KC Sinha Solution Class 12 Chapter 27 Vector or Cross product of Two vectors Exercise 27.1 (Q1-Q10)

Exercise 27.1

Type-1 :  दो सदिशों के सदिश गुणन पर आधारित प्रश्न :

Question 1

यदि दो सदिश $\vec{a}$ और $\vec{b}$ इस प्रकार हैं कि $|\vec{a}|=2,|\vec{b}|=7$ तथा $\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}$ तों $\vec{a}$ और $\vec{b}$ के बीच का कोण ज्ञात करे।

[If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=2,|\vec{b}|=7$ and $\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}$, find the angle between $\vec{a}$ and $\vec{b}$ ]]

Sol :

Given $|\vec{a}|=2,| \vec{b} |=7$ and  $\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}$

∵ $\vec{a} \times \vec{b}=3 \hat{i}+2 \hat{3}+6 \hat{k}$

∴ $|\vec{a} \times \vec{b}|=\sqrt{3^{2}+2^{2}+6^{2}}=\sqrt{49}=7$

Let θ be the angle between $\vec{a}\text{ and } \vec{b}$

Now ,
$\sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\frac{7}{2 \times 7}=\frac{1}{2}$

$\sin \theta=\frac{1}{2} \Rightarrow \sin \theta=\sin \frac{\pi}{6}$

∴ $\theta=\frac{\pi}{6}$

Hence , the angle between $\vec{a}$ and $\vec{b}=\frac{\pi}{6}$

Question 2

दिया है, $|\vec{a}|=10,|\vec{b}|=2$ तथा $\vec{a} \cdot \vec{b}=12$ तो $|\vec{a} \times \vec{b}|$ निकालें ।

Given $|\vec{a}|=10,|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=12$, find $|\vec{a} \times \vec{b}|$.]

Sol :
Given : $|\vec{a}|=10 ,|\vec{b}|=2$ and $\vec{a} \cdot \vec{b}=12$

Let θ be the angle between $\vec{a}$ and $\vec{b}$

∴  $\vec{a} \cdot \vec{b}$=abcosθ

12=10×2cosθ

$\Rightarrow \cos \theta=\frac{12}{20}=\frac{3}{5}$

$\Rightarrow \sin \theta =\sqrt{1-\cos ^2 \theta}$

$=\sqrt{1-{\left(\frac{3}{5}\right)}^2}$

$=\sqrt{1-\frac{9}{25}}=\sqrt{\frac{25-9}{25}}$

$\Rightarrow \sin \theta=\frac{4}{5}$

Now , $\vec{a} \times \vec{b}=a b \sin \theta \hat{n}$

$=10 \times 2 \times \dfrac{3}{5} \hat{n}$

$\vec{a} \times \vec{b}=16 \hat{n}$

$|\vec{a} \times \vec{b}|=16$

Question 3

$\vec{a} \cdot \vec{b}$ ज्ञात करें यदि $|\vec{a}|=2,|\vec{b}|=5,|\vec{a} \times \vec{b}|=8$.

[Find $\vec{a} \cdot \vec{b}$ if $|\vec{a}|=2,|\vec{b}|=5,|\vec{a} \times \vec{b}|=8 .]$

Sol :
Given : $|\vec{a}|=2 \quad|\vec{b}|=5$ and $|\vec{a} \times \vec{b}|=8$

Let θ be the angle between $\vec{a}$ and $\vec{b}$

Now , $|\vec{a} \times \vec{b}|=|\vec{a}|.|\vec{b} | \sin \theta$= 8

$\sin \theta=\frac{8}{|\overrightarrow{a}||\overrightarrow{b}|}=\frac{8}{2 \times 5}=\frac{4}{5}$

$\cos \theta=\frac{3}{5}$

∴ $\vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta$ $=\left(2 \times 5 \times \frac{3}{5}\right)=6$

