Question

Find the distance D from the point A(0,-8,8) to the plane 3x + 7y - 2z + 5 = 0

Answer 1

i did S(0, -8, 8)
P would then be P(0, -5/7, 0).... just because it's an easy point. the y intercept
\[\rightarrow PS\] = -51/7j + 8k

magnitude of 3x + 7y-2z = -5 would be \[\sqrt{60}\]


then just take dot product of ps and <3/\[\sqrt{60}\], 7/\[\sqrt{60}\], -2\[\sqrt{60}\]

would be 0 - 15/\[\sqrt{60}\] - 67/\[\sqrt{60}\]

Answer 2

er -15/sqr 60 - 16/sqrt 60 = -67/sqrt 60... absolute value so it would be 67 / sqrt 60