Vector Product (Part-2) | S N Dey

In the previous article we discussed Very Short Answer Type Questions from question no. 1 to 10 from the Chapter Product of Two Vectors of Chhaya Mathematics (Class XII) or alternatively S N Dey mathematics class 12 . In this article, we will solve few more problems . Here we will solve few more problem related to Vector Product . So, let's start.

11(i) Find the projection of the vector $~~(\vec{b}+\vec{c})~~$ on vector $~~\vec{a}~~$ where $~~\vec{a}=2\hat{i}-2\hat{j}+\hat{k}~,~~\vec{b}=\hat{i}+2\hat{j}-2\hat{k}~~$ and $~~ \vec{c}=2\hat{i}-\hat{j}+4\hat{k}.$

Solution.

$~ \vec{b}+\vec{c}\\=(\hat{i}+2\hat{j}-2\hat{k})+(2\hat{i}-\hat{j}+4\hat{k})\\=3\hat{i}+\hat{j}+2\hat{k}\\~~\\~~(\vec{b}+\vec{c}) \cdot \vec{a}\\=(3\hat{i}+\hat{j}+2\hat{k}) \cdot (2\hat{i}-2\hat{j}+\hat{k})\\=6-2+2\\=6\\~~\\~~|\vec{a}|=\sqrt{2^2+(-2)^2+1^2}=\sqrt{9}=3$

So, the projection of the vector $~~(\vec{b}+\vec{c})~~$ on vector $~~\vec{a}~~$ is given by :

$=\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac 63=2~~\text{unit}$

11(ii) Find the projection of the vector $~~(\vec{b}+\vec{c})~~$ on vector $~~\vec{a}~~$ where $~~\vec{a}=\hat{i}+2\hat{j}+\hat{k}~,~~\vec{b}=\hat{i}+3\hat{j}+\hat{k}~~$ and $~~\vec{c}=\hat{i}+\hat{k}.$

Solution.

$ \vec{b}+\vec{c}\\=(\hat{i}+3\hat{j}+\hat{k})+(\hat{i}+\hat{k})\\=2\hat{i}+3\hat{j}+2\hat{k} \\~~\\~~(\vec{b}+\vec{c}) \cdot \vec{a}\\=(2\hat{i}+3\hat{j}+2\hat{k}) \cdot (\hat{i}+2\hat{j}+\hat{k})\\=2+6+2\\=10\\~~\\~~|\vec{a}|=\sqrt{1^2+2^2+1^2}=\sqrt{6}$

So, the projection of the vector $~~(\vec{b}+\vec{c})~~$ on vector $~~\vec{a}~~$ is given by :

$=\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|}=\frac{10}{\sqrt{6}}=\frac{10\sqrt{6}}{6}=\frac{5\sqrt{6}}{3}~~\text{unit}$

12(i) If $~~|\vec{a}|~=\sqrt{3},~|\vec{b}|=2~~$ and $~~\vec{a}\cdot \vec{b}=\sqrt{6},~~$ then find the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}.$

Solution.

If $~\theta~$ is the the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b},~~$ then

$ \cos\theta=\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|} =\frac{\sqrt{6}}{\sqrt{3} \cdot 2} \\ \text{or,}~~ \cos\theta=\frac{\sqrt{3} \sqrt{2}}{2\sqrt{3}} \\ \text{or,}~~ \cos\theta=\frac{1}{\sqrt{2}}=\cos(\pi/4) \\ \therefore~\theta=\frac{\pi}{4} ~~\text{(ans.)}$

12(i) If $~~|\vec{a}|~=\sqrt{3},~|\vec{b}|=2~~$ and $~~\vec{a}\cdot \vec{b}=\sqrt{6},~~$ then find the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}.$

Solution.

If $~\theta~$ is the the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b},~~$ then

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(ii) Find the angle made by the vector $~~ \sqrt{2} \hat{i}+\hat{j}+\hat{k}~~$ with the $~~y$-axis.

