Class 12: Algebra of Vectors

$\displaystyle \textbf{Question 1: }\text{If the position vector of a point }(-4,-3)\text{ be }\vec{a},\text{ find }\lvert\vec{a}\rvert.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given a point }(-4,-3)\text{ such that its position vector }\overrightarrow{a} \\ \text{is given by }\overrightarrow{a}=-4\widehat{i}-3\widehat{j}$
$\displaystyle \text{Then,}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(-4)^{2}+(-3)^{2}}=\sqrt{16+9}=\sqrt{25}=5$

$\displaystyle \textbf{Question 2: }\text{If the position vector }\vec{a}\text{ of a point }(12,n)\text{ is such that }\lvert\vec{a}\rvert=13, \\ \text{ find the value(s) of }n.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given a position vector }\overrightarrow{a}\text{ of a point }(12,n)\text{ such that }\overrightarrow{a}=12\widehat{i}+n\widehat{j}. \\ \text{Then, }|\overrightarrow{a}|=\sqrt{12^{2}+n^{2}}$
$\displaystyle \text{Also, }|\overrightarrow{a}|=13\text{ (given)}$
$\displaystyle \text{Thus, we get,}$
$\displaystyle \sqrt{12^{2}+n^{2}}=13$
$\displaystyle \Rightarrow 12^{2}+n^{2}=169$
$\displaystyle \Rightarrow n^{2}=169-144$
$\displaystyle \Rightarrow n^{2}=25$
$\displaystyle \Rightarrow n=\pm 5$

$\displaystyle \textbf{Question 3: }\text{Find a vector of magnitude }4\text{ units which is parallel to the vector } \\ \sqrt{3}\,\hat{i}+\hat{j}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\sqrt{3}\widehat{i}+\widehat{j}$
$\displaystyle \text{Then,}$
$\displaystyle |\overrightarrow{a}|=\sqrt{(\sqrt{3})^{2}+1}=\sqrt{3+1}=\sqrt{4}=2$
$\displaystyle \text{A unit vector parallel to }\overrightarrow{a}\text{ is }\widehat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{1}{2}(\sqrt{3}\widehat{i}+\widehat{j})$
$\displaystyle \text{Hence, Required vector }=4\widehat{a}=4\times\frac{1}{2}(\sqrt{3}\widehat{i}+\widehat{j})=2\sqrt{3}\widehat{i}+2\widehat{j}$

$\displaystyle \textbf{Question 4: }\text{Express }\overrightarrow{AB}\text{ in terms of unit vectors }\hat{i}\text{ and }\hat{j},\text{ when the points are:}\\ \text{(i) }A(4,-1),\ B(1,3)\qquad \text{(ii) }A(-6,3),\ B(-2,-5).\\ \text{Find }\lvert\overrightarrow{AB}\rvert\text{ in each case.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Given: }A(4,-1)\text{ and }B(1,3)$
$\displaystyle \text{Then the position vector }\overrightarrow{AB}\text{ is given by }$
$\displaystyle \overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A=(\widehat{i}+3\widehat{j})-(4\widehat{i}-\widehat{j})$
$\displaystyle =\widehat{i}+3\widehat{j}-4\widehat{i}+\widehat{j}$
$\displaystyle =-3\widehat{i}+4\widehat{j}$
$\displaystyle \text{So,}$
$\displaystyle |\overrightarrow{AB}|=\sqrt{(-3)^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5$
$\displaystyle \text{(ii) }$
$\displaystyle \text{Given: }A(-6,3)\text{ and }B(-2,-5)$
$\displaystyle \text{Then, the position vector }\overrightarrow{AB}\text{ is given by }$
$\displaystyle \overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A=(-2\widehat{i}-5\widehat{j})-(-6\widehat{i}+3\widehat{j})$
$\displaystyle =-2\widehat{i}-5\widehat{j}+6\widehat{i}-3\widehat{j}$
$\displaystyle =4\widehat{i}-8\widehat{j}$
$\displaystyle \text{So,}$
$\displaystyle |\overrightarrow{AB}|=\sqrt{4^{2}+(-8)^{2}}=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}$

