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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.3 question 14

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.3  question 14

Answers (1)

Answer:

The answer of given question is the normal vector to the plane 2x+2y+2z=3 is equally inclined with the co-ordinate axes.

Hint:

Let α,β,γ be the angle that normal \vec{n} makes with the co-ordinate axes

Given:

2x+2y+2z=3

Solution:

The vector equation of the plane 2x+2y+2z=3 can be written as

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

The normal to this plane is

\vec{n}=2 \hat{\imath}+2 \hat{\jmath}+2 \hat{k} \ldots(i)

Direction ratio of  \vec{n}=2,2,2

Direction cosine of

\begin{aligned} &\vec{n}=\frac{2}{|\vec{n}|}, \frac{2}{|\vec{n}|}, \frac{2}{|\vec{n}|} \\ &\Rightarrow|\vec{n}|=\sqrt{(2)^{2}+(2)^{2}+(-2)^{2}} \\ &\Rightarrow|\vec{n}|=\sqrt{4+4+4} \\ &\Rightarrow|\vec{n}|=2 \sqrt{3} \end{aligned}

Direction cosine of

|\vec{n}|=\frac{2}{2 \sqrt{3}}, \frac{2}{2 \sqrt{3}}, \frac{2}{2 \sqrt{3}}=\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}

So,

l=\frac{1}{\sqrt{3}}, m=\frac{1}{\sqrt{3}}, n=\frac{1}{\sqrt{3}}

Let α,β,γ be the angle that normal \vec{n} makes with the co-ordinate axes respectively.

\begin{aligned} &l=\cos \alpha=\frac{1}{\sqrt{3}} \\ &\Rightarrow \alpha=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \ldots(i i) \\ &\Rightarrow \beta=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \ldots(i i i) \\ &\Rightarrow \gamma=\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right) \ldots(i v) \end{aligned}

Hence α = β = γ

So the normal vector to the plane 2x+2y+2z=3 is equally inclined with the co-ordinate axes.

Posted by

infoexpert26

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