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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.3 question 10

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.3  question 10

Answers (1)

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Answer:

The answer of given question is \frac{2}{3}, \frac{1}{3},-\frac{2}{3}

Hint:

Dividing by 3 on both sides.

Given:

The given equation of the plane is 2x+y-2z=3

Solution:

Dividing by 3 on both sides, we get

\begin{aligned} &\frac{2 x}{3}+\frac{y}{3}-\frac{2 z}{3}=\frac{3}{3} \\ &\Rightarrow \frac{\frac{x}{3}}{2}+\frac{y}{3}-\frac{\frac{z}{3}}{2}=1 \ldots (i) \end{aligned}

We know that if a,b,c are the intercepts by the plane on thee co-ordinates axes, then equation of the plane is

\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1

Comparing the equation (i) and (ii) we get

a=\frac{3}{2}, \quad b=3, \quad c=-\frac{3}{2}

Again the given equation of the plane is 2x+y-2z=3 writing this in vector form, we get

\begin{aligned} &\Rightarrow(x \hat{\imath}+y \hat{\jmath}+z \hat{k}) \cdot(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})=3 \\ &\Rightarrow \vec{r} \cdot(2 \hat{\imath}+\hat{\jmath}-2 \hat{k})=3 \end{aligned}

So vector normal to the plane is given by

\begin{aligned} &\vec{n}=2 \hat{\imath}+\hat{\jmath}-2 k \\ &\Rightarrow|\vec{n}|=\sqrt{(2)^{2}+(1)^{2}+(-2)^{2}} \\ &\Rightarrow|\vec{n}|=\sqrt{4+1+4} \\ &\Rightarrow|\vec{n}|=3 \end{aligned}

Direction vector of  \vec{n}=2,1,-2

Direction vector of  \vec{n}

\begin{aligned} &=\frac{2}{|\vec{n}|}, \frac{1}{|\vec{n}|}, \frac{-2}{|\vec{n}|} \\ &\Rightarrow \vec{n}=\frac{2}{3}, \frac{1}{3},-\frac{2}{3} \end{aligned}

So,

Intercepts by the plane on the co-ordinate axes are =\frac{2}{3}, \frac{1}{3},-\frac{2}{3}

Direction cosines of normal to the plane are =\frac{2}{3}, \frac{1}{3},-\frac{2}{3}

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