Verified answer

The parametric equations of the line are:

r(t):

x = 2 + 2t

y = 2 - t

z = - 3 - 3t

The directional vector v, of the line is:

v = <2, -1, -3>

A point Q, on the line can be found by setting parameter t = 0.

Q(2, 2, -3)

Calculate the vector QP.

QP = <P - Q> = <-1-2, 2-2, -1+3> = <-3, 0, 2>

Calculate the magnitude of the vectors.

|| v || = √[2² + (-1)² + (-3)²] = √(4 + 1 + 9) = √14

|| QP || = √[(-3)² + 0² + 2²] = √(9 + 0 + 4) = √13

Let

h = distance between point P and given line

θ = angle between vectors v and QP

sinθ = h / || QP ||

________

Calculate the magnitude of the cross product.

|| v X QP || = || <2, -1, -3> X <-3, 0, 2> || = || <-2, 5, -3> ||

= √[(-2)² + 5² + (-3)²] = √[4 + 25 + 9] = √38

The cross product can also be expressed another way.

|| v X QP || = || v || || QP || sinθ

= || v || || QP || (h / || QP ||) = || v || h

h = || v X QP || / || v || = √38 / √14 = √(19/7) ≈ 1.6475089

distance=sqrt(x-x0)^2+(y-y0)^2+(z-z0)^2)

=sqrt( 2+2t+1)^2+(2-t-2)^2+(-3-3t+1)^2)

=sqrt(9+4t^2+6t+t^2+4+9t^2+6t)

=sqrt(14 t^2+12t+13)

derivative of 14 t^2+12t+13 is 28t+12

hence 28t+12=0

t=-12/28

t=- 3/7

calculate sqrt(14 t^2+12t+13)

for t= -3/7