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  • Provide Solution for RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 16

Provide Solution for RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 16

Answers (1)

Answer:  \frac{1}{\sqrt{6}}  units

Hint: \mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}}

Given:  \vec{r} \cdot(\hat{i}+2 \hat{j}-\hat{k})=1 \text { and } \vec{r}=(-\hat{i}+\hat{j}+\hat{k})+\lambda(2 \hat{i}+\hat{j}+4 \hat{k})

Solution: We know that line \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\mathrm{kb}  and plane \vec{r} \cdot \vec{n}=\mathrm{d}  is parallel if  \vec{r} \cdot(\hat{i}+2 \hat{j}-\hat{k})=1

Given, equation of line \vec{r}=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{k})+k(2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{k}) and equation of plane is  \vec{r} \cdot(\hat{i}+2 \hat{j}-\hat{k})=1

So, \overrightarrow{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}+4 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{n}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}}

Now,

So, the line and plane are parallel

we know that the Distance ‘D’ of a plane  \vec{r} \cdot \vec{n}=d  from a point \vec{a} is given by

            \mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}}, \overrightarrow{\mathrm{a}}=(-\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})\mathrm{D}=\frac{1}{\sqrt{6}} \text { units }

            \begin{aligned} \Rightarrow D &=\frac{(-\hat{i}+\hat{j}+\hat{k})(\hat{i}+2 \hat{j}-\hat{k})-1}{\sqrt{(-1)^{2}+2^{2}+1^{2}}} \\ & \end{aligned}

            =\frac{-1+2-1-1}{\sqrt{1+4+1}}=\frac{-1}{\sqrt{6}}

We take the mod value,

            So,  \mathrm{D}=\frac{1}{\sqrt{6}} \text { units }

 

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