Question 4

दो सदिश $\vec{a}$ और $\vec{b}$ इस प्रकार हैं कि $|\vec{a}|=5,|\vec{b}|=4$ तथा $|\vec{a} \cdot \vec{b}|=10$ तो $\vec{a}$ और $\vec{b}$ के बीच का कोण ज्ञात करें तथा उससे $|\vec{a} \times \vec{b}|$ ज्ञात करें।

[If $\vec{a}$ and $\vec{b}$ are two vectors such that $|\vec{a}|=5,|\vec{b}|=4$ and $|\vec{a} \cdot \vec{b}|=10$, find the angle between $\vec{a}$ and $\vec{b}$ and hence find $|\vec{a} \times \vec{b}|$.]

Sol :
Given : $|\vec{a}|=5  , |\vec{b}|=4$ and $\vec{a} \cdot \vec{b}=10$

Let θ be the angle between $\vec{a}$ and $\vec{b}$

Now , $\vec{a} \cdot \vec{b}=a b \cos \theta$

10=5×4.cosθ

$\Rightarrow \cos \theta=\frac{10}{20}=\frac{1}{2}$

$\Rightarrow \cos \theta=\cos \frac{\pi}{3}$

$\Rightarrow \theta=\frac{\pi}{3}$

also , $\vec{\alpha} \times \vec{b}=a b \sin \theta \hat{n}$

$=5 \times 4 \times \frac{\sqrt{3}}{2} \hat{n}=10 \sqrt{3} \hat{n}$

$| \vec{a} \times \vec{b} |=10 \sqrt{3}$

Question 5

तोन सदिश $\vec{a}, \vec{b}, \vec{c}$ इस प्रकार हैं कि $\vec{a} \times \vec{b}=\vec{c}, \vec{b} \times \vec{c}=\vec{a}$. सिद्ध करें कि $\vec{a}, \vec{b}, \vec{c}$ परस्पर लम्ब हैं तथा $|\vec{b}|=1,|\vec{c}|=|\vec{a}|$

$[\vec{a}, \vec{b}, \vec{c}$ are three vectors such that $\vec{a} \times \vec{b}=\vec{c}, \vec{b} \times \vec{c}=\vec{a}$. Prove that $\vec{a}, \vec{b}, \vec{c}$ are mutually at right angles and $|\vec{b}|=1,|\vec{c}|=|\vec{a}|$.

Sol :
Given : $\vec{a} \times \vec{b}=\vec{c}$ , $\vec{b} \times \vec{c}=\vec{a}$

Now , $|\vec{a} \times \vec{b}|=|\vec{c}|$  , $|\vec{b} \times \vec{c}|=|\vec{a}|$

We know that
$|\vec{a} \times \vec{b}|=|\vec{a}||\vec{b}| \sin \theta$..(i)

$|\vec{b} \times \vec{c}|=|\vec{b}||\vec{c}| \sin \theta$..(ii)

From equation (i)
$\sin \theta=\frac{|\overrightarrow{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$..(iii)

From equation (ii)

$\sin \theta=\frac{|\vec{b} \times \vec{c}|}{|\vec{b}||\vec{c}|}$..(iv)

On comparing equation (iii) and (iv) , we got

$\sin \theta=\frac{|\overrightarrow{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}=\sin \theta=\frac{|\vec{b} \times \vec{c}|}{|\vec{b}||\vec{c}|}$

⇒ $\frac{|\vec{c}|}{|\vec{a}| | \vec{b}|}=\frac{|\vec{a}|}{|\vec{b}|| \vec{c} |}$

⇒ $|\vec{c}|^{2}=|\vec{a}|^{2}$

⇒ $|\vec{c}|=|\vec{a}|$

∵ $|\vec{c}|=|\vec{a} |$

∴$|\vec{b}|=1$

This condition is possible only when angle i.e. θ between the vectors $\vec{a}, \vec{b}, \vec{c} \text { is } 90^{\circ}$