Solution.

let $~~\vec{a}=\sqrt{2} \hat{i}+\hat{j}+\hat{k}~~$ and the unit vector along $~~y$-axis is $~~\hat{j}~~$ and let $~~\vec{b}=\hat{j}.$

$~\vec{a} \cdot \vec{b}=(\sqrt{2} \hat{i}+\hat{j}+\hat{k}) \cdot (\hat{j})=1, \\~~ |\vec{a}|=\sqrt{(\sqrt{2})^2+1^2+1^2}=\sqrt{4}=2, \\~~ |\vec{b}|=\sqrt{1^2}=1.$

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

$\cos\theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1}{2 \times 1} \\ \text{or,}~~ \cos\theta=\frac 12=\cos(\pi/3) \\ \therefore~ \theta=\frac{\pi}{3}~~\text{(ans.)}$

(iii) Find $~~|\vec{a}-\vec{b}|,~~$ if two vectors $~~\vec{a}~~$ and $~~\vec{b}~~$ are such that $~~|\vec{a}|=2,~~|\vec{b}|=3~~$ and $~~ \vec{a} \cdot \vec{b}=4.$

Solution.

$|\vec{a}-\vec{b}|^2\\=(\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b})\\=|\vec{a}|^2-\vec{a} \cdot \vec{b}-\vec{b}\cdot \vec{a}+|\vec{b}|^2\\=2^2-2\vec{a} \cdot \vec{b}+3^2\\=4-2 \times 4+9\\=4-8+9\\=5 \\ \therefore~|\vec{a}-\vec{b}|=\sqrt{5}~~\text{(ans.)}$

$13.~~$ Define the vector product of two vectors $~~\vec{a}~~\text{and}~~\vec{b}~~$. Give geometrical interpretation of $~~\vec{a} \times \vec{b}~.~$ Show that the vector product of two vectors is not commutative but it satisfies the distributive law.

Refer : Chhaya Mathematics (S N De)

14(i) If $~~|\vec{a}|=3,~~|\vec{b}|=4~~$ and $~~|\vec{a} \times \vec{b}|=6,~~$ then find the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}.$

Solution.

If $~~\theta~~$ is the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}~,~$ then

$~|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin\theta \\ \text{or,}~~ 6=3 \times 4 \times \sin\theta \\ \text{or,}~~ \sin\theta=\frac{6}{3 \times 4}=\frac 12 \\ \text{or,}~~ \sin\theta=\sin(\pi/6) \\ \therefore~ \theta=\frac{\pi}{6}~~\text{(ans.)}$

(ii) If $~~\vec{a},~~\vec{b}~~$ and $~~ \vec{a} \times \vec{b}~~$ are three unit vectors , find the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}.$

Solution.

If $~~\theta~~$ is the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}~,~$ then

$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin\theta \\ \text{or,}~~ 1=1 \times 1 \times \sin\theta \\ \text{or,}~~ \sin\theta=1=\sin(\pi/2) \\ \therefore~ \theta=\frac{\pi}{2}~~\text{(ans.)}$

(iii) If two vectors $~~\vec{a}~~$ and $~~\vec{a}~~$ are such that $~~|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}|~,~$ then find the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}.$

Solution.

$|\vec{a} \cdot \vec{b}|=|\vec{a} \times \vec{b}| \\ \text{or,}~~ |\vec{a}||\vec{b}| \cos\theta=|\vec{a}||\vec{b}|\sin\theta \\ \text{or,}~~ \cos\theta=\sin\theta \\ \text{or,}~~ \frac{\sin\theta}{\cos\theta}=1 \\ \text{or,}~~ \tan\theta=\tan(\pi/4) \\ \text{or,}~~ \theta=\frac{\pi}{4}~~\text{(ans.)} $

(iv) Two vectors $~~\vec{a}~~$ and $~~\vec{b}~~$ are such that $~~ |\vec{a}|=\sqrt{3},~~|\vec{b}|=\frac 23~~$ and the magnitude of $~~ \vec{a} \times \vec{b}~~$ is $~~1~;~$ ind the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}.$

Solution.

