The vector perpendicular to the vectors \(4 \mathbf{i}-\mathbf{j}+3 \mathbf{k}\) and \(-2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}\) whose magnitude
is 9

Let \(\mathbf{a}=4 \mathbf{i}-\mathbf{j}+3 \mathbf{k}, \mathbf{b}=-2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}\)
and \(\mathbf{c}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)
Given, \(\quad\) a \(\cdot \mathbf{c}=0\)
i.e., \(4 x-y+3 z=0\)...(i)
and b. \(\mathbf{c}=0\)
i.e., \(-2 x+y-2 z=0\)...(ii)
Also, \(|\mathbf{c}|=9\)
i.e., \(x^{2}+y^{2}+z^{2}=81\)...(iii)
Now, from Eqs. (i) and (ii), we get
\(
2 x+z=0 \Rightarrow z=-2 x
\)
On putting this value in Eq. (iii), we get
\(
\begin{aligned}
x^{2}+y^{2}+4 x^{2} &=81 \\
\Rightarrow \quad 5 x^{2}+y^{2} &=81...(iv)
\end{aligned}
\)
On multiplying Eq. (i) by 2 and Eq. (ii) by 3 and then adding, we get
\(
\begin{array}{r}
8 x-2 y+6 z=0 \\
\frac{-6 x+3 y-6 z=0}{2 x+y=0} \\
\Rightarrow \quad y=-2 x
\end{array}
\)
On putting this value in Eq. (iv), we get
\(
5 x^{2}+4 x^{2}=81
\)
\(
\begin{array}{lcc}
\Rightarrow & 9 x^{2}=81 \\
\Rightarrow & x^{2}=9 \\
\Rightarrow & x=\pm 3 \\
\therefore & y=\mp 6 & \text { and } z=\mp 6
\end{array}
\)
\(\therefore\) Required vector, \(\mathbf{c}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\)
\(
\begin{array}{l}
=\pm 3 \mathbf{i} \mp 6 \mathbf{j} \mp 6 \mathbf{k} \\
=3 \mathbf{i}-6 \mathbf{j}-6 \mathbf{k}
\end{array}
\)
or
\(
=-3 \mathbf{i}+6 \mathbf{j}+6 \mathbf{k}
\)