$\displaystyle \textbf{Question 5: }\text{Find the coordinates of the tip of the position vector which is equivalent} \\ \text{to }\overrightarrow{AB}, \text{ where the coordinates of }A\text{ and }B\text{ are }(-1,3)\text{ and }(-2,1)\text{ respectively.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }O\text{ be the origin. Let }P(x,y)\text{ be the required point. Then, } \\ \overrightarrow{P}\text{ is the tip of the position vector }\overrightarrow{OP}\text{ of the point }P.$
$\displaystyle \text{We have,}$
$\displaystyle \overrightarrow{OP}=x\widehat{i}+y\widehat{j}$
$\displaystyle \text{and, }\overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A$
$\displaystyle =(-2\widehat{i}+\widehat{j})-(-\widehat{i}+3\widehat{j})$
$\displaystyle =-2\widehat{i}+\widehat{j}+\widehat{i}-3\widehat{j}$
$\displaystyle =-\widehat{i}-2\widehat{j}$
$\displaystyle \text{Given that }\overrightarrow{OP}=\overrightarrow{AB}$
$\displaystyle \text{So, }x\widehat{i}+y\widehat{j}=-\widehat{i}-2\widehat{j}\Rightarrow x=-1,y=-2$
$\displaystyle \text{Hence, coordinates of the required point is }(-1,-2)$

$\displaystyle \textbf{Question 6: }ABCD\text{ is a parallelogram. If the coordinates of }A,B,C\text{ are }(-2,-1),(3,0) \\ \text{ and }(1,-2)\text{ respectively, find the coordinates of }D.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the coordinates of }D\text{ be }(x,y).$
$\displaystyle \text{ABCD is a parallelogram.}$
$\displaystyle AB=DC$
$\displaystyle \text{We have,}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{DC}$
$\displaystyle \Rightarrow 3\widehat{i}-(-2\widehat{i}-\widehat{j})=(\widehat{i}-2\widehat{j})-(x\widehat{i}+y\widehat{j})$
$\displaystyle \Rightarrow 5\widehat{i}+\widehat{j}=\widehat{i}(1-x)+\widehat{j}(-2-y)$
$\displaystyle \Rightarrow 1-x=5\text{ and }-2-y=1$
$\displaystyle \Rightarrow x=-4\text{ and }y=-3$
$\displaystyle \text{Hence, the coordinates of }D\text{ is }(-4,-3)$

$\displaystyle \textbf{Question 7: }\text{If the position vectors of the points }A(3,4),\ B(5,-6)\text{ and }C(4,-1) \\ \text{ are }\vec{a},\vec{b},\vec{c}\text{ respectively, compute }\vec{a}+2\vec{b}-3\vec{c}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ be the position vectors of the points }A(3,4),B(5,-6)\text{ and }C(4,-1)$
$\displaystyle \text{Then,}$
$\displaystyle \overrightarrow{a}=3\widehat{i}+4\widehat{j}$
$\displaystyle \overrightarrow{b}=5\widehat{i}-6\widehat{j}$
$\displaystyle \overrightarrow{c}=4\widehat{i}-\widehat{j}$
$\displaystyle \text{Therefore,}$
$\displaystyle \overrightarrow{a}+2\overrightarrow{b}-3\overrightarrow{c}$
$\displaystyle =3\widehat{i}+4\widehat{j}+2(5\widehat{i}-6\widehat{j})-3(4\widehat{i}-\widehat{j})$
$\displaystyle =3\widehat{i}+4\widehat{j}+10\widehat{i}-12\widehat{j}-12\widehat{i}+3\widehat{j}$
$\displaystyle =\widehat{i}-5\widehat{j}$

$\displaystyle \textbf{Question 8: }\text{If }\vec{a}\text{ be the position vector whose tip is }(5,-3),\text{ find the coordinates } \\ \text{of a point }B\text{ such that }\overrightarrow{AB}=\vec{a},\text{ the coordinates of }A\text{ being }(4,-1).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }O\text{ be the origin and let }P(5,-3)\text{ be the tip of the position vector }\overrightarrow{a}.$
$\displaystyle \text{Then, }\overrightarrow{a}=\overrightarrow{OP}=5\widehat{i}-3\widehat{j}.$
$\displaystyle \text{Let the coordinates of }B\text{ be }(x,y)\text{ and }A\text{ have coordinates }(4,-1).$
$\displaystyle \text{Therefore,}$
$\displaystyle \overrightarrow{AB}=\text{Position vector of }B-\text{Position vector of }A$
$\displaystyle =(x\widehat{i}+y\widehat{j})-(4\widehat{i}-\widehat{j})$
$\displaystyle =(x-4)\widehat{i}+(y+1)\widehat{j}$
$\displaystyle \text{Now,}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{a}$
$\displaystyle \Rightarrow (x-4)\widehat{i}+(y+1)\widehat{j}=5\widehat{i}-3\widehat{j}$
$\displaystyle \Rightarrow x-4=5\text{ and }y+1=-3$
$\displaystyle \Rightarrow x=9\text{ and }y=-4$
$\displaystyle \text{Hence, the coordinates of }B\text{ are }(9,-4).$

