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Please solve RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 13 maths textbook solution

Answers (1)

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Answer:

Option (b)   0<x<\frac{1}{2}

Hint:

Use the formula of finding the angle between two vectors   \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

Given:

\vec{a}=2 x^{2} \hat{i}+4 x \hat{j}+\hat{k} \text { and } \vec{b}=7 \hat{i}-2 \hat{j}+x \hat{k}  angle between them is obtuse and angle between  \vec{b} and z-axis is acute and less than  \frac{\pi }{6}

Solution:

\vec{a}=2 x^{2} \hat{i}+4 x \hat{j}+\hat{k}, \vec{b}=7 \hat{i}-2 \hat{j}+x \hat{k}

Let \cos A be the angle between \vec{a} \text { and } \vec{b}

Since, A is obtuse angle

\begin{aligned} &\cos A<0 \\\\ &\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}<0 \mid \quad[\because \vec{a} \cdot \vec{b}=|\vec{a}||\vec{b}| \cos \theta] \end{aligned}

\frac{\left(2 x^{2} \hat{i}+4 x \hat{j}+\hat{k}\right)(7 \hat{i}-2 \hat{j}+x \hat{k})}{\sqrt{\left(\left(2 x^{2}\right)^{2}+(4 x)^{2}+1^{2}\right)\left((7)^{2}+(-2)^{2}+x^{2}\right)}}<0

\begin{aligned} &\frac{14 x^{2}-8 x+x}{\sqrt{\left(4 x^{4}+16 x^{2}+1\right)\left(49+4+x^{2}\right)}}<0 \\\\ &14 x^{2}-7 x<0 \\\\ &7 x(2 x-1)<0 \\\\ &x<0 \; \; \; \; x<\frac{1}{2} \end{aligned}

Hence, option (b) is correct

 

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