KC Sinha: Exercise 27.1- Mathematics Solution Class 12 Chapter 27 Vector or Cross product of Two vectors

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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 $\vec{b}$ ]]

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

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

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

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

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

Question 11

दो सदिशों $\vec{a}$ और $\vec{b}$ के तल पर लम्ब इकाई सदिश निकालें, जहाँ

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

(ii) $\vec{a}=4 \hat{i}-\hat{j}+3 \hat{k}$ तथा (and) $\vec{b}=-2 \hat{i}+\hat{j}-2 \hat{k}$

Sol :

(i)

$\vec{a}=\hat{i}-\hat{j}$ , $\vec{b}=\hat{j}+\hat{k}$

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

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

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

$\hat{a} \times \hat{b}=-\hat{\imath}-\hat{j}+\hat{k}$

माना $\vec{c}=\vec{a} \times \vec{b}$

$|\vec{c}|=|\vec{a} \times \vec{b}|=\sqrt{(-1)^{2}+(-1)^{2}+1^{2}}=\sqrt{3}$

$=-\frac{1}{\sqrt{3}} \hat{i}-\frac{1}{\sqrt{3}} \hat{\jmath}+\frac{1}{\sqrt{3}} \hat{k}$

Question 12

निम्नलिखित सदिशों में से प्रत्येक पर लम्ब एक इकाई (मात्रक) सदिश निकालें। [Find unit vectors perpendicular to each of the following vectors.]

(i) $2 \hat{i}+3 \hat{j}-\hat{k}, \hat{i}+2 \hat{j}+3 \hat{k}$

(ii) $2 \hat{i}-\hat{j}-\hat{k}, 2 \hat{i}-\hat{j}+3 \hat{k}$

(iii) $4 \hat{i}-\hat{j}+3 \hat{k}, 2 \hat{i}+2 \hat{j}-\hat{k}$

Sol :

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

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{\imath} & \hat{\jmath} & \hat{k} \\ 2 & 3 & -1 \\ 1 & 2 & 3\end{array}\right|$

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

$=\hat{i}(9+2)-\hat{j}(6+1)+\hat{k}(4-3)$

$\vec{a} \times \vec{b}=11 \hat{i}-7 \hat{j}+\hat{k}$

$|\vec{a} \times \vec{b}|=\sqrt{11^{2}+(-7)^{2}+1^{2}}$

$=\sqrt{121+49+1}=\sqrt{171}$

$=\sqrt{19 \times  9 }=3\sqrt{19}$

$\vec{a}$ तथा $\vec{b}$  पर लंम्ब इकाई सदिश $=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

$=\frac{11 \hat{i}-7 \hat{j}+\hat{k}}{3 \sqrt{19}}$

$=\frac{7}{3 \sqrt{19}}(11 \hat{\imath}-7 \hat{\jmath}+\hat{k})$

Question 13

निम्नलिखित सदिशों में से प्रत्येक पर लम्ब एक सदिश निकालें।

[Find a vector which is perpendicular to each of the vectors in the following :]

(i) $\hat{i}-\hat{j}+\hat{k}$ तथा (and) $2 \hat{i}+3 \hat{j}-\hat{k}$

(ii) $\hat{i}+\hat{j}-2 \hat{k}$ तथा (and) $2 \hat{i}-2 \hat{j}+\hat{k}$

Sol :

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

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

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

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

$=-2 \hat{\imath}+3 \hat{j}+5 \hat{k}$

Question 14

सदिश $(\vec{a}+\vec{b})$ और $(\vec{a}-\vec{b})$ में से प्रत्येक के लंबवत् मात्रक सदिश ज्ञात कीजिए जहाँ $\vec{a}=\bar{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$ हैं।

Sol :

$\vec{a}=\bar{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+2 \hat{j}+3 \hat{k}$

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

$\vec{a}-\vec{b}=-\hat{j}-2 \hat{k}$

$(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & k \\ 2 & 3 & 4 \\ 0 & -1 & -3\end{array}\right|$

$=\hat{i}\left|\begin{array}{cc}3 & 4 \\ -1 & -2\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & 4 \\ 0 & -2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 3 \\ 0 & -1\end{array}\right|$

$=\hat{i}(-6+4)-\hat{j}(-4-0)+\hat{k}(-2-0)$

$=-2 \hat{i}+4 \hat{j}-2 \hat{k}$

$|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=\sqrt{(-2)^{2}+4^{2}+(-2)^{2}}$ $=\sqrt{4+16+4}=\sqrt{24}$

$=2 \sqrt{6}$

$\vec{a}+\vec{b}$ तथा $\vec{a}-\vec{b}$ पर लंब ईकाई सदिश$=\frac{(\vec{a}+\vec{b})\times\left(\vec{a}-\vec{b}\right)}{[(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})]}$

