Vector Product (Part-4) | S N Dey

In the previous article we have discussed 11 Short Answer Type Questions and their solutions of Vector Product . In this article, we will discuss the solutions of Short Answer Type Questions (12-22) in the chapter Product of Two Vectors as given in the Chhaya Publication Book of aforementioned chapter of S N De book. To download S N dey mathematics class 12 PDF, you can check out here. So, without wasting time, let’s start.

Vector Product | S N Dey mathematics class 12 Solutions of Ex-2A

$12.~~$ Let $~~~\vec{a}=2\hat{i}-2\hat{j} +\hat{k},~~\vec{b}=\hat{j}-\hat{k}~~\text{and}~~ \vec{c}=-\hat{i}+3\hat{j} +2\hat{k}~~$ be three given vectors.

Find $~~(i)~~\vec{a} \times \vec{b}.$

Solution.

$\vec{a} \times \vec{b} \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2 & ~-2 &~1 \\ 0 &~1&~-1 \\ \end{vmatrix} \\~~\\= (2-1)\hat{i}-(-2-0)\hat{j}+(2-0)\hat{k} \\= \hat{i}+2\hat{j}+2\hat{k}~~\text{(ans.)}$

Find $~~(ii)~~ \vec{c} \times (-\vec{a})$

Solution.

$\vec{c}=-\hat{i}+3\hat{j} +2\hat{k},~\\~~ -\vec{a}=-2\hat{i}+2\hat{j}-\hat{k}$

$\vec{c} \times (-\vec{a}) \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ -1 & ~3 &~2 \\ -2 &~2&~-1 \\ \end{vmatrix} \\= (-3-4)\hat{i}-(1+4)\hat{j} +(-2+6)\hat{k} \\= -7\hat{i}-5\hat{j} +4\hat{k}~~\text{(ans.)}$

Find $~~(iii)~~ (\vec{a}-2\vec{b}) \times \vec{c}$

Solution.

$\vec{a}-2\vec{b} \\= (2\hat{i}-2\hat{j} +\hat{k})-2(\hat{j}-\hat{k}) \\= 2\hat{i}-4\hat{j}+3\hat{k} ,\\~~\\~ \vec{c}=-\hat{i}+3\hat{j} +2\hat{k}$

$(\vec{a}-2\vec{b}) \times \vec{c} \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2 & -4 &~3 \\ -1 &~3&~2 \\ \end{vmatrix} \\ = (-8-9)\hat{i}-(4+3)\hat{j} +(6-4)\hat{k} \\= -17\hat{i}-7\hat{j} +2\hat{k} ~~~~\text{(ans.)}$

Find $~~(iv)~~ (\vec{a}+\vec{b}) \times (\vec{b}-\vec{c})$

Solution.

$\vec{a}+\vec{b}=(2\hat{i}-2\hat{j} +\hat{k})+(\hat{j}-\hat{k})=2\hat{i}-\hat{j}, \\~~~ \vec{b}-\vec{c} =(\hat{j}-\hat{k})-(-\hat{i}+3\hat{j} +2\hat{k})=\hat{i}-2\hat{j}-3\hat{k}.$

$~\text{So,}~~ (\vec{a}+\vec{b}) \times (\vec{b}-\vec{c})\\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2 & ~-1 &~0 \\ 1 &~-2&~-3 \\ \end{vmatrix} \\ = (3-0)\hat{i}-(-6-0)\hat{j} +(-4+1)\hat{k} \\= 3\hat{i}+6\hat{j} -3\hat{k}~~\text{(ans.)}$

Find $~~(v)~~$ angle between $~~\vec{a}~~\text{and}~~\vec{b}.$

Solution.

$\vec{a} \cdot \vec{b}=(2\hat{i}-2\hat{j} +\hat{k}) \cdot (\hat{j}-\hat{k}) \\ \text{or,} ~~ \vec{a} \cdot \vec{b}=0-2-1=-3 \rightarrow(1) \\ ~~|\vec{a}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}=3\rightarrow(2), \\~~ |\vec{b}|=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\rightarrow(3)$

If $~~\theta~~$ be the angle between $~~\vec{a}~~\text{and}~~\vec{b},~$

$\cos\theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{-3}{3 \sqrt{2}}~~[\text{By (1),(2),(3)}] \\ \text{or,}~~ \cos\theta=-\frac{1}{\sqrt{2}}=\cos(3\pi/4) \\ \text{or,}~~\theta=\frac{3\pi}{4} ~~\text{(ans.)}$

Find $~~(vi)~~$ sine of the angle between $~~\vec{a}~~\text{and}~~\vec{c}.$

Solution.

