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Question: If the points whose position, vectors are \(3\mathbf{i} - 2\mathbf{j} - \mathbf{k},\) \(2\mathbf{i} ...

If the points whose position, vectors are 3i−2j−k,3\mathbf{i} - 2\mathbf{j} - \mathbf{k}, 2i+3j−4k,2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k}, −i+j+2k- \mathbf{i} + \mathbf{j} + 2\mathbf{k}and 4i+5j+λk4\mathbf{i} + 5\mathbf{j} + \lambda\mathbf{k} lie on a plane, then λ=\lambda =

A

−14617- \frac{146}{17}

B

14617\frac{146}{17}

C

−17146- \frac{17}{146}

D

17146\frac{17}{146}

Answer

−14617- \frac{146}{17}

Explanation

Solution

Leta=3i−2j−k,b=2i+3j−4k,c=−i+j+2k\mathbf{a} = 3\mathbf{i} - 2\mathbf{j} - \mathbf{k},\mathbf{b} = 2\mathbf{i} + 3\mathbf{j} - 4\mathbf{k},\mathbf{c} = - \mathbf{i} + \mathbf{j} + 2\mathbf{k} and

d=4i+5j+λk.\mathbf{d} = 4\mathbf{i} + 5\mathbf{j} + \lambda\mathbf{k}.

Since the points are coplanar,

So, [dbc]+[dca]+[dab]=[abc]\lbrack\mathbf{dbc}\rbrack + \lbrack\mathbf{dca}\rbrack + \lbrack\mathbf{dab}\rbrack = \lbrack\mathbf{abc}\rbrack

4 & 5 & \lambda \\ 2 & 3 & - 4 \\ - 1 & 1 & 2 \end{matrix} \right| + \left| \begin{matrix} 4 & 5 & \lambda \\ - 1 & 1 & 2 \\ 3 & - 2 & - 1 \end{matrix} \right| + \left| \begin{matrix} 4 & 5 & \lambda \\ 3 & - 2 & - 1 \\ 2 & 3 & - 4 \end{matrix} \right|$$ $$= \left| \begin{matrix} 3 & - 2 & - 1 \\ 2 & 3 & - 4 \\ - 1 & 1 & 2 \end{matrix} \right|$$ $$\Rightarrow 40 + 5\lambda + 37 - \lambda + 94 + 13\lambda = 25 \Rightarrow \lambda = \frac{- 146}{17}.$$