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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.4 question 4 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.4 question 4 maths

Answers (1)

Answer:

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

Hint:

 You must know the rules of solving vector functions

Given:

 Reduce the equation

\vec{r}.\left ( \hat{i}-2\hat{j}+2\hat{k} \right )+6=0

to the normal form and hence,

find the length of perpendicular from the origin.

Solution:

We have

\vec{r}.\left ( \hat{i}-2\hat{j}+2\hat{k} \right )+6=0

\vec{r}.\left ( \hat{i}-2\hat{j}+2\hat{k} \right )=-6

The equation of a plane is in normal form is

\vec{r}.\hat{n}=d

[d= distance of plane from origin]

\begin{aligned} &\hat{n}=\hat{i}-2\hat{j}+2\hat{k}\\ &d=-6\\ &\left | \vec{n} \right |=\sqrt{(1)^{2}+(-2)^{2}+(2)^{2}}\\ &=\sqrt{1+4+4}=\sqrt{9}=3 \end{aligned}

For reducing the given equation to normal form
So, divide the given equation by

\begin{aligned} &\left | \vec{n} \right | \end{aligned}

We get

\begin{aligned} &\vec{r}.\left ( \frac{\hat{i}-2\hat{j}+2\hat{k}}{3} \right )=\frac{-6}{3}\\ &\vec{r}.\left ( \frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k} \right )=-2\\ &\vec{r}.\left ( -\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k} \right )=2 \qquad \qquad \rightarrow (1) \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 comparing equation (1) & (2)

we get d = 2 units
∴Length of perpendicular is d = 2 units

Posted by

Gurleen Kaur

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