In the previous article we discussed Very Short Answer Type Questions from question no. 11 to 17 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 problems related to Short Answer Type Questions of the aforementioned book . Here we will solve few more problem related to Vector Product . So, without wasting time , let’s start.
.png "Product of two Vectors")
1. Two vectors $~~\vec{a}~~$ and $~~\vec{b}~~$ are such that $~~|\vec{a}|=2,~~|\vec{b}|=1~~$ and $~~\vec{a} \cdot \vec{b}=1;~~$ then find the value of $~~(3\vec{a}-5\vec{b})\cdot (2\vec{a}+7\vec{b}).$
Solution.
$(3\vec{a}-5\vec{b})\cdot (2\vec{a}+7\vec{b}) \\= 6\vec{a}\cdot \vec{a}+21 \vec{a}\cdot \vec{b}-10\vec{b}\cdot \vec{a}-35 \vec{b}\cdot \vec{b}\\= 6|\vec{a}|^2+11\vec{a}\cdot \vec{b}-35|\vec{b}|^2 \\= 6 \times 2^2+11 \times 1-35 \times 1^2 \\= 24+11-35\\=35-35\\=0~~\text{(ans.)}$
2(i) If $~~|\vec{a}|=2,~~|\vec{b}|=3~~\text{and}~~\vec{a} \cdot \vec{b}=3,~~$ then find the projection of $~~\vec{b}~~$ on $~~\vec{a}.$
Solution.
The projection of $~~\vec{b}~~$ on $~~\vec{a}~~$ is given by :
$= \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}=\frac 32~~\text{(ans.)}$
$(ii)~~$ If projection of vector $~~\lambda \hat{i}-\hat{j}~~$ on vector $~~\hat{i}+\hat{j}~~$ is zero, find $~~\lambda.$
Solution.
According to the problem,
$\frac{(\hat{i}+\hat{j}) \cdot (\lambda \hat{i}-\hat{j})}{|\hat{i}+\hat{j}|}=0 \\ \text{or,}~~ \frac{\lambda (\hat{i} \cdot \hat{i})-\hat{i} \cdot \hat{j}+\lambda (\hat{j} \cdot \hat{i})-\hat{j} \cdot \hat{j}}{\sqrt{1^2+1^2}}=0 \\ \text{or,}~~ \lambda-1=0 ~~[~~ \because ~\hat{i} \cdot \hat{j}=\hat{j} \cdot \hat{i}=0 ~] \\ \therefore~\lambda=1 ~~\text{(ans.)}$
3. If $~~|\vec{a}|=4~~\text{and}~~|\vec{b}|=3,~~$ find the value of $~~\lambda~~$ for which the vectors $~~\vec{a}+\lambda \vec{b}~~$ and $~~\vec{a}-\lambda \vec{b}~~$ are perpendicular to each other.
Solution.
Since the vectors $~~\vec{a}+\lambda \vec{b}~~$ and $~~\vec{a}-\lambda \vec{b}~~$ are perpendicular to each other,
$ (\vec{a}+\lambda \vec{b}) \cdot (\vec{a}-\lambda \vec{b})=0 \\ \text{or,}~~ \vec{a} \cdot \vec{a}-\lambda \vec{a} \cdot \vec{b}+\lambda \vec{b} \cdot \vec{a}-\lambda^2 \vec{b} \cdot \vec{b}=0\\ \text{or,}~~ |\vec{a}|^2-\lambda^2|\vec{b}|^2=0 ~~(~\because ~~\vec{b} \cdot \vec{a}=\vec{a} \cdot \vec{b} ) \\ \text{or,}~~4^2-\lambda^2 \times 3^2=0 \\ \text{or,}~~ 16=9\lambda^2 \\ \text{or,}~~ \lambda=\pm \sqrt{\frac{16}{9}}=\pm \frac 43~~~\text{(ans.)} $
4. The vectors $~~\vec{a},~\vec{b},~\vec{c}~~$ are such that $~~ \vec{a}+\vec{b}+\vec{c}=\vec{0}. ~~$ If $~~|\vec{a}|=3,~~|\vec{b}|=4~~\text{and}~~ |\vec{c}|=5,~~$ show that $~~\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}=-25.$
Solution.
