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The plane — Maths STD 12 Science — Question

Maharashtra BoardEnglish MediumSTD 12 ScienceMathsThe plane2 Marks
Question
Write the distance of the plane $\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=12$ form the origin.
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Answer

The given equation of the plane is
$\vec{\text{r}}.(2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}})=12$ or $\vec{\text{r}}.\vec{\text{n}}=-6$, where $\vec{\text{n}}=2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{4+1+4}=3$
For reducing the given equation to normal form, we need to divide both sides by $|\vec{\text{n}}|$.
Then, we get $\vec{\text{r}}.\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{12}{|\vec{\text{n}}|}$
$=\vec{\text{r}}.\Big(\frac{2\hat{\text{i}}-\hat{\text{j}}+2\hat{\text{k}}}{3}\Big)=\frac{12}{3}$
$\Rightarrow\vec{\text{r}}.\Big(\frac{2}{3}\hat{\text{i}}+\frac{1}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}\Big)=4,\ ...(1)$
The eqution of the plane normal from is.
$\vec{\text{r}}.\text{n}=\text{d}\ ....(2)$
(where d is the distance of the plane from the origin)
Comparing (1) and (2),
Length of the perpendicular from the origin to the plane = d = 4 units.

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