Question

How do you find a unit vector perpendicular to a $3-D$ plane formed by points
$(1,0,1),(0,2,2)$ and $(3,3,0)$

Answer 1

Hint:A vector can be expressed as the unit vector when it is divided by its vector magnitude. The unit vector is also known as the normalized vector. Here we will first find the magnitude of the given vector. Then find the values for the unit vector and then place the values in the standard formula for the resultant value

Complete step by step solution:
Write the given points in the form of the vector.
$A(1,0,1)$
$B(0,2,2)$
$C(3,3,0)$
Find the distance between the points A and B and vector distance between the points B and C.
$\overrightarrow{AB}=(0-1)\hat{i}+(2-0)\hat{j}+(2-1)\hat{k}$
Simplify the above expression
$\overrightarrow{AB}=-\hat{i}+2\hat{j}+\hat{k}$ ….(i)
Similarly find the vector distance from the points B and C
$\overrightarrow{BC}=(3-0)\hat{i}+(3-2)\hat{j}+(0-2)\hat{k}$
Simplify the above expression
$\overrightarrow{BC}=3\hat{i}+\hat{j}-2\hat{k}$ ….(ii)
Take the cross product of the equation (i) and (II)
$\overrightarrow{AB}\times \overrightarrow{BC}=\left|\begin{array}{ccc}i& j& k\\ -1& 2& 1\\ 3& 1& -2\end{array}\right|$
Expand the above determinant –
$=i(-4-1)-j(2-3)+k(-1-6)$
Simplify the above equation, keeping sign convention the utmost carefully.
When there are two negative terms then you have to add both the values and sign of minus, whereas when you subtract a bigger number from the smaller number, then you have to do subtraction and sign of a bigger number.
$\overrightarrow{AB}\times \overrightarrow{BC}=i(-5)-j(-1)+k(-7)$
Product of plus and minus gives minus and product of minus into minus gives plus.
$\overrightarrow{AB}\times \overrightarrow{BC}=-5\hat{i}+\hat{j}-7\hat{k}$
Now the unit vector perpendicular to the place ABC can be given by –
$\hat{v}={\displaystyle \frac{\bar{v}}{\left|\bar{v}\right|}}$
Find the mode of the vector,
$\left|v\right|=\sqrt{5^2+1^2+7^2}$
Simplify the above equation:
$$\begin{gathered} \left|v\right|=\sqrt{25+1+49} \\ \left|v\right|=\sqrt{75} \end{gathered}$$
The above equation can be re-written as –
$\left|v\right|=\sqrt{25\times 3}$
Applying the concept of square and square-root cancel each other.
$\left|v\right|=\pm 5\sqrt{3}$
Place all the values in the standard equation:
$\hat{v}={\displaystyle \frac{-5\hat{i}+\hat{j}-7\hat{k}}{\pm 5\sqrt{3}}}$
This is the required solution.

Note: Be good in square and square-root and simplify the resultant value accordingly. Always remember that the square of negative numbers or the positive numbers always give positive value but the magnitude of any vector is always positive.