$~\because~~ \vec{a}+\vec{b}+\vec{c}=\vec{0}, \\ ~\therefore~~ (\vec{a}+\vec{b}+\vec{c})\cdot (\vec{a}+\vec{b}+\vec{c})=0 \\ \text{or,}~~|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2\\~+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\ \text{or,}~~ 3^2+4^2+5^2\\~~+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\ \text{or,}~~ 50 +2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\ \text{or,}~~2 \times [25+(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})]=0 \\ \text{or,}~~ 25+(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0 \\ \therefore~~ (\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=-25~~\text{(showed)} $
5. If $~~\vec{a}=\hat{i}-\hat{j} +2\hat{k},~~\vec{b}=\hat{i}+\hat{j} +\hat{k}~~$ and $~~ \vec{c}=2\hat{i}-\hat{j} +\hat{k},~~$ find the vector $~~\vec{r}~~$ satisfying the relations, $~~\vec{a} \cdot \vec{r}=1,~~\vec{b} \cdot \vec{r}=2~~$ and $~~\vec{c} \cdot \vec{r}=5.$
Solution.
let $~~\vec{r}=x\hat{i}+y\hat{j} +z\hat{k}$
$~ \vec{a} \cdot \vec{r}=1 \\ \text{or,}~~(\hat{i}-\hat{j} +2\hat{k}) \cdot (x\hat{i}+y\hat{j} +z\hat{k})=1 \\ \text{or,}~~x-y+2z=1\rightarrow(1) $
$~ \vec{b} \cdot \vec{r}=1 \\ \text{or,}~~(\hat{i}+\hat{j} +\hat{k}) \cdot (x\hat{i}+y\hat{j} +z\hat{k})=1 \\ \text{or,}~~x+y+z=1\rightarrow(2) $
$~ \vec{c} \cdot \vec{r}=1 \\ \text{or,}~~(2\hat{i}-\hat{j} +\hat{k}) \cdot (x\hat{i}+y\hat{j} +z\hat{k})=1 \\ \text{or,}~~2x-y+z=1\rightarrow(3) $
Adding $~~(1)~~\text{and}~~(2)~~$ we get,
$(x-y+2z)+(x+y+z)=1+2 \\ \text{or,}~~ 2x+3z-3=0\rightarrow(4)$
Adding $~~(2)~~\text{and}~~(3)~~$ we get,
$(x+y+z)+(2x-y+z)=2+5 \\ \text{or,}~~ 3x+2z-7=0\rightarrow(5)$
From $~~(4)~~\text{and}~~(5)~~$ we get,
$ \frac{x}{-21-(-6)}=\frac{z}{-9-(-14)}=\frac{1}{4-9} \\ \text{or,}~~ \frac{x}{-15}=\frac{z}{5}=\frac{1}{-5} \\ ~\therefore~ x=\frac{-15}{-5}=3, ~~y=\frac{5}{-5}=-1.$
JEE Advanced Mathematics - Algebra Paperback
$6.~~$ If $~~\vec{\alpha}~~$ and $~~\vec{\beta}~~$ are perpendicular to each other , show that
$(i)~ |\vec{\alpha}+\vec{\beta}|^2=|\vec{\alpha}|^2+|\vec{\beta}|^2\\~~~~~(ii)~~~|\vec{\alpha}+\vec{\beta}|^2=|\vec{\alpha}-\vec{\beta}|^2$
Solution.
Since $~~\vec{\alpha}~~$ and $~~\vec{\beta}~~$ are perpendicular to each other ,
$ \vec{\alpha} \cdot \vec{\beta}=0=\vec{\beta} \cdot \vec{\alpha}\rightarrow(1)$
$ |\vec{\alpha}+\vec{\beta}|^2\\~~~~=(\vec{\alpha}+\vec{\beta}) \cdot (\vec{\alpha}+\vec{\beta})\\~~~~=\vec{\alpha} \cdot \vec{\alpha} + \vec{\alpha} \cdot \vec{\beta}+\vec{\beta} \cdot \vec{\alpha}+\vec{\beta} \cdot \vec{\beta}\\~~~~=|\vec{\alpha}|^2+0+0+|\vec{\beta}|^2~~[\text{By (1)}]\\~~~~=|\vec{\alpha}|^2+ |\vec{\beta}|^2~~~\text{(showed)}$
Solution. $~~~(ii)$
$~|\vec{\alpha}+\vec{\beta}|^2-|\vec{\alpha}-\vec{\beta}|^2\\~~~~=(|\vec{\alpha}|^2+|\vec{\beta}|^2+2\vec{\alpha}\cdot \vec{\beta})-(|\vec{\alpha}|^2+|\vec{\beta}|^2-2\vec{\alpha} \cdot \vec{\beta})\\~~~~=4\vec{\alpha} \cdot \vec{\beta}\\~~~~=0~~[~\text{By (1)}] \\ \therefore~|\vec{\alpha}+\vec{\beta}|^2=|\vec{\alpha}-\vec{\beta}|^2~~\text{(showed)}$
$7(i)~~~ A(p,1,-1),~~B(2p,0,2)~~$ and $~~C(2+2p,p,p)~~$ are three points. Find the value of $~~p~~$ so that the vectors $~~\vec{AB}~~\text{and} ~~~\vec{BC}~~$ are perpendicular to each other.
