Let T be the plane 2x−3y = −2. Find the shortest distance d from the point P0=(−1, −2, 1) to T, and the point Q in T that is closest to P0. Use the square root symbol '√' where needed to give an exact value for your answer.

Question

I need help with this one question!!

Answer 1

Use the distance formula from point to plane:
distance from (−1, −2, 1) to 2x - 3y + 0z + 2 = 0 is
= |2(-1) - 3(-2) + 0(1) + 2 |/√(2^2 + 3^2 + 0^2)
= 6/√13

if you actually need to find Q:
normal to the plane is vector (2,-3,0), so the perpendicular line containing (−1, −2, 1) is
x = -1 + 2t
y = -2 - 3t
z = 1

sub those into the plane equation and solve for t
put the value of t into the parametric equations to get Q

Check my answer above by finding PQ using the formula for the distance between two points.

Answer 2

Q makes sense but d is calculated a different way

Answer 3

take each value of Q and subtract the respective value from P0
then square the 3 new values you have, add them together and then take the square root of their sum
Pythagorean theorem basically