Linear Algebra Question

Question

Find the shortest distance from the point P=(-3,-2,-3) to a point on the line given by 1: (x,y,z) = (-7t, 1t, 1t). Please help ASAP. Another person answered this before but got it wrong. I also need help in visualizing it.

Answer 1

Distance D = √((7t - 3)² + (t + 2)² + (t + 3)² = √51t² - 32t + 22; dD/dt = (51t - 16)/√(51t² - 32t + 22) = 0

51t - 16 = 0; t = 16/51; t = 16/51

min D = D(16/51) = √(112/51 - 3)² + (16/51 + 2)² + (16/51 + 3)²= 4.121

Answer 2

The distance between your line and the point is:

d = sqrt((-7t + 3)^2 + (t + 2)^2 + (t + 3)^2) = sqrt(49t^2 - 42t + 9 + t^2 +4t + 4 + t^2 + 6t + 9) = sqrt(51t^2 - 32t + 22)

(sqrt is the square root function.)

What you have is the square root of a quadratic function that opens up. The vertex of this quadratic function and this quadratic function after you take the square root out of it won't move.

The x coordinate of the vertex of the quadratic function is:

t = -b/2a = 32/(2*51) = 16/51

This is the same as the vertex of the square root of that quadratic function.

So the distance becomes:

d = sqrt(51(16 /51)^2 - 32(16/51) + 22) = 4.121