$=\frac{-2 \hat{i}+4 \hat{j}-2 \hat{k}}{2 \sqrt{6}}$

$=\frac{-2}{2 \sqrt{6}} \hat{i}+\frac{4}{2 \sqrt{6}} \hat{j}-\frac{2}{2 \sqrt{6}} \hat{k}$

$=-\frac{1}{\sqrt{6}} \hat{i}+\frac{2}{\sqrt{6}} \hat{j}-\frac{1}{\sqrt{6}} \hat{k}$

Question 15

सदिशों $\hat{i}+2 \hat{j}+\hat{k}$ तथा $3 \hat{i}+\hat{j}-\hat{k}$ के बीच का कोण ज्ञात करें साथ ही दोनों सदिशों में से प्रत्येक पर लम्ब एक इकाई सदिश भी ज्ञात करें।

Sol :

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

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{\imath} & \hat{J} & \hat{k} \\ 1 & 2 & 1 \\ 3 & 1 & -1\end{array}\right|$

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

$=\hat{i}(-2-1)-\hat{j}(-1-3)+\hat{k}(1-6)$

$=-3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k}$

$|\vec{a} \times \vec{b}|=\sqrt{(-3)^{2}+4^{2}+(-5)^{2}}=\sqrt{9+16+25}$

$=\sqrt{50}=5\sqrt{2}$

$\vec{a}$ तथा $\vec{b}$ दोनो पर लंब इकाई सदिश$=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

$=\frac{-3 \hat{\imath}+4 \hat{j}-5 \hat{k}}{5 \sqrt{2}}$

$=\frac{1}{5 \sqrt{2}}(-3 \hat{\imath}+4 \hat{\jmath}-5 \hat{k})$

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

$|\vec{a}|= \sqrt{1^{2}+2^{2}+1^{2}}=\sqrt{6}$ , $|\vec{b}|=\sqrt{3^{2}+1^{2}+(-1)^{2}}=\sqrt{9+1+1}=\sqrt{11}$

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

$\sin \theta=\frac{5 \sqrt{2}}{\sqrt{6}\times \sqrt{11}}$

$\theta=\sin ^{-1}\left(\frac{5}{\sqrt{33}}\right)$

Question 16

Question 17

$2 \hat{i}-\hat{j}+\hat{k}$ तथा $3 \hat{i}+4 \hat{j}-\hat{k}$ में से प्रत्येक पर लम्ब इकाई सदिश क्या है ? सिद्ध करें कि दोनों सदिशों के बीच के कोण का ज्या $\sqrt{\frac{155}{156}}$ है।

Sol :

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

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

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

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

$=-3 \hat{i}+5 \hat{j}+11 \hat{k}$

$|\vec{a} \times \vec{b}|=\sqrt{(-3)^{2}+5^{2}+11^{2}}=\sqrt{9+25+121}=\sqrt{155}$

$|\vec{a}|=\sqrt{2^{2}+(-1)^{2}+1^{2}}=\sqrt{6}$,$|\vec{b}|=\sqrt{3^{2}+4^{2}+(-1)^{2}}=\sqrt{9+16+1}=\sqrt{26}$

$\vec{a}$ तथा $\vec{b}$ पर लंब इकाई सदिश 

$=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{-3 \hat{\imath}+5 \hat{\jmath}+11 \hat{k}}{\sqrt{155}}$

$=\frac{1}{\sqrt{155}}(-3 \hat{i}+5 \hat{j}+11 \hat{k})$

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

$\sin \theta=\frac{\sqrt{155}}{\sqrt{6} \times \sqrt{26}}$

$\sin \theta=\frac{\sqrt{155}}{\sqrt{156}}$

$\sin \theta=\sqrt{\frac{155}{156}}$

Question 18

यदि बिन्दुएँ A, B, C क्रमश: (1,0,-1),(0,1,-1) तथा (-1,0,1) हैं तो रेखाओं AB तथा AC के बीच का कोण ज्ञात करें।

Sol :

$\overrightarrow{A B}=(0-1) \hat{i}+(1-0) \hat{\jmath}+(-1+1) \hat{k}$

$\overrightarrow{A B}=-\hat{i}+\hat{\jmath}$

$\overrightarrow{A C}=(-1-1) \hat{\imath}+(0-0) \hat{\jmath}+(1+1) \hat{k}$

$\overrightarrow{A C}=-2 \hat{\imath}+2 \hat{k}$

$\overrightarrow{A B} \times \overrightarrow{A C}=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ -1 & 1 & 0 \\ -2 & 0 & 2\end{array}\right|$

