Vector Algebra Question 6
Question 6 - 25 January - Shift 2
If the four points, whose position vectors are $3 \hat{i}-4 \hat{j}+2 \hat{k}, \hat{i}+2 \hat{j}-\hat{k},-2 \hat{i}-\hat{j}+3 \hat{k} \quad$ and $5 \hat{i}-2 \alpha \hat{j}+4 \hat{k}$ are coplanar, then $\alpha$ is equal to
(1) $\frac{73}{17}$
(2) $-\frac{107}{17}$
(3) $-\frac{73}{17}$
(4) $\frac{107}{17}$
Show Answer
Answer: (1)
Solution:
Formula: Scalar triple product : Condition for four vectors to be coplanar.
Let $A:(3,-4,2)$
B : $(1,2,-1)$
C : $(-2,-1,3)$
D : $(5,-2 \alpha, 4)$
A, B, C, D are coplanar points, then
$\Rightarrow \begin{vmatrix} 1-3 & 2+4 & -1-2 \\ -2-3 & -1+4 & 3-2 \\ 5-3 & -2 \alpha+4 & 4-2\end{vmatrix} =0$
$\Rightarrow \alpha=\frac{73}{17}$
