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Find the distance between the point P (6, 5, 9) and the plane determined by points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6). ?

Find the distance between the point P (6, 5, 9) and the plane determined by points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6). ?

Grade:12

1 Answers

SHAIK AASIF AHAMED
askIITians Faculty 74 Points
6 years ago
Hello student,
Please find the answer to your question below
The plane determined by points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6) is
\begin{vmatrix} x-3 & y+1 &z-2 \\ 2 &3 &2 \\ -6&-3 &2 \end{vmatrix}=0
Which gives 3x-4y+3z=19
distance between the point P (6, 5, 9) and the plane3x-4y+3z=19 is
distance=(3(6)-4(5)+3(9)-19)/sqrt(9+16+9)
=6/\sqrt{34}units

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