We have $~~\vec{a}=2\hat{i}-2\hat{j} +\hat{k} ~~\text{and }~~\vec{c}=-\hat{i}+3\hat{j} +2\hat{k}.$

$\text{So,}~~ \vec{a} \times \vec{c}\\=\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2 & ~-2 &~1 \\ -1 &~3&~2 \\ \end{vmatrix} \\= (-4-3)\hat{i}-(4+1)\hat{j} +(6-2)\hat{k} \\= -7\hat{i}-5\hat{j} +4\hat{k}\\~~ \\~~\therefore~~ |\vec{a} \times \vec{c}| \\= \sqrt{(-7)^2+(-5)^2+4^2} \\= \sqrt{49+25+16}\\= \sqrt{90}\\=\sqrt{9 \times 10}\\=3\sqrt{10} ,\\~~\\~~ |\vec{a}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}=3 , \\~~\\~~ |\vec{c}|=\sqrt{(-1)^2+3^2+2^2}=\sqrt{14}.$

$~~\therefore~~ \sin\theta =\frac{|\vec{a} \times \vec{c}|}{|\vec{a}||\vec{c}|}=\frac{3\sqrt{10}}{3\sqrt{14}}=\sqrt{\frac{10}{14}}=\sqrt{\frac 57}~~\text{(ans.)}$

$13.~~~$ In each of the following find a unit vector perpendicular to both $~~\vec{a}~~\text{and}~~\vec{b}.$

$~(i)~~ \vec{a}=\hat{i}+\hat{j}~~\text{and}~~ \vec{b}=-\hat{i}+\hat{k}.$

Solution.

$\vec{a} \times \vec{b} \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 1 & ~1 &~0 \\ -1 &~0&~1 \\ \end{vmatrix} \\ = (1-0)\hat{i}-(1-0)\hat{j} +(0+1)\hat{k} \\= \hat{i}-\hat{j} +\hat{k} ,\\~~\\~\therefore~~|\vec{a} \times \vec{b}|=\sqrt{1^2+(-1)^2+1^2}=\sqrt{3}.$

So, a unit vector perpendicular to both $~~\vec{a}~~\text{and}~~\vec{b}~$ is :

$\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{1}{\sqrt{3}} (\hat{i}-\hat{j} +\hat{k})~~\text{(ans.)}$

$(ii)~~ \vec{a}=2\hat{i}-2\hat{j} +\hat{k} ~~\text{and}~~~ \vec{b}=\hat{i}+2\hat{j} -2\hat{k}$

Solution.

$\vec{a} \times \vec{b} \\ = \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2 & -2 &~1 \\ 1 &~2&-2 \\ \end{vmatrix} \\ = (4-2)\hat{i}-(-4-1)\hat{j} +(4+2)\hat{k} \\= 2\hat{i}+5\hat{j} +6\hat{k}, \\~~\\~~\therefore~~ |\vec{a} \times \vec{b}| \\ = \sqrt{2^2+5^2+6^2}\\= \sqrt{4+25+36} \\= \sqrt{65}$

So, a unit vector perpendicular to both $~~\vec{a}~~\text{and}~~\vec{b}~~$ is :

$\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{1}{\sqrt{65}}(2\hat{i}+5\hat{j} +6\hat{k})~~\text{(ans.)}$

$(iii)~~\vec{a}=(2,1,1)~~~\text{and}~~~\vec{b}=(1,-1,2)$

Solution.