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

$=i(2-0)-\hat{\jmath}(-2+0)+\hat{k}(-0+2)$

$=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k}$

$|\overrightarrow{AB} \times \overrightarrow{A C}|=\sqrt{2^{2}+2^{2}+2^{2}}=\sqrt{12}=2 \sqrt{3}$

$|\overrightarrow{A B}|=\sqrt{(-1)^{2}+1^{2}}=\sqrt{2}$

$|AC|=\sqrt{(-2)^{2}+2^{2}}=\sqrt{8}$

$\sin \theta=\frac{|\overrightarrow{A B} \times \overrightarrow{A C}|}{|\overrightarrow{AB}||\overrightarrow{A C}|}$

$\sin \theta=\frac{2 \sqrt{3}}{\sqrt{2} \times \sqrt{8}}$

$\sin \theta=\frac{2 \sqrt{3}}{4}$

$\sin \theta=\frac{\sqrt{3}}{2}$

Question 19

इकाई परिमाण वाले सदिश का घटक निकालें जो सदिशों $2 \hat{i}+\hat{j}-4 \hat{k}$ तथा $3 \hat{i}+\hat{j}-\hat{k}$ पर लम्ब है।

Sol :

माना $\vec{a}=2 \hat{j}+\hat{j}-4 \hat{k}$

$\vec{b}=3 \hat{i}+\hat{j}-\hat{k}$

$\vec{a}$ तथा  $\vec{b}$ पर लंब सदिश $=\vec{a} \times \vec{b}$

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

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

$=\hat{i}(-1+4)-\hat{j}(-2+12)+\hat{k}(2-3)$

$=3 \hat{i}-10 \hat{j}-\hat{k}$

$|\vec{a} \times \vec{b}|=\sqrt{3^{2}+(-10)^{2}+(-1)^{2}}=\sqrt{9+100+1}=\sqrt{110}$

$\vec{a}$ तबा $\vec{b}$ पर लंब इकाई सदिश $=\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}=\frac{3 \hat{i}-10 \hat{j}-\hat{k}}{\sqrt{110}}$

$=\frac{1}{\sqrt{110}}(3 \hat{i}-10 \hat{j}-\hat{k})$

Question 20

यदि तीन बिन्दुओं $\mathrm{A}, \mathrm{B}, \mathrm{C}$ के स्थिति सदिश क्रमश: $2 \hat{i}+4 \hat{j}-\hat{k}, \hat{i}+2 \hat{j}-3 \hat{k}$ तथा $3 \hat{i}+\hat{j}+2 \hat{k}$ हैं, तो तल $\mathrm{ABC}$ पर लम्ब एक सदिश निकालें ।.

Sol :

$\overrightarrow{B C}=(3 \hat{i}+\hat{j}+2 \hat{k})-\left(\hat{i}+2 \hat{j}-3 \hat{k}\right)$

$=3 \hat{i}+\hat{j}+2 \hat{k}-\hat{i}-2 \hat{\jmath}+3 \hat{k}$

$\overrightarrow{B C}=2 \hat{i}-\hat{j}+5 \hat{k}$

$\overrightarrow{B A}=(2 \hat{i}+4 \hat{j}-\hat{k})-(\hat{i}+2 \hat{j}-3 \hat{k})$

$=2 \hat{i}+4 \hat{j}-\hat{k}-\hat{\imath}-2 \hat{\jmath}+3 \hat{k}$

$\overrightarrow{B A}=\hat{\imath}+2 \hat{j}+2 \hat{k}$

तल ABC पर लंब सदिश$=\overrightarrow{B A} \times \overrightarrow{B C}$

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

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

$=\hat{i}(10+2)-\hat{j}(5-4)+\hat{k}(-1-4)$☺

$=12 \hat{i}-\hat{j}-5 \hat{k}$

तल ABC पर लंब सदिश$=-(12 \hat{i}-\hat{\jmath}-5 \hat{k})$

$=-12 \hat{\imath}+\hat{j}+5 \hat{k}$

Question 21

दिया है (Given) $\vec{a}=\frac{1}{7}(2 \hat{i}+3 \hat{j}+6 \hat{k}), \vec{b}=\frac{1}{7}(3 \hat{i}-6 \hat{j}+2 \hat{k})$ तथा (and) $\vec{c}=\frac{1}{7}(6 \hat{i}+2 \hat{j}-3 \hat{k})$.

