Question

If a parallelogram is constructed on vectors \(\vec{a}=3\vec{p}-\vec{q}\), \(\vec{b}=\vec{p}+3\vec{q}\) and \(|\vec{p}|=|\vec{q}|=2\) and the angle between \(\vec{p}\) and \(\vec{q}\) is \(\frac{\pi }{3}\), then \(|\vec{a}|:|\vec{b}|=\ldots\)

• A. \(\sqrt{5}:\sqrt{13}\)
• B. \(\sqrt{7}:\sqrt{13}\)
• C. \(\sqrt{13}:\sqrt{7}\)
• D. \(\sqrt{13}:\sqrt{5}\)

Hint:

Find p dot q first, then compute the squared magnitudes of a and b.

Answer 1

The Correct Option is B

Step 1: Use the Square Formula
For $\vec{u}=m\vec{p}+n\vec{q}$, $|\vec{u}|^{2}=4m^{2}+4n^{2}+2mn(\vec{p}\cdot\vec{q})=4m^{2}+4n^{2}+4mn$.

Step 2: Substitute
For $\vec{a}$: $m=3,n=-1$, giving $36+4-12=28$. For $\vec{b}$: $m=1,n=3$, giving $4+36+12=52$.

Step 3: Compare
$\dfrac{|\vec{a}|^{2}}{|\vec{b}|^{2}}=\dfrac{28}{52}=\dfrac{7}{13}$, so the ratio is $\sqrt{7}:\sqrt{13}$.

Final Answer:
The magnitudes are in the ratio root 7 to root 13. \[ \boxed{\text{(B) }\sqrt{7}:\sqrt{13}} \]