Raju's Resource Hub
Google ad
Vectors constitute one of the several Mathematical systems which can be usefully employed to provide mathematical handling for certain types of problems in Geometry, Mechanics and other branches of Applied Mathematics. Vectors facilitate mathematical study of such physical quantities as possess Direction in addition to Magnitude. Velocity of a particle, for example, is one such quantity.
Physical quantities are broadly divided in two categories viz (a) Vector Quantities & (b) Scalar quantities.
( a ) Vector quantities :
Any quantity, such as velocity, momentum, or force, that has both magnitude and direction
Google ad
( b ) Scalar quantities :
A quantity, such as mass, length, time, density or energy, that has size or magnitude but does not involve the concept of direction is called scalar quantity.
Mathematical Description of Vector: A vector is a directed line segment. The length of the segment is called magnitude of the vector. The direction is indicated by an arrow joining the initial and final points of the line segment. The vector AB i.e, joining the initial point A and the final point B in the direction of AB is denoted as
\[\overrightarrow{AB}\]
\[\color {purple} {Example\ 1\ .}\ If\ \overrightarrow{a}\ =\ 5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\ and\ \overrightarrow{b}\ =\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k},\ \color {red} {find\ the\ value\ of\ 4\overrightarrow{a}\ +\ \overrightarrow{b}}\ \hspace{10cm}\]
\[\hspace{5cm}\ April\ 2024,\ October\ 2024\]
\[\color {blue} {Soln:}\ Given\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 4(5\overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k})\ +\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[=\ 20\overrightarrow{i}\ +\ 8\overrightarrow{j}\ -\ 12\overrightarrow{k}\ +\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j} +\ 5\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ \overrightarrow{b}\ =\ 23\overrightarrow{i}\ +\ 6\overrightarrow{j}\ -\ 7\overrightarrow{k}\ \hspace{5cm}\]
\[\color {purple} {Example\ 2\ .}\ If\ \overrightarrow{a}\ =\ 5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\ and\ \overrightarrow{b}\ = -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k},\ \color {red} {find\ 2\overrightarrow{a}\ +\ 6\overrightarrow{b}}\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue} {Soln:}\ Given\ \hspace{20cm}\]
\[\overrightarrow{a}= 5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k}\]
\[2\overrightarrow{a}\ +\ 6\overrightarrow{b}\ =\ 2(5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k})\ +\ 6( -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k})\]
\[=\ 10\overrightarrow{i}\ +\ 4\overrightarrow{j}\ -\ 6\overrightarrow{k}\ -\ 18\overrightarrow{i}\ -\ 12\overrightarrow{j}\ +\ 30\overrightarrow{k}\]
\[2\overrightarrow{a}\ +\ 6\overrightarrow{b}\ =\ -\ 8\overrightarrow{i}\ -\ 8\overrightarrow{j}\ +\ 24\overrightarrow{k}\ \hspace{5cm}\]
\[\color {purple} {Example\ 3\ .}\ If\ \overrightarrow{a}\ =\ 5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\ and\ \overrightarrow{b}\ = -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k},\ \color {red} {find\ 4\overrightarrow{a}\ +\ 6\overrightarrow{b}}\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2023\ Supp(25)\]
\[\color {blue} {Soln:}\ Given\ \hspace{20cm}\]
\[\overrightarrow{a}= 5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ 6\overrightarrow{b}\ =\ 4(5\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ 3\overrightarrow{k})\ +\ 6( -\ 3\overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 5\overrightarrow{k})\]
\[=\ 20\overrightarrow{i}\ +\ 8\overrightarrow{j}\ -\ 12\overrightarrow{k}\ -\ 18\overrightarrow{i}\ -\ 12\overrightarrow{j}\ +\ 30\overrightarrow{k}\]
\[4\overrightarrow{a}\ +\ 6\overrightarrow{b}\ =\ 2\overrightarrow{i}\ -\ 4\overrightarrow{j}\ +\ 18\overrightarrow{k}\ \hspace{5cm}\]
\[\color {purple} {Example\ 4\ .}\ \overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}\ and\ \overrightarrow{b}\ =\ \overrightarrow{j}\ – 2\ \overrightarrow{k},\ \color {red} {find\ \overrightarrow{a}\ -\ 2\overrightarrow{b}}\ \hspace{15cm}\]
\[\color {blue} {Soln:}\ Given\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{b}\ =\ \overrightarrow{j}\ – 2\ \overrightarrow{k}\]
\[\overrightarrow{a}\ -\ 2\overrightarrow{b}\ =\ 3\overrightarrow{i}\ + 2\overrightarrow{j} + \overrightarrow{k}\ -\ 2(\overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ +\ \overrightarrow{k}\ -\ 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[\overrightarrow{a}\ -\ 2\overrightarrow{b}\ =\ 3\overrightarrow{i}\ +\ 5\overrightarrow{k}\ \hspace{5cm}\]
Defn: If a is a vector, then
\[\color {green} {Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}}\]
\[\color {purple} {Example\ 5\ .}\ \color {red} {Find\ the\ Unit\ vector\ along\ the\ vector}\ 3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}\ \hspace{15cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\overrightarrow{a}= 3\overrightarrow{i} + 4\overrightarrow{j} – 5\overrightarrow{k}\]
\[\overrightarrow{|a|} = \sqrt{(3)^2 + (4)^2+(-5)^2 }\]
\[= \sqrt{(9 + 16 +25}\]
\[=\sqrt{50}\]
\[\overrightarrow{|a|}=\sqrt{50}\]
\[Unit\ vector\ along\ \overrightarrow{a}=\frac{\overrightarrow{a}}{\overrightarrow{|a|}}= \frac{3\overrightarrow{i}\ + 4\overrightarrow{j}- 5\overrightarrow{k}}{\sqrt{50}}\]
Position Vector: If P is any point in the space and 0 is the origin, then
\[\overrightarrow{OP}\]
is called the position vector of the point P.
