Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Solution

Let the equation of the plane be,

A ( x - x₁ ) + B ( y - y₁ ) + C ( z - z₁ ) = 0

Plane passes throughthe points ( -1, 3, 2 )

$∴$ A ( x + 1 ) + B ( y - 3 ) + C ( z - 2 ) = 0 ........(i)

Now applying the condition of perpendicularity to the plane (i) with planes

x + 2y + 3z = 5 and 3x + 3y + z = 0, We have,

A + 2B + 3C = 0

3A + 3B + C = 0

Solving we get

A + 2B + 3C = 0

9A + 9B + 3C = 0

By cross multiplication, we have,

$\frac{A}{2 x 3 - 9 x 3} = \frac{B}{9 x 3 - 1 x 3} = \frac{C}{1 x 9 - 2 x 9} \Rightarrow \frac{A}{6 - 27} = \frac{B}{27 - 3} = \frac{C}{9 - 18} \Rightarrow \frac{A}{- 21} = \frac{B}{24} = \frac{C}{- 9} \Rightarrow \frac{A}{- 7} = \frac{B}{8} = \frac{C}{- 3} \Rightarrow A = 7 λ; B = - 8 λ; C = 3 λ$

By substituting A and C in equation (i), we get,

Substituting the values of A, B and C in equation (i), we have,

$7 λ (x + 1) - 8 λ (y - 3) + 3 λ (z - 2) = 0 \Rightarrow 7 x + 7 - 8 y + 24 + 3 z - 6 = 0 \Rightarrow 7 x - 8 y + 3 z + 25 = 0$

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Three Dimensional Geometry

Find the equation of the plane passing through the point (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

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