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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:Upto college level

1 Answers

Latika Leekha
askIITians Faculty 165 Points
6 years ago
Hello student,
First of all our aim shall be to find out the equation of the plane determined by the points A (3, –1, 2), B (5, 2, 4) and C (–1, –1, 6).
The equation of the plane determined by the points (x1, y1, z1), (x2, y2, z2) and (x3, y3, z3) is given by
\begin{vmatrix} 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{vmatrix} = 0
So, \begin{vmatrix} x-3 & y+1 & z-2 \\ 5-3& 2+1 & 4-2 \\ -1-3& -1+1 & 6-2 \end{vmatrix}= 0
Hence, this gives (x-3)12 – (y+1)16 + (z-2)12 = 0
Hence, the eqaution of the plane is 12x-16y+12z = 76.
or, 3x – 4y + 3z = 19.
Now, the distance of the plane 3x – 4y + 3z = 19 from the point (6, 5, 9) is
\frac{|18-20+27-19|}{\sqrt{34}} = \frac{6}{\sqrt{34}}
Hence, this is the required answer.

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