If `(vec a xx vec b)^2 + (veca .vecb)^2` = 676 and `|vecb| = 2`, then `|veca|` is equal to

Options

A. 13
B. 26
C. 39
D. none of these

Answer

To solve the problem, we start with the given equation:

\[ (\vec{a} \times \vec{b})^2 + (\vec{a} \cdot \vec{b})^2 = 676 \]

We know the following vector identities:

1. The magnitude of the cross product:
\[ |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \]
Therefore,
\[ (\vec{a} \times \vec{b})^2 = (|\vec{a}| |\vec{b}| \sin \theta)^2 = |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta \]

2. The magnitude of the dot product:
\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]
Therefore,
\[ (\vec{a} \cdot \vec{b})^2 = (|\vec{a}| |\vec{b}| \cos \theta)^2 = |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta \]

Substituting these into the original equation gives:

\[ |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 676 \]

Factoring out \( |\vec{a}|^2 |\vec{b}|^2 \):

\[ |\vec{a}|^2 |\vec{b}|^2 (\sin^2 \theta + \cos^2 \theta) = 676 \]

Using the Pythagorean identity \( \sin^2 \theta + \cos^2 \theta = 1 \):

\[ |\vec{a}|^2 |\vec{b}|^2 = 676 \]

Given that \( |\vec{b}| = 2 \):

\[ |\vec{a}|^2 (2^2) = 676 \]

This simplifies to:

\[ |\vec{a}|^2 \cdot 4 = 676 \]

Dividing both sides by 4:

\[ |\vec{a}|^2 = \frac{676}{4} = 169 \]

Taking the square root of both sides gives:

\[ |\vec{a}| = \sqrt{169} = 13 \]

Thus, the magnitude of \( \vec{a} \) is:

\[ |\vec{a}| = 13 \]