Question

The position vector of a point at a distance of 3 root 11 units from i -j +2k on a line passing through the point i - j + 2k and parallel to the vector 3i + j + k is?

Answer 1

10i+2j+5k and -8i-4j-k are the possible position vectors of the point.

Step-by-step explanation:

we have to find the position vector of a point which is at a distance of $3\sqrt{11}$ from$i-j+2k$  i-j+2k and parallel to the vector $3i+j+k$ 3i+j+k.

let the position vector of required point be$Xi+Yj+Zk$

Xi+Yj+Zk

distance between $i-j+2k$ and $Xi+Yj+Zk$ is $3\sqrt{11}$

that is $\sqrt{(X-1)^{2} +(Y+1)^{2} +(Z-2)^{2} }$= $3\sqrt{11}$------------(1)

also the line $Xi+Yj+Zk-(i-j+2k)$ is parallel to $3i+j+k$.

That is $(X-1)i+(Y+1)j+(Z-2)k$ is parallel to $3i+j+k$

therefore, the co-ordinates are in proportion, they are $\frac{X-1}{3}=\frac{Y+1}{1} =\frac{Z-2}{1}$=λ

therefore, $X=$3λ+1, $Y=$λ-1, $Z=$λ+2------------(2)

substituting the values of equation 2 in equation 1.

we get, (3λ)²+λ²+λ²=99

⇒11λ²=99

⇒λ²=9

⇒λ=±3

therefore, (a) if λ=3, then $X=10,Y=2,Z=5$

(b) if λ= -3, then $X=-8,Y=-4, Z=-1$

therefore the position vectors are 10i+2j+5k  and -8i-4j-k