if `| vec(a) xx vec(b) |^(2) +| vec(a). vec(b)|^(2)= 144` and `| vec(a) | =4` then `|vec(b) |` is equal to

Options

A. 12
B. 3
C. 4
D. `-3`

Answer

To solve the problem, we need to use the given equation and the properties of vectors. Let's break it down step by step.

### Given:
1. \( |\vec{a} \times \vec{b}|^2 + |\vec{a} \cdot \vec{b}|^2 = 144 \)
2. \( |\vec{a}| = 4 \)

### Step 1: Use the properties of vector operations
We know that:
- \( |\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin \theta \)
- \( |\vec{a} \cdot \vec{b}| = |\vec{a}| |\vec{b}| \cos \theta \)

where \( \theta \) is the angle between the vectors \( \vec{a} \) and \( \vec{b} \).

### Step 2: Substitute the expressions into the equation
Substituting the expressions for the cross product and dot product into the given equation:

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

### Step 3: Simplify the equation
This simplifies to:

\[ |\vec{a}|^2 |\vec{b}|^2 \sin^2 \theta + |\vec{a}|^2 |\vec{b}|^2 \cos^2 \theta = 144 \]

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

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

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

\[ |\vec{a}|^2 |\vec{b}|^2 = 144 \]

### Step 4: Substitute the known value of \( |\vec{a}| \)
Now, substituting \( |\vec{a}| = 4 \):

\[ (4)^2 |\vec{b}|^2 = 144 \]

This simplifies to:

\[ 16 |\vec{b}|^2 = 144 \]

### Step 5: Solve for \( |\vec{b}|^2 \)
Dividing both sides by 16:

\[ |\vec{b}|^2 = \frac{144}{16} \]

Calculating the right side:

\[ |\vec{b}|^2 = 9 \]

### Step 6: Take the square root to find \( |\vec{b}| \)
Taking the square root of both sides:

\[ |\vec{b}| = 3 \]

### Final Answer:
Thus, \( |\vec{b}| = 3 \).

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