Vectors — Maths STD 12 Science — Question

$\bar{a}=\hat{i}-\hat{j}+\hat{k}, \bar{b}=-\hat{i}+\hat{j}+\hat{k}, \bar{c}=\hat{i}+\hat{j}+\hat{k}, \bar{d}=2 \hat{i}-3 \hat{j}+4 \hat{k}$

$\therefore \overline{\mathrm{AB}}=\bar{b}-\bar{a}=(-\hat{i}+\hat{j}+\hat{k})-(\vec{i}-\vec{j}+\hat{k})$

$=-2 \hat{i}+2 \hat{j}$

$\overline{\mathrm{AC}}=\bar{c}-\bar{a}=(\hat{i}+\hat{j}+\hat{k})-(\hat{i}-\hat{j}+\hat{k})=2 \hat{j}$

and $\overline{\mathrm{AD}}=\bar{d}-\bar{a}=(2 \hat{i}-3 \hat{j}+4 \hat{k})-(\hat{i}-\hat{j}+\hat{k})$

$=\hat{i}-2 \hat{j}+3 \hat{k}$

If $\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}$ are coplanar, then there exist scalars $x, y$ such that

$\overline{\mathrm{AB}}=x \cdot \overline{\mathrm{AC}}+y \cdot \overline{\mathrm{AD}}$

$\begin{aligned} & \therefore-2 \hat{i}+2 \hat{j}=x(2 \hat{j})+y(\hat{i}-2 \hat{j}+3 \hat{k}) \\ & \therefore-2 \hat{i}+2 \hat{j}=y \hat{i}+(2 x-2 y) \hat{j}+3 y \hat{k}\end{aligned}$

By equality of vectors,
y = -2 ….(1)
2x – 2y = 2 … (2)
3y = 0 … (3)
From (1), y = -2
From (3), y = 0 This is not possible.
Hence, the points A, B, C, D are not coplanar.