Question

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

• A. Parallelogram
• B. Rectangle
• C. Trapezium
• D. Square

Answer 1

The correct option is C

Trapezium

Explanation for the correct option

The given points are $\hat{i}+\hat{j}+\hat{k},2\hat{i}+3\hat{j},3\hat{i}+5\hat{j}-2\hat{k},\hat{k}-\hat{j}$

$$\begin{gathered} \overrightarrow{AB}=\left(2\hat{i}+3\hat{j}\right)-\left(\hat{i}+\hat{j}+\hat{k}\right)=\hat{i}+2\hat{j}-\hat{k} \\ \therefore \left|\overrightarrow{AB}\right|=\sqrt{1^2+2^2+{\left(-1\right)}^2}=\sqrt{6} \\ \overrightarrow{BC}=\left(3\hat{i}+5\hat{j}-2\hat{k}\right)-\left(2\hat{i}+3\hat{j}\right)=\hat{i}+2\hat{j}-2\hat{k} \\ \therefore \left|\overrightarrow{BC}\right|=\sqrt{1^2+2^2+{\left(-2\right)}^2}=\sqrt{9}=3 \\ \overrightarrow{CD}=\left(\hat{k}-\hat{j}\right)-\left(3\hat{i}+5\hat{j}-2\hat{k}\right)=-3\hat{i}-6\hat{j}+3\hat{k} \\ \therefore \left|\overrightarrow{CD}\right|=\sqrt{{\left(-3\right)}^2+{\left(-6\right)}^2+{\left(3\right)}^2}=\sqrt{54} \\ \overrightarrow{DA}=\left(\hat{i}+\hat{j}+\hat{k}\right)-\left(\hat{k}-\hat{j}\right)=\hat{i}+2\hat{j} \\ \therefore \left|\overrightarrow{DA}\right|=\sqrt{{\left(1\right)}^2+{\left(2\right)}^2+{\left(0\right)}^2}=\sqrt{5} \\ \therefore \overrightarrow{AB}\parallel \overrightarrow{CD} \end{gathered}$$

So the given figure has only two parallel sides and the length of the sides are not equal

So, the points make a trapezium

Hence, option (C) is the correct answer.