Question

find the vector equation of line passing through the point whose position vector is 3 i + J - K and which is parallel to vector to vector 2i-j+2k.
if P is a point on his line such that length of AP is equal to 15 find the position vector of point P

Answer 1

Since, i have used the formula of equation

Answer 2

Therefore the coordinate of P are (9,-7,13) and (-9.8,2.4,-5.8)

Step-by-step explanation:

The equation of line which passes through the point A(3,1,-1) and parallel to the vector 2i-j+2k is

r= 3i+j-k+λ(2i-j+2k)

Therefore any point on the line be (2t-3,-t-1,2t+1)

Let the coordinate of P be ( 2t - 3,-t-1,2t+1)

The distance between A and P is $\sqrt{(2t-3-3)^2+(-t-1-1)^2+(2t+1+1)^2}$

According To Problem,

$\sqrt{(2t-3-3)^2+(-t-1-1)^2+(2t+1+1)^2}$=15

⇒9$t^{2}$-12t-181=0

⇒t=6,-3.4

Therefore the coordinate of P are (9,-7,13) and (-9.8,2.4,-5.8)