Question

The equation of the plane in normal form which passes through the points $(-2,1,3)$, $(1,1,1)$ and $(2,3,4)$ is

• A. $\left(\frac{2}{3}\right)x+\left(-\frac{2}{3}\right)y+\left(\frac{1}{3}\right)z=\frac{1}{3}$
• B. $\left(-\frac{2}{3}\right)x+\left(\frac{2}{3}\right)y+\left(-\frac{1}{3}\right)z=\frac{1}{3}$
• C. $\left(\frac{-4}{\sqrt{173}}\right)x+\left(\frac{11}{\sqrt{173}}\right)y+\left(\frac{-6}{\sqrt{173}}\right)z=\frac{1}{\sqrt{173}}$
• D. $\left(\frac{4}{\sqrt{173}}\right)x+\left(\frac{-11}{\sqrt{173}}\right)y+\left(\frac{6}{\sqrt{173}}\right)z=\frac{1}{\sqrt{173}}$

Answer 1

Answer: (C)

Given points are $A(-2,1,3),B(1,1,1)$ and $C(2,3,4)$.
$\therefore A=-2\hat{i}+\hat{j}+3\hat{k},B=\hat{i}+\hat{j}+\hat{k}$ and
$C=2\hat{i}+3\hat{j}+4\hat{k}$
Now, $AB=(\hat{i}+\hat{j}+\hat{k})-(-2\hat{i}+\hat{j}+\hat{k})=3\hat{i}-2\hat{k}$
$BC=(2\hat{i}+3\hat{j}+4\hat{k})-(\hat{i}+\hat{j}+\hat{k})=\hat{i}+2\hat{j}+3\hat{k}$
The normal of the plane $n=AB\times BC$
$\therefore AB\times BC=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 0& -2\\ 1& 2& 3\end{array}\right|$
$=\hat{i}(0+4)-\hat{j}(9+2)+\hat{k}(6-0)$
$n=4\hat{i}-11\hat{j}+6\hat{k}$
Now, equation of plane
$\Rightarrow 4(x-2)-11(y-3)+6(z-4)=0$
$\Rightarrow 4x-8-11y+33+6z-24=0$
$\Rightarrow 4x-11y+6z+1=0$
As we know that, if $ax+by+cz+d=0$
is the equation of plane, then the normal form is
$\left(\frac{a}{\sqrt{a^2+b^2+c^2}}\right)x+\left(\frac{b}{\sqrt{a^2+b^2+c^2}}\right)y$
$+\left(\frac{c}{\sqrt{a^2+b^2+c^2}}\right)z+\left(\frac{d}{\sqrt{a^2+b^2+c^2}}\right)=0$
$\therefore$ Required normal form is
$\left(\frac{4}{\sqrt{4^2+11^2+6^2}}\right)x+\left(\frac{-11}{\sqrt{4^2+11^2+6^2}}\right)y$
$+\left(\frac{6}{\sqrt{4^2+11^2+6^2}}\right)z+\left(\frac{1}{\sqrt{4^2+11^2+6^2}}\right)=0$
$\Rightarrow \frac{4x}{\sqrt{16+12l+36}}-\frac{11y}{\sqrt{16+12l+36}}$
$+\frac{6z}{\sqrt{16+121+36}}+\frac{1}{\sqrt{16+121+36}}=0$
$\Rightarrow \frac{4x}{\sqrt{173}}-\frac{11y}{\sqrt{173}}+\frac{6z}{\sqrt{173}}+\frac{1}{\sqrt{173}}=0$
$\Rightarrow \frac{4x}{\sqrt{173}}-\frac{11y}{\sqrt{173}}+\frac{6z}{\sqrt{173}}=-\frac{1}{\sqrt{173}}$
$\Rightarrow -\frac{4x}{\sqrt{173}}+\frac{11y}{\sqrt{173}}-\frac{6z}{\sqrt{173}}=\frac{1}{\sqrt{173}}$
$\Rightarrow \left(-\frac{4}{\sqrt{173}}\right)x+\left(\frac{11}{\sqrt{173}}\right)y+\left(\frac{-6}{\sqrt{173}}\right)z=\frac{1}{\sqrt{173}}$