Question

If the vectors $6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k},2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}$ and $3\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}$ form a triangle, then it is

• A. Right angled
• B. Obtuse angled
• C. Equilteral
• D. Isosceles

Answer 1

Answer: (B)

Obtuse angled

Answer 2

$\overset{\rightarrow}{AB}$= Position vector of $\overset{\rightarrow}{B}$– Position vector of $\overset{\rightarrow}{A}$

$= (2\mathbf{i} + 3\mathbf{j} - 6\mathbf{k}) - (6\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}) = - 4\mathbf{i} + 5\mathbf{j} - 9\mathbf{k}$

⇒ $|\overset{\rightarrow}{AB}| = \sqrt{16 + 25 + 81} = \sqrt{122}$, $\overset{\rightarrow}{BC} = \mathbf{i} + 3\mathbf{j} + 4\mathbf{k}$

⇒$|\overset{\rightarrow}{BC}| = \sqrt{1 + 9 + 16} = \sqrt{26}$ and $\overset{\rightarrow}{AC} = - 3\mathbf{i} + 8\mathbf{j} - 5\mathbf{k}$

⇒ $|\overset{\rightarrow}{AC}| = \sqrt{98}$

Therefore, $AB^{2} = 122$, $BC^{2} = 26$ and $AC^{2} = 98$.

$\Rightarrow AB^{2} + BC^{2} = 26 + 122 = 148$

Since $AC^{2} < AB^{2} + BC^{2}$, therefore $\Delta ABC$ is an obtuse-angled triangle.