Question

If A=2i +3j-k and B= - i+3j+4k,then what will be a unit vector perpendicular to both A and B? ​

Answer 1

Given:

Two vectors-

$A=2\hat{i}+3\hat{j}-\hat{k}$

$B=-\hat{i}+3\hat{j}+4\hat{k}$

To Find:

A unit vector perpendicular to both A and B

Concept to be Used:

A vector perpendicular to two non-parallel vectors is the cross product of given two vectors  

Solution:

We know that,

• Unit vector of a vector is given by

$\pink{\boxed{\bf{\hat{A}=\dfrac{\vec{A}}{\mid A\mid}}}}$

$\rule{190}{1}$

Now, firstly we have to find cross product of A and B

$\longrightarrow\rm{A\times B}$

$\longrightarrow\rm{\left|\begin{array}{ccc}\hat{i} &\hat{j}&\hat{k}\\2&3&-1\\-1&3&4\end{array}\right|}$

$\longrightarrow\rm{\hat{i}(12+3)-\hat{j}(8-1)+\hat{k}(6+3)}$

$\longrightarrow\rm{\hat{i}(15)-\hat{j}(7)+\hat{k}(9)}$

$\longrightarrow\rm{15\hat{i}-7\hat{j}+9\hat{k}}$

$\rule{190}{1}$

Let $\rm{\vec{C}=15\hat{i}-7\hat{j}+9\hat{k}}$

Now, we have to find the unit vector along $\vec{C}$

So,

$\longrightarrow\rm{\hat{C}=\dfrac{\vec{C}}{\mid C\mid}}$

$\longrightarrow\rm{\hat{C}=\dfrac{15\hat{i}-7\hat{j}+9\hat{k}}{\sqrt{(15)^{2}+(-7)^{2}+(9)^{2}}}}$

$\longrightarrow\rm{\hat{C}=\dfrac{15\hat{i}-7\hat{j}+9\hat{k}}{\sqrt{225+49+81}}}$

$\longrightarrow\rm{\hat{C}=\dfrac{15\hat{i}-7\hat{j}+9\hat{k}}{\sqrt{355}}}$

$\longrightarrow\rm\green{\hat{C}=\dfrac{15\hat{i}}{\sqrt{355}}-\dfrac{7\hat{j}}{\sqrt{355}}+\dfrac{9\hat{k}}{\sqrt{355}}}$

Hence, $\bf{\hat{C}}$ is the required unit vector.