Exercise 27.1
Question 1
यदि दो सदिश $\vec{a}$ और $\vec{b}$ इस प्रकार हैं कि $|\vec{a}|=2,|\vec{b}|=7$ तथा $a \times \vec{b}=3 \hat{i}+2 \hat{j}+6 \hat{k}$ तो $\vec{a}$ और $\vec{b}$ के बीच का कोण ज्ञात करें।
Ans:
$\begin{aligned}|\vec{a}|=2,|\vec{b}|=7, \quad \vec{a} \times \vec{b} &=3 \hat{i}^{2}+2 \hat{j}+6 k \\|\vec{a} \times \vec{b}| &=\sqrt{3^{2}+2^{2}+6^{2}} \\ &=\sqrt{9+4+31}=\sqrt{49}=7 \end{aligned}$
$\sin \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}||\vec{b}|}$
$\sin 7=\frac{x}{2 \times 7}$
$\sin \theta=\frac{1}{2}$
Question 2
दिया है, $|\vec{a}|=10,|\vec{b}|=2$ तथा $\vec{a} \cdot \vec{b}=12$ तो $|\vec{a} \times \vec{b}|$ निकालें ।
Ans: $\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\cos \theta=\frac{12}{10 \times 2}$
$\begin{aligned} \cos \theta &=\frac{3}{5} \\ \sin \theta &=\sqrt{1-\cos ^{2} \theta} \end{aligned}$
$=\sqrt{1-\left(\frac{3}{5}\right)^{2}}$
$=\sqrt{1-\frac{9}{25}}=$$\sqrt{\frac{25-9}{25}}$
$\sin \theta=\frac{4}{5}$
$\frac{|\vec{a} \times \vec{b}|}{|\vec{a}| \cdot|\vec{b}|}=\frac{4}{5}$
$\frac{|\vec{a} \times \vec{b}|}{10 \times 2}=\frac{4}{5}$
$|\vec{a} \times \vec{b}|=\frac{4}{5} \times 20$
=16
Question 3
$\vec{a} \cdot \vec{b}$ ज्ञात करें यदि $|\vec{a}|=2,|\vec{b}|=5,|\vec{a} \times \vec{b}|=8$.
Ans: $|\vec{a}|=2,|\vec{b}|=5$
$|\vec{a} \times \vec{b}|=8$
$\sin \theta=\frac{(\vec{a} \times \vec{b} \mid}{|\vec{a}||\vec{b}|}$
$\sin θ =\frac{8}{2 \times 5}$
$\sin \theta=\frac{4}{5}$
$\begin{aligned} \cos \theta &=\sqrt{1-\sin ^{2} \theta} \\ \cos \theta &=\sqrt{1-\left(\frac{4}{5}\right)^{2}} \\ &=\sqrt{1-\frac{16}{25}}=\sqrt{\frac{25-16}{25}} \end{aligned}$
$\begin{aligned} \cos \theta &=\frac{3}{5} \\ \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} &=\frac{3}{5} \end{aligned}$
$\frac{\vec{a} \cdot \vec{b}}{2 \times 5}=\frac{3}{5} \Rightarrow \vec{a}-\vec{b}$
=$\frac{3}{5} \times 2 \times 5$
= $\vec{a} \cdot \vec{b}= ± 6$
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।
Ans: $|\vec{a}|=5,|\vec{b}|=4,|\vec{a} \cdot \vec{b}|=10$
$\cos \theta=\frac{\hat{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$
$\operatorname{Cos} \theta=\frac{10}{5 \times 4}$
=$\cos \theta=\frac{1}{2}$
$\theta=\frac{\pi}{3}$
$\operatorname{son} \theta=\frac{|\vec{a} \times \vec{b}|}{|\vec{a}|\left|\vec{b}\right|}$
$\frac{|\vec{a} \times \vec{b}|}{5 \times 4}=\sin \frac{\pi}{3}$
$\frac{|\vec{a} \times \vec{b}|}{20}=\frac{\sqrt{3}}{2}$
$|\vec{a} \times \vec{b}|=10 \sqrt{3}$
Question 5
तीन 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 $|\vec{b}|=1,|\vec{c}|=|\vec{a}|$
Ans: $\vec{a} \times \vec{b}=\vec{c}, \vec{b} \times \vec{c}=\vec{a}$
$\vec{c} \perp \vec{a}, \vec{c} \perp \vec{b} \quad$ तबा $\vec{a} \perp \vec{b}, \vec{a} \perp \vec{c}$
$\because \vec{a} \times \vec{b}=\vec{c} \quad$ तथा $\vec{b} \times \vec{c}=\vec{a}$
