Question

If \(|\vec{a}\times\vec{b}|+|\vec{a}.\vec{b}|^2\)=144 and \(|\vec{a}|\) = 6, then \(|\vec{b}|\) is equal to

• A. 6
• B. 3
• C. 2
• D. 4

Answer 1

The Correct Option is C

If \(|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144\) and \(|\vec{a}| = 6\), then we need to find \(|\vec{b}|\).

We know that \(|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta\) and \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta\), where \(\theta\) is the angle between \(\vec{a}\) and \(\vec{b}\).

So, \(|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta\) and \(|\vec{a} \cdot \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta\).

We are given that \(|\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144\). Substituting the expressions for the squared magnitudes, we get:

\(|\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 144\)

\(|\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta) = 144\)

Since \(\sin^2 \theta + \cos^2 \theta = 1\), we have:

\(|\vec{a}|^2 |\vec{b}|^2 = 144\)

We are given \(|\vec{a}| = 6\), so:

\(6^2 |\vec{b}|^2 = 144\)

\(36 |\vec{b}|^2 = 144\)

\(|\vec{b}|^2 = \frac{144}{36}\)

\(|\vec{b}|^2 = 4\)

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

Therefore, \(|\vec{b}| = 2\).

Thus, the correct option is (C) 2.