दिखाएँ कि $\vec{a}, \vec{b}, \vec{c}$ परस्पर लम्ब इकाई सदिश हैं तथा $\vec{a} \times \vec{b}=\vec{c}$

Sol :

$\vec{a}=\frac{2}{7} \hat{\imath}+\frac{3}{7} \hat{j}+\frac{6}{7} \hat{k}$

$\vec{b}=\frac{3}{7} \hat{i}-\frac{6}{7} \hat{j}+\frac{2}{7} \hat{k}$

$\vec{c}=\frac{6}{7} \hat{\imath}+\frac{2}{7} \hat{j}-\frac{3}{7} \hat{k}$

$|\vec{a}|=\sqrt{\left(\frac{2}{7}\right)^{2}+\left(\frac{3}{7}\right)^{2}+\left(\frac{6}{7}\right)^{2}}=$

$=\sqrt{\frac{4}{49}+\frac{9}{49}+\frac{36}{49}}=\sqrt{1}$=1

$|\vec{b}|=\sqrt{\left(\frac{3}{7}\right)^{2}+\left(-\frac{6}{7}\right)^{2}+\left(\frac{2}{7}\right)^{2}}$

=1

$|\vec{c}|=\sqrt{\left(\frac{6}{7}\right)^{2}+\left(\frac{2}{7}\right)^{2}+\left(\frac{-3}{7}\right)^{2}}=1$

$\vec{a} \cdot \vec{b}=\frac{6}{49}-\frac{18}{49}+\frac{12}{49}=\frac{6-18+12}{49}=0$

$\therefore \vec{a} \perp \vec{b}$

$\vec{b} \cdot \vec{c}=\frac{18}{49}-\frac{12}{49}-\frac{6}{49}=0$

∴$\vec{b} \perp \vec{c}$

$\vec{c} \cdot \vec{0}=\frac{12}{49}+\frac{6}{49}-\frac{18}{49}=0$

∴$\vec{c} \perp \vec{a}$

∴$\vec{a}, \vec{b}$ तथा $\vec{c}$ परसपर लंब इकाई सदिश है।

अब,$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ \frac{2}{7} & \frac{3}{7} & \frac{6}{7} \\ \frac{3}{7} & \frac{-6}{7} & \frac{2}{7}\end{array}\right|$

$=\frac{1}{49}\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 3 & -6 & 2\end{array}\right|$

$=\frac{1}{49}\left[\hat{i}\left|\begin{array}{cc}3 & 6 \\ -6 & 2\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}2 & 6 \\ 3 & 2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 3 \\ 3 & -6\end{array}\right|\right]$

$=\frac{1}{49}[\hat{i}(6+36)-\hat{j}(4-18)+\hat{k}(-12-9]]$

$=\frac{1}{49}[42 \hat{i}+14 \hat{j}-21 \hat{k}]$○

$=\frac{1}{49} \times 7[6 \hat{\jmath}+2 \hat{j}-3 \hat{k}]$

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

Question 22

यदि (If) $\vec{a}=7 \hat{i}+3 \hat{j}-5 \hat{k}, \vec{b}=2 \hat{i}+5 \hat{j}-\hat{k}$ तथा (and)

$\vec{c}=-\hat{i}+2 \hat{j}+4 \hat{k}$ तो सत्यापित करें कि (Then verify that)

$\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$

Sol :

LHS

$\vec{a} \times(\vec{b}+\vec{c})=(7 \hat{i}+3 \hat{j}-5 \hat{k}) \times(\hat{i}+7 \hat{j}+3 \hat{k})$

$=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 7 & 3 & -5 \\ 1 & 7 & 3\end{array}\right|$

RHS

$\vec{a} \times \vec{b}+\vec{a} \times \vec{c}=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 7 & 3 & -5 \\ 2 & 5 & -1\end{array}\right|+\left|\begin{array}{ccc}\hat{i} & \hat{\jmath} & \hat{k} \\ 7 & 3 & -5 \\ -1 & 2 & 4\end{array}\right|$

$=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & -5 \\ 2+(-1) & 5+2 & -1+4\end{array}\right|=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 7 & 3 & -5 \\ 1 & 7 & 3\end{array}\right|$

$\therefore \vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$

Question 23

माना कि (Let) $\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}, \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}$ तथा (and) $\vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$. तो दिखाएँ कि (Then show that) $\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c} .$

Sol :

LHS

$\vec{a} \times(\vec{b}+\vec{c})=\left(a_{1} \hat{i}+a_{2} \hat{\jmath}+a_{3} \hat{k}\right) \times\left[\left(b_{1}+c_{1}\right) \hat{i}+\left(b_{2}+c_{2}\right) \hat{j}+(b_3+c_3)\hat{k}\right]$

$=\left|\begin{array}{ccc}\hat{i} & \hat{\jmath} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3}\end{array}\right|$

$=\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|+\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right|$

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

Question 24

यदि (If) $\vec{a}=2 \hat{i}+5 \hat{j}-7 \hat{k}, \vec{b}=-3 \hat{i}+4 \hat{j}+\hat{k}$ तथा (and)

$\vec{c}=\hat{i}-2 \hat{j}-3 \hat{k}$, दिखाएँ कि (show that) $(\vec{a} \times \vec{b}) \times \vec{c}$ तथा (and)