Solution.
According to the problem, suppose that the position vectors of
$A=p\hat{i}+\hat{j} -\hat{k}, ~~B= 2p\hat{i}+2\hat{k},\\~~~~~C=(2+2p)\hat{i}+p\hat{j} +p\hat{k}$
$\text{So,}~~\vec{AB}\\=(2p\hat{i}+2\hat{k})-(p\hat{i}+\hat{j} -\hat{k})\\=(2p-p)\hat{i}-\hat{j} +(2+1)\hat{k}\\=p\hat{i}-\hat{j} +3\hat{k}$
$\text{Again,}~~\vec{BC}\\=[(2+2p)\hat{i}+p\hat{j} +p\hat{k}]-(2p\hat{i}+2\hat{k})\\=(2+2p-2p)\hat{i}+p\hat{j} +(p-2)\hat{k} \\=2\hat{i}+p\hat{j} +(p-2)\hat{k}$
$ \because~~\vec{AB} \perp \vec{BC},\\~~\vec{AB} \cdot \vec{BC}=0 \\ \text{or,}~~(p\hat{i}-\hat{j} +3\hat{k}) \cdot [2\hat{i}+p\hat{j} +(p-2)\hat{k}]=0 \\ \text{or,}~~ 2p-p+3(p-2)=0 \\ \text{or,}~~ p+3p-6=0 \\ \text{or,}~~ 4p-6=0 \\ \text{or,}~~ 4p=6 \\ \text{or,}~~ p=\frac 64=\frac 32~~\text{(ans.)} $
$(ii)~~$ If $~~ \vec{\alpha}=\hat{i}+2\hat{j} -3\hat{k}~~\text{and}~~\vec{\beta}=3\hat{i}-\hat{j} +2\hat{k},~~$ find the cosine of the angle between the vectors $~~ (2\vec{\alpha}+\vec{\beta})~~\text{and} ~~ (\vec{\alpha}+2\vec{\beta}).$
Solution.
$(2\vec{\alpha}+\vec{\beta}) \\=2(\hat{i}+2\hat{j} -3\hat{k})+(3\hat{i}-\hat{j} +2\hat{k}) \\= 5\hat{i}+3\hat{j} -4\hat{k} \\ ~\therefore~~ |2\vec{\alpha}+\vec{\beta}| \\= \sqrt{5^2+3^2+(-4)^2}\\=\sqrt{25+9+16}\\=\sqrt{50}\rightarrow(1) \\~~~\\~~~ (\vec{\alpha}+2\vec{\beta})\\= (\hat{i}+2\hat{j} -3\hat{k})+2(3\hat{i}-\hat{j} +2\hat{k}) \\=7\hat{i} +\hat{k}\\ ~~\\~~\therefore~ |\vec{\alpha}+2\vec{\beta}|=\sqrt{7^2+1^2}=\sqrt{50}\rightarrow(2)$
$(2\vec{\alpha}+\vec{\beta}) \cdot (\vec{\alpha}+2\vec{\beta})\\=(5\hat{i}+3\hat{j} -4\hat{k}) \cdot (7\hat{i} +\hat{k})\\=35-4\\=31 \rightarrow(3)$
If $~~\theta~~$ be the angle $~~ (2\vec{\alpha}+\vec{\beta})~~\text{and}~~ (\vec{\alpha}+2\vec{\beta})~~$ then
$ \cos\theta\\=\frac{(2\vec{\alpha}+\vec{\beta}) \cdot (\vec{\alpha}+2\vec{\beta})}{|2\vec{\alpha}+\vec{\beta}||\vec{\alpha}+2\vec{\beta}|}\\=\frac{31}{\sqrt{50} \sqrt{50}}~~~[\text{By (1),(2),(3)}]\\=\frac{31}{50}~~\text{(ans.)} $
JEE Advanced Mathematics - Coordinated Geometry Paperback
$8(i)~~$ Show that the vector $~~\vec{a}~~$ is perpendicular to the vector $~~\vec{b}-\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}~\vec{a}.$
Solution.
