Explain solution RD Sharma class 12 chapter Vector or Cross Product exercise 24.1 question 32 maths

Answers (1)

Answer:

 No, take any two collinear vectors

Hint:

To solve this we let

\hat{i}+\hat{j}+\hat{k}=\overrightarrow{0} ; 2 \hat{i}+2 \hat{j}+2 \hat{k}=\overrightarrow{0}

Given: \vec{a}=0 ; \vec{b}=0 \text { then } \vec{a} \times \vec{b}=0

Solution:

Statement:

If either \vec{a}=\vec{b} \text { or } \vec{b}=\overrightarrow{0} \Rightarrow \vec{a} \times \vec{b}=0

Converse: \vec{a}=0 ; \vec{b}=0 \text { then } \vec{a} \times \vec{b}=0

Let 

\begin{aligned} &\vec{a}=\hat{i}+\hat{j}+\hat{k}=(1,1,1) \\\\ &\vec{b}=2 \hat{i}+2 \hat{j}+2 \hat{k}=(2,2,2) \end{aligned}

\begin{aligned} &\vec{a} \times \vec{b}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 2 & 2 \end{array}\right| \\\\ &=\hat{i}(2-2)+\hat{j}(2-2)+\hat{k}(2-2) \\\\ &=\vec{o} \end{aligned}

But  \vec{a} \neq 0 \text { and } \vec{b} \neq 0

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