$|\vec{a} \times \vec{b}|=|\vec{c}|, \quad|\vec{b} \times \vec{c}|=|\vec{a}|$
$|\vec{a}||\vec{b}| \cdot \sin \frac{\pi}{2}=|\vec{c}| \quad,|\vec{b}||\vec{c}| \cdot \sin \frac{\pi}{2}=|\vec{a}|$
$|\vec{a}||\vec{b}|=|\vec{c}|$,...........(1)
$|\vec{b}||\vec{c}|=|\vec{a}|$.......(2)
1 मे 2 से गु़णा करने पर
$|\vec{a}||\vec{b}||\vec{b}||\vec{c}|=|\vec{c}||\vec{a}|$
$\begin{aligned}|\vec{b}|^{2} &=1 \\|\vec{b}| &=1 \end{aligned}$
समीकरण 1 से
$|\vec{a}||\vec{b}|=|\vec{c}|$
$|\vec{a}| \cdot 1=|\vec{c}| \Rightarrow|\vec{c}|=|\vec{a}|$
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) $\vec{a}=\hat{i}-7 \hat{j}+7 \hat{k}$ तथा (and) $\vec{b}=3 \hat{i}-2 \hat{j}+2 \hat{k}$
Ans: (i) $\vec{a}=2 \hat{i}+\hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+5 \hat{j}-2 k$
$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 3 \\ 3 & 5 & -2\end{array}\right|$
$=\hat{i}\left|\begin{array}{cc}1 & 3 \\ 5 & -2\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & 3 \\ 3 & -2\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & 1 \\ 3 & 5\end{array}\right|$
$\begin{aligned} &=\hat{i}(-1-15)-\hat{j}(-4-9)+\hat{k}(10-3) \\ \vec{a} \times b &=-17 \hat{i}+13 \hat{j}+7 \hat{k} \end{aligned}$
$\begin{aligned}(\vec{a} \times \vec{b}) &=\sqrt{(-17)^{2}+(13)^{2}+7^{2}} \\ &=\sqrt{289+169+49}= \end{aligned}$ $\sqrt{507}$
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 कै जो $\vec{a}$ और $\vec{b}$ दोनों पर लम्ब है।
Ans: $\vec{o}=2 \hat{i}-\hat{\jmath}+\hat{k}, \quad \vec{b}=3 \hat{i}+4 \hat{j}-\hat{k}$
$\vec{a} \times \vec{b}=\left|\begin{array}{rrr}\hat{i} & \hat{j} & \hat{k} \\ 2 & -1 & 1 \\ 3 & 4 & -1\end{array}\right|$
$=\hat{i}\left|\begin{array}{cc}-1 & 1 \\ 4 & -1\end{array}\right|-\hat{j}\left|\begin{array}{cc}2 & 1 \\ 3 & -1\end{array}\right|+\hat{k}\left|\begin{array}{cc}2 & -1 \\ 3 & 4\end{array}\right|$
$\begin{aligned} &=\hat{j}(1-4)-\hat{j}(-2-3)+\hat{k}(8+3) \\ \vec{a} \times \vec{b} &=-3 \hat{i}+5 \hat{j}+11 \hat{k} \end{aligned}$
$\vec{a} \cdot(\vec{a} \times \vec{b})=(2 \hat{\imath}-\hat{\jmath}+\hat{k}) \cdot\left(-3 \hat{i}^{2}+5 \hat{j}+11 \hat{k}^{2}\right)=-6-5+11=0$
$\vec{b} \cdot(\vec{a} \times \vec{b})=(3 \hat{i}+4 \hat{j}-\hat{k}) \cdot(-3 \hat{i}+5 \hat{j}+11 \hat{k})=-9+20-11=0$
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 ।
Ans: $\vec{a}=7 \hat{\imath}+3 \hat{j}-5 \hat{k}, \vec{b}=2 \hat{u}+5 j^{7}-k, \vec{c}=-\hat{\imath}+2 \hat{j}+4 \hat{k}$
$(\vec{a}-\vec{b}) \times(\vec{c}-\vec{b})=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 5 & -2 & -4 \\ -3 & -3 & 5\end{array}\right|$
$=\hat{j}\left|\begin{array}{cc}-2 & -4 \\ -3 & 5\end{array}\right|-\hat{j}\left|\begin{array}{cc}5 & -4 \\ -3 & 5\end{array}\right|+\hat{k}\left|\begin{array}{cc}5 & -2 \\ -3 & -3\end{array}\right|$
$=\hat{i}(-10-12)-\hat{\jmath}(25-12)+\hat{k}(-15-6)$
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