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

Answer: (B)

Let $\vec{c}=c_1\hat{i}+c_2\hat{j}+c_3\hat{k}$ be the unit vector which is perpendicular to both $\vec{a}$ and $\vec{b}$, then
$\vec{a}\cdot \vec{c}=0\Rightarrow 2c_1+c_2+c_3=0...\left(i\right)$
and $\vec{b}\cdot \vec{c}=0\Rightarrow c_2+c_3=0...\left(ii\right)$
$\vec{c}$ is unit vector
$\Rightarrow c_1^2+c_2^2+c_3^2=1...\left(iii\right)$
From eq. $\left(i\right)$ and $\left(ii\right)$, we get
$2c_1+c_3=0...\left(iv\right)$
From eq. $\left(ii\right)$ and $\left(iii\right)$, we get
$c_1^2+2c_3^2=1...\left(v\right)$
Now, from eq. $\left(iv\right)$ and $\left(v\right)$
$c_1^2+2{\left(-2c_1\right)}^2=1$
$\Rightarrow c_1^2+8c_1^2=1$
$\Rightarrow c_1^2=\frac{1}{9}$
$\Rightarrow c_1=\pm \frac{1}{3}$
Putting the value of $c_1$ in eq. $\left(iv\right)$, we get
$c_3=\mp \frac{2}{3}$
Now, from eq. $\left(i\right)$
$c_2=-2\left(c_1+c_3\right)$
If $c_1=-\frac{1}{3}$, then $c_3=\frac{-2}{3}$
$\therefore c_2=-2\left[\frac{-1}{2}+\frac{2}{3}\right]=-2\left[\frac{1}{3}\right]=\frac{-2}{3}$
Hence, vector $\vec{c}=\frac{1}{3}\hat{i}+\frac{2}{3}\hat{j}-\frac{2}{3}\hat{k}$ and $\vec{c}=-\frac{1}{3}\hat{i}-\frac{2}{3}\hat{j}+\frac{2}{3}\hat{k}$
Hence, there are two unit vectors perpendicular to the given vectors $\vec{a}\&\vec{b}$.