JEE MAIN - Mathematics (2023 - 29th January Morning Shift - No. 18)
If the vectors $$\overrightarrow a = \lambda \widehat i + \mu \widehat j + 4\widehat k$$, $$\overrightarrow b = - 2\widehat i + 4\widehat j - 2\widehat k$$ and $$\overrightarrow c = 2\widehat i + 3\widehat j + \widehat k$$ are coplanar and the projection of $$\overrightarrow a $$ on the vector $$\overrightarrow b $$ is $$\sqrt {54} $$ units, then the sum of all possible values of $$\lambda + \mu $$ is equal to :
24
0
18
6
Explanation
$\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}, \vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$
Now, $\vec{a} \cdot \vec{b}=\sqrt{54} \Rightarrow \frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}$
$\Rightarrow-2 \lambda+4 \mu-8=36$
$\Rightarrow 2 \mu-\lambda=22\quad...(i)$
and $\left|\begin{array}{ccc}\lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{array}\right|=0$
$10 \lambda-2 \mu-56=0 \quad...(ii)$
By (i) & (ii) $\lambda=\frac{78}{9}, \mu=\frac{138}{9}$
$\therefore \mu+\lambda=24$
Now, $\vec{a} \cdot \vec{b}=\sqrt{54} \Rightarrow \frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}$
$\Rightarrow-2 \lambda+4 \mu-8=36$
$\Rightarrow 2 \mu-\lambda=22\quad...(i)$
and $\left|\begin{array}{ccc}\lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{array}\right|=0$
$10 \lambda-2 \mu-56=0 \quad...(ii)$
By (i) & (ii) $\lambda=\frac{78}{9}, \mu=\frac{138}{9}$
$\therefore \mu+\lambda=24$
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