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  • Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 21

Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 21

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Answer:  \overrightarrow{\mathrm{r}}=(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\mathrm{m}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-5 \hat{\mathrm{k}})

Hint:  \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{kb}

Given:  \vec{r} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})+9=0 \text { and }(1,2,3)

Solution: We know that equation of line passing through point  \vec{a} and parallel to vector \vec{b}  is given by  \vec{r}=\vec{a}+k \vec{b}                                 …………. (1)

Given that, the line is passing through  \left ( 1,2,3 \right )

            So,  \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}

It is given that line is perpendicular to plane  \vec{r} \cdot(\hat{i}+2 \hat{j}-5 \hat{k})+9=0

So, normal to plane \left ( \vec{n} \right ) is parallel to  \vec{b}

So, Let  \vec{b}=\vec{n}=1(\hat{i}+2 \hat{j}-5 \hat{k})

Putting \vec{a} \& \vec{b}  in (1), equation of line is

            \begin{aligned} &\vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+k\{1(\hat{i}+2 \hat{j}-5 \hat{k})\} \\ & \end{aligned}

            \Rightarrow \vec{r}=(\hat{i}+2 \hat{j}+3 \hat{k})+m(\hat{i}+2 \hat{j}-5 \hat{k})

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