Question

Find the equation of a plane that is perpendicular to the vector 2i −2j +3k and passing through the point (3,4,−2)

The plane is given by:

Answer 1

$\\*\text{The equation of a plane perpendicular to the vector }\vec n=a \hat i+b\hat j+c \hat k \\*\text{and passing through the point }(x_1,y_1,z_1)\text{ is:} \\*\text{\quad\quad\quad\quad\quad\quad} a ( x - x_1) + b (y - y_1) + c (z -z_1) = 0$.

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Here, ñ = 2î – 2ị + 3k = a=2, b= -2, c= 3 (x1, 41, 21) = (3, 4, -2) = x1 = 3, Y1 = 4, 21 = -2

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Hence, the required plane is 2(x – 3) - 2(y – 4) + 3(z + 2) = 0 2x – 6 – 2y + 8 + 3z + 6 = 0 2x – 2y + 3z + 8 = 0