Algebra> Find the distance between the point P (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).

Manvendra Singh chahar , 11 Years ago

Grade Upto college level

1 Answers

Latika Leekha

Last Activity: 9 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
$\frac{|18-20+27-19|}{\sqrt{34}}$
= $\frac{6}{\sqrt{34}}$