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Find the shortest distance from the point (1, 0, −2) to the

plane 2x+4y+2z = 8

User Hubert
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1 Answer

5 votes
To find the shortest distance from a point to a plane, we can use the formula:

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

Here, the equation of the plane is 2x + 4y + 2z = 8, which can be rewritten as 2x + 4y + 2z - 8 = 0.

Comparing this to the general form of the equation ax + by + cz + d = 0, we have:
a = 2, b = 4, c = 2, and d = -8.

Now, let's substitute the values of the point (1, 0, -2) into the formula:

Distance = |(2 * 1) + (4 * 0) + (2 * -2) - 8| / √((2^2) + (4^2) + (2^2))

Simplifying:
Distance = |2 - 4 - 4 - 8| / √(4 + 16 + 4)
Distance = |-14| / √24
Distance = 14 / √24
Distance ≈ 2.85 (rounded to two decimal places)

Therefore, the shortest distance from the point (1, 0, -2) to the plane 2x + 4y + 2z = 8 is approximately 2.85 units.
User Bill White
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