Question 64

Mathematics Scalar and Vector Products Easy

Each of the angle between vectors \( \vec{a}, \vec{b} \) and \( \vec{c} \) is equal to \( 60^{\circ} \) |f \( |\vec{a}|=4,|\vec{b}|=2 \) and \( |\vec{c}|=6 \) then the modulus of \( \vec{a}+\vec{b}+\vec{c} \) is

(A) 10
(B) 15
(C) 12
(D) 20
View Dynamic Solution & Explanation
Correct Solution: Option A

Step-by-step Solution:

Solution: We use the identity: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 + 2(\vec{a} \cdot \vec{b} + \vec{b} \cdot \vec{c} + \vec{c} \cdot \vec{a}) \] Step 1: Compute Dot Products Given that the angle between each pair of vectors is \( \frac{\pi}{3} \), we use: \[ \vec{x} \cdot \vec{y} = |\vec{x}| |\vec{y}| \cos \frac{\pi}{3} \] Since \( \cos \frac{\pi}{3} = \frac{1}{2} \), we find: \[ \vec{a} \cdot \vec{b} = 4 \times 2 \times \frac{1}{2} = 4 \] \[ \vec{b} \cdot \vec{c} = 2 \times 6 \times \frac{1}{2} = 6 \] \[ \vec{c} \cdot \vec{a} = 6 \times 4 \times \frac{1}{2} = 12 \] Step 2: Compute Magnitude Given \( |\vec{a}| = 4, |\vec{b}| = 2, |\vec{c}| = 6 \), substitute in the identity: \[ |\vec{a} + \vec{b} + \vec{c}|^2 = 4^2 + 2^2 + 6^2 + 2(4 + 6 + 12) \] \[ = 16 + 4 + 36 + 2(22) \] \[ = 100 \] \[ \Rightarrow |\vec{a} + \vec{b} + \vec{c}| = \sqrt{100} = 10 \]