$\vec{a} \times(\vec{b} \times \vec{c})$. समान नहीं हैं। (are not same)

Sol :

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 5 & -7 \\ -3 & 4 & 1\end{array}\right|=\hat{i}\left|\begin{array}{cc}5 & -7 \\ 4 & 1\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & -7 \\ -3 & 1\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 &5 \\ -3 & 4\end{array}\right|$

$=\hat{i}(5+28)-\hat{j}(2-21)+\hat{k}(8+15)$

$=33 \hat{i}+19 \hat{j}+23 \hat{k}$

$\vec{b} \times \vec{c}=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ -3 & 4 & 1 \\ 1 & -2 & -3\end{array}\right|$

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

$=\hat{i}(-12+2)-\hat{j}(9-1)+\hat{k}(6-4)$

$=-10 \hat{i}-8 \hat{j}+2 \hat{k}$

$(\vec{a} \times \vec{b}) \times \vec{c}=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 33 & 19 & 23 \\ 1 & -2 & -3\end{array}\right|$

$=\hat{i}\left|\begin{array}{cc}19 & 23 \\ -2 & -3\end{array}\right|-\hat{j}\left|\begin{array}{cc}33 & 23 \\ 1 & -3\end{array}\right|+\hat{k}\left|\begin{array}{cc}33 & 19 \\ 1 & -2\end{array}\right|$

$=\hat{i}(-57+46)-\hat{j}(-99-23)+\hat{k}(-66-19)$

$=-11 \hat{i}+122 \hat{\jmath}-85 \hat{k}$

$\vec{a} \times(\vec{b} \times \vec{c})=\left|\begin{array}{ccc}\hat{\hat{\imath}} & \hat{\jmath} & \hat{k} \\ 2 & 5 & -7 \\ -10 & -8 & 2\end{array}\right|=\hat{\imath}\left|\begin{array}{cc}5 & -7 \\ -8 & 2\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}2 & -7 \\ -10 & 2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 5 \\ -10 & -8\end{array}\right|$

$=\hat{i}(10-56)-\hat{\jmath}(4-70)+\hat{k}(-16+50)$

$=-46 \hat{i}+66\hat{\jmath}+34 \hat{k}$

$\therefore(\vec{a} \times \vec{b}) \times \vec{c} \neq \vec{a} \times(\vec{b} \times \vec{c})$

Question 25

यदि (If) $\vec{a}=2 \hat{i}+2 \hat{j}-\hat{k}$, तथा (and) $\vec{b}=3 \hat{i}-\hat{j}-\hat{k}$ तथा (and) $\vec{c}=\hat{i}+2 \hat{j}-3 \hat{k}$. तो सत्यापित करें कि (then verify that)

$\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$.

Sol :

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

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

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

$=\hat{i}(3+2)-\hat{j}(-9+1)+\hat{k}(6+1)$

$=5 \hat{i}+8 \hat{j}+7 \hat{k}$

$\vec{a} \cdot \vec{c}$

=2+4+3=9,

$ \vec{a} \cdot \vec{b}$

=6-2+1=5

LHS

$\vec{a} \times\left(\vec{b} \times \vec{c}\right)=\left|\begin{array}{ccc}\hat{\imath} & \hat{j} & \hat{k} \\ 2 & 2 & -1 \\ 5 & 8 & 7\end{array}\right|=\hat{i}\left|\begin{array}{cc}2 & -1 \\ 8 & 7\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & -1 \\ 5 & 7\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 2 \\ 5 & 8\end{array}\right|$

$=\hat{\imath}(14+8)-\hat{j}(14+5)+\hat{k}(16-10)$

$=22 \hat{i}-19 \hat{j}+6 \hat{k}$

RHS

$(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=9 \cdot(3 \hat{i}-\hat{j}-\hat{k})-5(\hat{i}+2 \hat{j}-3 \hat{k})$

$=27 \hat{i}-9\hat{j}-9 \hat{k}-5 \hat{i}-10 \hat{j}+15 \hat{k}$○

$=22 \hat{i}-19 \hat{j}+6 \hat{k}$

$\therefore \vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \overrightarrow{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}$

Question 26

$\mathrm{P}(-\hat{i}+2 \hat{j}+6 \hat{k})$ से $\mathrm{A}(2 \hat{i}+3 \hat{j}-4 \hat{k})$ तथा $\mathrm{B}(8 \hat{i}+6 \hat{j}-8 \hat{k})$ को मिलाने वाली रेखा पर खींचे गये लम्ब की लम्बाई प्राप्त करें।

Sol :

