If `vecA=4hati+6hatj` and `vecB=2hati+3hatj` . Then

Options

A. `vecA.vecB=29`
B. `vecAxxvecB=vec0`
C. `(|vecB|)/(|vecA|)=2/1`
D. angle between `vecA` and `vecB` is `30^@`

Explanation

To solve the problem involving the vectors \(\vec{A} = 4\hat{i} + 6\hat{j}\) and \(\vec{B} = 2\hat{i} + 3\hat{j}\), we will calculate the following:

1. The dot product of the two vectors.
2. The cross product of the two vectors.
3. The magnitude ratio of the two vectors.
4. The angle between the two vectors.

### Step 1: Calculate the Dot Product

The dot product of two vectors \(\vec{A}\) and \(\vec{B}\) is given by:

\[ \vec{A} \cdot \vec{B} = A_x B_x + A_y B_y \]

Where \(A_x\) and \(A_y\) are the components of \(\vec{A}\), and \(B_x\) and \(B_y\) are the components of \(\vec{B}\).

Substituting the values:

\[ \vec{A} \cdot \vec{B} = (4)(2) + (6)(3) = 8 + 18 = 26 \]

### Step 2: Calculate the Cross Product

The cross product of two vectors can be calculated using the determinant of a matrix formed by the unit vectors and the components of the vectors:

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 6 & 0 \\ 2 & 3 & 0 \end{vmatrix} \]

Calculating the determinant:

\[ \vec{A} \times \vec{B} = \hat{i}(6 \cdot 0 - 0 \cdot 3) - \hat{j}(4 \cdot 0 - 0 \cdot 2) + \hat{k}(4 \cdot 3 - 6 \cdot 2) \]

This simplifies to:

\[ \vec{A} \times \vec{B} = 0\hat{i} - 0\hat{j} + (12 - 12)\hat{k} = 0\hat{i} + 0\hat{j} + 0\hat{k} = \vec{0} \]

### Step 3: Calculate the Magnitudes of the Vectors

The magnitude of a vector \(\vec{A}\) is given by:

\[ |\vec{A}| = \sqrt{A_x^2 + A_y^2} \]

Calculating the magnitudes:

\[ |\vec{A}| = \sqrt{4^2 + 6^2} = \sqrt{16 + 36} = \sqrt{52} \]

\[ |\vec{B}| = \sqrt{2^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

### Step 4: Calculate the Magnitude Ratio

The magnitude ratio of \(\vec{B}\) to \(\vec{A}\) is:

\[ \text{Magnitude Ratio} = \frac{|\vec{B}|}{|\vec{A}|} = \frac{\sqrt{13}}{\sqrt{52}} = \frac{\sqrt{13}}{2\sqrt{13}} = \frac{1}{2} \]

### Step 5: Calculate the Angle Between the Vectors

Using the dot product and magnitudes, the cosine of the angle \(\theta\) between the vectors is given by:

\[ \vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos(\theta) \]

Substituting the values we calculated:

\[ 26 = \sqrt{52} \cdot \sqrt{13} \cdot \cos(\theta) \]

Calculating \(\sqrt{52} = 2\sqrt{13}\):

\[ 26 = (2\sqrt{13})(\sqrt{13}) \cos(\theta) = 2 \cdot 13 \cos(\theta) = 26 \cos(\theta) \]

Thus:

\[ \cos(\theta) = 1 \implies \theta = 0^\circ \]

### Summary of Results

1. Dot Product: \(26\)
2. Cross Product: \(\vec{0}\) (null vector)
3. Magnitude Ratio: \(\frac{1}{2}\)
4. Angle: \(0^\circ\)