Find the equation of the plane determined by the point A( 3, - 1, 2 ), B( 5, 2, 4 ) and C( -1, -1, 6 ) and hence find the distance between the plane and the point P( 6, 5, 9 ).

Solution

We know that, equation of a plane passing through 3 points,

$|\begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} \\ x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ x_{3} - x_{1} & y_{3} - y_{1} & z_{3} - z_{1} \end{matrix}| = 0 \Rightarrow |\begin{matrix} x - 3 & y + 1 & z - 2 \\ 2 & 3 & 2 \\ - 4 & 0 & 4 \end{matrix}| = 0 \Rightarrow (x - 3) (12 - 0) - (y + 1) (8 + 8) + (z - 2) (0 + 12) = 0 \Rightarrow 12 x - 36 - 16 y - 16 + 12 z - 24 = 0 \Rightarrow 12 x - 16 y + 12 z - 76 = 0 \Rightarrow 3 x - 4 y + 3 z - 19 = 0$

Also, perpendicular distance of P ( 6, 5, 9 ) to the plane 3 x - 4 y + 3 z - 19 = 0

$= \frac{|3 \times 6 - 4 \times 5 + 3 \times 9 - 19|}{\sqrt{9 + 16 + 9}} = \frac{6}{\sqrt{34}} units$

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Vector Algebra

Find the equation of the plane determined by the point A( 3, - 1, 2 ), B( 5, 2, 4 ) and C( -1, -1, 6 ) and hence find the distance between the plane and the point P( 6, 5, 9 ).

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