$\vec{a} \times \vec{b}\\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2 & 1 &~1 \\ 1 &-1&2 \\ \end{vmatrix} \\ = (2+1)\hat{i}-(4-1)\hat{j} +(-2-1)\hat{k} \\= 3\hat{i}-3\hat{j} -3\hat{k} \\~~\\~\therefore~|\vec{a} \times \vec{b}| \\= \sqrt{3^2+(-3)^2+(-3)^2}\\=\sqrt{9+9+9}\\=\sqrt{27}\\= \sqrt{3 \times 3 \times 3}\\=3\sqrt{3}$

So, a unit vector perpendicular to both $~~\vec{a}~~\text{and}~~\vec{b}~~$ is :

$\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} =\frac{1}{3\sqrt{3}}(3\hat{i}-3\hat{j} -3\hat{k})~~\text{(ans.)}$

$14.~~$ If $~~\vec{a}=\hat{i}+\hat{j} -\hat{k},~~\vec{b}=2\hat{i}-2\hat{j} +\hat{k}~~\text{and}~~\vec{c}=3\hat{i}+2\hat{j} -2\hat{k}~~$ show that,

$(i)~~ \vec{a} \cdot (\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c} \\~~~(ii)~~ \vec{a} \times (\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a}\times\vec{c}.$

Solution. $~~(i)$

$\vec{b}+\vec{c}\\=(2\hat{i}-2\hat{j} +\hat{k})+(3\hat{i}+2\hat{j} -2\hat{k})\\=5\hat{i}-\hat{k} \\ ~~\\~~\therefore ~~\vec{a} \cdot (\vec{b}+\vec{c})\\=(\hat{i}+\hat{j} -\hat{k}) \cdot (5\hat{i}-\hat{k})\\=5+0+1\\=6~\rightarrow(1)\\~~\\~~ \vec{a} \cdot \vec{b}=(\hat{i}+\hat{j} -\hat{k}) \cdot (2\hat{i}-2\hat{j} +\hat{k})\\~\therefore~\vec{a} \cdot \vec{b} =2-2-1=-1 \rightarrow(2) \\~~\\~~~~ \vec{a} \cdot \vec{c}=(\hat{i}+\hat{j} -\hat{k}) \cdot (3\hat{i}+2\hat{j} -2\hat{k})\\~ \therefore~\vec{a} \cdot \vec{c}=3+2+2=7\rightarrow(3)$

So, by $~(2),(3)~~$ we get,

$\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=-1+7=6 \rightarrow(4)$

Hence , by $~(1),~(4)~~$ the result follows.

Solution. $~~(ii)$

$\vec{a} \times (\vec{b}+\vec{c}) \\ =\begin{vmatrix} \hat{i}&\hat{j} &\hat{k}\\ 1 & 1 & -1 \\ 5 &0&-1 \\ \end{vmatrix} \\ = (-1-0)\hat{i}-(-1+5)\hat{j} +(0-5)\hat{k} \\= -\hat{i}-4\hat{j} -5\hat{k}\rightarrow(5)$

$\vec{a} \times \vec{b} \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 1 & 1 & -1 \\ 2 &-2&1 \\ \end{vmatrix}\\=(1-2)\hat{i}-(1+2)\hat{j} +(-2-2)\hat{k}\\=-\hat{i}-3\hat{j} -4\hat{k}\rightarrow(6)$

$\vec{a} \times \vec{c}\\=\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 1 & 1 & -1 \\ 3 &2&-2 \\ \end{vmatrix}\\=(-2+2)\hat{i}-(-2+3)\hat{j} +(2-3)\hat{k}\\=- \hat{j} -\hat{k} \rightarrow(7)$

So, by $~(6),(7)~~$ we get,

$\vec{a} \times \vec{b}+\vec{a} \times \vec{c}\\=(-\hat{i}-3\hat{j} -4\hat{k}) +(-\hat{j} -\hat{k}) \\= -\hat{i}-4\hat{j} -5\hat{k} \rightarrow(8)$

Hence , by $~(5),~(8)~~$ the result follows.

$15.~~$ By vector method show that the points $~~(2,-3,4),~(-2,1,0)~~\text{and}~~ (1,-2,3)~~$ are collinear.

Solution.

$\text{let}~~A=(2,-3,4)=2\hat{i}-3\hat{j} +4\hat{k},\\~~~B=(-2,1,0)=-2\hat{i}+\hat{j},\\~~~C=(1,-2,3)=\hat{i}-2\hat{j} +3\hat{k}.$

$\vec{AB}=(-2\hat{i}+\hat{j})-(2\hat{i}-3\hat{j} +4\hat{k})=-4\hat{i}+4\hat{j}-4\hat{k}\rightarrow(1),\\~~~~\vec{BC}=(\hat{i}-2\hat{j} +3\hat{k})-(-2\hat{i}+\hat{j})=3\hat{i}-3\hat{j}+3\hat{k}\rightarrow(2)$

Hence, by $~(1),~(2)~~$ we get,

$\vec{BC}=-3(-\hat{i}+\hat{j}-\hat{k}), ~~~\vec{AB}=4(-\hat{i}+\hat{j}-\hat{k})\\~~\therefore ~\frac{\vec{BC}}{\vec{AB}}=-\frac 34 \Rightarrow \vec{AB}=-\frac 43 \vec{BC} \rightarrow(3)$

So, by $~(3)~$ we can conclude that three given points are collinear.

