MCQ
The equation of the plane through $(-1,1,2)$, whose normal makes equal acute angles with co-ordinate axes is
- A$\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=2$
- B
- C
- D
(a) : The equation of plane passing through $A(\vec{a})$ and $\perp$ to $\vec{n}$ is $\vec{r} \cdot \vec{n}=\vec{a} \cdot \vec{n}$ ...(i)
Let $\alpha$ be the equal acute angle that the normal makes with coordinate axes.
So, $l=m=n=\cos \alpha$
Now, $l^2+m^2+n^2=1$
$
\Rightarrow \cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1 \Rightarrow \cos \alpha=\frac{1}{\sqrt{3}}
$
So, $l=m=n=\frac{1}{\sqrt{3}}$
Hence, dr's of $\vec{n}$ are $< 1,1,1>$
Now, $\vec{a}=-\hat{i}+\hat{j}+2 \hat{k}$ and $\vec{n}=\hat{i}+\hat{j}+\hat{k}$
So, from (i), we get
$
\begin{aligned}
\quad \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k}) & =(-\hat{i}+\hat{j}+2 \hat{k}) \cdot(\hat{i}+\hat{j}+\hat{k}) \\
\Rightarrow \quad & \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=-1+1+2=2
\end{aligned}
$
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