The vector equation of the plane which is at a distance of 8 units from the origin and which is normal to the vector $2\hat{i} + \hat{j} + 2 \hat{k},$ is
Answer & explanation
Correct answer: option 2
Here, d = 8 and $\vec{n} = 2\hat{i} + \hat{j} + 2 \hat{k}.$
$∴ \hat{n} =\frac{\vec{n}}{|\vec{n}|}=\frac{2\hat{i} + \hat{j} + 2 \hat{k}}{\sqrt{4+1+4}}=\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}$
Hence, the required equation of the plane is
$\vec{r}.\left(\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}\right)=8 ⇒92\hat{i} + \hat{j} + 2 \hat{k})=24.$