Question

Consider the points $A(4\hat{i} + \hat{j} + 3\hat{k})$, $B(2\hat{j})$ and $C(-4\hat{i} + 3\hat{j} - 3\hat{k})$. Which of the following statements is TRUE?

• A. $A, B$ and $C$ are collinear.
• B. $\vec{AB} + 3\vec{BC}$ is perpendicular to $\vec{AC}$.
• C. $\vec{AB} \times \vec{BC} = \hat{i} + \hat{j} + \hat{k}$.
• D. $\vec{AB}, \vec{BC}$ and $\vec{CA}$ are mutually perpendicular.

Hint:

To quickly check for collinearity of three points $A$, $B$, and $C$, compute the components of $\vec{AB}$ and $\vec{BC}$.
If their respective components are proportional, i.e., $\frac{x_1}{x_2} = \frac{y_1}{y_2} = \frac{z_1}{z_2}$, the points are collinear.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Points are collinear if the vectors formed between them are parallel ($\vec{u} = \lambda \vec{v}$).

Step 2: Detailed Explanation:
1. Calculate vector $\vec{AB$:}
$\vec{AB} = \vec{B} - \vec{A} = (0 - 4)\hat{i} + (2 - 1)\hat{j} + (0 - 3)\hat{k} = -4\hat{i} + \hat{j} - 3\hat{k}$.
2. Calculate vector $\vec{BC$:}
$\vec{BC} = \vec{C} - \vec{B} = (-4 - 0)\hat{i} + (3 - 2)\hat{j} + (-3 - 0)\hat{k} = -4\hat{i} + \hat{j} - 3\hat{k}$.
3. Comparison:
$\vec{AB} = \vec{BC}$ (Scalar $\lambda = 1$).
Since the vectors are identical and share a common point (B), the points A, B, and C lie on the same straight line.

Step 3: Final Answer:
The points are collinear.
This matches option (A).