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Q.
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
Vector Algebra
Solution:
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 eq. $\left(i\right)$ and $\left(ii\right)$, we get
$ 2c_{1} + c_{3} = 0 \quad...\left(iv\right)$
From eq. $\left(ii\right)$ and $\left(iii\right)$, we get
$c_1^{2} +2c_3^{2} =1 \quad ...\left(v\right)$
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 $\vec{a}\, \&\, \vec{b}$.