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  • Provide solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28 .9 Question 2 textbook solution.

Provide solution for RD Sharma maths Class 12 Chapter 28 The Plane Exercise 28.9 Question 2 textbook solution.

Answers (1)

Answer : The given points are equidistant from the plane

Hint : \text { Distance }=\frac{\left|A x_{1}+B y_{1}+C z_{1}+D\right|}{\sqrt{A^{2}+B^{2}+C^{2}}}

Given :

Two points (\hat{i}-\hat{j}+3 \hat{k}) and (3 \hat{i}+3 \hat{j}+3 \hat{k}), plane \vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0

Solution :

Points given by the equation

\vec{a}=\hat{i}-\hat{j}+3 \hat{k}, \vec{b}=3 \hat{i}+3 \hat{j}+3 \hat{k}

Plane given by the equation

\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0, where the normal is \vec{n}=5 \hat{i}+2 \hat{j}-7 \hat{k}

We know, the distance of \vec{a} from the plane \vec{r} \cdot \vec{n}-d=0 Is given by

p=\left|\frac{\vec{a} \cdot \vec{n}-d}{|\vec{n}|}\right|

Distance of \hat{i}-\hat{j}+3 \hat{k} from the plane

\begin{aligned} &=\left|\frac{(\hat{i}-\hat{j}+3 \hat{k}) \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9}{|5 \hat{i}+2 \hat{j}-7 \hat{k}|}\right| \\ &=\frac{9}{\sqrt{78}} \text { Units } \end{aligned}

And,

Distance of 3 \hat{i}+3 \hat{j}+3 \hat{k} from the plane

=\left|\frac{(3 \hat{i}+3 \hat{j}+3 \hat{k}) \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9}{|5 \hat{i}+2 \hat{j}-7 \hat{k}|}\right|

= \frac{9}{\sqrt{78}} Units

The point \hat{i}-\hat{j}+3 \hat{k} \; \text {and} \; 3 \hat{i}+3 \hat{j}+3 \hat{k} are equidistant from the plane

\vec{r} \cdot(5 \hat{i}+2 \hat{j}-7 \hat{k})+9=0

Posted by

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