If $~\theta~$ is the angle between the vectors $~~\vec{a}~~$ and $~~\vec{b}~,~$ then

$\sin\theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{1}{\sqrt{3} \times \frac 23} \\ \text{or,}~~ \sin\theta=\frac{1}{\frac{2}{\sqrt{3}}}=\frac{\sqrt{3}}{2} \\ \text{or,}~~ \sin\theta=\sin(\pi/3) \\ \therefore~ \theta=\frac{\pi}{3} ~~\text{(ans.)} $

15(i) If vectors $~~ \vec{a}=p\hat{i}+8\hat{j}+6\hat{k}~~$ and $~~\vec{b}=-3\hat{i}+4\hat{j}+q\hat{k}~~$ are parallel, find $~~p~~$ and $~~q.$

Solution.

We know if two vectors $~~ \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k},~~ \vec{b}=b_1\vec{i}+b_2\hat{j}+b_3\hat{k}~~$ are parallel if

$~\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3} .$

Hence , by the given question, we have

$~\frac{p}{-3}=\frac 84=\frac 6q \rightarrow(1)$

So, by $~~(1)~~$ we get,

$~~\frac{p}{-3}=\frac 84=2\\~~~ \Rightarrow p=-3 \times 2=-6$

Again, by $~~(1)~~$ we get,

$~~ \frac 6q=\frac 84=2\\ ~~~\Rightarrow q=\frac 62=3.$

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(ii) If $~~(2\hat{i}+6\hat{j}+27\hat{k}) \times (\hat{i}+3\hat{j}+p\hat{k})=\vec{0}, ~~$ find $~~p.$

Solution.

$ (2\hat{i}+6\hat{j}+27\hat{k}) \times (\hat{i}+3\hat{j}+p\hat{k})=\vec{0} \\ \text{or,}~~\begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 2& 6 & 27 \\ 1& 3 & p \\ \end{vmatrix}=0 \\ \text{or,}~~ (6p-81)\hat{i}-(2p-27)\hat{j}+(6-6)\hat{k}=0 \\ \text{or,}~~ (6p-81)\hat{i}-(2p-27)\hat{j} =0 \\ \therefore ~~ 6p-81=0,~~ 2p-27=0 \\ \Rightarrow p=\frac{27}{2} ~~\text{(ans.)}$

(iii) If $~~(2\hat{i}+6\hat{j}+14\hat{k}) \times (\hat{i}-\lambda \hat{j}+7\hat{k})=\vec{0},~~$ find $~~\lambda .$

Solution.

$ (2\hat{i}+6\hat{j}+14\hat{k}) \times (\hat{i}-\lambda \hat{j}+7\hat{k})=\vec{0} \\ \text{or,}~~\begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 2& 6 & 14 \\ 1& -\lambda & 7 \\ \end{vmatrix}=0\\ \text{or,}~~ (42+14\lambda)\hat{i}-(14-14)\hat{j}+(-2\lambda-6)\hat{k}=0 \\ \text{or,}~~ (42+14\lambda)\hat{i}-(2\lambda+6)\hat{k} =0 \\ \text{or,} ~~ 42+14\lambda)=0,~~2\lambda+6=0 \\ \therefore ~~\lambda=-\frac{42}{14}=-\frac 62=-3~~\text{(ans.)} $

16. Find the area of the parallelogram whose

(i) adjacent sides are $~~ \vec{a}=3\hat{i}+\hat{j}+4\hat{k}~~$ and $~~\vec{b}=\hat{i}-\hat{j}+\hat{k}. $

(ii) whose vertices are the points $~~ (0,-3,-1),~~(2,1,-1),~~(3,-3,2)~~$ and $~~ (1,-7,2)~~$ taken in order.

