Question

If\(|\vec {a}+\vec {b}|=15, |\vec {a}-\vec{b}| =10,|\vec a|=\frac{11}{2}\) then the value of |\(\vec b\)| is/are:

• A. \(\frac{23}{2}\)
• B. \(± \frac{23}{2}\)
• C. \(±23\)
• D. \(\frac{23}{\sqrt2}\)

Answer 1

The Correct Option is A

Given:   \( |\vec{a} + \vec{b}| = 15 \), \( |\vec{a} - \vec{b}| = 10 \), \( |\vec{a}| = \frac{11}{2} \)

Using vector identities:

\( |\vec{a} + \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b} \quad \text{(1)} \)

\( |\vec{a} - \vec{b}|^2 = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b} \quad \text{(2)} \)

From the given magnitudes:

\( |\vec{a} + \vec{b}|^2 = 15^2 = 225 \)

\( |\vec{a} - \vec{b}|^2 = 10^2 = 100 \)

Subtracting equation (2) from (1):

\( 225 - 100 = (|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a} \cdot \vec{b}) - (|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a} \cdot \vec{b}) \)

\( 125 = 4\vec{a} \cdot \vec{b} \)

\( \vec{a} \cdot \vec{b} = \frac{125}{4} \)

Now substitute into equation (1):

\( 225 = \left(\frac{11}{2}\right)^2 + |\vec{b}|^2 + 2 \cdot \frac{125}{4} \)

\( 225 = \frac{121}{4} + |\vec{b}|^2 + \frac{250}{4} \)

\( 225 = \frac{371}{4} + |\vec{b}|^2 \)

Solve for \( |\vec{b}|^2 \):

\( |\vec{b}|^2 = 225 - \frac{371}{4} = \frac{900 - 371}{4} = \frac{529}{4} \)

\( |\vec{b}| = \sqrt{\frac{529}{4}} = \frac{\sqrt{529}}{2} = \frac{23}{2} \)

Final Answer: \( |\vec{b}| = \frac{23}{2} \)