Question

Let \(\overset{⃗}{P} = \hat{i}+\hat{j}+\hat{k}\) and \(\overset{⃗}{Q} = -(\hat{i}+\hat{j}+\hat{k})\). The angle between \((\overset{⃗}{P}-\overset{⃗}{Q})\) and \(\overset{⃗}{P}\) is

• A. \(90^{\circ}\)
• B. \(60^{\circ}\)
• C. \(0^{\circ}\)
• D. \(30^{\circ}\)

Hint:

Since Q = -P, the vector P - Q equals 2P, which points along P.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The angle between two vectors follows from \(\cos\theta = \dfrac{\vec A\cdot\vec B}{|\vec A||\vec B|}\). Two vectors in the same direction have an angle of 0.

Step 2: Compute P - Q:
\(\vec Q = -\vec P\), so \(\vec P - \vec Q = \vec P + \vec P = 2\vec P = 2(\hat i + \hat j + \hat k)\).

Step 3: Find the angle:
\[ \cos\theta = \frac{2\vec P\cdot\vec P}{|2\vec P||\vec P|} = \frac{2|\vec P|^2}{2|\vec P|^2} = 1 \Rightarrow \theta = 0^\circ \]

Step 4: Check:
Numerically: \(2\vec P = (2,2,2)\), \(\vec P = (1,1,1)\), dot product is 6, magnitudes \(2\sqrt3\) and \(\sqrt3\), product 6. So \(\cos\theta = 1\). Option (C).

Final Answer:
P - Q is twice P, so the angle is 0 degrees. \[ \boxed{\text{(C) }0^{\circ}} \]