#### Provide solution for RD Sharma Maths Class 12 Chapter 24 Vector or Cross Product exercise Very Short Answer Question, question 13.

ANSWER: $\overrightarrow{0}$

HINT:

Cross product the value and then do dot product

GIVEN:

Two vectors $\overrightarrow{a}\; \text{and}\: \overrightarrow{b}$ is given

SOLUTION:

Let

\begin{aligned} &\vec{a}=a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k} \\ &\vec{b}=a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k} \end{aligned}

Now,

\begin{aligned} &\operatorname{For}(\vec{a} \times \vec{b}) \cdot \vec{b} \\ &\begin{aligned} (\vec{a} \times \vec{b}) &=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right| \\ &=\hat{i}\left(b_{1} c_{2}-b_{2} c_{1}\right)-\hat{j}\left(a_{1} c_{2}-a_{2} c_{1}\right)+\hat{k}\left(a_{1} b_{2}-a_{2} b_{1}\right) \end{aligned} \end{aligned}

Now for

\begin{aligned} (\vec{a} \times \vec{b}) \cdot \vec{b} &=\left[\hat{i}\left(b_{1} c_{2}-b_{2} c_{1}\right)-\hat{j}\left(a_{1} c_{2}-a_{2} c_{1}\right)+\hat{k}\left(a_{1} b_{2}-a_{2} b_{1}\right)\right] \cdot\left[a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right] \\ &=a_{2} b_{1} c_{2}-a_{2} b_{2} c_{1}-a_{1} b_{2} c_{2}+a_{2} b_{2} c_{1}+a_{1} b_{2} c_{2}-a_{2} b_{1} c_{2} \\ &=0 \\ & \therefore(\vec{a} \times \vec{b}) \cdot \vec{b}=0 \end{aligned}

Hence, $\overrightarrow{0}$

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