Let P be a point in a Three dimensional Space. Let 0 be the origin and i, j and k the unit vectors along the x ,yand z axes . Then if P is (x, y, z) , the position vector of the point P is
\[\overrightarrow{OP}= x\overrightarrow{i}\ + y\overrightarrow{j}+ z\overrightarrow{k}\]
\[OP =\overrightarrow{|OP|} = \sqrt{x^2 + y^2 + z^2 }\]
Triangle Law of Addition of Two vectors:
\[\overrightarrow{OA} = \overrightarrow{a}\] \[\overrightarrow{AB} = \overrightarrow{b}\]\[\overrightarrow{OA}+ \overrightarrow{AB} = \overrightarrow{OB}\]\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[\color {purple} {Example\ 6\ .}\ If\ position\ vectors\ of\ the\ points\ A\ and\ B\ are\ \hspace{8cm}\]\[\overrightarrow{i}\ + 2\overrightarrow{j}- 3\overrightarrow{k}\ and\ 2\overrightarrow{i}\ – \overrightarrow{j},\ \color {red} {find\ \overrightarrow{|AB|}}\ \hspace{6cm}\]
\[\color {blue} {Soln\ :}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}= \overrightarrow{i}\ + 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{OB}= 2\overrightarrow{i}\ – \overrightarrow{j}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=2\overrightarrow{i}\ – \overrightarrow{j}- (\overrightarrow{i}\ + 2\overrightarrow{j}- 3\overrightarrow{k})\]
\[=2\overrightarrow{i}\ – \overrightarrow{j}- \overrightarrow{i}\ – 2\overrightarrow{j} + 3\overrightarrow{k}\]
\[\overrightarrow{AB}= \overrightarrow{i} – 3\overrightarrow{j} + 3\overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(1)^2 + (-3)^2 +(3)^2 }\]
\[ = \sqrt{(1 + 9 +9 })\]
\[ = \sqrt{19}\]
\[\underline{\color {red} {Condition\ for\ three\ position\ vectors\ \overrightarrow{OA}\ ,\ \overrightarrow{OB}\ and\ \overrightarrow{OC}\ to\ be}}\ \hspace{10cm}\]
\[i)\ \color {green} {collinear \ if \overrightarrow{BC} = K\overrightarrow{AB}} \]
\[ii) \color {green} {Equilateral\ triangle\ if\ \overrightarrow{|AB|}=\overrightarrow{|BC|}=\overrightarrow{|AC|}} \]
\[iii) \color {green} {Isosceles\ triangle\ if\ \overrightarrow{|AB|}=\overrightarrow{|BC|}\not=\overrightarrow{|AC|} or\ \overrightarrow{|BC|}=\overrightarrow{|AC|}\not=\overrightarrow{|AB|}}\]
\[iv) \color {green} {Right\ angled\ triangle\ if\ {AB}^2\ =\ {BC}^2\ +\ {AC}^2\ (OR)\ {BC}^2\ =\ {AB}^2\ +\ {AC}^2}\]\[\color {green}{(OR)\ {AC}^2\ =\ {AB}^2\ +\ {BC}^2}\]
\[\color {purple} {Example\ 7\ .}\ \color {red} {Show\ that\ the\ points\ whose\ position\ vectors}\ \hspace{12cm}\]\[2\overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ 5\overrightarrow{k},\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 2\overrightarrow{k}\ and\ 6\overrightarrow{i}\ -\ 5 \overrightarrow{j}\ +\ 7\overrightarrow{k}\ \color {red} {are\ collinear}\ \hspace{5cm}\]
\[\color {blue} {Soln\ :}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}= 2\overrightarrow{i}\ + 3\overrightarrow{j}- 5\overrightarrow{k}\]
\[\overrightarrow{OB}= 3\overrightarrow{i}\ +\overrightarrow{j} – 2\overrightarrow{k}\]
\[\overrightarrow{OC}= 6\overrightarrow{i}\ – 5\overrightarrow{j} + 7\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=3\overrightarrow{i}+ \overrightarrow{j} – 2\overrightarrow{k}- (2\overrightarrow{i}\ + 3\overrightarrow{j}- 5\overrightarrow{k})\]
\[=3\overrightarrow{i}+ \overrightarrow{j} – 2\overrightarrow{k}- 2\overrightarrow{i}\ – 3\overrightarrow{j}+ 5\overrightarrow{k}\]
\[\overrightarrow{AB}= \overrightarrow{i} – 2\overrightarrow{j} + 3\overrightarrow{k}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=6\overrightarrow{i}- 5 \overrightarrow{j} + 7\overrightarrow{k}- (3\overrightarrow{i}\ + \overrightarrow{j}- 2\overrightarrow{k})\]
\[=6\overrightarrow{i}-5 \overrightarrow{j} +7\overrightarrow{k}- 3\overrightarrow{i}\ – \overrightarrow{j}+ 2\overrightarrow{k}\]
\[\overrightarrow{BC}= 3\overrightarrow{i} – 6\overrightarrow{j} + 9\overrightarrow{k}\]
\[=\ 3( \overrightarrow{i}\ – 2\overrightarrow{j}\ +\ 3\overrightarrow{k})\]
\[\overrightarrow{BC}\ =\ 3\ \overrightarrow{AB}\]
\[\overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC} \ are\ collinear\]
\[\color {purple} {Example\ 8\ .}\ \color {red} {Prove\ that\ the\ points}\ \hspace{15cm}\]\[5\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 7\overrightarrow{k}, 6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k} and\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ \color {red} {form\ an\ equilateral\ triangle}\]
\[\hspace{5cm}\ October\ 2023\ October\ 2024\]
\[\color {blue} {Soln\ :}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}\ =\ 5\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 7\overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k}\ -\ (5\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 7\overrightarrow{k})\]
\[=\ 6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k}\ -\ 5\overrightarrow{i}\ -\ 6\overrightarrow{j}\ – 7\overrightarrow{k}\]
\[\overrightarrow{AB}\ =\ \overrightarrow{i}\ +\ \overrightarrow{j}\ – 2\overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(1)^2 + (1)^2 +(-2)^2 }\]
\[ = \sqrt{1 + 1 +4 }\]
\[AB = \sqrt{6}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ -\ (6\overrightarrow{i}\ +\ 7\overrightarrow{j}\ +\ 5\overrightarrow{k})\]
\[=\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ – 6\overrightarrow{i}\ – 7\overrightarrow{j}\ – 5\overrightarrow{k}\]
\[\overrightarrow{BC}= \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ \overrightarrow{k}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (-2)^2 +(1)^2 }\]
\[ = \sqrt{1 + 4 + 1}\]
\[BC = \sqrt{6}\]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ -\ (5\overrightarrow{i}\ + 6\overrightarrow{j}\ +\ 7\overrightarrow{k})\]
\[=\ 7\overrightarrow{i}\ +\ 5 \overrightarrow{j}\ +\ 6\overrightarrow{k}\ – 5\overrightarrow{i}\ -\ 6\overrightarrow{j}\ -\ 7\overrightarrow{k}\]
\[\overrightarrow{AC}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(2)^2 + (-1)^2 +(-1)^2 }\]
\[ = \sqrt{4 + 1 + 1}\]
\[AC = \sqrt{6}\]
\[AB = BC = AC = \sqrt{6}\]
\[The\ given\ triangle\ is\ an\ equilateral\ triangle\]
\[\color {purple} {Example\ 9\ .}\ \color {red} {Prove\ that\ the\ points}\ \hspace{15cm}\]\[4\overrightarrow{i}\ + 2\overrightarrow{j}+ 3\overrightarrow{k}, 2\overrightarrow{i}\ + 3\overrightarrow{j}+ 4\overrightarrow{k} and\ 3\overrightarrow{i}\ +4 \overrightarrow{j}+ 2\overrightarrow{k}\ \color {red} {form\ an\ equilateral\ triangle}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln\ :}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}= 4\overrightarrow{i}\ + 2\overrightarrow{j}+ 3\overrightarrow{k}\]
