Find the distance from the point to the given plane. (1, −7, 8), 3x + 2y + 6z = 5

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dominic24 dominic24 · Answer 1 · 2018-03-09T18:16:30+01:00

3x + 2y + 6z = 5??
then
distance = |3(1) + 2(-2) + 6(4) - 5| / √(3^2 + 2^2 + 6^2)
the plane is ax + by + cz - d = 0 while the point is (x1, y1, z1)

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topeadeniran2 topeadeniran2 · Answer 2 · 2022-06-21T22:56:45+02:00

The distance from the point to the given plane will be 32/7.

The distance from the point to the given plane will be:

L = (ax1 + by¹ + cz¹ - d)/(✓a² ✓b² + ✓c²)

L = [3(1) + 2(-7) + 6(8) - 5]/✓(3² + 2² + 6²)

L = (3 - 14 + 48 - 5)/(✓9 + 4 + 36)

L = 32/✓49

L = 32/7

In conclusion, the distance from the point to the given plane will be 32/7.

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