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Please solve RD Sharma Class 12 Chapter 24 Vector or Cross Product Exercise Very Short Answer Question, question 10 Maths textbook solution.

Answers (1)

ANSWER:

  0

HINT:

  Find the cross product and then the dot product

GIVEN:

 Two vectors \overrightarrow{a} and \overrightarrow{b}                                              

SOLUTION:

Let

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

Now,

For, \overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{a})

We calculate

\begin{aligned} (\vec{b} \times \vec{a}) &=\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}

Now,

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

Thus, 

\overrightarrow{a}.(\overrightarrow{b}\times \overrightarrow{a})=0 

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