Question

The angle subtended at the point $(1, 2, 3)$ by the points $P(2, 4, 5)$ and $Q(3, 3, 1)$, is:

• A. $90^\circ$
• B. $60^\circ$
• C. $30^\circ$
• D. $0^\circ$
• E. $45^\circ$

Hint:

Always check the dot product first. If it results in zero, you instantly know the angle is $90^\circ$ without having to calculate the magnitudes of the vectors.

Answer 1

The Correct Option is A

Concept: The angle subtended at point $A$ by points $P$ and $Q$ is the angle between the vectors $\vec{AP}$ and $\vec{AQ}$. We can find this angle using the dot product formula: $\cos \theta = \frac{\vec{AP} \cdot \vec{AQ}}{|\vec{AP}| |\vec{AQ}|}$.

Step 1: Find the vectors $\vec{AP}$ and $\vec{AQ}$.
Let $A = (1, 2, 3)$, $P = (2, 4, 5)$, and $Q = (3, 3, 1)$. \[ \vec{AP} = (2-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = \hat{i} + 2\hat{j} + 2\hat{k} \] \[ \vec{AQ} = (3-1)\hat{i} + (3-2)\hat{j} + (1-3)\hat{k} = 2\hat{i} + \hat{j} - 2\hat{k} \]

Step 2: Calculate the dot product $\vec{AP} \cdot \vec{AQ}$.
\[ \vec{AP} \cdot \vec{AQ} = (1)(2) + (2)(1) + (2)(-2) \] \[ \vec{AP} \cdot \vec{AQ} = 2 + 2 - 4 = 0 \]

Step 3: Determine the angle.
Since the dot product is zero, the vectors are perpendicular. \[ \cos \theta = 0 \quad \Rightarrow \quad \theta = 90^\circ \]