Question

Find the unit vector perpendicular to each of the vectors $6\hat{i}+2\hat{j}+3\hat{k}$ and $3\hat{i}-2\hat{k}$

Answer 1

We know that vectors are quantities which have both direction and magnitude. We also know that two or more vectors can be added by vector addition and multiplied by vector cross product. Clearly in both the cases, we get another vector which has its own direction and magnitude.

Complete step-by-step solution:
We know that two vectors which make an angle $\theta$ with each other can be multiplied to obtain a resultant vector , this is called a vector cross product. It is mathematically denoted as
$\vec{A}\times\vec{ B}=A.B.sin\theta\hat{n}$, where $\vec{A}$ and $\vec{B}$ are vectors while $A$ and $B$ are the magnitude of the vectors respectively . Also $\hat n$ is a unit vector perpendicular to both $\vec{A}$ and $\vec{B}$.
This means to say that the resultant of two vectors which is cross multiplied is always perpendicular to both the parent vectors, $\vec{A}$ and $\vec{B}$.
We can also express the same in terms of matrix as
$\vec{A}\times \vec{B}=\left[\begin{matrix} i&j;&k;\\\ a_1&a;_2&a;_3\\\ b_1&b;_2&b;_3\end{matrix}\right]$, where $\vec{A}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$ and $\vec{B}=b_1\hat{i}+b_2\hat{j}+b_3\hat{k}$.
Then the resultant is given as $\vec{A}\times \vec{B}=\\{(a_2b_3)-(a_3b_2)\\} \hat{i}+\\{(a_3b_1)-(a_1b_3)\\}\hat{j}+\\{(a_1b_2)-(a_2b_1)\\}\hat{k}$
Here, we have $\vec{A}=6\hat{i}+2\hat{j}+3\hat{k}$ and $\vec{B}=3\hat{i}-2\hat{k}$, representing the same in matrix, we have
$\vec{A}\times \vec{B}=\left[\begin{matrix} i&j;&k;\\\ 6&2&3\\\ 3&0&-2\end{matrix}\right]$ $\implies \vec{A}\times \vec{B}=(-4-0)\hat{i}+(9+12)\hat{j}+(0-6)\hat{k}$
$\implies \vec{A}\times \vec{B}=(-4)\hat{i}+(21)\hat{j}+(-6)\hat{k}$
The unit vector along $\vec{A}\times \vec{B}$can be written as
$\hat{n}=\dfrac{\vec{A}\times \vec{B}}{||\vec{A}\times \vec{B}||}$
$\implies \hat{n}=\dfrac{(-4)\hat{i}+(21)\hat{j}+(-6)\hat{k}}{\sqrt{(-4)^2+(21)^2+(-6)^2}}$
$\implies \hat{n}=\dfrac{(-4)\hat{i}+(21)\hat{j}+(-6)\hat{k}}{\sqrt{16+441+36}}$
$\therefore \hat{n}=\dfrac{(-4)\hat{i}+(21)\hat{j}+(-6)\hat{k}}{\sqrt{493}}$
Thus the required unit vector is $\dfrac{(-4)\hat{i}+(21)\hat{j}+(-6)\hat{k}}{\sqrt{493}}$

Note: There are two other types of vector multiplication as well, namely the vector dot product, where the vector is either multiplied by a scalar or another vector and the resultant is a scalar, and hence is also called as the scalar multiplication. And the other is the scalar triple product or the box product, which again results in a scalar, here two vectors are cross multiplied and the resultant is again scalar multiplied with the other vector.