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Three Dimensional Geometry — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsThree Dimensional Geometry5 Marks
Question
Reduce the equation $\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})+6=0$ to the normal form and, hence, find the length of the perpendicular from the origin to the plane.
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Answer

The given equation of the plane is,
$\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})+6=0$
$\Rightarrow\vec{\text{r}}\cdot(\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}})=-6$ or $\vec{\text{r}}\cdot\vec{\text{n}}=-6,$ where $\vec{\text{n}}=\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}$
$|\vec{\text{n}}|=\sqrt{1+4+4}=3$
For reducing the given equation to normal form, we need to divide it by $|\vec{\text{n}}|$ Then, we get
$|\vec{\text{r}}|\cdot\frac{\vec{\text{n}}}{|\vec{\text{n}}|}=\frac{-6}{|\vec{\text{n}}|}$
$\Rightarrow\vec{\text{r}}\cdot\Big(\frac{\hat{\text{i}}-2\hat{\text{j}}+2\hat{\text{k}}}{3}\Big)=\frac{-6}{3}$
$\Rightarrow\vec{\text{r}}\cdot\Big(\frac{1}{3}\hat{\text{i}}-\frac{2}{3}\hat{\text{j}}+\frac{2}{3}\hat{\text{k}}\Big)=-2$
Dividing both sides by -1 we get
$\vec{\text{r}}\cdot\Big(-\frac{1}{3}\hat{\text{i}}+\frac{2}{3}\hat{\text{j}}-\frac{2}{3}\hat{\text{k}}\Big)=2\ ...(\text{i})$
The equation of the plane in normal form is
$\vec{\text{r}}\cdot\vec{\text{n}}=\text{d}\ ...(\text{ii})$
(where d is distance of the plane from the origin)
Comparing (i) and (ii)
length of the perpendicular from the origin to the plane = d = 2 units

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