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  • Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 15

Explain solution RD Sharma class 12 Chapter 28 The Plane exercise 28.11 question 15

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Answer: Therefore, the line \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}  and plane \vec{r} \cdot \vec{n}=d are parallel. And line

\overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\lambda(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})  is parallel to \vec{r} \cdot(2 \hat{j}+\hat{k})=3 .. The distance between lines and plane is \frac{1}{\sqrt{5}}  units.

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

Given:  \vec{r}=\hat{i}+\hat{j}+\lambda(3 \hat{i}-\hat{j}+2 \hat{k}) \text { and } \vec{r} \cdot(2 \hat{j}+\hat{k})=3

Solution: We know that line \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{b}}  and plane \vec{r} \cdot \vec{n}=d  is parallel

if   \vec{b} \cdot \vec{n}=0                                            ……………… (1)

Given, equation of line \overrightarrow{\mathrm{r}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})   and equation of plane is \vec{r} \cdot(2 \hat{j}+\hat{k})=3

So, \overrightarrow{\mathrm{b}}=3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}} \text { and } \overrightarrow{\mathrm{n}}=2 \hat{\mathrm{j}}+\hat{\mathrm{k}}

Now,  \overrightarrow{\mathrm{b}} \cdot \vec{n}=(3 \hat{\mathrm{i}}-\hat{\mathrm{j}}+2 \hat{\mathrm{k}})(2 \hat{\mathrm{j}}+\hat{\mathrm{k}})=-2+2=0

So, the line and plane are parallel.

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

\begin{aligned} &\mathrm{D}=\frac{\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{n}}-\mathrm{d}}{\overrightarrow{\mathrm{n}}} \\ & \end{aligned}

\overrightarrow{\mathrm{a}}=(\hat{\mathrm{i}}+\hat{\mathrm{j}})

\begin{aligned} D &=\frac{(\hat{i}+\hat{j})(2 \hat{j}+\hat{k})-3}{\sqrt{2^{2}+1^{2}}} \\ & \end{aligned}

=\frac{2-3}{\sqrt{4+1}}=\frac{-1}{\sqrt{5}}

We take the mod value

            D=\frac{1}{\sqrt{5}}

 

Posted by

infoexpert27

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