Question

Given \( |\vec{a}|=|\vec{b}|=1 \) and \( |\vec{a}+\vec{b}|=\sqrt{3} \). If \( \vec{c} \) is a vector such that \( \vec{c}-\vec{a}-2\vec{b}=3(\vec{a}\times\vec{b}) \), then \( \vec{c}\cdot\vec{b} \) is equal to

• A. \(-\frac{1}{2}\)
• B. \(\frac{1}{2}\)
• C. \(\frac{3}{2}\)
• D. \(\frac{5}{2}\)

Hint:

Find \( \vec{a}\cdot\vec{b} \) from \( |\vec{a}+\vec{b}|^2 \), then dot the given relation with \( \vec{b} \); the scalar triple product term vanishes.

Answer 1

The Correct Option is D

Since \( |\vec{a}+\vec{b}|^2=|\vec{a}|^2+|\vec{b}|^2+2\vec{a}\cdot\vec{b} \), we get \( 3=1+1+2\vec{a}\cdot\vec{b} \), so \( \vec{a}\cdot\vec{b}=\frac{1}{2} \).
From the given relation, \( \vec{c}=\vec{a}+2\vec{b}+3(\vec{a}\times\vec{b}) \).
Taking dot product with \( \vec{b} \): \( \vec{c}\cdot\vec{b}=\vec{a}\cdot\vec{b}+2|\vec{b}|^2+3(\vec{a}\times\vec{b})\cdot\vec{b} \).
Now \( (\vec{a}\times\vec{b})\cdot\vec{b}=0 \) because \( \vec{a}\times\vec{b} \) is perpendicular to \( \vec{b} \).
Hence \( \vec{c}\cdot\vec{b}=\frac{1}{2}+2+0=\frac{5}{2} \).
Answer: (4) \( \frac{5}{2} \)