If `vec(A) = 3hat(i) - 4hat(j) + hat(k)` and `vec(B) = 4hat(j) + phat(i) + hat(k)` for what value of p, `vec(A)` and `vec(B)` will ve collinear ?

Options

A. 3
B. `-5`
C. `-(16)/(3)`
D. `vec(A)` and `vec(B)` cannot be collinear

Answer

To determine the value of \( p \) for which the vectors \( \vec{A} \) and \( \vec{B} \) are collinear, we can use the property that two vectors are collinear if their cross product is zero.

Given:
\[ \vec{A} = 3\hat{i} - 4\hat{j} + \hat{k} \]
\[ \vec{B} = p\hat{i} + 4\hat{j} + \hat{k} \]

### 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 \( \vec{A} \) and \( \vec{B} \).

\[ \vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 1 \\ p & 4 & 1 \end{vmatrix} \]

### Step 2: Calculate the determinant
To calculate the determinant, we expand it as follows:

\[ \vec{A} \times \vec{B} = \hat{i} \begin{vmatrix} -4 & 1 \\ 4 & 1 \end{vmatrix} - \hat{j} \begin{vmatrix} 3 & 1 \\ p & 1 \end{vmatrix} + \hat{k} \begin{vmatrix} 3 & -4 \\ p & 4 \end{vmatrix} \]

Calculating each of the 2x2 determinants:

1. For \( \hat{i} \):
\[ \begin{vmatrix} -4 & 1 \\ 4 & 1 \end{vmatrix} = (-4)(1) - (1)(4) = -4 - 4 = -8 \]

2. For \( -\hat{j} \):
\[ \begin{vmatrix} 3 & 1 \\ p & 1 \end{vmatrix} = (3)(1) - (1)(p) = 3 - p \]

3. For \( \hat{k} \):
\[ \begin{vmatrix} 3 & -4 \\ p & 4 \end{vmatrix} = (3)(4) - (-4)(p) = 12 + 4p \]

Putting it all together:
\[ \vec{A} \times \vec{B} = -8\hat{i} - (3 - p)\hat{j} + (12 + 4p)\hat{k} \]

### Step 3: Set the cross product equal to zero
For the vectors to be collinear, the cross product must equal zero:
\[ -8\hat{i} - (3 - p)\hat{j} + (12 + 4p)\hat{k} = 0 \]

This gives us three equations:
1. From the \( \hat{i} \) component: \( -8 = 0 \) (This cannot be satisfied)
2. From the \( \hat{j} \) component: \( 3 - p = 0 \)
3. From the \( \hat{k} \) component: \( 12 + 4p = 0 \)

### Step 4: Solve the equations
From the second equation:
\[ 3 - p = 0 \implies p = 3 \]

From the third equation:
\[ 12 + 4p = 0 \implies 4p = -12 \implies p = -3 \]

### Conclusion
We have two values for \( p \) from the equations, \( p = 3 \) and \( p = -3 \). However, since the \( \hat{i} \) component gives a constant value of -8, which cannot be zero, we conclude that the vectors \( \vec{A} \) and \( \vec{B} \) cannot be collinear for any value of \( p \).

### Final Answer
The vectors \( \vec{A} \) and \( \vec{B} \) cannot be collinear for any value of \( p \).

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