Question

The distance between a point and a plane is given by the formula: d = |Ax1 + By1 + Cz1 + D| / √(A^2 + B^2 + C^2) In this case, the point P is (-7, -3, -4) and the plane is given by -8x + 2y + 3z = 2. Therefore, the distance between the point P and the plane is: d = |-8(-7) + 2(-3) + 3(-4) - 2| / √((-8)^2 + 2^2 + 3^2)

          The distance between a point and a plane is given by the formula:

d = |Ax1 + By1 + Cz1 + D| / √(A^2 + B^2 + C^2)

In this case, the point P is (-7, -3, -4) and the plane is given by -8x + 2y + 3z = 2.

Therefore, the distance between the point P and the plane is:

d = |-8(-7) + 2(-3) + 3(-4) - 2| / √((-8)^2 + 2^2 + 3^2)
        
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Added by Ana G.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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The distance between a point and a plane is given by the formula: d = |Ax1 + By1 + Cz1 + D| / √(A^2 + B^2 + C^2) In this case, the point P is (-7, -3, -4) and the plane is given by -8x + 2y + 3z = 2. Therefore, the distance between the point P and the plane is: d = |-8(-7) + 2(-3) + 3(-4) - 2| / √((-8)^2 + 2^2 + 3^2)
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Transcript

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00:02 Given that a plane passes through the origin 1 ,1 ,2 and 0, -1 now the equation of the plane is a of x -x1 plus b of y -y1 plus c of z -z1 which is equal to 0 now as the normal vector to the plane, the plane is the cross product of b1 ,b2 b1 cross b2 this one is our b1, this one is our b2 now b1 cross b2 is 1 ,1 ,2 cross 2 ,0, -1 which is equal to 5 -2 so the equation of plane is equation of plane is equation of plane is now let the closest…
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