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Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 1 sub question 2 maths textbook solution

Answers (1)

Answer:

              \cos ^{-1}\left ( -\frac{4}{21} \right )

Hint:

           \cos \theta=\frac{\overrightarrow{n_{1}} \cdot \overrightarrow{n_{2}}}{\left|\overrightarrow{n_{1}}\right|\left|\overrightarrow{n_{2}}\right|}

Given:

              \overrightarrow{\boldsymbol{r}} \cdot(\mathbf{2} \hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}}+2 \widehat{\boldsymbol{k}})=\mathbf{6} \text { And } \vec{r} \cdot(3 \hat{\imath}+6 \hat{\jmath}-2 \hat{k})=9

Solution:

\begin{aligned} &\vec{n}_{1}=2 \hat{\imath}-\hat{\jmath}+2 \hat{k} \\ &\vec{n}_{2}=3 \hat{\imath}+6 \hat{\jmath}-2 \hat{k} \\ &\cos \theta=\frac{\vec{n}_{1} \cdot \vec{n}_{2}}{\left|\vec{n}_{1}\right| \cdot\left|\vec{n}_{2}\right|}=\frac{(2 \hat{\imath}-\hat{\jmath}+2 \hat{k}) \cdot(3 \hat{\imath}+6 \hat{\jmath}-2 \hat{k})}{|2 \hat{\imath}-\hat{\jmath}+2 \hat{k}| \cdot|3 \hat{\imath}+6 \hat{\jmath}-2 \hat{k}|} \\ &=\frac{6-6-4}{\sqrt{2^{2}+(-1)^{2}+2^{2}} \cdot \sqrt{3^{2}+6^{2}+(-2)^{2}}}=\frac{-4}{3 \times 7}=-\frac{4}{21} \\ &\theta=\cos ^{-1}\left(-\frac{4}{21}\right) \end{aligned}

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