Question

The component vector of $4\overline{i}-4\overline{j}-7\overline{k}$ perpendicular to $\overline{i}-\overline{2j}-\overline{k}$ is

• A. $\frac{5\overline{i}+14\overline{j}-23\overline{k}}{6}$
• B. $5\overline{i}+14\overline{j}-23\overline{k}$
• C. $\frac{-5\overline{i}-14\overline{j}+23\overline{k}}{3}$
• D. $3\overline{i}-4\overline{j}+6\overline{k}$

Answer 1

The correct option is A $\frac{5\overline{i}+14\overline{j}-23\overline{k}}{6}$

Here,

the component vector is, $4\overline{i}-4\overline{j}-7\overline{k}$

and , the perpendicular is $\overline{i}-2\overline{j}-\overline{k}$,

$\begin{array}{c}\therefore \overline{a}=\frac{\overline{a}.\overline{b}}{\overline{b}.\overline{b}},\overline{b}=\frac{\overline{(a}.\overline{b)}.\overline{b}}{\overline{b}.\overline{b}}\\ \overline{a}.\overline{b}=\left|\begin{array}{c}ijk\\ 1-2-1\\ 4-2-7\end{array}\right|\\ =i\left|\begin{array}{c}-2-1\\ -2-7\end{array}\right|-j\left|\begin{array}{c}1-1\\ 4-7\end{array}\right|+k\left|\begin{array}{c}1-2\\ 4-2\end{array}\right|\\ =i(14-4)-j(-7+4)+k(-4+8)\end{array}$

$\begin{array}{c}=10i+3j+4k\\ and\\ (b.a)\times b=\left|\begin{array}{c}ijk\\ 103-4\\ 1-2-1\end{array}\right|\\ =i\left|\begin{array}{c}3-4\\ -2-1\end{array}\right|-j\left|\begin{array}{c}10-4\\ 1-1\end{array}\right|+k\left|\begin{array}{c}103\\ 1-2\end{array}\right|\\ =i(-3+8)-j(-10+4)+k(-20-3)\\ =5i+6j-23k\\ \overline{b}.\overline{b}=(i-2j-k)(i-2j-k)\\ =1+4+1=6\\ thenthevectorcomponentis=\frac{5\overline{i}+6\overline{j}-23\overline{k}}{6}\end{array}$

so,that the correct option is A.