$\displaystyle \textbf{Question 9: }\text{Show that the points }2\hat{i},\ -\hat{i}-4\hat{j}\text{ and }-\hat{i}+4\hat{j}\text{ form an isosceles triangle.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given: The points }A,B,C\text{ with position vectors }\overrightarrow{a},\overrightarrow{b},\overrightarrow{c}\text{ respectively.}$
$\displaystyle \text{Also,}$
$\displaystyle \overrightarrow{a}=2\widehat{i}$
$\displaystyle \overrightarrow{b}=-\widehat{i}-4\widehat{j}$
$\displaystyle \overrightarrow{c}=-\widehat{i}+4\widehat{j}$
$\displaystyle \text{Then,}$
$\displaystyle \overrightarrow{AB}=\overrightarrow{b}-\overrightarrow{a}$
$\displaystyle \Rightarrow \overrightarrow{AB}=(-\widehat{i}-4\widehat{j})-2\widehat{i}$
$\displaystyle \Rightarrow \overrightarrow{AB}=-3\widehat{i}-4\widehat{j}$
$\displaystyle \text{Now, }|\overrightarrow{AB}|=\sqrt{(-3)^{2}+(-4)^{2}}=\sqrt{9+16}=\sqrt{25}=5$
$\displaystyle \overrightarrow{BC}=\overrightarrow{c}-\overrightarrow{b}$
$\displaystyle \Rightarrow \overrightarrow{BC}=(-\widehat{i}+4\widehat{j})-(-\widehat{i}-4\widehat{j})$
$\displaystyle \Rightarrow \overrightarrow{BC}=-\widehat{i}+4\widehat{j}+\widehat{i}+4\widehat{j}$
$\displaystyle \Rightarrow \overrightarrow{BC}=8\widehat{j}$
$\displaystyle \text{and}$
$\displaystyle \overrightarrow{AC}=\overrightarrow{c}-\overrightarrow{a}$
$\displaystyle \Rightarrow \overrightarrow{AC}=(-\widehat{i}+4\widehat{j})-2\widehat{i}$
$\displaystyle \Rightarrow \overrightarrow{AC}=-3\widehat{i}+4\widehat{j}$
$\displaystyle \text{Now, }|\overrightarrow{AC}|=\sqrt{(-3)^{2}+4^{2}}=\sqrt{9+16}=\sqrt{25}=5$
$\displaystyle \text{Since, the magnitude of }AB\text{ and }AC\text{ is equal.}$
$\displaystyle \text{Hence, the points }2\widehat{i},-\widehat{i}-4\widehat{j}\text{ and }-\widehat{i}+4\widehat{j}\text{ form an isosceles triangle.}$

$\displaystyle \textbf{Question 10: }\text{Find a unit vector parallel to the vector }\hat{i}+\sqrt{3}\,\hat{j}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }\overrightarrow{a}=\widehat{i}+\sqrt{3}\widehat{j}$
$\displaystyle \text{Then,}$
$\displaystyle |\overrightarrow{a}|=\sqrt{1^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=\sqrt{4}=2$
$\displaystyle \text{Unit vector parallel to }\overrightarrow{a}\text{ is }$
$\displaystyle \hat{a}=\frac{\overrightarrow{a}}{|\overrightarrow{a}|}=\frac{1}{2}(\widehat{i}+\sqrt{3}\widehat{j})=\frac{1}{2}\widehat{i}+\frac{\sqrt{3}}{2}\widehat{j}$

