The figure formed by the four points $(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}),(2…
The figure formed by the four points $(\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}),(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}),(5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ and $(\hat{\mathbf{k}}-\hat{\mathbf{j}})$ is
trapezium
rectangle
parallelogram
quadrilateral
Solution
Let $\mathbf{O A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$
$\begin{aligned} \mathbf{O B} & =2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}} \\ \mathbf{O C} & =5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ \mathbf{O D} & =\hat{\mathbf{k}}-\hat{\mathbf{j}}\end{aligned}$
$\therefore \quad \mathbf{A B}=O B-O A$
$=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$
$\begin{aligned} \Rightarrow \quad|\mathbf{A B}| & =\sqrt{1^2+2^2+1^2}=\sqrt{6} \\ \mathbf{B C} & =\mathbf{O C}-\mathbf{O B}=-2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\end{aligned}$
$\begin{aligned} \Rightarrow \quad|\mathbf{B C}| & =\sqrt{(-2)^2+2^2+(-2)^2}=2 \sqrt{3} \\ \mathbf{C D} & =\mathbf{O D}-\mathbf{O C}=-6 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}\end{aligned}$
$\begin{aligned} \Rightarrow \quad|\mathbf{C D}| & =3 \sqrt{5} \\ \mathbf{D A} & =\mathbf{O A}-\mathbf{O D}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\end{aligned}$
$\Rightarrow \quad|\mathbf{D A}|=3$
$\because$ All the sides has different length and no two sides are equal and parallel.
Hence, it forms only quadrilateral.