(iii) whose diagonals are the vectors $~~3\hat{i}+\hat{j} -2\hat{k}~~$ and $~~ \hat{i}-3\hat{j} +4\hat{k}.$

Solution. $~~(i)$

$~ \vec{a} \times \vec{b} \\=\begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 3& 1 & 4 \\ 1& -1 & 1 \\ \end{vmatrix} \\ =(1+4)\hat{i}-(3-4)\hat{j}+(-3-1)\hat{k} \\ =5\hat{i}+\hat{j}-4\hat{k} $

So, the area of the parallelogram is given by :

$|\vec{a} \times \vec{b}|\\=\sqrt{5^2+1^2+(-4)^2}\\=\sqrt{25+1+16}\\=\sqrt{42}~~\text{sq. unit.}$

Solution $~~(ii)$

Let the vertices of the parallelogram be denoted by $~~ A,~~B,~~C,~~D~~$ respectively where $~~ A= -3\hat{j}-\hat{k} ,~~ B= 2\hat{i}+\hat{j}-\hat{k},~~\\ ~~ C= 3\hat{i}-3\hat{j}+2\hat{k}, ~~D= \hat{i}-7\hat{j}+2\hat{k}.$

$\text{Now,}~~ \vec{AB}\\=(2\hat{i}+\hat{j}-\hat{k})-(-3\hat{j}-\hat{k})\\=2\hat{i}+4\hat{j},\\~~\\~~\vec{AD}\\=(\hat{i}-7\hat{j}+2\hat{k})-(-3\hat{j}-\hat{k})\\=\hat{i}-4\hat{j}+3\hat{k}$

$~\therefore~~ \vec{AB} \times \vec{AD} \\~~~~ = \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 2& 4 & 0 \\ 1& -4 & 3 \\ \end{vmatrix} \\~~~~= (12-0)\hat{i}-(6-0)\hat{j}+(-8-4)\hat{k} \\ ~~~~=12\hat{i}-6\hat{j}-12\hat{k}$

So, the area of the parallelogram is :

$~~~=|\vec{AB} \times \vec{AD}|\\~=\sqrt{12^2+(-6)^2+(-12)^2}\\~=\sqrt{144+36+144}\\~=\sqrt{324}\\~=18~~\text{sq. unit}$

Solution $~~(iii)$

$\text{let}~~\vec{a}=3\hat{i}+\hat{j} -2\hat{k} , ~~ \vec{b}=\hat{i}-3\hat{j} +4\hat{k} \\ \text{so that }~~ |\vec{a} \times \vec{b}| \\= \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 3& 1 & -2 \\ 1& -3 & 4 \\ \end{vmatrix} \\= (4-6)\hat{i}-(12+2)\hat{j} +(-9-1)\hat{k} \\= -2\hat{i}-14\hat{j} -10\hat{k} \\~~\\~\therefore~ |\vec{a} \times \vec{b}|\\=\sqrt{(-2)^2+(-14)^2+(-10)^2}\\=\sqrt{4+196+100}\\=\sqrt{300}\\=\sqrt{3\times 100}=10\sqrt{3} $

So, the area of the parallelogram is :

$~~=\frac 12|\vec{a} \times \vec{b}|\\~=\frac 12 \times 10\sqrt{3}\\~=5\sqrt{3}~~\text{sq. unit}$

17. Find the area of the triangle

(i) drawn on the vectors $~~ \vec{a}=6\hat{i}+2\hat{j} -3\hat{k}~~$ and $~~\vec{b}=4\hat{i}-\hat{j} -2\hat{k}. $

(ii) whose vertices have position vectors $~~ \hat{i}+\hat{j} +2\hat{k}, ~~ 2\hat{i}+2\hat{j} +3\hat{k}~~~~$ and $~~~~3\hat{i}-\hat{j} -\hat{k}.$

(iii) whose vertices are the points $~~(1,2,3),~~(2,3,1),~~$ and $~~ (1,1,1).$

Solution. (i)