$\vec{a} \cdot \left(\vec{b}-\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}~\vec{a}\right)\\~~=\vec{a} \cdot \vec{b}-\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}(\vec{a} \cdot \vec{a})\\~~=\vec{a} \cdot \vec{b}-\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2} |\vec{a}|^2\\~~=\vec{a} \cdot \vec{b}-\vec{a}\cdot \vec{b}\\~~=0\rightarrow(1)$
Hence by $~(1)~$ , the result follows.
$(ii)~~$ For any two vectors $~~\vec{a}~~\text{and}~~\vec{b}~~$ show that $~~|\vec{a} \cdot \vec{b}| \leq |\vec{a}||\vec{b}|.$
Solution.
If $~~\theta~~$ is the angle between any two vectors $~~\vec{a}~~\text{and}~~\vec{b},$
$ \vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}| \cos\theta \\ \text{or,}~~ \cos\theta =\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\rightarrow(1) $
$\text{Now,}~~ |\cos\theta| \leq 1 \rightarrow(2) $
So, by $~~(1)~~\text{and}~~(2),~~$ we get
$ \left|\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\right| \leq 1 \Rightarrow |\vec{a} \cdot \vec{b}| \leq |\vec{a}||\vec{b}|.$
9. If $~~\vec{a},~~\vec{b},~~\vec{c}~~$ are non-coplanar vectors $~~\vec{r} \cdot \vec{a}=\vec{r} \cdot \vec{b}=\vec{r} \cdot \vec{c}=0,~~$ show that $~~\vec{r}~~$ is a zero vector.
Solution.
let $~~ \vec{r}=x\vec{a}+y\vec{b} +z\vec{c}~~$ where $~~x,y,z~~$ are scalars.
$\vec{r} \cdot \vec{a}=0 \\ \Rightarrow x|\vec{a}|^2+y~\vec{a} \cdot \vec{b}+z~ \vec{a} \cdot \vec{c}=0\rightarrow(1) \\~~~ \vec{r} \cdot \vec{b}=0 \\ \Rightarrow x~ \vec{a} \cdot \vec{b}+y|\vec{b}|^2+z~ \vec{b} \cdot \vec{c}=0\rightarrow(2) \\~~~ \vec{r} \cdot \vec{c}=0 \\ \Rightarrow x~\vec{a} \cdot \vec{c}+y~\vec{b} \cdot \vec{c}+z |\vec{c}|^2=0\rightarrow(3)$
$ \begin{vmatrix} |\vec{a}|^2 &\vec{a} \cdot \vec{b} & \vec{a} \cdot \vec{c} \\ ~\vec{a} \cdot \vec{b} & |\vec{b}|^2 & \vec{b} \cdot \vec{c} \\ ~\vec{a} \cdot \vec{c} &\vec{b} \cdot \vec{c} & |\vec{c}|^2 \\ \end{vmatrix}=[\vec{a}~~\vec{b}~~\vec{c}] \neq 0$
Since $~~\vec{a},~~\vec{b},~~\vec{c}~~$ are non-coplanar vectors, the only solution of $~(1),~(2),~(3)~$ is :
$ x=y=z=0 \Rightarrow \vec{r}=\vec{0} ~~$ and hence the result follows.
10(i) For the vectors $~~\vec{a}~~\text{and}~~\vec{b}~~$ if $~~|\vec{a}+\vec{b}|=|\vec{a}-\vec{b}|,~~$ show that $~~\vec{a}~~\text{and}~~\vec{b}~~$ are perpendicular.
Solution.
