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 :
Solution
$\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}, \vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}$
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Now, $\vec{a} \cdot \vec{b}=\sqrt{54} \Rightarrow \frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}$
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$\Rightarrow-2 \lambda+4 \mu-8=36$
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$\Rightarrow 2 \mu-\lambda=22\quad...(i)$
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and $\left|\begin{array}{ccc}\lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{array}\right|=0$
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$10 \lambda-2 \mu-56=0 \quad...(ii)$
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By (i) & (ii) $\lambda=\frac{78}{9}, \mu=\frac{138}{9}$
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$\therefore \mu+\lambda=24$
About this question
Subject: Mathematics · Chapter: Vector Algebra · Topic: Types of Vectors
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