Get Answers to all your Questions

header-bg qa

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

Answers (1)

best_answer

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

View full answer

Crack CUET with india's "Best Teachers"

  • HD Video Lectures
  • Unlimited Mock Tests
  • Faculty Support
cuet_ads