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  • please solve RD sharma class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 4 maths textbook solution

please solve RD sharma class 12 chapter 23 Scalar or dot product exercise Fill in the blanks question 4 maths textbook solution

Answers (1)

best_answer

 \pi

Hint:

Angle between \vec{a} and \vec{c}.

Given:

The non-zero vector \vec{a},\vec{b},\vec{c} are related by \vec{a}=8b and \vec{c}=-7b then angle between \vec{a} and \vec{c} is

Solution:

We have,

\vec{a}=8b and \vec{c}=-7b

Let,
   \begin{aligned} &\vec{b}=b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k} \\ \end{aligned}

Then,

\begin{aligned} &\vec{a}=8\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right) \\ \end{aligned}

And  \begin{aligned} &\vec{c}=-7\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right) \\ \end{aligned}

Consider angle between \vec{a} and \vec{c} is \theta.

Now, we know the rule of dot product.

\begin{aligned} &\vec{a} . \vec{c}=|\vec{a} \| \vec{c}| \cos \theta \\ &\therefore \cos \theta=\frac{\vec{a} \cdot \vec{c}}{|\vec{a}||\vec{c}|} \\ \end{aligned}

\begin{aligned} &\cos \theta=\frac{\left[\left(8\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right) \cdot\left(-7\left(b_{1} \hat{\imath}+b_{2} \hat{\jmath}+b_{3} \hat{k}\right)\right)\right)\right]}{|-7| \times|8| \cdot \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}} \cdot \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}} \\ \end{aligned}

\begin{aligned} &\cos \theta=\frac{-56\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)}{|-56|\left(b_{1}^{2}+b_{2}^{2}+b_{3}^{2}\right)} \quad\left[\begin{array}{l} \imath{\imath} . \hat{\imath}=1, \hat{\jmath} \cdot \hat{\jmath}=1, \\ \hat{k} \cdot \hat{k}=1 \end{array}\right] \\ \end{aligned}

\begin{aligned} &\cos \theta=\frac{-56}{56} \\ &\cos \theta=-1 \end{aligned}

Therefore,
\begin{aligned} \theta=\pi &&& \quad[\because \cos \pi=-1] \end{aligned}

Hence, angle between \vec{a} and \vec{c} is \pi.

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