Question

Find the distance of point -2i+3j-4k from the line r=i+2j-k+¥[i+3j-9k]measured parallel to the plane x-y+2z-3=0

Answer 1

Answer:

Step-by-step explanation:

Call the given point P, so P = (-2, 3, -4).

We need to get the distance from P to a point X on the given line, such that PX is parallel to the given plane.

As X is on the line, it takes the form

X = (1, 2, -1) + (1, 3, -9)u, for some parameter u.

As PX is parallel to the given plane, it is perpendicular to the normal to the plane.  The normal is n = (1, -1, 2), given by the coefficients of x, y, z.  To have PX perpendicular to n means the dot product n.(P-X) is zero.,  So n.X = n.P.

Putting in the above expression for X, we have

n . [ (1, 2, -1) + (1, 3, -9)u ] = n . P

and putting in the known values of n and P, this becomes

-3 - 20u = -13  =>  20u = 10  =>  u = 1/2.

So X = (1.5, 3.5, -5.5).

The required distance d is then the distance from P to X.  So

d² = (3.5)² + (0.5)² + (1.5)² = 14.75

and so d = √14.75, which is approximately 3.84.