\[\overrightarrow{OB}= 2\overrightarrow{i}\ +3\overrightarrow{j} + 4\overrightarrow{k}\]
\[\overrightarrow{OC}= 3\overrightarrow{i}\ + 4\overrightarrow{j} + 2\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=2\overrightarrow{i}+ 3\overrightarrow{j} + 4\overrightarrow{k}- (4\overrightarrow{i}+ 2\overrightarrow{j}+ 3\overrightarrow{k})\]
\[=2\overrightarrow{i}+ 3\overrightarrow{j} + 4\overrightarrow{k}- 4\overrightarrow{i}- 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{AB}= -2\overrightarrow{i} +\overrightarrow{j} + \overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(-2)^2 + (-1)^2 +(1)^2 }\]
\[ = \sqrt{(4 + 1 +1 }\]
\[AB = \sqrt{6}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- (2\overrightarrow{i}+ 3\overrightarrow{j}+ 4\overrightarrow{k})\]
\[=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- 2\overrightarrow{i}- 3\overrightarrow{j}- 4\overrightarrow{k}\]
\[\overrightarrow{BC}= \overrightarrow{i} +\overrightarrow{j} -2\overrightarrow{k}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (1)^2 +(-2)^2 }\]
\[ = \sqrt{(1 + 1 + 4}\]
\[BC = \sqrt{6}\]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- (4\overrightarrow{i}\ + 2\overrightarrow{j}+ 3\overrightarrow{k})\]
\[=3\overrightarrow{i}+ 4\overrightarrow{j} + 2\overrightarrow{k}- 4\overrightarrow{i}- 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{AC}= -\overrightarrow{i} +2\overrightarrow{j} -\overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(-1)^2 + (2)^2 +(-1)^2 }\]
\[ = \sqrt{(1 + 4 + 1}\]
\[AC = \sqrt{6}\]
\[AB = BC = AC = \sqrt{6}\]
\[The\ given\ triangle\ is\ an\ equilateral\ triangle\]
\[\color {purple} {Example\ 10\ .}\ \color {red} {Prove\ that\ the\ points}\ \hspace{15cm}\]\[3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k}\ ,\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k} and\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ \color {red} {form\ an\ isosceles\ triangle}\]
\[\hspace{5cm}\ October\ 2024\ Supp\ June\ 25\]
\[\color {blue} {Soln\ :}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}\ =\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ – (3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ -\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[\overrightarrow{AB}\ =\ 2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\ \overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(2)^2\ +\ (2)^2\ +\ (-1)^2 }\]
\[ = \sqrt{4\ +\ 4\ +\ 1}\]
\[ =\ \sqrt{9}\]
\[AB =\ 3 \]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 3\overrightarrow{k})\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 5\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 3\overrightarrow{k}\]
\[\overrightarrow{BC}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(1)^2\ +\ (-2)^2\ +\ (2)^2 }\]
\[ = \sqrt{1\ +\ 4\ +\ 4}\]
\[ =\ \sqrt{9}\]
\[BC =\ 3 \]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ – (3\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 6\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ \overrightarrow{k}\ -\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{AC}\ =\ 4\overrightarrow{i}\ +\ \overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(4)^2\ +\ (0)^2\ +\ (1)^2 }\]
\[ = \sqrt{16\ +\ 0\ +\ 1}\]
\[AC = \sqrt{17}\]
\[\boxed{AB = BC\ \neq\ AC}\]
\[\color {purple} {Example\ 11\ .}\ \color {red} {Prove\ that\ the\ points\ whose\ position\ vectors\ are}\ \hspace{15cm}\]\[3\overrightarrow{i}\ – \overrightarrow{j}+ 6\overrightarrow{k}, 5\overrightarrow{i}\ – 2\overrightarrow{j}+ 7\overrightarrow{k} and\ 6\overrightarrow{i}\ -5 \overrightarrow{j}+ 2\overrightarrow{k}\ \color {red} {form\ a\ right\ angled\ triangle}\]
\[\color {blue} {Soln\ :}\ Given\ \hspace{17cm}\]
\[\overrightarrow{OA}= 3\overrightarrow{i} – \overrightarrow{j}+ 6\overrightarrow{k}\]
\[\overrightarrow{OB}= 5\overrightarrow{i}\ -2\overrightarrow{j} + 7\overrightarrow{k}\]
\[\overrightarrow{OC}= 6\overrightarrow{i}\ – 5\overrightarrow{j} + 2\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=5\overrightarrow{i}- 2\overrightarrow{j} + 7\overrightarrow{k}- (3\overrightarrow{i}- \overrightarrow{j}+ 6\overrightarrow{k})\]
\[=5\overrightarrow{i}- 2\overrightarrow{j} + 7\overrightarrow{k}- 3\overrightarrow{i}+ \overrightarrow{j}- 6\overrightarrow{k}\]
\[\overrightarrow{AB}= 2\overrightarrow{i} -\overrightarrow{j} + \overrightarrow{k}\]
\[AB =\overrightarrow{|AB|} = \sqrt{(2)^2 + (-1)^2 +(1)^2 }\]
\[ = \sqrt{(4 + 1 +1 }\]
\[AB = \sqrt{6}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- (5\overrightarrow{i}- 2\overrightarrow{j}+ 7\overrightarrow{k})\]
\[=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- 5\overrightarrow{i}+ 2\overrightarrow{j}- 7\overrightarrow{k}\]
\[\overrightarrow{BC}= \overrightarrow{i} -3\overrightarrow{j} -5\overrightarrow{k}\]
\[BC =\overrightarrow{|BC|} = \sqrt{(1)^2 + (-3)^2 +(-5)^2 }\]
\[ = \sqrt{(1 + 9 + 25}\]
\[BC = \sqrt{35}\]
\[\overrightarrow{AC} = \overrightarrow{OC}-\overrightarrow{OA}\]
\[=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- (3\overrightarrow{i}- \overrightarrow{j}+ 6\overrightarrow{k})\]
\[=6\overrightarrow{i}- 5\overrightarrow{j} + 2\overrightarrow{k}- 3\overrightarrow{i}+\overrightarrow{j}- 6\overrightarrow{k}\]
\[\overrightarrow{AC}= 3\overrightarrow{i} -4\overrightarrow{j} -4\overrightarrow{k}\]
\[AC =\overrightarrow{|AC|} = \sqrt{(3)^2 + (-4)^2 +(-4)^2 }\]
\[ = \sqrt{(9 + 16 + 16}\]
\[AC = \sqrt{41}\]
\[AB^2 = 6, BC^2 = 35, AC^2 = 41 \]
\[AB^2 + BC^2 = 6 + 35 = 41=AC^2\]
\[AB^2 + BC^2 = AC^2\]
The given triangle is an right angled triangle.
Direction Cosines and Direction ratios
When a directed line OP passing through the origin makes α, β and γ angles with the x, y and z axis respectively with O as the reference, these angles are referred as the direction angles of the line and the cosine of these angles give us the direction cosines. These direction cosines are usually represented as l, m and n.
\[\overrightarrow{OP}= x\overrightarrow{i}\ + y\overrightarrow{j}+ z\overrightarrow{k}\]
\[OP =\overrightarrow{|OP|} = \sqrt{x^2 + y^2 + z^2 }= r\]