माना PD⟂AB है।

$\begin{aligned} \overrightarrow{A B} &=(8 \hat{i}+6 \hat{j}-8 \hat{k})-(2 \hat{i}+3 \hat{j}-4 \hat{k}) \\ &=8 \hat{i}+6 \hat{j}-8 \hat{k}-2 \hat{i}-3 \hat{j}+4 \hat{k} \\ \overrightarrow{A B} &=6 \hat{i}+3 \hat{j}-4 \hat{k} \end{aligned}$

$\begin{aligned} \overrightarrow{A P} &=(-\hat{\imath}+2 \hat{j}+6 \hat{k})-(2 \hat{\jmath}+3 \hat{j}-4 \hat{k}) \\ &=-\hat{i}+2 \hat{j}+6 \hat{k}-2 \hat{i}-3 \hat{j}+4 \hat{k} \\ \overrightarrow{A P} &=-3 \hat{i}-\hat{j}+10 \hat{k} \end{aligned}$

$\overrightarrow{A B} \times \overrightarrow{A P}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 6 & 3 & -4 \\ -3 & -1 & 10\end{array}\right|$

$=\hat{i}\left|\begin{array}{cc}3 & -4 \\ -1 & 10\end{array}\right|-\hat{j}\left|\begin{array}{cc}6 & -4 \\ -3 & 10\end{array}\right|+\hat{k}\left|\begin{array}{cc}6 & 3 \\ -3 & -1\end{array}\right|$

$=\hat{i}(30-4)-\hat{j}(60-12)+\hat{k}(-6+9)$

$=26 \hat{i}-48 \hat{j}+3 \hat{k}$

$|\overrightarrow{A B} \times \overrightarrow{AP}|=\sqrt{26^{2}+(-48)^{2}+3^{2}}$

$=\sqrt{676+2304+9}$

$=\sqrt{2989}=\sqrt{7 \times 7 \times 61}=7\sqrt{61}$

$|\overrightarrow{A B}|=\sqrt{6^{2}+3^{2}+(-4)^{2}}$

$=\sqrt{36+3+16}=\sqrt{61}$

$P D=\frac{|\overrightarrow{AB} \times \overrightarrow{A P}|}{|\overrightarrow{AB}|}=\frac{7 \sqrt{61}}{\sqrt{61}}$=7

Question 27

माना कि $\vec{a}=(3,-1,0)$ तथा $\vec{b}=\left(\frac{1}{2}, \frac{3}{2}, 1\right) \cdot \vec{a} \times \vec{c}=4 \vec{b}$ तथा $\vec{a} \cdot \vec{c}=1$ को संतुष्ट करता हुआ सदिश $\vec{c}$ ज्ञात करें।

Sol :

माना $\vec{a}=3 \hat{\imath}-\hat{\jmath}$ , $\vec{b}=\frac{1}{2} \hat{i}+\frac{3}{2} \hat{j}+\hat{k}$

$\vec{z}=x \hat{i}+y \hat{j}+z \hat{k}$

$\vec{a} \times \vec{c}=4 \vec{b}$

$\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 0 \\ x & y & z\end{array}\right|=4\left(\frac{1}{2} \hat{i}+\frac{3}{2} \hat{\jmath}+\hat{k}\right)$

$i\left|\begin{array}{cc}-1 & 0 \\ y & z\end{array}\right|-\hat{\jmath}\left|\begin{array}{ll}3 & 0 \\ x & z\end{array}\right|+\hat{k}\left|\begin{array}{cc}3 & -1 \\ x & y\end{array}\right|=2 \hat{i}+6 \hat{j}+4 \hat{k}$

$\hat{i}(-z-0)-\hat{\jmath}(3 z-0)+\hat{k}(3{y}+x)=2 \hat{i}+6 \hat{j}+4 \hat{k}$

$-z \hat{i}-3 z \hat{j}+(x+3 y) \hat{k}=2 \hat{i}+6 \hat{j}+4 \hat{k}$

-z=2⇒z=-2, x+3y=4...(i)

∵$\vec{a} \cdot \vec{c}=1$

$(3 \hat{i}-\hat{\jmath}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})=1$

3x-y=1....(ii)×3

समीकरण (i) तथा (ii) 

$\begin{aligned}x+3y&=4\\9x-3y&=3 \\ \hline 10x&=7\end{aligned}$

$x=\frac{7}{10} \Rightarrow y=\frac{11}{10}$

∴$\vec{c}=\frac{7}{10} \hat{i}+\frac{11}{10} \hat{j}+(-2) \hat{k}$

$=\frac{7 \hat{i}+10 \hat{j}-20 \hat{k}}{10}$

$=\frac{1}{10}(7 \hat{i}+10 \hat{\jmath}-20 \hat{k})$

Question 28

यदि $\vec{a}=(0,1,-1)$ तथा $\vec{c}=(1,1,1)$ दिए गए सदिश हैं तो एक सदिश $\vec{b}$ ज्ञात करें जो $\vec{a} \times \vec{b}+\vec{c}=0$ तथा $\vec{a} \cdot \vec{b}=3$ को संतुष्ट करता है।

