A unit vector which is perpendicular to the vector $2 \hat{i}-\hat{j}+2 \hat{k}$ and is coplanar with the…
A unit vector which is perpendicular to the vector $2 \hat{i}-\hat{j}+2 \hat{k}$ and is coplanar with the vectors $\hat{i}+\hat{j}-\hat{k}$ and $2 \hat{i}+2 \hat{j}-\hat{k}$ is
$\frac{2 \hat{j}+\hat{k}}{\sqrt{5}}$
$\frac{3 \hat{i}+2 \hat{j}-2 \hat{k}}{\sqrt{17}}$
$\frac{3 \hat{i}+2 \hat{j}+2 \hat{k}}{\sqrt{17}}$
$\frac{2 \hat{i}+2 \hat{j}-\hat{k}}{3}$
Solution
Let $x \hat{i}+y \hat{j}+z \hat{k}$ be the required unit vector.
Since $\hat{a}$ is perpendicular to $(2 \hat{i}-\hat{j}+2 \hat{k})$.
$
\therefore \quad 2 x-y+2 z=0
$
Since vector $x \hat{i}+y \hat{j}+z \hat{k}$ is coplanar with the vector $\hat{i}+\hat{j}-\hat{k}$ and $2 \hat{i}+2 \hat{j}-\hat{k}$.
$
\begin{aligned}
\therefore \quad x \hat{i} & +y \hat{j}+z \hat{k} \\
& =p(\hat{i}+\hat{j}-\hat{k})+q(2 \hat{i}+2 \hat{j}-\hat{k}),
\end{aligned}
$
where $p$ and $q$ are some scalars.
$
\begin{aligned}
& \Rightarrow x \hat{i}+y \hat{j}+z \hat{k} \\
& =(p+2 q) \hat{i}+(p+2 q) \hat{j}-(p+q) \hat{k} \\
& \Rightarrow x=p+2 q, y=p+2 q, z=-p-q
\end{aligned}
$
Now from equation (i),
$
\begin{aligned}
& 2 p+4 q-p-2 q-2 p-2 q=0 \\
\Rightarrow & -p=0 \Rightarrow p=0 \\
\therefore & x=2 q, y=2 q, z=-q
\end{aligned}
$
Since vector $x \hat{i}+y \hat{j}+z \hat{k}$ is a unit vector, therefore
$
\begin{aligned}
& |x \hat{i}+y \hat{j}+z \hat{k}|=1 \\
\Rightarrow & \sqrt{x^2+y^2+z^2}=1 \\
\Rightarrow & x^2+y^2+z^2=1 \\
\Rightarrow & 4 q^2+4 q^2+q^2=1
\end{aligned}
$
$
\Rightarrow 9 q^2=1 \Rightarrow q=\pm \frac{1}{3}
$
When $q=\frac{1}{3}$, then $x=\frac{2}{3}, y=\frac{2}{3}$,
$
z=-\frac{1}{3}
$
When $q=-\frac{1}{3}$, then $x=-\frac{2}{3}, y=-\frac{2}{3}$,
$
z=\frac{1}{3}
$
Here required unit vector is $\frac{2}{3} \hat{i}+\frac{2}{3} \hat{j}-\frac{1}{3} \hat{k}$ or $-\frac{2}{3} \hat{i}-\frac{2}{3} \hat{j}+\frac{1}{3} \hat{k}$