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Find the distance from the point to the given plane. (–6, 3, 5), x – 2y – 4z = 8

User Robince
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To find the distance from a point to a plane, we can use the formula:

Distance = |ax + by + cz + d| / sqrt(a^2 + b^2 + c^2)

In this case, the point is (-6, 3, 5) and the equation of the plane is x - 2y - 4z = 8.

Let's substitute the values into the formula:

Distance = |(-6) - 2(3) - 4(5) + 8| / sqrt(1^2 + (-2)^2 + (-4)^2)

Distance = |-6 - 6 - 20 + 8| / sqrt(1 + 4 + 16)

Distance = |-24| / sqrt(21)

Distance = 24 / sqrt(21)

Therefore, the distance from the point (-6, 3, 5) to the plane x - 2y - 4z = 8 is 24 / sqrt(21).
User Simekadam
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