Question

Find the perpendicular distance of the point (2,1,-1) from the plane passing through (1,1,1), (1,-2,3),
(-3,1,2)

Answer 1

Coordinate Geometry (3D)

Formula to find the distance of a point from a plane:

Let the equation of a plane be

$\quad\quad ax+by+cz+d=0$

Then the distance of a point $(x_{1},\:y_{1},\:z_{1})$ from the plane is given by

$\quad\quad \frac{|ax_{1}+by_{1}+cz_{1}+d|}{\sqrt{a^{2}+b^{2}+c^{2}}}$

Solution:

The given plane passes through the points $(1,\:1,\:1)$, $(1,\:-2,\:3)$ and $(-3,\:1,\:2)$.

The equation of the plane passing through the point $(1,\:1,\:1)$ is

$\quad A(x-1)+B(y-1)+C(z-1)=0\quad ...(i)$

If it passes through the points $(1,\:-2,\:3)$ and $(-3,\:1,\:2)$, then we have

$\quad 0.A-3B+2C=0\quad ...(ii)$

$\quad -4A+0.B+C=0\quad ...(iii)$

Eliminating $A$, $B$ and $C$ from $(i)$, $(ii)$ and $(iii)$, we get the equation of the required plane as

$\quad$$\left|\begin{array}{ccc}x-1&y-1&z-1\\0&-3&2\\-4&0&1\end{array}\right|=0$

$\Rightarrow (x-1)(-3-0)-(y-1)(0+8)+(z-1)(0-12)=0$

$\quad$(Expanding along $R_{1}$)

$\Rightarrow -3x+3-8y+8-12z+12=0$

$\Rightarrow 3x+8y+12z-23=0$

$\therefore$ the distance of the point $(2,\:1,\:-1)$ from the given plane is

$\quad \frac{|3(2)+8(1)+12(-1)-23|}{\sqrt{3^{2}+8^{2}+12^{2}}}$

$=\frac{|6+8-12-23|}{\sqrt{217}}$

$=\frac{21}{\sqrt{217}}$

Answer: The perpendicular distance of the point $(2,\:1,\:-1)$ from the plane passing through the points $(1,\:1,\:1)$, $(1,\:-2,\:3)$, $(-3,\:1,\:2)$ is $\frac{21}{\sqrt{217}}$ units.

Answer 2

Answer:

2/rt(217)

Step-by-step explanation:

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