$\vec{a} \times \vec{b} \\=\begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 6& 2 & -3 \\ 4& -1 & -2 \\ \end{vmatrix} \\= (-4-3)\hat{i}-(-12+12)\hat{j} +(-6-8)\hat{k} \\= -7\hat{i}-14\hat{k}$

So, the area of the triangle :

$~=\frac 12|\vec{a} \times \vec{b}|\\~=\frac 12 |-7\hat{i}-14\hat{k} |\\~=\frac 72 |-\hat{i}-2\hat{k}|\\~=\frac 72 \times \sqrt{(-1)^2+(-2)^2}\\~=\frac{7\sqrt{5}}{2}~~\text{sq. unit}$

Solution. (ii)

let the vertices of the triangle be denoted by

$~~A,~~B,~~C~~$ where $~~ A=\hat{i}+\hat{j} +2\hat{k}, ~~ B=2\hat{i}+2\hat{j} +3\hat{k}, ~~\\ C=3\hat{i}-\hat{j} -\hat{k}.$

$~~\vec{AB}=(2\hat{i}+2\hat{j} +3\hat{k})-(\hat{i}+\hat{j} +2\hat{k}) \\ \therefore~~\vec{AB}=\hat{i}+\hat{j}+\hat{k}.$

$~~\vec{AC}=(3\hat{i}-\hat{j} -\hat{k})-(\hat{i}+\hat{j} +2\hat{k}), \\ \therefore~~\vec{AC}=2\hat{i}-2\hat{j}-3\hat{k}.$

$\therefore~~ \vec{AB} \times \vec{AC} \\~= \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 1& 1 & 1 \\ 2& -2 & -3 \\ \end{vmatrix} \\~= (-3+2)\hat{i}-(-3-2)\hat{j} +(-2-2)\hat{k} \\~= -\hat{i}+5\hat{j} -4\hat{k} $

So, the area of the triangle :

$=\frac 12| \vec{AB} \times \vec{AC}|\\~=\frac 12 \sqrt{(-1)^2+5^2+(-4)^2}\\~=\frac 12 \sqrt{1+25+16}\\~=\frac 12 \sqrt{42}\\~=\sqrt{\frac{42}{4}}=\sqrt{\frac{21}{2}}~~\text{sq. unit}$

Solution. (iii)

let the vertices of the triangle be denoted by $~~A(1,2,3),~~B(2,3,1)~~$ and $~~ C(1,1,1).$

$\text{So,}~~ A=\hat{i}+2\hat{j} +3\hat{k}, ~~ B=2\hat{i}+3\hat{j} +\hat{k},\\~~C=\hat{i}+\hat{j} +\hat{k}.$

$\vec{AB}=(2\hat{i}+3\hat{j} +\hat{k})-(\hat{i}+2\hat{j} +3\hat{k})\\ \text{or,}~~\vec{AB}=\hat{i}+\hat{j} -2\hat{k}$

$\vec{AC}=(\hat{i}+\hat{j} +\hat{k})-(\hat{i}+2\hat{j} +3\hat{k})\\ \text{or,}~~\vec{AC}=-\hat{j}-2\hat{k}.$

$\therefore~~ \vec{AB} \times \vec{AC} \\ ~~~~~=\begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 1& 1 & -2 \\ 0& -1 & -2 \\ \end{vmatrix} \\ ~~~~~= (-2-2)\hat{i}-(-2-0)\hat{j} +(-1-0)\hat{k} \\~~~~~= -4\hat{i}+2\hat{j} -\hat{k}.$

$\therefore~~ $ the area of the triangle :

$~=\frac 12|\vec{AB} \times \vec{AC}|\\~=\frac 12 \sqrt{(-4)^2+2^2+(-1)^2}\\~=\frac 12\sqrt{16+4+1}\\~=\frac 12 \sqrt{21}~~\text{sq. unit}$