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  • Please solve RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 21 maths textbook solution

Please solve RD Sharma class 12 chapter 23 Scaler and Dot Product exercise Multiple choice question 21 maths textbook solution

Answers (1)

best_answer

Answer:

Option (c) R-[-4, 7]

Hint:

You must know the formula for finding angle between two vectors  \cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}

Given:

x \hat{i}+3 \hat{j}-7 \hat{k} \text { and } \hat{x} \hat{i}-x \hat{j}+4 \hat{k}, angle between them is acute.

Solution:

\begin{aligned} &\frac{(\hat{x} \hat{i}+3 \hat{j}-7 \hat{k})(\hat{x}-x \hat{j}+4 \hat{k})}{\sqrt{\left(x^{2}+9+49\right)\left(2 x^{2}+16\right)}}=\cos \theta \\\\ &\frac{x^{2}-3 x-28}{\sqrt{\left(x^{2}+58\right)\left(2 x^{2}+16\right)}}=\cos \theta \end{aligned}                \left[\because \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}=\cos \theta\right]

Angle between the vectors is acute

\begin{aligned} &\theta<\frac{\pi}{2} \Rightarrow \cos \theta>0 \\\\ &\frac{x^{2}-3 x-28}{\sqrt{\left(x^{2}+56\right)\left(2 x^{2}+16\right)}}>0 \end{aligned}

\begin{aligned} &x^{2}-3 x-28>0 \\ &x^{2}-7 x+4 x-28>0 \\ &x(x-7)+4(x-7)>0 \\ &(x-7)(x+4)>0 \\ &x \in R-[-4,7] \end{aligned}

Hence, option (c) is correct

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