Question

The points $(5,-4,2),(4,-3,1),(7,-6,4)$ and $(8,-7,5)$ are the vertices of

• A. A rectangle
• B. A square
• C. A parallelogram
• D. None of these

Answer 1

The correct option is D

None of these

Explanation of the correct option.

Let the vertices are $A(5,-4,2),B(4,-3,1),C(7,-6,4)$and $D(8,-7,5)$ .

$$\begin{gathered} \overrightarrow{\mathrm{AB}}=(4-5)\hat{\mathrm{i}}+(-3+4)\hat{\mathrm{j}}+(1-2)\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{AB}}=-1\hat{\mathrm{i}}+1\hat{\mathrm{j}}-1\hat{\mathrm{k}} \end{gathered}$$

Magnitude of $\overrightarrow{\mathrm{AB}}$,

$$\begin{gathered} \left|\overrightarrow{\mathrm{AB}}\right|=\sqrt{{(-1)}^2+{\left(1\right)}^2+{(-1)}^2} \\ \Rightarrow \left|\overrightarrow{\mathrm{AB}}\right|=\sqrt{3} \end{gathered}$$

$$\begin{gathered} \overrightarrow{\mathrm{BC}}=(7-4)\hat{\mathrm{i}}+(-6+3)\hat{\mathrm{j}}+(4-1)\hat{\mathrm{k}} \\ \Rightarrow \overrightarrow{\mathrm{BC}}=3\hat{\mathrm{i}}-3\hat{\mathrm{j}}+3\hat{\mathrm{k}} \end{gathered}$$

Magnitude of $\overrightarrow{\mathrm{BC}}$

,$$\begin{gathered} \left|\overrightarrow{BC}\right|=\sqrt{{\left(3\right)}^2+{(-3)}^2+{\left(3\right)}^2} \\ \Rightarrow \left|\overrightarrow{BC}\right|=\sqrt{9+9+9} \\ \Rightarrow \left|\overrightarrow{BC}\right|=3\sqrt{3} \end{gathered}$$

Since $\overrightarrow{\mathrm{AB}}.\overrightarrow{\mathrm{BC}}=\left|\overrightarrow{\mathrm{AB}}\right|\left|\overrightarrow{\mathrm{BC}}\right|\cos \left(\mathrm{\theta }\right)$,

$$\begin{gathered} -3-3-3=\left(\sqrt{3}\right)\left(3\sqrt{3}\right)\cos \left(\mathrm{\theta }\right) \\ \Rightarrow -9=9\cos \left(\mathrm{\theta }\right) \\ \Rightarrow -1=\cos \left(\mathrm{\theta }\right) \end{gathered}$$

Since $\cos \left(\mathrm{\pi }\right)=-1$

Therefore, $\theta =\mathrm{\pi }$.

Since angle between the two sides is $\pi$.

Therefore it can not be a rectangle, square or a parallelogram.

Hence, option $\left(\mathrm{D}\right)$ is the correct answer.