∴ So, these vectors ate mutually perpendicular to each other

proved

TYE-II : $\hat{i}, \hat{j}, \hat{k}$ के पदों में व्यक्त सदिशों के सदिश गुणनफल पर आधारित प्रश्न :

Question 6

$\vec{a} \times \vec{b}$ तथा $|\vec{a} \times \vec{b}|$ ज्ञात करें यदि [Find $\vec{a} \times \vec{b}$ and $|\vec{a} \times \vec{b}|$ if ]

(i) $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}$ तथा (and) $\vec{b}=3 \hat{i}+5 \hat{j}-2 \hat{k}$

(ii) $a=\hat{i}-7 \hat{j}+7 \hat{k}$ तथा (and) $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$

Sol :

(i) $\vec{a}=2 \hat{\jmath}+\hat{j}+3 \hat{k}, \vec{b}=3 \hat{\imath}+5 \hat{j}-2 \hat{k}$
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2\end{array}\right|$

$=\hat{i}\left|\begin{array}{cc}1 & 3 \\ 5 & -2\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & 3 \\ 3 & -2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 1 \\ 3 & 5\end{array}\right|$

$=\hat{i}(-2-15)-\hat{j}(-4-9)+\hat{k}(10-3)$

$\vec{a} \times \vec{b}=-17 \hat{i}+13 \hat{j}+7 \hat{k}$

$\begin{aligned}|\vec{a} \times \overrightarrow{b}| &=\sqrt{(-17)^{2}+(13)^{2}+7^{2}} \\ &=\sqrt{289+169+49}\\&=\sqrt{507}\end{aligned}$

Question 7

यदि $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ तथा $\vec{b}=3 \hat{i}+4 \hat{j}-\hat{k}$ तो सिद्ध करें कि $\vec{a} \times \vec{b}$ एक सदिश है जो $\vec{a}$ और $\vec{b}$ दोनों पर लम्ब है ।

If $\vec{a}=2 \hat{i}-\hat{j}+\hat{k}$ and $\vec{b}=3 \hat{i}+4 \hat{j}-\hat{k}$, prove that $\vec{a} \times \vec{b}$ represents a vector which is perpendicular to both $\vec{a}$ and $\vec{b}$.]

Sol :

$\vec{a}=2 \hat{i}-\hat{\jmath}+\hat{k},\vec{b}=3 \hat{\imath}+4 \hat{\jmath}-\hat{k}$

$\vec{a} \times \vec{b}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 4 & -1\end{array}\right|$

$=\hat{i}\left|\begin{array}{cc}-1 & 1 \\ 4 & -1\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & 1 \\ 3 & -1\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right|$

$=\hat{i}(1-4)-\hat{j}(-2-3)+\hat{k}(8+3)$

$\vec{a} \times \vec{b}=-3 \hat{i}+5 \hat{j}+11 \hat{k}$

$\vec{a} \cdot(\vec{a} \times \vec{b})=(2 \hat{i}-\hat{\jmath}+\hat{k}) \cdot\left(-3 \hat{i}+5 \hat{j}+11 \hat{k}\right)$

=-6-5+11=0

$\vec{b} \cdot(\vec{a} \times \vec{b})=(3 \hat{i}+4 \hat{i}-\hat{k}) \cdot(-3 \hat{i}+5 \hat{j}+11 \hat{k})$=-9+20-11=0

अतः $\vec{a} \times \vec{b}$ एक सदिश है, जो $\vec{a}$ तथा  $\vec{b}$ दोनो पर लंब है।

Question 8

यदि $\vec{a}=7 \hat{i}+3 \hat{j}-5 \hat{k}, \vec{b}=2 \hat{i}+5 \hat{j}-\hat{k}$ तथा $\vec{c}=-\hat{i}+2 \hat{j}+4 \hat{k}$ तो $(\vec{a}-\vec{b}) \times(\vec{c}-\vec{b})$ ज्ञात करें ।

Sol :

