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  • Explain Solution R.D.Sharma Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 46 Maths Textbook Solution.

Explain Solution R.D.Sharma Class 12 Chapter 23 Scalar or Dot Products Exercise 23.1 Question 46 Maths Textbook Solution.

Answers (1)

Answer:  xt\left ( -3,-2 \right )\: \: U\left ( 2,3 \right )

Hints: You must know the rules of solving vectors.

Given: \begin{aligned} &\vec{a}=x^{2} \hat{\imath}+2 \hat{\jmath}-2 \hat{k} \\ \end{aligned},\begin{aligned} &\vec{b}=\hat{\imath}-\hat{\jmath}+\hat{k} \\ \end{aligned},\begin{aligned} &\vec{c}=x^{2} \hat{\imath}+5 \hat{\jmath}-4 \hat{k} \end{aligned}

Solution: We have,

\begin{aligned} &\vec{a}=x^{2} \hat{\imath}+2 \hat{\jmath}-2 \hat{k} \\ &\vec{b}=\hat{\imath}-\hat{\jmath}+\hat{k} \\ &\vec{c}=x^{2} \hat{\imath}+5 \hat{\jmath}-4 \hat{k} \end{aligned}

Let \theta _{1} be the angle between \vec{a}and \vec{b}

      \theta _{2}      be the angle between \vec{b} and \vec{c}

Given that \theta _{1} is acute and  \theta _{2} is obtuse

\begin{aligned} &\Rightarrow \cos \theta_{1}>0 \quad \text { and } \theta_{2}<0 \\ \end{aligned}

\begin{aligned} &\Rightarrow \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}>0 \text { and } \frac{\vec{b} \cdot \vec{c}}{|\vec{b}||\cdot \vec{c}|}<0 \\ \end{aligned}

\begin{aligned} &\Rightarrow \frac{x^{2}-4}{\sqrt{x^{4}+4+4} \sqrt{1+1+1}}>0 \quad \text { and } \frac{x^{2}-9}{\sqrt{x^{4}+25+16} \sqrt{1+1+1}}<0 \\ \end{aligned}

\begin{aligned} &\Rightarrow x^{2}-4>0 \quad \text { and } x^{2}-9<0 \end{aligned}

\begin{aligned} &\Rightarrow x t(-\infty,-2) U(2, \infty) \text { and } x t(-3,3) \\ \end{aligned}

\begin{aligned} &\Rightarrow x t(-3,-2) U(2,3) \end{aligned}

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