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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}}
=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}

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