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)

two

Answer 2

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 $\vec{a} \cdot \vec{c} = 0 \Rightarrow 2c_{1} +c_{2} +c_{3} = 0 \quad...\left(i\right)$ and $\vec{b} \cdot \vec{c} = 0 \Rightarrow c_{2} + c_{3} = 0 \quad...\left(ii\right)$ $\vec{c}$ is unit vector $\Rightarrow c_{1}^{2} + c_{2}^{2} + c_{3}^{2} = 1 \quad...\left(iii\right)$ From e $\left(i\right)$ and $\left(ii\right)$, we get $2c_{1} + c_{3} = 0 \quad...\left(iv\right)$ From e $\left(ii\right)$ and $\left(iii\right)$, we get $c_1^{2} +2c_3^{2} =1 \quad ...\left(v\right)$ Now, from e $\left(iv\right)$ 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 e $\left(iv\right)$, we get $c_{3} = \mp\frac{2}{3}$ Now, from e $\left(i\right)$ $c_{2} = -2 \left(c_{1} +c_{3}\right)$ 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 $\vec{a}\, \&\, \vec{b}$.