Functions Ans 10
Question 10 - 29 January - Shift 1
Let $\overrightarrow{\mathrm{a}}, \mathrm{b}$ and $\overrightarrow{\mathrm{c}}$ be three non-zero non-coplanar vectors. Let the position vectors of four points 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 $\overrightarrow{A B}$, $\overrightarrow{\mathrm{AC}}$ and $\overrightarrow{\mathrm{AD}}$ are coplanar, then $\lambda$ is :
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Answer: (2)
Solution:
Formula: Scalar Triple Product (8.5)
$ \begin{aligned} & \overline{A B}=(\lambda-1) \bar{a}-2 \bar{b}+3 \bar{c} \\ & \overline{A C}=2 \bar{a}+3 \bar{b}-4 \bar{c} \\ & \overline{A D}=\bar{a}-3 \bar{b}+5 \bar{c} \text{ athonges } \\ & \begin{vmatrix} \lambda-1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix} =0 \\ & \Rightarrow(\lambda-1)(15-12)+2(-10+4)+3(6-3)=0 \end{aligned} $
