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Find the equation of the plane through P0(5,5,−7) with normal vector ⃗ =[ 2 −1 4]

User Wajih
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1 Answer

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Final answer:

The equation of the plane through P0(5,5,-7) with normal vector (2,-1,4) is 2x - y + 4z = -43.

Step-by-step explanation:

To find the equation of the plane, we can use the formula:

Ax + By + Cz = D

where (A, B, C) is the normal vector and (x, y, z) are the coordinates of a point on the plane.

In this case, the normal vector is (2, -1, 4), and a point on the plane is (5, 5, -7). Plugging these values into the equation, we get:

2x - y + 4z = D

To find the value of D, we can substitute the coordinates of the given point:

2*5 - (-1)*5 + 4*(-7) = D

Simplifying, we get D = -43. Therefore, the equation of the plane is:

2x - y + 4z = -43

User David Guan
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