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Find An Equation Of The Plane That Passes Through The Point P(−6,−7,2) And Has The Vector N=⟨−7,2,5⟩ As A Normal.

User Cortwave
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The equation of the plane can be found using the formula of a plane, which is Ax + By + Cz = D, where A, B and C are the coefficients of the normal vector and D can be found by substituting the point into the equation.

Step 1: Let's define the normal vector N=⟨−7,2,5⟩. The coefficients A, B, and C of the normal vector become -7, 2, and 5, respectively.

So, A = -7, B = 2, and C = 5.

Step 2: Substitute the coordinates of point P(-6, -7, 2) into the equation Ax + By + Cz = D to find the value of D.

Here, x = -6, y = -7, and z = 2. So, substituting these values, we get:

D = A*(-6) + B*(-7) + C*2 = -7*-6 + 2*-7 + 5*2 = 42 - 14 + 10 = 38.

So, D = 38.

Step 3: Insert A, B, C, and D into the equation of the plane giving us the following equation:

-7x + 2y + 5z = 38.

So, the equation of the plane that passes through the point P(-6, -7, 2) and has the normal vector N=⟨−7,2,5⟩ is -7x + 2y + 5z = 38.

User Himanshu Mori
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