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
Answers
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.