Respuesta :
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)
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)
The distance from the point to the given plane will be 32/7.
How to calculate the distance?
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.
Learn more about distance on:
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