Question

If $(\vec{a} \times \vec{b})^2+(\vec{a}.\vec{b})^2=144$ and $|\vec{a}|$ then $ |\vec{b}|$=

• A. 16
• B. 8
• C. 3
• D. 12

Answer 1

The Correct Option is C

The correct answer is C:3
Given, \(\left(a \times b\right)^{2} + \left(a.b\right)^{2} = 144\)
\(|a|=4, |b|=?\)
Let us consider the angle between \(\vec{a}\) and \(\vec{b}\) is \(\theta\) then;
\(|\vec{a}|^2|\vec{b}|^2sin^2\theta+|\vec{a}|^2|\vec{b}|^2.cos^2\theta=144\)
⇒\(|\vec{a}|^2|\vec{b}|^2=144 (sin^2\theta+cos^2\theta=1)\)
⇒\(4|\vec{b}|^2=144\)
⇒\(|\vec{b}|^2=\frac{144}{4}\)
⇒\(|\vec{b}|=\frac{12}{4}\)
⇒\(|b|=3\)