Question

A unit vector $n$ perpendicular to the plane determined by the points $A(0,-2,1),B(1,-1,-2)$and
$C(-1,1,0)$

• A. $\frac{1}{3}(2i+j+2k)$
• B. $\frac{1}{\sqrt{6}}(2i+j+k)$
• C. $\frac{1}{\sqrt{3}}(i-j+k)$
• D. $\frac{1}{\sqrt{14}}(3i+j+k)$

Answer 1

Answer: (B)

$\frac{1}{\sqrt{6}}(2i+j+k)$

eq of plane from three points is

$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-1\end{array}\right|=0$

Given points A(0,−2,1),B(1,−1,−2) and C(−1,1,0)

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

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

$8x-(y+2)(-4)+(z-1)\left(4\right)=0$

$8x+4y+8+4z-4=0$

$8x+4y+4z+4=0$

$2x+y+z+1=0$

eq of plane

$\overrightarrow{r}\cdot (2\hat{i}+\hat{j}+\hat{k})+1=0$

where $\overrightarrow{n}$ is perpendicular to plane

$\hat{n}=\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{2^2+1^2+1^2}}$

$\hat{n}=\frac{2\hat{i}+\hat{j}+\hat{k}}{\sqrt{6}}$

$\hat{n}=\frac{1}{\sqrt{6}}(2\hat{i}+\hat{j}+\hat{k})$