Q is somewhere on the vector
<5,5,-3> + c<3,0,-2> = <5+3c,5,-3-2c>
But that point must lie in the plane, so
3(5+3c) - 2(-3-2c) = 14
c = -7/13
Now you know Q = (5,5,-3) - 7/13 (3,0,-2)
The distance is, as usual
|3*5 + 0*5 - 2(-3)-14|/√(3^2+0^2+2^2) = 7/√13
Let T be the plane 3x−2z = 14. Find the shortest distance d from the point P0=(5, 5, −3) 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.
d=?
Q=(?,?,?)
2 answers
Thanks!!!!