\[cos\ α\ =\ \frac{x}{r},\ cos\ β\ =\ \frac{y}{r},\ cos\ γ\ =\ \frac{z}{r}\]
\[ Direction\ cosines\ are \frac{x}{r}, \frac{y}{r}, \frac{z}{r} \]
\[\color {purple} {Example\ 12\ .}\ \color {red} {Find\ the\ Direction\ cosines}\ of\ the\ vector\ \overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2023,\ April\ 2024\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\overrightarrow{a}\ =\ \overrightarrow{i}\ + 2\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[r =\overrightarrow{|a|} = \sqrt{(1)^2 + (2)^2 + (-3)^2 }\]
\[ = \sqrt{(1\ +\ 4\ +\ 9) }\]
\[r =\sqrt{14}\]
\[ Direction\ cosines\ are \frac{1}{\sqrt(14)}, \frac{2}{\sqrt(14)}, \frac{-3}{\sqrt(14)} \]
\[\color {purple} {Example\ 13\ .}\ \color {red} {Find\ the\ Direction\ cosines\ of}\ \overrightarrow{i}\ + 3\overrightarrow{j}\ -\ 3\overrightarrow{k}\ \hspace{5cm}\]
\[\hspace{5cm}\ April\ 2025\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\overrightarrow{a}\ =\ \overrightarrow{i}\ + 3\overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[r =\overrightarrow{|a|} = \sqrt{(1)^2 + (3)^2 + (-3)^2 }\]
\[ = \sqrt{(1\ +\ 9\ +\ 9) }\]
\[r =\sqrt{19}\]
\[ Direction\ cosines\ are \frac{1}{\sqrt(19)}, \frac{3}{\sqrt(19)}, \frac{-3}{\sqrt(19)} \]
\[\color {purple} {Example\ 14\ .}\ \color {red} {Find\ the\ Direction\ cosines\ of}\ 5\overrightarrow{i}\ – 3\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \hspace{5cm}\]
\[\hspace{5cm}\ Supp\ June\ 2025\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\overrightarrow{a}\ =\ 5\overrightarrow{i}\ – 3\overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[r =\overrightarrow{|a|} = \sqrt{(5)^2 + (-3)^2 + (4)^2 }\]
\[ = \sqrt{(25\ +\ 9\ +\ 16) }\]
\[r =\sqrt{50}\]
\[ Direction\ cosines\ are \frac{5}{\sqrt(50)}, \frac{-3}{\sqrt(50)}, \frac{4}{\sqrt(50)} \]
\[\color {purple} {Example\ 15\ .}\ \color {red} {Find\ the\ Direction\ cosines\ of}\ 3\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2024\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\overrightarrow{a}\ =\ 3\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[r =\overrightarrow{|a|} = \sqrt{(3)^2 + (2)^2 + (4)^2 }\]
\[ = \sqrt{(9\ +\ 4\ +\ 16) }\]
\[r =\sqrt{29}\]
\[ Direction\ cosines\ are \frac{3}{\sqrt(29)}, \frac{2}{\sqrt(29)}, \frac{4}{\sqrt(29)} \]
PRODUCT OF VECTORS
SCALAR OR DOT PRODUCT
Definition:
\[Let \overrightarrow{a}\ and \overrightarrow{b}\ be\ two\ non\ zero\ vectors\ inclined\ at\ an\ angle\ \theta.\] \[Then\ the\ scalar\ product\ of\ \overrightarrow{a} and \overrightarrow{b} is\ denoted\ by\]\[\overrightarrow{a}.\overrightarrow{b} and\ is\ defined\ as\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{|a|}\overrightarrow{|b|}\cos\theta \]
Properties of Scalar Product:
\[1. \overrightarrow{a}\ and\overrightarrow{b}are\ perpendicular\ vectors\ if\ and\ only\ if \overrightarrow{a}.\overrightarrow{b}=0. \]
\[2. \overrightarrow{i},\overrightarrow{j}\ and\ \overrightarrow{k}\ are\ the\ unit\ vectors\ along\ the\ x, y, z\ axis\ respectively \]
\[Then\ i) \overrightarrow{i}.\overrightarrow{i}= \overrightarrow{j}.\overrightarrow{j}=\overrightarrow{k}.\overrightarrow{k}=1 \]
\[ ii) \overrightarrow{i}.\overrightarrow{j}= \overrightarrow{j}.\overrightarrow{k}=\overrightarrow{k}.\overrightarrow{i}= \overrightarrow{j}.\overrightarrow{i}= \overrightarrow{k}.\overrightarrow{j}=\overrightarrow{i}.\overrightarrow{k}=0 \]
\[3. \ Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[4. \ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[If\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ and\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k}\ . Then\]
\[ \overrightarrow{a}.\overrightarrow{b}= a_1b_1 + a_2b_2 + a_3b_3\ (using\ property\ 2)\]
\[\color {purple} {Example\ 16\ .}\ \color {red} {Find\ the\ scalar\ product\ of}\ 2\overrightarrow{i} -4\overrightarrow{j}\ + 8\overrightarrow{k}\ and\ \overrightarrow{i}\ +6\overrightarrow{j}+ 12\overrightarrow{k}\ \hspace{10cm}\]
\[\color {blue} {Soln\ :}\ \hspace{18cm}\]
\[\overrightarrow{a}\ =\ 2\overrightarrow{i} -4\overrightarrow{j}\ + 8\overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}\ +\ 6\overrightarrow{j}+ 12\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (2\overrightarrow{i} -4\overrightarrow{j}\ + 8\overrightarrow{k}) .( \overrightarrow{i} +6\overrightarrow{j}+ 12\overrightarrow{k})\]
= 2(1) – 4 (6) + 8 (12)
= 2 – 24 + 96
= 74
\[ \boxed{\overrightarrow{a}.\overrightarrow{b}= 74}\]
\[\color {purple} {Example\ 17\ .}\ \color {red} {Find\ the\ Dot\ product\ of}\ \overrightarrow{i}+\overrightarrow{j}\ +\ \overrightarrow{k}\ ,\ \overrightarrow{i}\ + 3\overrightarrow{k}\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i}+\overrightarrow{j}\ +\ \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}\ + 3\overrightarrow{k}\]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+\overrightarrow{j}\ +\ \overrightarrow{k}) .(\overrightarrow{i}\ + 3\overrightarrow{k})\]
\[=\ 1(1)\ +\ 1(0)\ +\ 1(3)\]
\[=\ 1\ +\ 0\ +\ 3\]
\[=\ 4\]
\[ \boxed{\overrightarrow{a}.\overrightarrow{b}\ =\ 4}\]
\[\color {purple} {Example\ 18\ .}\ \color {red} {Show\ that\ the\ vectors}\ \overrightarrow{i}\ -\ 3\overrightarrow{j}\ + 5\overrightarrow{k} and\ -\ 2\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 4\overrightarrow{k} are\ perpendicular\]
\[\hspace{5cm}\ October\ 2023,\ April\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k} \]
\[\overrightarrow{b}= – 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i} – 3\overrightarrow{j} + 5\overrightarrow{k}) .(- 2\overrightarrow{i}+ 6\overrightarrow{j}+4\overrightarrow{k})\]
\[=\ 1(-2)\ +\ -3(6)\ +\ 5 (4)\]
\[=\ -\ 2\ -\ 18\ +\ 20\]
\[=\ 0\]
\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[\color {purple} {Example\ 19\ .}\ \color {red} {Find\ the\ value\ of\ m}\ if\ the\ vectors\ 3\overrightarrow{i} -\overrightarrow{j} + 5\overrightarrow{k} and\ -6\overrightarrow{i}+ m\overrightarrow{j}+4\overrightarrow{k} are\ perpendicular\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= 3\overrightarrow{i} -\overrightarrow{j} + 5\overrightarrow{k}\]
\[ \overrightarrow{b}= -6\overrightarrow{i}+ m\overrightarrow{j}+4\overrightarrow{k} \]
\[Given\ \overrightarrow{a} and\ \overrightarrow{b} are\ perpendicular\ to\ each\ other \]
\[i.e\ \overrightarrow{a}.\overrightarrow{b}= 0\]
