Practicing Success

Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Three-dimensional Geometry

Question:

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

Options:

$\vec{r}.(2\hat{i} + \hat{j} + 2 \hat{k})=8$

$\vec{r}.(2\hat{i} + \hat{j} + 2 \hat{k})=24$

$\vec{r}.(2\hat{i} + \hat{j} + 2 \hat{k})=4$

none of these

Correct Answer:

$\vec{r}.(2\hat{i} + \hat{j} + 2 \hat{k})=24$

Explanation:

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 

$\vec{r}.\left(\frac{2}{3}\hat{i} + \frac{1}{3}\hat{j} + \frac{2}{3}\hat{k}\right)=8 ⇒92\hat{i} + \hat{j} + 2 \hat{k})=24.$