Question

If $\vec {a}= {2\hat{i}}+3\hat{j}-\hat k,\vec b = \hat i+2 \hat j-5 \hat k ,\vec{ c} =3 \hat i+ 5\hat j-\hat k,$ then a vector perpendicular to $\vec{a}$ and in the plane containing $\vec {b}$ and $\vec {c}$ is

• A. $17 \hat i+21 \hat{j} -123 \hat k$
• B. $-17 \hat i+21 \hat{j} -97 \hat k$
• C. $-17 \hat i-21 \hat{j} -97 \hat k$
• D. $-17 \hat i-21 \hat{j} +97 \hat k$

Answer 1

Answer: (C)

We know that a vector perpendicular to a and in the plane containing $\vec{b}$ and $\vec{c}$ is given by
$\vec{a} \times (\vec{b} \times \vec{c}).$
Given $\vec{a} = 2\hat{i} + 3 \hat{j} - \hat{k}$
$\vec{b} = \hat{i} + 2 \hat{j} - 5 \hat{k} $ and $\vec{c} = 3 \hat{i } + 5 \hat{j} - \hat{k} $
$\therefore \vec{b} \times\vec{c} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 1 &2&-5\\ 3&5&-1\end{vmatrix} = 23 \hat{i} - 14 \hat{j} - \hat{k} $
Now, $ \vec{a} \times\left( \vec{b} \times\vec{c}\right) = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\ 2&3&-1\\ 23&-14&-1\end{vmatrix} $
$ = (-3 - 14) \hat{ i} - \hat{j} (-2+ 23) + \hat{k} (-28 - 69)$
$= -17 \hat{i} - 21 \hat{ j} - 9 7 \hat{k}$
Which is the required vector.