KC Sinha Solution Class 12 Chapter 27 Vector or Cross product of Two vectors Exercise 27.1 (Q31-Q34)

Exercise 27.1

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 :

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

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

Sol :

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

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

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

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