$\vec{a}=7 \hat{i}+3 \hat{j}-5 \hat{k}, \vec{b}=2 \hat{i}+5 \hat{j}-\hat{k}$, $\vec{c}=-\hat{i}+2 \hat{j}+4 \hat{k}$ 

$\vec{a}-\vec{b}=5 \hat{i}-2 \hat{j}-4 \hat{k}$ , $\vec{c}-\vec{b}=-3 \hat{i}-3 \hat{j}+5 \hat{k}$

$(\vec{a}-\vec{b}) \times(\vec{c}-\vec{b})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 5 & -2 & -4 \\ -3 & -3 & 5\end{array}\right|$

$=\hat{i}\left|\begin{array}{cc}-2 & -4 \\ -3 & 5\end{array}\right|-\hat{j}\left|\begin{array}{cc}5 & -4 \\ -3 & 5\end{array}\right|+\hat{k}\left|\begin{array}{cc}5 & -2 \\ -3 & -3\end{array}\right|$

$=\hat{i}(-10-12)-\hat{\jmath}(25-12)+\hat{k}(-15-6)$

$=-22\hat{i}-13\hat{j}-21 \hat{k}$

Question 9

सदिरा $\overrightarrow{\mathrm{A}}$ और $\overrightarrow{\mathrm{B}}$ प्राप्त होते हैं। सदिश $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}$ का परिमाण तथा इसकी दिक्-कोज्याएँ ज्ञात करें।

Sol :

$\vec{A}=\hat{i}-\hat{k}, \vec{B}=-\hat{i}+\hat{j}+\hat{k}$

$\vec{A} \times \vec{B}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & -1 \\ -1 & 1 & 1\end{array}\right|$

$=\hat{\imath}\left|\begin{array}{ll}0 & -1 \\ 1 & 1\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}1 & -1 \\ -1 & 1\end{array}\right|+\hat{k}\left|\begin{array}{cc}1 & 0 \\ -1 & 0\end{array}\right|$

$=\hat{\imath}(0+1)-\hat{j}(1-1)+\hat{k}(1+0)$

$\vec{A} \times \vec{B}=\hat{i}+\hat{k}$

$|\vec{A} \times \vec{B}|=\sqrt{1^{2}+1^{2}}= \sqrt{2}$

माना $\hat{C}, \vec{A} \times \vec{B}$ का इकाई सदिश है।

$=\frac{1}{\sqrt{2}}(\hat{i}+\hat{k})=\frac{1}{\sqrt{2}} \hat{i}+\frac{1}{\sqrt{2}} \hat{k}$

दिक् कोज्याएँ : $\frac{1}{\sqrt{2}}, 0, \frac{1}{\sqrt{2}}$

Question 10

यदि $\overrightarrow{\mathrm{A}}=2 \hat{i}-3 \hat{j}+\hat{k}$ तथा $\overrightarrow{\mathrm{B}}=3 \hat{i}+2 \hat{j}$ तो $\overrightarrow{\mathrm{A}} \cdot \overrightarrow{\mathrm{B}}$ और $\overrightarrow{\mathrm{A}} \times \overrightarrow{\mathrm{B}}$ निकालें ।

Sol :

$\vec{A}=2 \hat{i}-3 \hat{j}+\hat{k}, \vec{B}=3 \hat{i}+2 \hat{j}$

$\vec{A} \cdot \vec{B}=$6-6=0

$\vec{A} \times \vec{B}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 1 \\ 3 & 2 & 0\end{array}\right|$

$=\hat{i}\left|\begin{array}{rr}-3 & 1 \\ 2 & 0\end{array}\right|+\hat{j}\left|\begin{array}{cc}2 & 1 \\ 3 & 0\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & -3 \\ 3 & 2\end{array}\right|$

$=\hat{i}(0-2)-\hat{j}(0-3)+\hat{k}(4+9)$

$\vec{A} \times \vec{B}=-2 \hat{\imath}+3 \hat{\jmath}+13 \hat{k}$