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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 2

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 2

Answers (1)

Answer:

 \vec{r}.\left ( \frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \right )=5

Hint:

 You must know the rules of solving vector functions

Given:

Find the vector equation of a plane which is at a distance of 5 units from origin

and normal to the vector

\hat{i}-2\hat{j}-2\hat{k}

Solution:

We have

Normal vector,

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

Now,

\begin{aligned} &\hat{n}=\frac{\vec{n}}{\left | \vec{n} \right |}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{(1)^{2}+(-2)^{2}+(-2)^{2}}}\\ &=\frac{\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{1+4+4}}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{\sqrt{9}}=\frac{\hat{i}-2\hat{j}-2\hat{k}}{3}\\ &\therefore \hat{n}=\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \end{aligned}

The equation of a plane is in normal form is

\begin{aligned} &\vec{r}.\hat{n}=d \qquad \qquad \rightarrow (1) \end{aligned}

[d= distance of plane from origin]
By putting

\begin{aligned} &\hat{n}=\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \end{aligned}

d = 5units in (1) relation
We get,
\vec{r}.\left ( \frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \right )=5

This is the required vector equation

Posted by

Gurleen Kaur

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