$ |\vec{a}+\vec{b}|=|\vec{a}-\vec{b}| \\ \text{or,}~~ |\vec{a}+\vec{b}|^2=|\vec{a}-\vec{b}|^2 \\ \text{or,}~~ (\vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b})=(\vec{a}-\vec{b}) \cdot (\vec{a}-\vec{b}) \\ \text{or,}~~ |\vec{a}|^2+|\vec{b}|^2+2\vec{a} \cdot \vec{b}=|\vec{a}|^2+|\vec{b}|^2-2\vec{a} \cdot \vec{b} \\ \text{or,}~~ 2\vec{a}\cdot \vec{b}=-2\vec{a} \cdot \vec{b} \\ \text{or,}~~ 4\vec{a} \cdot \vec{b}=0 \\ \text{or,}~~ \vec{a} \cdot \vec{b}=0 \\ \therefore~~\vec{a} \perp \vec{b}. $
$(ii)~~~$ Prove that $~~( \vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2,~~$ if and only if $~~\vec{a},~~\vec{b}~~$ are perpendicular , given $~~ \vec{a} \neq \vec{0},~~ \vec{b} \neq \vec{0}.$
Solution.
$( \vec{a}+\vec{b}) \cdot (\vec{a}+\vec{b})=|\vec{a}|^2+|\vec{b}|^2 \\ \text{or,}~~ \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{b}\cdot \vec{a}+\vec{b} \cdot \vec{b}=|\vec{a}|^2+|\vec{b}|^2 \\ \text{or,}~~ |\vec{a}|^2+2\vec{a} \cdot \vec{b}+|\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2 \\ \text{or,}~~ 2\vec{a} \cdot \vec{b}=0 \\ \text{or,}~~ \vec{a} \cdot \vec{b}=0 \\ \therefore~~ \vec{a} \perp \vec{b}~~~\text{(proved )}$
$11.~~$ If $~~\hat{i},~~\hat{j}~~\text{and}~~\hat{k}~~$ are unit vectors along three mutually perpendicular axes and $~~ \vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}~ ; ~~\vec{b}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}~~\text{and}~~ \vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}~~$ prove that
$(i)~ \vec{a} \cdot (\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c} ~~(ii)~~(\vec{b}+\vec{c}) \times \vec{a}=\vec{b} \times \vec{a}+\vec{c} \times \vec{a}$
Solution. $~~(i)$
$(\vec{b}+\vec{c})\\~~~=(b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}$
$\vec{a} \cdot(\vec{b}+\vec{c}) \\= (a_1\hat{i}+a_2\hat{j}+a_3\hat{k}) \cdot [(b_1+c_1)\hat{i}+(b_2+c_2)\hat{j}+(b_3+c_3)\hat{k}]\\=a_1(b_1+c_1)+a_2(b_2+c_2)+a_3(b_3+c_3) \\= (a_1b_1+a_2b_2+a_3b_3)+(a_1c_1+a_2c_2+a_3c_3)\rightarrow(1) $
$ \vec{a} \cdot \vec{b}\\=(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot (b_1\hat{i}+b_2\hat{j}+b_3\hat{k})\\=a_1b_1+a_2b_2+a_3b_3 \rightarrow(2)$
$ \vec{a} \cdot \vec{c}\\=(a_1\hat{i}+a_2\hat{j}+a_3\hat{k})\cdot (c_1\hat{i}+c_2\hat{j}+c_3\hat{k})\\=a_1c_1+a_2c_2+a_3c_3 \rightarrow(3)$
Hence, by $~(1),~(2),~(3)~$ we can conclude that
$\vec{a} \cdot (\vec{b}+\vec{c})=\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}$
$(ii)~~(\vec{b}+\vec{c}) \times \vec{a} \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ (b_1+c_1) & ~(b_2+c_2) &~(b_3+c_3) \\ a_1 &~a_2 &~a_3 \\ \end{vmatrix}\\~~~ \\= \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ b_1 & ~b_2 &~b_3 \\ a_1 &~a_2 &~a_3 \\ \end{vmatrix} + \begin{vmatrix} \hat{i}&\hat{j} &\hat{k} \\ c_1 & ~c_2 &~c_3 \\ a_1 &~a_2 &~a_3 \\ \end{vmatrix} \\= \vec{b} \times \vec{a}+\vec{c} \times \vec{a} ~~\text{(proved)}$
Read More :
Vector Product (Part-4) | S N Dey | Class 12
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