\[(3\overrightarrow{i} -\overrightarrow{j} + 5\overrightarrow{k}).(-6\overrightarrow{i}+ m\overrightarrow{j}+4\overrightarrow{k}) = 0 .\]
3(-6) – 1(m) + 5 (4) = 0
-18 – m + 20 = 0
-m + 2 = 0
\[\boxed{m\ =\ 2}\]
\[\color {purple} {Example\ 20\ .}\ \color {red} {Prove\ that}\ \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k},\ \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}\ and\ 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k} are\ mutually\ orthogonal.\ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2023\ October\ 2024\ Supp(June)\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= \overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k}\]
\[ \overrightarrow{b}= \overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k} \]
\[\overrightarrow{c}= 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k}) .(\overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k})\]
= 1(1) + 2(1) + 1(-3)
= 0
\[\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors.\]
\[ \overrightarrow{b}.\overrightarrow{c}= (\overrightarrow{i} + \overrightarrow{j}- 3\overrightarrow{k}) .( 7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k})\]
= 1(7) + 1(-4) – 3(1)
= 7 – 4 – 3
= 0
\[\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors.\]
= 7(1) -4 (2) + 1 (1)
\[ \overrightarrow{c}.\overrightarrow{a}= (7\overrightarrow{i}-4\overrightarrow{j}+\overrightarrow{k}) .(\overrightarrow{i}+2\overrightarrow{j}+ \overrightarrow{k})\]
= 7 – 8 + 1
= 0
\[\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors.\]
\[ The\ three\ vectors\ are\ mutually\ perpendicular.\]
\[\color {purple} {Example\ 21\ .}\ \color {red} {Show\ that}\ the\ vectors\ 2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ + \overrightarrow{k},\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ + 2\overrightarrow{k}\ \hspace{10cm}\\ and\
2\overrightarrow{i}\ -\overrightarrow{j}\ -\ 2\overrightarrow{k} are\ mutually\ perpendicular.\ \hspace{5cm}\]
\[\hspace{5cm}\ April\ 2024\ April\ 2025\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ + \overrightarrow{k}\]
\[\overrightarrow{b}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ + 2\overrightarrow{k}\]
\[\overrightarrow{c}\ =\ 2\overrightarrow{i}\ -\overrightarrow{j}\ -\ 2\overrightarrow{k}\]
\[ \overrightarrow{a}.\overrightarrow{b}= (2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ + \overrightarrow{k}) .(\overrightarrow{i}\ -\ 2\overrightarrow{j}\ + 2\overrightarrow{k})\]
\[=\ 2(1)\ +\ 2(- 2)\ +\ 1(2)\]
\[=\ 2\ -\ 4\ +\ 2\]
\[=\ 0\]
\[\boxed{\overrightarrow{a}\ and\ \overrightarrow{b}\ are\ perpendicular\ vectors}\]
\[ \overrightarrow{b}.\overrightarrow{c}= (\overrightarrow{i}\ -\ 2\overrightarrow{j}\ + 2\overrightarrow{k}) .(2\overrightarrow{i}\ -\overrightarrow{j}\ -\ 2\overrightarrow{k})\]
\[=\ 1(2)\ +\ -\ 2(-1)\ +\ 2(-2)\]
\[=\ 2\ +\ 2 -\ 4\]
\[=\ 0\]
\[\boxed{\overrightarrow{b} and\ \overrightarrow{c}\ are\ perpendicular\ vectors}\]
\[ \overrightarrow{c}.\overrightarrow{a}= (2\overrightarrow{i}\ -\overrightarrow{j}\ -\ 2\overrightarrow{k}) .(2\overrightarrow{i}\ +\ 2\overrightarrow{j}\ + \overrightarrow{k})\]
\[=\ 2(2)\ +\ -\ 1(2)\ +\ -\ 2(1)\]
\[=\ 4\ -\ 2 -\ 2\]
\[=\ 0\]
\[\boxed{\overrightarrow{c} and\ \overrightarrow{a}\ are\ perpendicular\ vectors}\]
\[\therefore\ The\ three\ vectors\ are\ mutually\ perpendicular.\]
\[\underline{Projection\ of\ \overrightarrow{a}\ on\ \overrightarrow{b}}\]
\[\color {purple} {Example\ 22\ .}\ \color {red} {Find\ the\ projection\ of\ the\ vector}\ 2\overrightarrow{i}\ + \overrightarrow{j}\ -\ 2\overrightarrow{k} on\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 2\overrightarrow{k} \ \hspace{5cm}\]
\[\hspace{5cm}\ October\ 2024\ April\ 25\ Supp(June)\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 2\overrightarrow{k} \]
\[\overrightarrow{b}\ =\ \overrightarrow{i}\ -\ 2 \overrightarrow{j}\ -\ 2\overrightarrow{k} \]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[=\ \frac{(2\overrightarrow{i}\ + \overrightarrow{j}\ -\ 2\overrightarrow{k}).(\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 2\overrightarrow{k})}{\sqrt{(1)^2 + (-2)^2 + (-2)^2 }}\]
\[=\ \frac{2(1)\ +\ 1(-2)\ +\ (-2) (-2)}{\sqrt{(1 + 4 + 4 }}\]
\[= \frac{2\ -\ 2\ +\ 4}{\sqrt{9}}\]
\[\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{4}{3}}\]
\[\color {purple} {Example\ 23\ .}\ \color {red} {Find\ the\ projection\ of\ the\ vector}\ 3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k} on\ the\ vector\ \overrightarrow{i}+ 2\overrightarrow{j}+2\overrightarrow{k} \ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k} \]
\[\overrightarrow{b}= \overrightarrow{i}+2 \overrightarrow{j}+ 2\overrightarrow{k} \]
\[Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|b|}}\]
\[=\frac{(3\overrightarrow{i}+ 4\overrightarrow{j}- 5\overrightarrow{k}).(\overrightarrow{i}+2\overrightarrow{j}+ 2\overrightarrow{k})}{\sqrt{(1)^2 + (2)^2 + (2)^2 }}\]
\[= \frac{3(1)+ 4(2)- 5(2)}{\sqrt{(1 + 4 + 4 }}\]
\[= \frac{3+ 8- 10}{\sqrt{9}}\]
\[\boxed{Projection\ of\ \overrightarrow{a} on \overrightarrow{b} =\frac{1}{3}}\]
Angle between two vectors using scalar product
\[\underline{Angle\ between\ two\ vectors\ \overrightarrow{a} and\ \overrightarrow{b}}: \ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[\color {purple} {Example\ 24\ .}\ \color {red} {Find\ the\ angle\ between\ vectors}\ 3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k} and\ 2\overrightarrow{i} -3\overrightarrow{j}- 5\overrightarrow{k}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= 3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k} \]
\[\overrightarrow{b}= 2\overrightarrow{i}-3 \overrightarrow{j}- 5\overrightarrow{k} \]
\[ \overrightarrow{a}.\overrightarrow{b}= (3\overrightarrow{i}+ 4\overrightarrow{j}- 2\overrightarrow{k}) .(2\overrightarrow{i}-3 \overrightarrow{j}- 5\overrightarrow{k})\]
= 3 ( 2 ) + 4 ( – 3 ) – 2 ( – 5 )
= 6 – 12 + 10
= 4
\[\boxed{ \overrightarrow{a}.\overrightarrow{b}= 4}\]
\[\overrightarrow{|a|} = \sqrt{(3)^2 + (4)^2 + (-2)^2 }=\sqrt{(9 + 16 +4 }=\sqrt{29}\]
\[\overrightarrow{|b|} = \sqrt{(2)^2 + (-5)^2 + (-3)^2 }=\sqrt{(4 + 25 +9 }=\sqrt{38}\]
\[\ cos\ \theta =\frac{\overrightarrow{a}.\overrightarrow{b}}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[= \frac{4}{\sqrt{29}\sqrt{38}}\]
\[\boxed{\theta = \cos ^-1 ( \frac{4}{\sqrt{29}\sqrt{38}})}\]
