Vector Algebra Question 8
Question 8 - 29 January - Shift 1
If the vectors $\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-2 \hat{k}$ and $\vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$ are coplanar and the projection of $\vec{a}$ on the vector $\vec{b}$ is $\sqrt{54}$ units, then the sum of all possible values of $\lambda+\mu$ is equal to
(1) 0
(2) 6
(3) 24
(4) 18
Show Answer
Answer: (3)
Solution:
Formula: Scalar product of two vectors; projection of a vector on the another vector, Scalar triple product; Condition for four vectors to be coplanar
$ \begin{vmatrix} \lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{vmatrix} =0$
$\Rightarrow\lambda(10)-\mu\cdot2+4\cdot(-14)=0$
$\Rightarrow 10\lambda-2 \mu=56$
$\Rightarrow 5 \lambda-\mu=28$
$\frac{\overrightarrow{{}a} \cdot \overrightarrow{{}b}}{|\overrightarrow{{}b}|}=\sqrt{54}$
$\Rightarrow \frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}$
$\Rightarrow -2 \lambda+4 \mu-8=\sqrt{54 \times 24}$
By solving equation (1) and (2)
$\Rightarrow \lambda+\mu=24$
