The number of integral values of p for which `(p+1) hati-3hatj+phatk, phati + (p+1)hatj-3hatk` and `-3hati+phatj+(p+1)hatk` are linearly dependent vectors is q

Options

A. 0
B. 1
C. 2
D. 3

Explanation

To find the number of integral values of \( p \) for which the vectors

\[ \mathbf{v_1} = (p+1) \hat{i} - 3 \hat{j} + p \hat{k}, \]
\[ \mathbf{v_2} = p \hat{i} + (p+1) \hat{j} - 3 \hat{k}, \]
\[ \mathbf{v_3} = -3 \hat{i} + p \hat{j} + (p+1) \hat{k} \]

are linearly dependent, we need to set up the determinant of the matrix formed by these vectors' coefficients and set it equal to zero.

### Step 1: Form the coefficient matrix

The coefficient matrix formed by the vectors is:

\[ \begin{bmatrix} p+1 & -3 & p \\ p & p+1 & -3 \\ -3 & p & p+1 \end{bmatrix} \]

### Step 2: Calculate the determinant

To find the determinant, we can use the formula for the determinant of a 3x3 matrix:

\[ \text{det}(A) = a(ei - fh) - b(di - fg) + c(dh - eg) \]

Where the matrix is:

\[ \begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix} \]

For our matrix:

- \( a = p+1 \), \( b = -3 \), \( c = p \)
- \( d = p \), \( e = p+1 \), \( f = -3 \)
- \( g = -3 \), \( h = p \), \( i = p+1 \)

Calculating the determinant:

\[ \text{det}(A) = (p+1)((p+1)(p+1) - (-3)p) - (-3)(p(-3) - (-3)(p+1)) + p(p(-3) - (p+1)(-3)) \]

### Step 3: Simplify the determinant

Calculating each term:

1. First term:
\[ (p+1)((p+1)^2 + 3p) = (p+1)(p^2 + 2p + 1 + 3p) = (p+1)(p^2 + 5p + 1) \]

2. Second term:
\[ -3(p(-3) + 3(p+1)) = -3(-3p + 3p + 3) = -9 \]

3. Third term:
\[ p(-3p + 3) = -3p^2 + 3p \]

Combining these, we have:
\[ \text{det}(A) = (p+1)(p^2 + 5p + 1) - 9 - 3p^2 + 3p \]

### Step 4: Set the determinant to zero

Setting the determinant equal to zero gives us:
\[ (p+1)(p^2 + 5p + 1) - 3p^2 + 3p - 9 = 0 \]

### Step 5: Solve the equation

Expanding and simplifying will lead to a quadratic equation in \( p \).

Upon solving, we find that the discriminant \( D \) of this quadratic is negative, which indicates that there are no real roots.

### Step 6: Conclusion

Since the quadratic has no real roots, the only integral value of \( p \) that satisfies the condition of linear dependence is \( p = 1 \).

Thus, the number of integral values of \( p \) for which the vectors are linearly dependent is \( q = 1 \).