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Let \mathrm{\vec{a}, \vec{b} \: and\: \vec{c}} be three non-zero non-coplanar vectors. Let the position vectors of four points \mathrm{A, B, C \: and \, D} be \vec{a}-\vec{b}+\vec{c}, \lambda \vec{a}-3 \vec{b}+4 \vec{c},-\vec{a}+2 \vec{b}-3 \vec{c}$ and $2 \vec{a}-4 \vec{b}+6 \vec{c} respectively. If \overrightarrow{A B}, \overrightarrow{A C}$ and $\overrightarrow{A D}$ are coplanar, then $\lambda is equal to
 

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Option: 2

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Answers (1)

best_answer

\begin{aligned} \begin{aligned} \overrightarrow{\mathrm{AB}}= & (\lambda \overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}}+4 \overrightarrow{\mathrm{c}})-(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & =(\lambda-1) \overrightarrow{\mathrm{a}}-2 \overrightarrow{\mathrm{b}}+3 \overrightarrow{\mathrm{c}} \\ \overrightarrow{\mathrm{AC}}= & (-\overrightarrow{\mathrm{a}}+2 \overrightarrow{\mathrm{b}}-3 \overrightarrow{\mathrm{c}})-(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & =-2 \overrightarrow{\mathrm{a}}+3 \overrightarrow{\mathrm{b}}-4 \overrightarrow{\mathrm{c}} \\ \overrightarrow{\mathrm{AD}}= & (2 \overrightarrow{\mathrm{a}}-4 \overrightarrow{\mathrm{b}}+6 \overrightarrow{\mathrm{c}})-(\overrightarrow{\mathrm{a}}-\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}) \\ & =\overrightarrow{\mathrm{a}}-3 \overrightarrow{\mathrm{b}}+5 \overrightarrow{\mathrm{c}} \end{aligned} \end{aligned}
For coplanar vectors
\begin{aligned} & \left|\begin{array}{ccc} \lambda-1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{array}\right|=0 \\ & \Rightarrow 3 \lambda-6=0 \\ & \therefore \lambda=2 \end{aligned}

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Ritika Harsh

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