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  • Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 11

Need solution for RD Sharma maths class 12 chapter The Plane exercise 28.4 question 11

Answers (1)

Answer:

 \frac{6}{\sqrt{29}}

Hint:

 You must know the rules of solving vector functions

Given:

 Find the distance of the plane 2x - 3y + 4z - 6 = 0 from the origin

Solution:

We have

\begin{aligned} &2x-3y+4z-6=0\\ &2x-3y+4=6 \qquad \qquad \dots (1)\\ &\vec{r}(2\hat{i}-3\hat{j}+4\hat{k})=6 \end{aligned}

Now,

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

Divide equation (1) by \begin{aligned} &\sqrt{29} \end{aligned}

We get,

\begin{aligned} &\frac{2}{\sqrt{29}}x-\frac{3}{\sqrt{29}}y+\frac{4}{\sqrt{29}}z=\frac{6}{\sqrt{29}} \end{aligned}

Which is the normal of the plane

Therefore the length of perpendicular from the origin to the plane is

\begin{aligned} &\frac{6}{\sqrt{29}} \end{aligned}

Posted by

Gurleen Kaur

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