$16.~~$ If $~~\vec{\alpha}=2\hat{i}+\hat{j} -3\hat{k}~~\text{and}~~ \vec{\beta}=\hat{i}-2\hat{j} +\hat{k},~~$ find a vector of magnitude $~~5~~$ perpendicular to both $~~\vec{\alpha}~~\text{and}~~\vec{\beta}.$

Solution.

$\vec{\beta} \times \vec{\alpha}\\=\begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 1 &-2&1 \\ 2 & 1 & -3 \\ \end{vmatrix} \\= (6-1) \hat{i}-(-3-2)\hat{j} +(1+4)\hat{k} \\= 5\hat{i}+5\hat{j} +5\hat{k}\\=5(\hat{i}+\hat{j} +\hat{k} )$

So, a vector of magnitude $~~5~~$ perpendicular to both $~~\vec{\alpha}~~\text{and}~~\vec{\beta}~~$ is :

$5 \cdot \frac{\vec{\beta} \times \vec{\alpha}}{|\vec{\beta} \times \vec{\alpha}|}\\=5 \cdot \frac{5(\hat{i}+\hat{j} +\hat{k} )}{\sqrt{5^2+5^2+5^2}}\\=5 \cdot \frac{5(\hat{i}+\hat{j} +\hat{k} )}{\sqrt{25 \times 3}}\\=5 \cdot \frac{5(\hat{i}+\hat{j} +\hat{k} )}{5\sqrt{3}}\\=\frac{5}{\sqrt{3}}(\hat{i}+\hat{j} +\hat{k})~~\text{(ans.)}$

$17.~~$ If $~~\vec{a}+\vec{b}+\vec{c}=\vec{0}~~$ and $~~|\vec{a}|=6,~~|\vec{b}|=4~~\text{and}`~|\vec{c}|=3,~~$ find the cosine of the angle between the vectors $~~\vec{b}~~\text{and}~~\vec{c}.$

Solution.

$\vec{a}+\vec{b}+\vec{c}=\vec{0} \\ \text{or,}~~ \vec{b}+\vec{c} =-\vec{a} \\ \text{or,}~~ (\vec{b}+\vec{c}) \cdot (\vec{b}+\vec{c}) =(-\vec{a}) \cdot (-\vec{a}) \\ \text{or,}~~\vec{b}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{b}+\vec{c}\cdot \vec{c}=|\vec{a}|^2 \\ \text{or,}~~ |\vec{b}|^2+2\vec{b} \cdot \vec{c}+|\vec{c}|^2=|\vec{a}|^2 \\ \text{or,}~~ 4^2+2\vec{b} \cdot \vec{c}+3^2=6^2 \\ \text{or,}~~ 25+2\vec{b} \cdot \vec{c}=36 \\ \text{or,}~~ 2\vec{b} \cdot \vec{c} =36-25 \\ \text{or,}~~ \vec{b} \cdot \vec{c} =\frac{11}{2} \rightarrow(1)$

If $~~\theta~~$ is the angle between $~~\vec{b} ~~\text{and}~~ \vec{c}~~$ then

$\cos\theta=\frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\vec{c}|}=\frac{11/2}{4 \times 3} \\ \text{or,}~~ \cos\theta=\frac{11}{2} \times \frac{1}{12}=\frac{11}{24} ~~\text{(ans.)}$

$18.~~$ Prove that,

$(i)~~(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})=2(\vec{a} \times \vec{b})$

Solution.

$(\vec{a}-\vec{b}) \times (\vec{a}+\vec{b})\\=\vec{a} \times \vec{a}+\vec{a} \times \vec{b}-\vec{b} \times \vec{a}-\vec{b} \times \vec{b}\\=0+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}-0~~[*]\\=2(\vec{a} \times \vec{b})~~\text{(proved)}$

Note [*] : $~~\vec{a} \times \vec{b}=-\vec{b} \times \vec{a}$

$(ii)~~(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2=|\vec{a}|^2 |\vec{b}|^2$

Solution.

