Question

The points $A,B$ and $C$ with position vectors, $\mathrm{a}=3\mathrm{i}-4\mathrm{j}-4\mathrm{k}$, vector $\mathrm{b}=2\mathrm{i}-\mathrm{j}+\mathrm{k}$ and vector $\mathrm{c}=\mathrm{i}-3\mathrm{j}-5\mathrm{k}$ respectively form the vertices of ?

• A. Right angled triangle
• B. Isosceles triangle
• C. Equilateral triangle
• D. None of the above

Answer 1

The correct option is A

Right angled triangle

Explanation of the correct option

Given : Position vectors of $\mathrm{a}=3\mathrm{i}-4\mathrm{j}-4\mathrm{k}$, $\mathrm{b}=2\mathrm{i}-\mathrm{j}+\mathrm{k}$ and $\mathrm{c}=\mathrm{i}-3\mathrm{j}-5\mathrm{k}$

$$\begin{gathered} {\left|AB\right|}^2={\left(2-3\right)}^2+{\left(-1+4\right)}^2+{(1+4)}^2 \\ \Rightarrow {\left|AB\right|}^2=35 \end{gathered}$$

$$\begin{gathered} {\left|BC\right|}^2={\left(1-2\right)}^2+{\left(-3+1\right)}^2+{(-5-1)}^2 \\ \Rightarrow {\left|BC\right|}^2=41 \end{gathered}$$

$$\begin{gathered} {\left|CA\right|}^2={\left(3-1\right)}^2+{\left(-4+3\right)}^2+{(-4+5)}^2 \\ \Rightarrow {\left|CA\right|}^2=6 \end{gathered}$$

Since, ${\left|BC\right|}^2={\left|AB\right|}^2+{\left|CA\right|}^2$

Therefore, according to the Pythagoras theorem, it is a right-angled triangle.

Hence, option $\left(A\right)$ is the correct option.