Sol :

माना $\vec{b}=x \hat{i}+y \hat{\jmath}+z \hat{k}$

$\vec{a}=\hat{\jmath}-\hat{k}, \vec{c}=\hat{i}+\hat{j}+\hat{k}$

$\vec{a} \times \vec{b}+\vec{c}=\vec{0}$

$\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & -1 \\ x & y & z\end{array}\right|+(\hat{i}+\hat{j}+\hat{k})=\overrightarrow{0}$

$\hat{i}\left|\begin{array}{cc}1 & -1 \\ y & z\end{array}\right|-\hat{\jmath}\left|\begin{array}{cc}0 & -1 \\ x & z\end{array}\right|+\hat{k}\left|\begin{array}{ll}0 & 1 \\ x & y\end{array}\right|+\hat{i}+\hat{j}+\hat{k}=\overrightarrow{0}$

$\hat{\imath}(z+y)-\hat{j}(0+x)+\hat{k}(0-x)+\hat{i}+\hat{j}+\hat{k}=\overrightarrow{6}$

$\hat{i}(z+y)-\hat{j}(0+x)+\hat{k}(0-x)+\hat{\imath}+\hat{j}+\hat{k}=\vec{0}$

$(1+y+z) \hat{i}+(1-x) \hat{j}+(1-x) \hat{k}=0 \hat{i}+0 \hat{j}+0 \cdot \hat{k}$

1+y+z=0 , 1-x=0

y+z=-1...(i) , x=1

$\vec{a} \cdot \vec{b}=3 \Rightarrow(\hat{j}-\hat{k}) \cdot(x \hat{i}+y \hat{j}+z \hat{k})=3$

y-z=3

समीकरण (i) तथा (ii) से,

$\begin{aligned}y+z&=-1 \\y-z&=3 \\ \hline2 y&=2\end{aligned}$

y=1 , z=-2

∴$\vec{b}=1 \cdot \hat{i}+1 \cdot \hat{j}+(-2) \hat{k} \Rightarrow \vec{b}=\hat{i}+\hat{\jmath}-2 \hat{k}$

Question 29

दिखाएँ कि $(\vec{a}-\vec{d}) \times(\vec{b}-\vec{c})+(\vec{b}-\vec{d}) \times(\vec{c}-\vec{a})+(\vec{c}-\vec{d}) \times(\vec{a}-\vec{b}), \vec{d}$ से स्वतंत्र है ।

Sol :

$(\vec{a}-\vec{d}) \times(\vec{b}-\vec{c})+(\vec{b}-\vec{d}) \times(\vec{c}-\vec{a})+(\vec{c}-\vec{d}) \times(\vec{a}-\vec{b})$

$= \vec{a} \times \vec{b}-\vec{a} \times \vec{c}-\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{d} \times \vec{c}$ $+\vec{d} \times \vec{a}+\vec{c} \times \vec{a}-\vec{c} \times \vec{b}-\vec{d} \times \vec{a}+\vec{d} \times \vec{b}$

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

$=2 \vec{a} \times \vec{b}+2 \vec{b} \times \vec{c}+2 \vec{c} \times \vec{a}$

$=2(\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}), \vec{d}$ से स्वतंत्र है।

Question 30 

(i) सिद्ध करें कि (Prove that)

$(\vec{a}+3 \vec{b}) \times(\vec{a}+\vec{b})+(3 \vec{a}-5 \vec{b}) \times(\vec{a}-\vec{b})=\overline{0}$

Sol :

LHS

$(\vec{a}+3 \vec{b}) \times(\vec{a}+\vec{b})+(3 \vec{a}-5 \vec{b})\times(\vec{a}-\vec{b})$

$=\vec{a} \times \vec{a}+\vec{a} \times \vec{b}+3 \vec{b} \times \vec{a}+3 \vec{b} \times \vec{b}+3 \vec{a} \times \vec{a}$ $-3 \vec{a} \times \vec{b}-5 \vec{b} \times \vec{a}+5 \vec{b} \times \vec{b}$

$=\overrightarrow{0}-2 \vec{a} \times \vec{b}-2 \vec{b} \times \vec{a}+3 \times \vec{0}+3 \times \overrightarrow{0}+5 \times \overrightarrow{0}$

$=-2 \vec{a} \times \vec{b}+2 \vec{a} \times \vec{b}$

=0

(ii) सिद्ध करें कि (Prove that) $|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=2 a b$ यदि (if) $\vec{a} \perp \vec{b}$.