Vector Product of Two Vectors
Definition
\[Let \overrightarrow{a}\ and \overrightarrow{b}\ be\ two\ non\ zero\ vectors\ inclined\ at\ an\ angle\ \theta.\] \[Then\ the\ vector\ product\ of\ \overrightarrow{a} and \overrightarrow{b} is\ denoted\ by\]\[\overrightarrow{a}×\overrightarrow{b} and\ is\ defined\ as\overrightarrow{a}×\overrightarrow{b}=\overrightarrow{|a|}\overrightarrow{|b|}\sin\theta\ n^\wedge \]
Properties of Vector Product:
\[1. \overrightarrow{a}\ and\overrightarrow{b}are\ parellel\ vectors\ if\ and\ only\ if \overrightarrow{a}× \overrightarrow{b}= 0. \]
\[2.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ any\ two\ vectors\ then \overrightarrow{a}×\overrightarrow{b} = -\overrightarrow{b}×\overrightarrow{a}\]
\[3. \overrightarrow{i},\overrightarrow{j}\ and\ \overrightarrow{k}\ are\ the\ unit\ vectors\ along\ the\ x, y, z\ axis\ respectively \]
\[Then\ i) \overrightarrow{i}×\overrightarrow{i}= \overrightarrow{j}×\overrightarrow{j}=\overrightarrow{k}×\overrightarrow{k}=0 \]
\[ ii) \overrightarrow{i}×\overrightarrow{j}= \overrightarrow{k};\overrightarrow{j}×\overrightarrow{k}= \overrightarrow{i};\overrightarrow{k}×\overrightarrow{i}=\overrightarrow{j}\]
\[ iii) \overrightarrow{j}×\overrightarrow{i}= -\overrightarrow{k};\overrightarrow{k}×\overrightarrow{j}= -\overrightarrow{i};\overrightarrow{i}×\overrightarrow{k}= -\overrightarrow{j}\]
4. Vector product in determinant form
\[Let\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ and\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k}\ . Then\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
\overrightarrow{a_1} & \overrightarrow{a_2} & \overrightarrow{a_3}\\
\overrightarrow{b_1} & \overrightarrow{b_2} & \overrightarrow{b_3}\\
\end{vmatrix}\]
\[5.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ parellelogram. Then\ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|\]
\[6.\ If \overrightarrow{d_1}\ and \overrightarrow{d_2}\ are\ two\ diagonals\ of\ a\ parellelogram. Then\ Area\ of \ parellelogram = \frac{1}{2}|\overrightarrow{d_1} × \overrightarrow{d_2}|\]
\[7.\ If \overrightarrow{a}\ and \overrightarrow{b}\ are\ two\ adjacent\ sides\ of\ a\ triangle. Then\ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{a} × \overrightarrow{b}|\]
\[8.\ Area\ of\ the\ triangle\ formed\ by\ the\ points\ whose\ position\ vectors\ \overrightarrow{OA},\overrightarrow{OB}\ and\ \overrightarrow{OC}\ is\ \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[9.\ \ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}\]
\[10.\ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}\]
Scalar triple Product
\[11.\ Let\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} be\ any\ three\ vectors,\]\[their\ scalar\ triple\ product\ is\ denoted\ by\ [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}]\]
\[12.\ Let\ \overrightarrow{a}= a_1\overrightarrow{i}\ + a_2\overrightarrow{j}+ a_3\overrightarrow{k}\ ,\ \overrightarrow{b}= b_1\overrightarrow{i}\ + b_2\overrightarrow{j}+ b_3\overrightarrow{k} and\ \overrightarrow{c}= c_1\overrightarrow{i}\ + c_2\overrightarrow{j}+ c_3\overrightarrow{k}\]
\[\underline{\color {red} {Condition\ for\ three\ vectors\ \overrightarrow{a}\, \overrightarrow{b}\ and\ \overrightarrow{c}\ to\ be\ coplanar\ (or)\ lie\ on\ the\ same\ plane}}\]\[if\ \color {green} {\begin{vmatrix}
{a_1} &{a_2} & {a_3}\\
{b_1} & {b_2} & {b_3}\\
{c_1} & {c_2} & {c_3}\\
\end{vmatrix}\ =\ 0}\]
\[\color {purple} {Example\ 25\ .}\ \color {red} { Find\ the\ values\ of}\ \overrightarrow{i}\ . \ \overrightarrow{j}\ and\ \overrightarrow{i}\ \times\ \overrightarrow{j}\ \hspace{18cm}\]
\[\hspace{5cm}\ October\ 2023\]
\[\color {blue} {Soln:}\ \overrightarrow{i}\ . \ \overrightarrow{j}\ =\ 0\ and\ \overrightarrow{i}\ \times\ \overrightarrow{j}\ =\ \overrightarrow{k}\ \hspace{15cm}\]
\[\color {purple} {Example\ 26\ .}\ \color {red} { Find}\ (i)\ \overrightarrow{k}\ . \ \overrightarrow{i}\ and\ (ii)\ \overrightarrow{k}\ \times\ \overrightarrow{i}\ \hspace{18cm}\]
\[\hspace{5cm}\ April\ 2024\ April\ 25\]
\[\color {blue} {Soln:}\ (i)\ \overrightarrow{k}\ . \ \overrightarrow{i}\ =\ 0\ and\ (ii)\ \overrightarrow{k}\ \times\ \overrightarrow{i}\ =\ \overrightarrow{j}\ \hspace{15cm}\]
\[\color {purple} {Example\ 27\ .}\ If\ \overrightarrow{a}\ =\ \overrightarrow{i}\ +\ \overrightarrow{j}\ + \overrightarrow{k} and \overrightarrow{b}= 2\overrightarrow{i}- \overrightarrow{j}+ 3\overrightarrow{k}\ \color {red} {Find\ \overrightarrow{a}× \overrightarrow{b}}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{18cm}\]
\[\overrightarrow{a}= \overrightarrow{i}+ \overrightarrow{j}+\overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
1 & 1 & 1\\
2 & -1 & 3\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( 3 + 1) -\overrightarrow{j}(3-2)+\overrightarrow{k}(-1-2)\]
\[ = \overrightarrow{i}(4) -\overrightarrow{j}(1)+\overrightarrow{k}(-3)\]
\[\boxed{ \overrightarrow{a}× \overrightarrow{b}= 4\overrightarrow{i}-\overrightarrow{j}-3\overrightarrow{k}}\]
\[\color {purple} {Example\ 28\ .}\ \color {red} {Show\ that\ the\ vectors}\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 6\overrightarrow{k}\ and\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\ \color {red} {are\ parallel}.\ \hspace{10cm}\]
\[\hspace{5cm}\ October\ 2024,\ Supp\ June\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[To\ show\ \overrightarrow{a}×\overrightarrow{b} =\ 0\ \hspace{10cm}\]
\[\overrightarrow{a}\ =\ 4\overrightarrow{i}\ -\ 2\overrightarrow{j}\ -\ 6\overrightarrow{k} \]
\[\overrightarrow{b}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ -\ 3\overrightarrow{k}\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
4 & -2 & -6\\
2 & -1 & -3\\
\end{vmatrix}\]
\[ = \overrightarrow{I}(6\ -\ 6)\ -\overrightarrow{j}(-12\ +\ 12)\ +\ \overrightarrow{k}(-4\ +\ 4)\]
\[ = \overrightarrow{i}(0) -\overrightarrow{j}(0)+\overrightarrow{k}(0)\]
\[\boxed{\overrightarrow{a}×\overrightarrow{b} =\ 0}\ \hspace{7cm}\]
\[\therefore\ The\ given\ vectors\ \overrightarrow{a}\ and\ \overrightarrow{b}\ are\ parallel\]
\[\color {purple} {Example\ 29\ .}\ \color {red} {Find\ the\ area\ of\ the\ parellelogram}\ whose\ adjacent\ sides\ are\ \hspace{10cm}\]\[-\ \overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\ and\ \overrightarrow{i}\ – \overrightarrow{j}\ -\ \overrightarrow{k}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}=-\ \overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}\ – \overrightarrow{j}\ -\ \overrightarrow{k}\]
