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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 10

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 10

Answers (1)

Answer:

\vec{r}\left ( \frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \right )=\frac{6}{\sqrt{29}} \text { and }2x-3y+4z=6 

Hint:

 You must know the rules of solving vector functions

Given:

 find the vector equation of the plane which is at a distance of

\frac{6}{\sqrt{29}}

from the origin and it’s normal vector from the origin is

2\hat{i}-3\hat{j}+4\hat{k}

Solution:

We have

normal vector =

\vec{n}=2\hat{i}-3\hat{j}+4\hat{k}

Unit vector to the plane ,

\begin{aligned} &\hat{n}=\frac{\vec{n}}{\left | \vec{n} \right |}\\ &=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{(2)^2+(-3)^2+(4)^2}}\\ &=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{4+9+16}}\\ &=\frac{2\hat{i}-3\hat{j}+4\hat{k}}{\sqrt{29}}\\ &\hat{n}=\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \end{aligned}

Equation of plane in normal form is

\begin{aligned} &\vec{r}.\hat{n}=d\\ &\text { Put }\hat{n}=\frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k}\text { and }d=\frac{6}{\sqrt{29}}\\ &\vec{r} \left ( \frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \right ) =\frac{6}{\sqrt{29}}\\ \end{aligned}

This is the required vector equation.

For finding Cartesian form

Put

\begin{aligned} &\vec{r}=\left ( x\hat{i}+y\hat{j}+z\hat{k} \right )\\ &\left ( x\hat{i}+y\hat{j}+z\hat{k} \right ) \left ( \frac{2}{\sqrt{29}}\hat{i}-\frac{3}{\sqrt{29}}\hat{j}+\frac{4}{\sqrt{29}}\hat{k} \right ) =\frac{6}{\sqrt{29}}\\ &\frac{2x-3y+4z}{\sqrt{29}}=\frac{6}{\sqrt{29}}\\ &2x-3y+4z=\frac{6(\sqrt{29})}{\sqrt{29}}\\ &2x-3y+4z=6 \end{aligned}

 

 

Posted by

Gurleen Kaur

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