$\vec{a} \times \vec{b} =|\vec{a}||\vec{b}|\sin\theta ,~~\vec{a} \cdot \vec{b}= |\vec{a}||\vec{b}|\cos\theta$

$\text{So,}~~(\vec{a} \times \vec{b})^2+(\vec{a} \cdot \vec{b})^2 \\= |\vec{a}|^2|\vec{b}|^2\sin^2\theta+|\vec{a}|^2|\vec{b}|^2\cos^2\theta \\= |\vec{a}|^2|\vec{b}|^2 (\sin^2\theta+\cos^2\theta) \\= |\vec{a}|^2|\vec{b}|^2~~\text{(proved)}$

$(iii)~~ \vec{a} \times (\vec{b}+\vec{c})+\vec{b} \times (\vec{c}+\vec{a})+\vec{c} \times (\vec{a}+\vec{b})=\vec{0}.$

Solution.

$\vec{a} \times (\vec{b}+\vec{c})+\vec{b} \times (\vec{c}+\vec{a})+\vec{c} \times (\vec{a}+\vec{b})\\=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}+\vec{b} \times \vec{a}+\vec{c} \times \vec{a}+\vec{c} \times \vec{b}\\=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}-\vec{a} \times \vec{b}-\vec{a} \times \vec{c}-\vec{b} \times \vec{c}\\=0~~\text{(proved)}$

$(iv)~~$ Given that $~~\vec{a} \cdot \vec{b}=0~~$ and $~~\vec{a} \times \vec{b}=\vec{0}.~~$ What can you conclude about the vector $~~\vec{a}~~\text{and}~~\vec{b}~$ ?

Solution.

If possible let $~~|\vec{a}| \neq 0~~~\text{and}~~|\vec{b}| \neq 0.$

$\text{So,}~~\vec{a} \cdot \vec{b}=0 \\ \text{or,}~~ |\vec{a}||\vec{b}| \cos\theta=0 \\ \text{or,}~~\cos\theta=0=\cos(\pi/2)~~[\because~~|\vec{a}| \neq 0~,|\vec{b}| \neq 0] \\ \text{or,}~~ \theta=\frac{\pi}{2}\rightarrow(1)$

$\text{Now,}~~ \vec{a} \times \vec{b}=\vec{0} ~~(\text{given}) \\ \text{or,} ~~ |\vec{a}||\vec{b}| \sin\theta=0 \\ \text{or,}~~ \sin\theta=0 \\ \text{or,}~~\theta=0\rightarrow(2)$

Hence , the value of $~~\theta~~$ obtained from $~~(1)~~$ and $~~(2)~~$ are contradictory to each other and so our assumption is wrong and so either $~~|\vec{a}|=0~~\text{or,}~~ |\vec{b}|=0.$

$19.~~$ If $~~\vec{a}+\vec{b}+\vec{c}=\vec{0},~~$ show that, $~~\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}.$

Solution.

$\vec{a}+\vec{b}+\vec{c} =\vec{0} \\ \text{or,}~~ \vec{a}\times (\vec{a}+\vec{b}+\vec{c})=\vec{a} \times \vec{0} \\ \text{or,}~~ \vec{a} \times \vec{a}+\vec{a} \times \vec{b}+\vec{a} \times \vec{c}=\vec{0} \\ \text{or,}~~ \vec{0}+ \vec{a} \times \vec{b}+\vec{a} \times \vec{c}=\vec{0} \\ \text{or,}~~ \vec{a} \times \vec{b}=-\vec{a} \times \vec{c} \\ \text{or,}~~ \vec{a} \times \vec{b}=\vec{c} \times \vec{a}\rightarrow(1)$

$\vec{a}+\vec{b}+\vec{c}=\vec{0} \\ \text{or,}~~ \vec{b}\times (\vec{a}+\vec{b}+\vec{c})=\vec{b} \times \vec{0} \\ \text{or,}~~ \vec{b} \times \vec{a}+\vec{b} \times \vec{b}+\vec{b} \times \vec{c}=\vec{0} \\ \text{or,}~~ \vec{b} \times \vec{a}+\vec{0}+\vec{b} \times \vec{c}=\vec{0} \\ \text{or,}~~ -\vec{a} \times \vec{b}=-\vec{b} \times \vec{c} \\ \text{or,}~~ \vec{a} \times \vec{b}=\vec{b} \times \vec{c}\rightarrow(2)$

Hence by $~~(1)~~\text{and}~~(2),~~$ we get,

$\vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a}.$

$20.~~$ Find the value of $~~[(\vec{k} \times \vec{j})\cdot \vec{i}+\vec{j} \cdot \vec{k}]$

Solution.

