Question

09. Find the point on the line (x+2)/3=y+1)/42-3)/2 at a distance of 32 from the point(1,2,3) .​

Answer 1

Answer:

Let

3

x−2

=

4

y+1

=

12

z−2

=t

Any point on the line can be written in the parametric form as (3t+2,4t−1,12t+2)

To find the point of intersection, let us substitute the point in the equation of the plane.

⇒3t+2−4t+1+12t+2=16

⇒11t=11

⇒t=1

Hence, the point of intersection is (5,3,14)

The distance of (5,3,14) from (1,0,2) =

16+9+144

=13