Question

The position vectors of the points A, B, C are $(2\mathbf{i} + \mathbf{j} - \mathbf{k}),$

$(3\mathbf{i} - 2\mathbf{j} + \mathbf{k})$ and $(\mathbf{i} + 4\mathbf{j} - 3\mathbf{k})$ respectively. These points

• A. Form an isosceles triangle
• B. Form a right-angled triangle
• C. Are collinear
• D. Form a scalene triangle

Answer 1

Answer: (C)

Are collinear

Answer 2

$\overset{\rightarrow}{AB} = (3 - 2)\mathbf{i} + ( - 2 - 1)\mathbf{j} + (1 + 1)\mathbf{k} = \mathbf{i} - 3\mathbf{j} + 2\mathbf{k}$

$\mathbf{r} = \frac{\mathbf{b} \times (\mathbf{c} \times \mathbf{a})}{|\mathbf{b} \times (\mathbf{c} \times \mathbf{a})|}$

$\overset{\rightarrow}{CA} = (2 - 1)\mathbf{i} + (1 - 4)\mathbf{j} + ( - 1 + 3)\mathbf{k} = \mathbf{i} - 3\mathbf{j} - 2\mathbf{k}$

$|\overset{\rightarrow}{AB}| = \sqrt{1 + 9 + 4} = \sqrt{14}$

$|\overset{\rightarrow}{BC}| = \sqrt{4 + 36 + 16} = \sqrt{56} = 2\sqrt{14}$

$|\overset{\rightarrow}{CA}| = \sqrt{1 + 9 + 4} = \sqrt{14}$

So, $|\overset{\rightarrow}{AB}| + |\overset{\rightarrow}{AC}| = |\overset{\rightarrow}{BC}|$ and angle between AB and BC is 180°.

∴ Points A, B, C can not form an isosceles triangle.

Hence A, B, C are collinear.