Question

find unit vector perpendicular to plane of a=2i-2j+k and b=i+2j+2k

Answer 1

Given:

Two vectors-

$\longrightarrow\rm{\overrightarrow{a}=2\hat{i}-2\hat{j}+\hat{k}}$

$\longrightarrow\rm{\overrightarrow{b}=\hat{i}+2\hat{j}+2\hat{k}}$

To Find:

Unit vector perpendicular to plane having given vectors

Concept to be Used:

We have to find the cross product a and b because direction of a×b is perpendicular to both a and b and then we have to find a unit vector in the direction of a×b

Solution:

Let $\rm{\overrightarrow{p}=\overrightarrow{a}\times\overrightarrow{b}}$

So,

$\longrightarrow\rm{\overrightarrow{p}=\left|\begin{array}{ccc}\hat{i}&\hat{j}&\hat{k}\\2&-2&1\\1&2&2\end{array}\right|}$

$\longrightarrow\rm{\overrightarrow{p}=\hat{i}(-4-2)-\hat{j}(4-1)+\hat{k}(4+2)}$

$\longrightarrow\rm{\overrightarrow{p}=-6\hat{i}-3\hat{j}+6\hat{k}}$

Now, let the unit vector in the direction of $\overrightarrow{p}$ be $\hat{p}$

So,

$\longrightarrow\rm{\hat{p}=\dfrac{-6\hat{i}-3\hat{j}+6\hat{k}}{\sqrt{(-6)^{2}+(-3)^{2}+(6)^{2}}}}$

$\longrightarrow\rm{\hat{p}=\dfrac{-6\hat{i}-3\hat{j}+6\hat{k}}{\sqrt{81}}}$

$\longrightarrow\rm{\hat{p}=\dfrac{-6\hat{i}-3\hat{j}+6\hat{k}}{9}}$

$\longrightarrow\rm{\hat{p}=-\dfrac{6}{9}\hat{i}-\dfrac{3}{9}\hat{j}+\dfrac{6}{9}\hat{k}}$

$\longrightarrow\rm\green{\hat{p}=-\dfrac{2}{3}\hat{i}-\dfrac{1}{3}\hat{j}+\dfrac{2}{3}\hat{k}}$

Hence, the unit vector perpendicular to plane of a and  b is $\rm\pink{-\dfrac{2}{3}\hat{i}-\dfrac{1}{3}\hat{j}+\dfrac{2}{3}\hat{k}}$.