Sol :

यदि $\vec{a} \perp \vec{b} \Rightarrow|\vec{a} \times \vec{b}|=a b \sin 90^{\circ}$

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

LHS

$|(\vec{a}+\vec{b}) \times(\vec{a}-\vec{b})|=\mid \vec{a} \times \vec{a}-\vec{a} \times \vec{b}+\vec{b} \times \vec{a}-\vec{b} \times \vec{b}|$

$=|\vec{0}-\vec{a} \times \vec{b}-\vec{a} \times \vec{b}-\vec{0}|$

$=|-2 \vec{a} \times \vec{b}|$

$=2|\vec{a} \times \vec{b}|$

=2ab

Question 31

$\vec{a}, \vec{b}, \vec{c}$ शून्येत्तर सदिश हैं । यदि $\vec{a} \times \vec{b}=\vec{a} \times \vec{c}$ तथा $\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}$ तो दिखाएँ कि $\vec{b}=\vec{c}$.

Sol :

$\begin{aligned} \vec{a} \times \vec{b}=\vec{a} & \times \vec{c} \\ \vec{a} \times \vec{b}-\vec{a} \times \vec{c} &=\overrightarrow{0} \\ \vec{a} \times(\vec{b}-\vec{c}) &=\vec{b}  \end{aligned}$

$\vec{b}-\vec{c} =\vec{o}$ $(\because \vec{a} \neq \vec{b})$

$\vec{b}=\vec{c}$ का $\vec{a} \|(\vec{b}-\vec{c})$

CASE-I

$\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}$

$\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c}=0$

$\vec{a} \cdot(\vec{b}-\vec{c})=0$ $(\because \vec{a} \neq \vec{o})$

$\vec{b}-\vec{c}=0$

$\vec{b}=\vec{c}$ या $\vec{a} \perp(\vec{b}-\vec{c})$

$\therefore \vec{b}=\vec{c}$

Question 32

निम्नलिखित का मान ज्ञात करें [Find the value of]

(i) $|(\hat{i}+\hat{j}) \times(\hat{i}+2 \hat{j}+\hat{k})|$

Sol :

$(\hat{i}+\hat{j}) \times(\hat{\imath}+2 \hat{\jmath}+\hat{k})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 0 \\ 1 & 2 & 1\end{array}\right|$

$=\hat{\imath}\left|\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right|-\hat{j}\left|\begin{array}{ll}1 & 0 \\ 1 & 1\end{array}\right|+\hat{k}\left|\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right|$

$=\hat{i}(1-0)-\hat{j}(1-0)+\hat{k}(2-1)$

$=\hat{i}-\hat{\jmath}+\hat{k}$

$|(\hat{i}+\hat{j}) \times(\hat{i}+2 \hat{\jmath}+\hat{k})|$

$=\sqrt{1^{2}+(-1)^{2}+1^{2}}=\sqrt{1+1+1}=\sqrt{3}$

(ii) $|(3 \hat{i}+\hat{j}) \times(2 \hat{i}-\hat{j})|$

Sol :

Question 33

(i) $|\hat{i} \times(\hat{i}+\hat{j}+\hat{k})|$

Sol :

$|\hat{i} \times(\hat{i}+\hat{j}+\hat{k})|=|\hat{i} \times \hat{i}+\hat{j} \times \hat{j}+\hat{i} \times \hat{k}|$

$=|\overrightarrow{0}+\hat{k}-\hat{j}|$

$=|\hat{k}-\hat{j}|$

$=\sqrt{1^{2}+(-1)^{2}}=\sqrt{1+1}=\sqrt{2}$

(ii) $|\hat{i} \times \hat{j}|+|\hat{j} \times \hat{k} \mid$.

Sol :

$=|\hat{k}|+|\hat{i}|=\sqrt{1^{2}}+\sqrt{1^{2}}$

=2

Question 34

सिद्ध करें कि [Prove that]

(i) $(2 \hat{i}+3 \hat{j}) \times(\hat{i}+2 \hat{j})=\hat{k}$

Sol :

LHS

$(2 \hat{\imath}+3 \hat{j}) \times(\hat{i}+2 \hat{j})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 0 \\ i & 2 & 0\end{array}\right|$

$=\hat{k}\left|\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right|=\hat{k}(4-3)=\hat{k}$

(ii) $(2 \vec{a}-\vec{b}) \times(\vec{a}+2 \vec{b})=5 \vec{a} \times \vec{b}$.

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

$=2 \vec{a} \times \vec{a}+4 \vec{a} \times \vec{b}-\vec{b} \times \vec{a}-2 \vec{b} \times \vec{b}$

$=2(\vec{0})+4 \vec{a} \times \vec{b}+\vec{a} \times \vec{b}-2(\vec{0})$

$=5 \vec{a} \times \vec{b}$