Find the point Q on the plane 2x 3y z 4 that is closest t

Find the point Q on the plane 2x ? 3y + z = 4 that is closest to the point P = (5, ?12, 10), and the distance from P to the plane in two ways:

(a) Using projections.

(b) By noticing that Q will be the point of intersection between the plane and the line L that passes through the point P in the direction of the normal vector to the plane.

2. Find the point Q on the plane 2x-3y+ z = 4 that is closest to the point P- (5,-12, 10), and the distance from P to the plane in two ways: (a) Using projections b) By noticing that Q will be the point of intersection between the plane and the b) By noticing that 2 will be the point of intersection between the plane and the line L that passes through the point P in the direction of the normal vector to the plane.

Solution

First let us find the equation of the line L passing through Q and the point P(5, -12, 10)

The line L passes through P(5, -12, 10) and is parallel to <2, -3, 1> ( the co efficients of the plane equation)

So we have,

(x - 5)/2 = (y + 12)/(-3) = (z - 10)/1 = t

So the parametric equations of L are,

x = 2t + 5

y = -3t - 12

z = t + 10

This line L intersects the plane 2x - 3y + z = 4 if and only if

2(2t + 5) -3(-3t - 12) + (t + 10) = 4

That is, 4t + 10 + 9t + 36 + t + 10 = 4

That is, 14t + 56 = 4. So 14t = 4 - 56 = -52

14 t = -52, So t = - 52/14 = -26/7

t = -26/7

Substituting t as -26/7 in the parametric equations of L gives the point Q as

x = 2t + 5 = 2(-26/7) + 5 = -(52/7) + 5 = -17/7

y = -3t - 12 = -3(-26/7) - 12 = (78/7) - 12 = -6/7

z = t + 10 = -(26/7) + 10 = 44/7

So, the point Q is (-17/7, -6/7, 44/7)