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

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

Answers (1)

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

 

Posted by

infoexpert26

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