Given `vec(A)=2hat(i)+phat(j)+qhat(k)` and `vec(B)=5hat(i)+7hat(j)+3hat(k)`. If `vec(A)||vec(B)`, then the values of p and q are, respectively,

Options

A. `14/5and6/5`
B. `14/3 and 6/5`
C. `6/5 and 1/3`
D. `3/4 and 1/4`

Explanation

To solve the problem, we need to find the values of \( p \) and \( q \) such that the vectors \( \vec{A} \) and \( \vec{B} \) are parallel. Two vectors are parallel if their cross product is equal to zero.

Given:
\[ \vec{A} = 2\hat{i} + p\hat{j} + q\hat{k} \]
\[ \vec{B} = 5\hat{i} + 7\hat{j} + 3\hat{k} \]

### Step 1: Set up the cross product
The cross product \( \vec{A} \times \vec{B} \) can be calculated using the determinant of a matrix formed by the unit vectors \( \hat{i}, \hat{j}, \hat{k} \) and the components of the vectors \( \vec{A} \) and \( \vec{B} \):

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & p & q \\ 5 & 7 & 3 \end{vmatrix} \]

### Step 2: Calculate the determinant
Using the determinant formula, we can expand this as follows:

\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} p & q \\ 7 & 3 \end{vmatrix} - \hat{j} \begin{vmatrix} 2 & q \\ 5 & 3 \end{vmatrix} + \hat{k} \begin{vmatrix} 2 & p \\ 5 & 7 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} p & q \\ 7 & 3 \end{vmatrix} = p \cdot 3 - q \cdot 7 = 3p - 7q \]

2. For \( \hat{j} \):
\[ \begin{vmatrix} 2 & q \\ 5 & 3 \end{vmatrix} = 2 \cdot 3 - q \cdot 5 = 6 - 5q \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 2 & p \\ 5 & 7 \end{vmatrix} = 2 \cdot 7 - p \cdot 5 = 14 - 5p \]

Putting it all together, we have:
\[ \vec{A} \times \vec{B} = (3p - 7q)\hat{i} - (6 - 5q)\hat{j} + (14 - 5p)\hat{k} \]

### Step 3: Set the cross product to zero
For the vectors to be parallel, the cross product must equal zero:
\[ 3p - 7q = 0 \quad (1) \]
\[ 6 - 5q = 0 \quad (2) \]
\[ 14 - 5p = 0 \quad (3) \]

### Step 4: Solve the equations
From equation (2):
\[ 6 - 5q = 0 \implies 5q = 6 \implies q = \frac{6}{5} \]

Substituting \( q \) into equation (1):
\[ 3p - 7\left(\frac{6}{5}\right) = 0 \]
\[ 3p - \frac{42}{5} = 0 \implies 3p = \frac{42}{5} \implies p = \frac{42}{15} = \frac{14}{5} \]

From equation (3):
\[ 14 - 5p = 0 \implies 5p = 14 \implies p = \frac{14}{5} \]

### Final Result
Thus, the values of \( p \) and \( q \) are:
\[ p = \frac{14}{5}, \quad q = \frac{6}{5} \]