Question

The figure formed by the four points $\overbrace{i}+\overbrace{j}-\overbrace{k},2\overbrace{i}+3\overbrace{j},5\overbrace{j}-2\overbrace{k}$ and $\hat{k}-\hat{j}$is?

• A. Trapezium
• B. Rectangle
• C. Parallelogram
• D. None of these

Answer 1

The correct option is D

None of these

Explanation for the correct option

Given, four points, say $P,Q,R,S$

$$\begin{gathered} \overrightarrow{PQ}=\left(2\hat{i}+3\hat{j}\right)-\left(\hat{i}+\hat{j}-\hat{k}\right)=\hat{i}+2\hat{j}+\hat{k} \\ \therefore \left|\overrightarrow{PQ}\right|=\sqrt{1^2+2^2+1^2}=\sqrt{6} \\ \overrightarrow{QR}=\left(5\hat{j}-2\hat{k}\right)-\left(2\hat{i}+3\hat{j}\right)=-2\hat{i}+2\hat{j}-2\hat{k} \\ \therefore \left|\overrightarrow{QR}\right|=\sqrt{{\left(-2\right)}^2+2^2+{\left(-2\right)}^2}=\sqrt{12} \\ \overrightarrow{RS}=\left(\hat{k}-\hat{j}\right)-\left(5\hat{j}-2\hat{k}\right)=-6\hat{j}-3\hat{k} \\ \therefore \left|\overrightarrow{RS}\right|=\sqrt{{\left(0\right)}^2+{\left(-6\right)}^2+{\left(-3\right)}^2}=\sqrt{45} \\ \overrightarrow{SP}=\left(\hat{i}+\hat{j}-\hat{k}\right)-\left(\hat{k}-\hat{j}\right)=\hat{i}+2\hat{j}-2\hat{k} \\ \therefore \left|\overrightarrow{SP}\right|=\sqrt{1^2+2^2+{\left(-2\right)}^2}=\sqrt{9}=3 \end{gathered}$$

The magnitude of the sides of the shape does not satisfy the conditions of the quadrilateral, trapezium and rectangle .

So the figure does not show any of the shape given in option

Hence, option(D) is the correct answer.