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

Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28 point 6 question 1 sub question 1 maths textbook solution

Answers (1)

Answer:

\theta=\cos ^{-1} \frac{-5}{\sqrt{58}}             

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}}-3 \hat{\boldsymbol{\jmath}}+\mathbf{4} \widehat{\boldsymbol{k}})=\mathbf{1} \text { And } \vec{r} \cdot(-\hat{\imath}+\hat{\jmath})=4

Solution:

\begin{aligned} &\vec{r} \cdot(2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k})=1 \\ &\vec{n}_{1}=2 \hat{\imath}-3 \hat{\jmath}+4 \hat{k} \\ &\vec{r} \cdot(-\hat{\imath}+\hat{\jmath})=4 \\ &\vec{n}_{2}=-\hat{\imath}+\hat{\jmath} \\ &\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}-3 \hat{\jmath}+4 \hat{k}) \cdot(-\hat{\imath}+\hat{\jmath})}{\sqrt{2^{2}+(-3)^{2}+(4)^{2}} \cdot \sqrt{(-1)^{2}+\left ( 1 \right )^{2}}} \\ &=\frac{-2-3}{\sqrt{29 \times \sqrt{2}}}=\frac{-5}{\sqrt{58}} \\ \end{aligned}

\theta=\cos ^{-1}\frac{-5}{\sqrt{58}}

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