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Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.2 question 2 sub question (i)

Answers (1)

Answer:

Intercepts form of the given equation is \frac{x}{3}+\frac{y}{4}+\frac{z}{-2}=1 & its intercepts on co-ordinate axes are 3,-4  &  -2

Hint:

First divide the given equation by 12 then compare it with  \frac{x}{p}+\frac{y}{q}+\frac{z}{r}=1,   where p, qrintercepts on co-ordinate axes.

Given:

Given equation is 4x+3y-6z-12=0

Explanation:

It is given that,

equation of plane is  4x+3y-6z-12=0                                         ….....(i)

Divide equation (i) by 12

        \begin{aligned} \Rightarrow & \frac{4 x+3 y-6 z-12}{12}=0 \\ \end{aligned}

        \Rightarrow \frac{x}{3}+\frac{y}{4}-\frac{z}{2}=1 \\

        \Rightarrow \frac{x}{3}+\frac{y}{4}+\frac{z}{(-2)}=1

This is in the form of  \frac{x}{l}+\frac{y}{m}+\frac{z}{n}=1

Here, l=3, m=4, n=-2

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