Vector Algebra Question 9
Question 9 - 29 January - Shift 1
Let $\vec{a}, b$ and $\vec{c}$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A$, $B$, $C$ and $D$ be $\vec{a}-\vec{b}+\vec{c}, \quad \lambda \vec{a}-3 \vec{b}+4 \vec{c}$, $-\vec{a}+2 \vec{b}-3 \vec{c}$ and $2 \vec{a}-4 \vec{b}+6 \vec{c}$ respectively. If $\overrightarrow{{}A B}$, $\overrightarrow{{}AC}$ and $\overrightarrow{{}AD}$ are coplanar, then $\lambda$ is :
Show Answer
Answer: (2)
Solution:
Formula: Scalar triple product; Condition for three vectors to be coplanar or non-coplanar
$ \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} \\ & \therefore \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} $