\[ Area\ of \ parellelogram = |\overrightarrow{a} × \overrightarrow{b}|\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
– 1 & 2 & 4\\
1 & -1 & -1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}(-\ 2 +\ 4)\ -\ \overrightarrow{j}(1\ -\ 4)\ +\ \overrightarrow{k}(1\ -\ 2)\]
\[ = \overrightarrow{i}(2)\ -\overrightarrow{j}(-\ 3)\ +\ \overrightarrow{k}(- 1)\]
\[ \overrightarrow{a}× \overrightarrow{b}= 2\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ -\ \overrightarrow{k}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(2)^2\ +\ (3)^2 + (-1)^2 }=\sqrt{(4\ +\ 9\ +\ 1}=\sqrt{14}\]
\[\boxed{Area\ of \ parellelogram = \sqrt{14} sq.units}\]
\[\color {purple} {Example\ 30\ .}\ \color {red} {Find\ the\ area\ of\ the\ parellelogram}\ whose\ diagonals\ are\ represented\ by\ \hspace{10cm}\]\[3\ \overrightarrow{i}\ + \overrightarrow{j}\ -\ 2\overrightarrow{k}\ and\ \overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 4\ \overrightarrow{k}\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{d_1}\ =\ 3\ \overrightarrow{i}\ + \overrightarrow{j}\ -\ 2\overrightarrow{k}\]
\[\overrightarrow{d_2}\ =\ \overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 4\ \overrightarrow{k}\]
\[\overrightarrow{d_1}\ ×\ \overrightarrow{d_2}\ =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
3 & 1 & -2\\
1 & -3 & 4\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( 4\ -\ 6) -\overrightarrow{j}(12\ +\ 2)+\overrightarrow{k}(-9\ -\ 1)\]
\[ = \overrightarrow{i}(-2) -\overrightarrow{j}(14)\ +\ \overrightarrow{k}(- 10)\]
\[\overrightarrow{d_1}\ ×\ \overrightarrow{d_2}\ =\ -\ 2\overrightarrow{i}\ -\ 14\ \overrightarrow{j}\ -\ 10\overrightarrow{k}\]
\[ Area\ of \ parellelogram =\ \frac{1}{2} |\overrightarrow{d_1}\ ×\ \overrightarrow{d_2}|\]
\[=\ \frac{1}{2} \sqrt{(-2)^2 + (-14)^2 + (-10)^2 }\]
\[=\ \frac{1}{2}\ \sqrt{(4\ +\ 196\ +\ 100)}\]
\[=\ \frac{1}{2}\ \sqrt{300}\]
\[\boxed{Area\ of \ parellelogram =\ \frac{1}{2}\ \sqrt{300} sq.units}\]
\[\color {purple} {Example\ 31\ .}\ \color {red} {Find\ the\ area\ of\ the\ triangle}\ whose\ adjacent\ sides\ are\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}\ \hspace{8cm}\\ and\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\ \hspace{15cm}\]
\[\hspace{5cm}\ April\ 2024\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 3\overrightarrow{i}\ +\ \overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ \overrightarrow{i}\ -\ 2\overrightarrow{j}\ +\ 4\overrightarrow{k}\]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
3 & 1 & 2\\
1 & -2 & 4\\
\end{vmatrix}\]
\[ = \overrightarrow{i}(1 . 4\ -\ 2(-2))\ -\ \overrightarrow{j}(3 . 4\ -\ 2(1))\ +\ \overrightarrow{k}(3 . (-2)\ -\ 1(1))\]
\[ = \overrightarrow{i}(4\ +\ 4)\ -\ \overrightarrow{j}(12\ -\ 2)\ +\ \overrightarrow{k}(-6\ -\ 1)\]
\[ = \overrightarrow{i}(8)\ -\ \overrightarrow{j}(10)\ +\ \overrightarrow{k}(-7)\]
\[ \overrightarrow{a}× \overrightarrow{b}\ =\ 8\overrightarrow{i}\ -\ 10\overrightarrow{j}\ -\ 7\overrightarrow{k}\]
\[ Area\ of \ triangle\ =\ \frac{1}{2} |\overrightarrow{a}\ ×\ \overrightarrow{b}|\]
\[=\ \frac{1}{2} \sqrt{((8)^2\ +\ (-10)^2 + (-7)^2 }\]
\[=\ \frac{1}{2}\ \sqrt{(64\ +\ 100\ +\ 49)}\]
\[=\ \frac{1}{2}\ \sqrt{213}\]
\[\color {purple} {Example\ 32\ .}\ \color {red} {Find\ the\ area\ of\ the\ triangle}\ formed\ by\ the\ points\ whose\ position\ vectors\ are\ \hspace{8cm}\]\[ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}\ , 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ and\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ \hspace{7cm}\]
\[\hspace{5cm}\ October\ 2023\ October\ 2024\ October\ 25,\ (Supp)\ June\ 25\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{OA}\ =\ \overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}\]
\[\overrightarrow{OB}\ =\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\]
\[\overrightarrow{OC}\ =\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ – (\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k})\]
\[=\ 2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k}\ -\ \overrightarrow{i}\ – 3\overrightarrow{j}\ – 2\overrightarrow{k}\]
\[\boxed{\overrightarrow{AB}\ =\ \overrightarrow{i}\ – 4\overrightarrow{j}\ -\ \overrightarrow{k}}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – (2\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ \overrightarrow{k})\]
\[=\ -\overrightarrow{i}\ + 2\overrightarrow{j}\ +\ 3\overrightarrow{k}\ – 2\overrightarrow{i}\ +\ \overrightarrow{j}\ -\overrightarrow{k}\]
\[\boxed{\overrightarrow{BC}\ =\ -\ 3\overrightarrow{i}\ +\ 3\overrightarrow{j}\ +\ 2\overrightarrow{k}}\]
\[\overrightarrow{AB}×\overrightarrow{BC} =\ \begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\
1 & -4 & -1\\
-3 & 3 & 2\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( -8 + 3)\ -\ \overrightarrow{j}(2 – 3)\ +\ \overrightarrow{k}(3 – 12)\]
\[ = \overrightarrow{i}(-5) -\overrightarrow{j}(-1)+\overrightarrow{k}(-9)\]
\[\boxed{\overrightarrow{AB}× \overrightarrow{BC}\ =\ -5\overrightarrow{i}\ +\ \overrightarrow{j}\ -\ 9\overrightarrow{k}}\]
\[|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-5 )^2\ +\ (1)^2\ + (-9)^2 }=\sqrt{25 + 1 + 81 }=\sqrt{107}\]
\[ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[\boxed{Area\ of \ triangle\ =\frac{\sqrt{107}}{2}\ sq. units}\]
\[\color {purple} {Example\ 33\ .}\ \color {red} {Find\ the\ area\ of\ the\ triangle}\ formed\ by\ the\ points\ whose\ position\ vectors\ \hspace{8cm}\]\[ 3\overrightarrow{i}\ +\ 2\overrightarrow{j}\ -\overrightarrow{k}\ , 2\overrightarrow{i}\ -\ 3\overrightarrow{j}\ +\ \overrightarrow{k}\ and\ 5\overrightarrow{i}\ + \overrightarrow{j}\ +\ 3\overrightarrow{k}\ \hspace{7cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{OA}= 3\overrightarrow{i}\ + 2\overrightarrow{j}- \overrightarrow{k}\]
\[\overrightarrow{OB}= 2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}\]
\[\overrightarrow{OC}= 5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}\]
\[\overrightarrow{AB} = \overrightarrow{OB}-\overrightarrow{OA}\]
\[=2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}- (3\overrightarrow{i} + 2\overrightarrow{j}- \overrightarrow{k})\]
