Question

The number of vectors of unit length perpendicular to the vectors $\vec{a}=2\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$ is
A. One
B. Two
C. Three
D. Infinite

Answer 1

Hint: At first, we find the cross product of the two given vectors according to the formula $\vec{a}\times \vec{b}=\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|$ . We then find the unit vector in this direction. This vector is perpendicular to the plane containing a and b, and so it is perpendicular to both a and b. Similarly, we find $\vec{b}\times \vec{a}$ and thus get two vectors.

Complete step by step answer:
Vectors are notations of a quantity which has both magnitude as well as direction. They are represented by a letter or a number with an arrow on top of it. For example, $\vec{2}$ means a vector of magnitude $2$ and having a certain direction.
Now, we know that the cross product of two vectors $\vec{a}={{a}_{1}}\hat{i}+{{a}_{2}}\hat{j}+{{a}_{3}}\hat{k}$ and $\vec{b}={{b}_{1}}\hat{i}+{{b}_{2}}\hat{j}+{{b}_{3}}\hat{k}$ can be written as,
$\vec{a}\times \vec{b}=\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|$
It is given as the two vectors are $\vec{a}=2\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$ . So, doing the cross product gives,
$\vec{a}\times \vec{b}=\left| \begin{matrix}
   {\hat{i}} & {\hat{j}} & {\hat{k}} \\
   2 & 1 & 2 \\
   0 & 1 & 1 \\
\end{matrix} \right|=\left( 1-2 \right)\hat{i}-\left( 2-0 \right)\hat{j}+\left( 2-0 \right)\hat{k}=-\hat{i}-2\hat{j}+2\hat{k}$
This vector is perpendicular to both the vectors a and b. The unit vector in this direction is,
$\hat{n}=\dfrac{\vec{a}\times \vec{b}}{\left| {\vec{a}} \right|\left| {\vec{b}} \right|}=\dfrac{-\hat{i}-2\hat{j}+2\hat{k}}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}}=\dfrac{-\hat{i}-2\hat{j}+2\hat{k}}{3}=-\dfrac{1}{3}\hat{i}-\dfrac{2}{3}\hat{j}+\dfrac{2}{3}\hat{k}$
Similarly, $\vec{b}\times \vec{a}$ is also perpendicular to both a and b and the unit vector in its direction is $\dfrac{1}{3}\hat{i}+\dfrac{2}{3}\hat{j}-\dfrac{2}{3}\hat{k}$ .
Thus, we can conclude that there are two vectors of unit length perpendicular to the vectors $\vec{a}=2\hat{i}+\hat{j}+2\hat{k}$ and $\vec{b}=\hat{j}+\hat{k}$ .

So, the correct answer is “Option B”.

Note: We should understand the fact that the number of vectors perpendicular to a is infinite and the number of vectors perpendicular to b are also infinite. But, since a and b lie on the same plane, the vector which is mutually perpendicular to both of them is just a single vector (perpendicular to the plane) and another vector opposite to it. The answer is not infinite.