First slide

Planes in 3D

Question

The centre of the circle given by $\vec{r} \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k}) = 15$ and $| \vec{r} - (\hat{j} + 2 \hat{k}) | = 4 is$

Moderate

A

(0,1,2)
B

(1,3,4)
C

(- 1,3,4)
D

none of these

Solution

The equation of the line through the centre $\hat{j} + 2 \hat{k}$ and normal to the given plane is
$\vec{r} = \hat{j} + 2 \hat{k} + λ (\hat{i} + 2 \hat{j} + 2 \hat{k})$ -------(i)
This meets the plane for which
$\begin{matrix} \hat{[j} + 2 \hat{k} + λ (\hat{i} + 2 \hat{j} + 2 \hat{k})] \cdot (\hat{i} + 2 \hat{j} + 2 \hat{k}) = 15 \\ or 6 + 9 λ = 15 or λ = 1 \end{matrix}$
Putting in (i), we get
$\vec{r} = \hat{j} + 2 \hat{k} + (\hat{i} + 2 \hat{j} + 2 \hat{k}) = \hat{i} + 3 \hat{j} + 4 \hat{k}$
Hence, centre is (1 ,3,4).

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App