$\displaystyle \textbf{Question 11: }\text{The position vectors of points }A,B\text{ and }C\text{ are } \\ \lambda\hat{i}+3\hat{j},\ 12\hat{i}+\mu\hat{j} \text{ and }11\hat{i}-3\hat{j}\text{ respectively. If }C\text{ divides the line segment }\text{joining } \\ A\text{ and }B \text{ in the ratio }3:1,\text{ find the values of }\lambda\text{ and }\mu.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The position vectors of points }A,B\text{ and }C\text{ are }\lambda\widehat{i}+3\widehat{j},\,12\widehat{i}+\mu\widehat{j}\text{ and }11\widehat{i}-3\widehat{j}\text{ respectively.}$
$\displaystyle \text{It is given that }C\text{ divides the line segment joining }A\text{ and }B\text{ in the ratio }3:1.$
$\displaystyle 11\widehat{i}-3\widehat{j}=\frac{3(12\widehat{i}+\mu\widehat{j})+1(\lambda\widehat{i}+3\widehat{j})}{3+1}$
$\displaystyle \Rightarrow 11\widehat{i}-3\widehat{j}=\frac{(36+\lambda)\widehat{i}+(3\mu+3)\widehat{j}}{4}$
$\displaystyle \Rightarrow 44\widehat{i}-12\widehat{j}=(36+\lambda)\widehat{i}+(3\mu+3)\widehat{j}$
$\displaystyle \text{Equating the corresponding components, we get }36+\lambda=44$
$\displaystyle \Rightarrow \lambda=44-36=8$
$\displaystyle \text{and}$
$\displaystyle 3\mu+3=-12$
$\displaystyle \Rightarrow 3\mu=-15$
$\displaystyle \Rightarrow \mu=-5$
$\displaystyle \text{Thus, the values of }\lambda\text{ and }\mu\text{ are }8\text{ and }-5\text{ respectively.}$

$\displaystyle \textbf{Question 12: }~\text{Find the components along the coordinate axes of the position } \\ \text{vector of each of the following points:}\\ \text{(i) }P(3,2)\qquad \text{(ii) }Q(-5,1)\qquad \text{(iii) }R(-11,-9)\qquad \text{(iv) }S(4,-3).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(i) }$
$\displaystyle \text{Let }O\text{ be the origin.}$
$\displaystyle \text{The position vector of point }P(3,2)\text{ is}$
$\displaystyle \overrightarrow{OP}=3\widehat{i}+2\widehat{j}$
$\displaystyle \text{Component of }\overrightarrow{OP}\text{ along the }x\text{-axis is a vector of magnitude }3\text{ having its direction along} \\ \text{the positive direction of the }x\text{-axis.}$
$\displaystyle \text{Component of }\overrightarrow{OP}\text{ along the }y\text{-axis is a vector of magnitude }2\text{ having its direction along} \\ \text{the positive direction of the }y\text{-axis.}$
$\displaystyle \text{(ii) }$
$\displaystyle \text{The position vector of point }Q(-5,1)\text{ is}$
$\displaystyle \overrightarrow{OQ}=-5\widehat{i}+\widehat{j}$
$\displaystyle \text{Component of }\overrightarrow{OQ}\text{ along the }x\text{-axis is a vector of magnitude }5\text{ having its direction along} \\ \text{the negative direction of the }x\text{-axis.}$
$\displaystyle \text{Component of }\overrightarrow{OQ}\text{ along the }y\text{-axis is a vector of magnitude }1\text{ having its direction along} \\ \text{the positive direction of the }y\text{-axis.}$
$\displaystyle \text{(iii) }$
$\displaystyle \text{The position vector of point }R(-11,-9)\text{ is}$
$\displaystyle \overrightarrow{OR}=-11\widehat{i}-9\widehat{j}$
$\displaystyle \text{Component of }\overrightarrow{OR}\text{ along the }x\text{-axis is a vector of magnitude }11\text{ having its direction along} \\ \text{the negative direction of the }x\text{-axis.}$
$\displaystyle \text{Component of }\overrightarrow{OR}\text{ along the }y\text{-axis is a vector of magnitude }9\text{ having its direction along} \\ \text{the negative direction of the }y\text{-axis.}$
$\displaystyle \text{(iv) }$
$\displaystyle \text{The position vector of point }S(4,-3)\text{ is}$
$\displaystyle \overrightarrow{OS}=4\widehat{i}-3\widehat{j}$
$\displaystyle \text{Component of }\overrightarrow{OS}\text{ along the }x\text{-axis is a vector of magnitude }4\text{ having its direction along} \\ \text{the positive direction of the }x\text{-axis.}$
$\displaystyle \text{Component of }\overrightarrow{OS}\text{ along the }y\text{-axis is a vector of magnitude }3\text{ having its direction along} \\ \text{the negative direction of the }y\text{-axis.}$