\[=2\overrightarrow{i}\ -3\overrightarrow{j} +\overrightarrow{k}- 3\overrightarrow{i}- 2\overrightarrow{j}+ \overrightarrow{k}\]
\[\boxed{\overrightarrow{AB}= -\overrightarrow{i} – 5\overrightarrow{j} + 2\overrightarrow{k}}\]
\[\overrightarrow{BC} = \overrightarrow{OC}-\overrightarrow{OB}\]
\[=5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}- (2\overrightarrow{i} -3\overrightarrow{j} +\overrightarrow{k})\]
\[=5\overrightarrow{i}\ +\overrightarrow{j} + 3\overrightarrow{k}- 2\overrightarrow{i} + 3\overrightarrow{j} -\overrightarrow{k}\]
\[\boxed{\overrightarrow{BC}= 3\overrightarrow{i} + 4\overrightarrow{j} + 2\overrightarrow{k}}\]
\[\overrightarrow{AB}×\overrightarrow{BC} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} &\overrightarrow{k}\\
-1 & -5 & 2\\
3 & 4 & 2\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( -10 – 8) -\overrightarrow{j}(-2 – 6)+\overrightarrow{k}(-4 + 15)\]
\[ = \overrightarrow{i}(-18) -\overrightarrow{j}(8)+\overrightarrow{k}(11)\]
\[\boxed{\overrightarrow{AB}× \overrightarrow{BC}= -18\overrightarrow{i}\ +\ 8\overrightarrow{j}+11\overrightarrow{k}}\]
\[|\overrightarrow{AB} × \overrightarrow{BC}| = \sqrt{(-18 )^2 + (-8)^2 + (11)^2 }=\sqrt{(324 + 64 + 121 }=\sqrt{509}\]
\[ Area\ of \ triangle = \frac{1}{2}|\overrightarrow{AB} × \overrightarrow{BC}|\]
\[\boxed{Area\ of \ triangle\ =\frac{\sqrt{509}}{2}\ sq. units}\]
\[\color {purple} {Example\ 34\ .}\ \color {red} {Find\ the\ unit\ vector}\ perpendicular\ to\ each\ of\ the\ vectors\ 2\overrightarrow{i} -\overrightarrow{j}+\overrightarrow{k} and\ 3\overrightarrow{i}+ 4\overrightarrow{j} -\overrightarrow{k}.\]\[\color {red} {Also\ find\ the\ sine\ of\ the\ angle}\ between\ the\ vectors .\]
\[\overrightarrow{a}= 2\overrightarrow{i} -\overrightarrow{j}+\overrightarrow{k} \]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{b}=3\overrightarrow{i}+ 4\overrightarrow{j} -\overrightarrow{k} \]
\[\overrightarrow{a}×\overrightarrow{b} =\begin{vmatrix}
\overrightarrow{i} & \overrightarrow{j} & \overrightarrow{k}\\
2 & -1 & 1\\
3 & 4 & -1\\
\end{vmatrix}\]
\[ = \overrightarrow{i}( 1 – 4) -\overrightarrow{j}(-2 – 3)+\overrightarrow{k}(8 + 3)\]
\[ = \overrightarrow{i}(-3) -\overrightarrow{j}(-5)+\overrightarrow{k}(11)\]
\[\boxed{\overrightarrow{a}× \overrightarrow{b}= -3\overrightarrow{i} +5\overrightarrow{j}+11\overrightarrow{k}}\]
\[|\overrightarrow{a} × \overrightarrow{b}| = \sqrt{(-3 )^2 + (5)^2 + (11)^2 }=\sqrt{9 + 25 + 121}=\sqrt{155}\]
\[ n^\wedge =\frac{\overrightarrow{a}× \overrightarrow{b}}{|\overrightarrow{a} × \overrightarrow{b}|}= \frac{-3\overrightarrow{i}\ + 5\overrightarrow{j}+ 11\overrightarrow{k}}{\sqrt{155}} \]
\[\boxed{n^\wedge = \frac{-3\overrightarrow{i}\ + 5\overrightarrow{j}+ 11\overrightarrow{k}}{\sqrt{155}}}\]
\[\overrightarrow{|a|} = \sqrt{(2)^2 + (-1)^2 + (1)^2 }=\sqrt{(4 + 1 + 1 }=\sqrt{6}\]
\[\overrightarrow{|b|} = \sqrt{(3)^2 + (4)^2 + (-1)^2 }=\sqrt{(9 + 16 + 1 }=\sqrt{26}\]
\[\ sin\ \theta =\frac{|\overrightarrow{a} × \overrightarrow{b}|}{\overrightarrow{|a|}\overrightarrow{|b|}}= \frac{\sqrt{155}}{\sqrt{6}\sqrt{26}}\]
\[\boxed{sin\ \theta =\frac{\sqrt{155}}{\sqrt{6}\sqrt{26}}}\]
\[\color {purple} {Example\ 35\ .}\ \color {red} {Show\ that}\ 5\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 7\overrightarrow{k},\ 7\overrightarrow{i}\ -\ 8\overrightarrow{j}\ +\ 9\overrightarrow{k}\ and\ 3\overrightarrow{i}\ +\ 20\overrightarrow{j}\ +\ 5\overrightarrow{k}\ \color {red} {are\ coplanar}.\ \hspace{10cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 5\overrightarrow{i}\ +\ 6\overrightarrow{j}\ +\ 7\overrightarrow{k}\]
\[\overrightarrow{b}\ =\ 7\overrightarrow{i}\ -\ 8\overrightarrow{j}\ +\ 9\overrightarrow{k}\]
\[\overrightarrow{c}\ =\ 3\overrightarrow{i}\ +\ 20\overrightarrow{j}\ +\ 5\overrightarrow{k}\]
\[ [\overrightarrow{a}\ \overrightarrow{b} \overrightarrow{c}] =\begin{vmatrix}
5 & 6 & 7\\
7 & -8 & 9\\
3 & 20 & 5\\
\end{vmatrix}\]
\[=\ 5(-\ 40\ -\ 180)\ -\ 6(35\ -\ 27)\ +\ 7(140\ +\ 24)\]
\[=\ 5(-220)\ -\ 6(8)\ +\ 7(164)\]
\[=\ -\ 1100\ -\ 48\ +\ 1148\]
\[=\ 0\]
\[\color {purple} {Example\ 36\ .}\ \color {red} {Find\ the\ value\ of\ ‘m’\ so\ that\ the\ vectors}\ \hspace{10cm}\]\[2\ \overrightarrow{i}\ – \overrightarrow{j}\ +\ \overrightarrow{k},\ \overrightarrow{i}\ +\ 2\ \overrightarrow{j}\ -\ 3\ \overrightarrow{k}\ ,\ 3\overrightarrow{i}\ +\ m\ \overrightarrow{j}\ +\ 5\ \overrightarrow{k}\ coplanar\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}= 2\overrightarrow{i}- \overrightarrow{j}+ \overrightarrow{k}\]
\[\overrightarrow{b}= \overrightarrow{i}+ 2\overrightarrow{j}- 3\overrightarrow{k}\]
\[\overrightarrow{c}= 3\overrightarrow{i}+ m\overrightarrow{j}+ 5\overrightarrow{k}\]
\[Given\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c} are\ coplanar\ \implies [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0\]
\[\begin{vmatrix}
2 &- 1 & 1\\
1 & 2 & -3\\
3 & m & 5\\
\end{vmatrix}=0\]
\[2(10\ +\ 3\ m)\ +\ 1(5\ +\ 9)\ +\ 1(m\ -\ 6)\ =\ 0\]
\[20\ +\ 6m\ +\ 14\ +\ m\ -\ 6\ =\ 0\]
\[28\ +\ 7m\ =\ 0\]
\[7m\ =\ -\ 28\]
\[m\ =\ -\ 4\]
\[\color {purple} {Example\ 37\ .}\ \color {red} {Find\ the\ value\ of\ ‘p’\ such\ that\ the\ vectors}\ \hspace{10cm}\]\[2\ \overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 5\ \overrightarrow{k},\ p\ \overrightarrow{i}\ +\ 2\ \overrightarrow{j}\ -\ \ \overrightarrow{k}\ and\ \ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\ \overrightarrow{k}\ lie\ on\ the\ same\ plane\ \hspace{5cm}\]
\[\color {blue} {Soln:}\ \hspace{20cm}\]
\[\overrightarrow{a}\ =\ 2\ \overrightarrow{i}\ -\ 3\ \overrightarrow{j}\ +\ 5\ \overrightarrow{k}\]
\[\overrightarrow{b}\ =\ p\ \overrightarrow{i}\ +\ 2\ \overrightarrow{j}\ -\ \ \overrightarrow{k}\]
\[\overrightarrow{c}\ =\ 3\overrightarrow{i}\ -\ \overrightarrow{j}\ +\ 4\ \overrightarrow{k}\]
\[Given\ \overrightarrow{a},\overrightarrow{b} and\ \overrightarrow{c}\ lie\ on\ the\ same\ plane\ \implies\ [\overrightarrow{a}\ \overrightarrow{b}\ \overrightarrow{c}] = 0\]
\[\begin{vmatrix}
2 &- 3 & 5\\
p & 2 & -1\\
3 & -1 & 4\\
\end{vmatrix}=0\]
\[2(8\ -\ 1)\ +\ 3(4\ p\ +\ 3)\ +\ 5(-\ p\ -\ 6)\ =\ 0\]
\[14\ +\ 12\ p\ +\ 9\ +\ m\ -\ 5\ p\ -\ 30\ =\ 0\]
\[-\ 7\ +\ 7\ p\ =\ 0\]
\[7\ p\ =\ 7\]
\[p\ =\ 1\]
📚 Related Posts
Google ad