We know, $~~~~~\vec{k} \times \vec{j}=-\hat{i},~~\vec{j} \cdot \vec{k}=0 \rightarrow(1)$

$[(\vec{k} \times \vec{j})\cdot \vec{i}+\vec{j} \cdot \vec{k}]\\=[-\hat{i} \cdot \hat{i}+0]~~[\text{By (1)}]\\=-1+0 =-1~~~\text{(ans.)}$

$22(i)~~$ Find a unit vector perpendicular to both the vectors $~~\vec{a}=2\hat{i}+\hat{j} -2\hat{k}~~$ and $~~ \vec{b}=3\hat{i}-\hat{j} +\hat{k}.$

Solution.

$\vec{a} \times \vec{b} \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 2 & 1 &-2 \\ 3 &-1&1 \\ \end{vmatrix} \\= (1-2)\hat{i}-(2+6)\hat{j} +(-2-3)\hat{k} \\= -\hat{i}-8\hat{j} -5\hat{k}, \\~~\\~~\therefore~~|\vec{a} \times \vec{b}| \\= \sqrt{(-1)^2+(-8)^2+(-5)^2} \\= \sqrt{1+64+25} \\= \sqrt{90}\\= \sqrt{9 \times 10} \\= 3\sqrt{10}.$

Clearly, $~~\vec{a} \times \vec{b}~~$ is a vector which is perpendicular to both $~~\vec{a}~~\text{and}~~\vec{b}.$

So, a unit vector perpendicular to both the given vectors is given by :

$\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{1}{3\sqrt{10}}(-\hat{i}-8\hat{j} -5\hat{k})~~\text{(ans.)}$

$(ii)~~$ Find a unit vector perpendicular to each of the vectors $~~\vec{a}+\vec{b}~~\text{and}~~ \vec{a}-\vec{b}~~$ where $~~\vec{a}=3\hat{i}+2\hat{j} +2\hat{k} ~~\text{and}~~ \vec{b}=\hat{i}+2\hat{j} -2\hat{k}.$

Solution.

$(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) \\= \vec{a} \times \vec{a}-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}-\vec{b} \times \vec{b} \\= 0-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}-0 \\= -\vec{a} \times \vec{b}-\vec{a} \times \vec{b} \\= -2(\vec{a} \times \vec{b})\rightarrow(1)$

$~\therefore~~ \vec{a} \times \vec{b} \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ 3 & 2 &~2 \\ 1 &~2&-2 \\ \end{vmatrix} \\= (-4-4)\hat{i}-(-6-2)\hat{j} +(6-2)\hat{k} \\= -8\hat{i}+8\hat{j} +4\hat{k}\rightarrow(2)$

From $~~(1)~~\text{and}~~(2),~~$ we get

$(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b}) \\ =-2 \times(-8\hat{i}+8\hat{j} +4\hat{k}) \\= 16\hat{i}-16\hat{j} -8\hat{k}\\=2(8\hat{i}-8\hat{j} -4\hat{k})$

So, a unit vector perpendicular to each of the vectors

$\vec{a}+\vec{b}~~\text{and}~~ \vec{a}-\vec{b}~~$ is given by :

$\pm\frac{(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})}{|(\vec{a}+\vec{b}) \times (\vec{a}-\vec{b})|}\\=\pm \frac{2(8\hat{i}-8\hat{j} -4\hat{k})}{2\sqrt{8^2+(-8)^2+(-4)^2}}\\=\pm \frac{8\hat{i}-8\hat{j} -4\hat{k}}{\sqrt{144}}\\=\pm \frac{8\hat{i}-8\hat{j} -4\hat{k}}{12}~~\text{(ans.)}$

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Vector Product (Part-4) | S N Dey | Class 12

Vector Product | S N Dey mathematics class 12 Solutions of Ex-2A

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