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 Question:
We are given three points in three-dimensional space: $A$, $B$, and $C$, with their position vectors specified.
We need to evaluate the properties of the vectors formed by these points (such as collinearity, perpendicularity, and cross product) to find the true statement.

Step 2: Key Formula or Approach:
Let the position vectors of the points be $\vec{OA}$, $\vec{OB}$, and $\vec{OC}$.
We can compute the displacement vectors:
\[ \vec{AB} = \vec{OB} - \vec{OA} \]
\[ \vec{BC} = \vec{OC} - \vec{OB} \]
Two vectors $\vec{u}$ and $\vec{v}$ are collinear if $\vec{u} = \lambda \vec{v}$ for some scalar $\lambda$.
If they share a common point (like $B$), then the three points $A$, $B$, and $C$ are collinear.

Step 3: Detailed Explanation:

• Let us write down the given position vectors:
\[ \vec{OA} = 4\hat{i} + \hat{j} + 3\hat{k} \]
\[ \vec{OB} = 2\hat{j} \]
\[ \vec{OC} = -4\hat{i} + 3\hat{j} - 3\hat{k} \]

• Next, we calculate the vector $\vec{AB}$:
\[ \vec{AB} = \vec{OB} - \vec{OA} = (0 - 4)\hat{i} + (2 - 1)\hat{j} + (0 - 3)\hat{k} \]
\[ \vec{AB} = -4\hat{i} + \hat{j} - 3\hat{k} \]

• Now, we calculate the vector $\vec{BC}$:
\[ \vec{BC} = \vec{OC} - \vec{OB} = (-4 - 0)\hat{i} + (3 - 2)\hat{j} + (-3 - 0)\hat{k} \]
\[ \vec{BC} = -4\hat{i} + \hat{j} - 3\hat{k} \]

• Comparing the two vectors $\vec{AB}$ and $\vec{BC}$:
\[ \vec{AB} = \vec{BC} \]
This means the direction of both vectors is identical ($\lambda = 1$), and since they share the point $B$, the points $A$, $B$, and $C$ must lie on the same straight line.
Thus, the points $A$, $B$, and $C$ are collinear.

• Let us verify why the other options are incorrect:
Since $\vec{AB}$ and $\vec{BC}$ are parallel:
The cross product $\vec{AB} \times \vec{BC}$ must be the zero vector $\vec{0}$, which makes option (C) incorrect.
Collinear vectors cannot be mutually perpendicular, which makes option (D) incorrect.
$\vec{AB} + 3\vec{BC} = 4\vec{AB}$ and $\vec{AC} = 2\vec{AB}$. Since both are along the same direction, they are parallel, not perpendicular, which makes option (B) incorrect.

Step 4: Final Answer:
Therefore, the statement "A, B and C are collinear" is true.