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  • Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 5

Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 5

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Answer: Therefore, the given line is parallel to given plane is showed. And distance is \frac{7}{\sqrt{3}} unit

Hint: Use properties of vector and formula d=\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|}

Given:  \overrightarrow{\mathrm{r}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}+\lambda(\hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \text { and } \overrightarrow{\mathrm{r}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=7

Solution: The given plane passes through the point with position vector  \vec{a}=2 \hat{i}+5 \hat{j}+7 \hat{k}  and is parallel to the vector \vec{b}=\hat{i}+3 \hat{j}+4 \hat{k}

The given plane is  \vec{r} \cdot(\hat{i}+\hat{j}-\hat{k})=7

So, the normal vector  \vec{n}=\hat{i}+\hat{j}-\hat{k} \text { and } d=7

Now, \vec{b} \cdot \vec{n}=(\hat{i}+3 \hat{j}+4 \hat{k}) \cdot(\hat{i}+\hat{j}-\hat{k})=1+3-4=0

So, \vec{b}  is perpendicular to \vec{n}

So, the given line is parallel to the given plane.

The distance between the line and the parallel plane.

Then, d = length of perpendicular from the point \overrightarrow{\mathrm{a}}=2 \hat{\mathrm{i}}+5 \hat{\mathrm{j}}+7 \hat{\mathrm{k}}  to the plane \vec{r} \cdot \vec{n}=d

            \begin{aligned} d &=\frac{|\vec{a} \cdot \vec{n}-d|}{|\vec{n}|} \\ & \end{aligned}

                =\frac{|(2 \hat{i}+5 \hat{j}+7 \hat{k})(\hat{i}+\hat{j}-\hat{k})-7|}{|\hat{i}+\hat{j}-\hat{k}|}

                \begin{aligned} &=\frac{|2+5-7-7|}{\sqrt{1+1+1}} \\ & \end{aligned}

                =\frac{7}{\sqrt{3}} \text { units }

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