Question

Let \( \mathbf{a} = \mathbf{i} - 2\mathbf{j} \), \( \mathbf{b} = 2\mathbf{j} + 3\mathbf{k} \), \( \mathbf{c} = p\mathbf{i} + q\mathbf{j} \) and \( \mathbf{d} = p\mathbf{j} - q\mathbf{k} \) be four vectors. If \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = 3 \) and \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d} = 0 \), then \( 3p + q = \)

• A. \( 0 \)
• B. \( 3 \)
• C. \( -2 \)
• D. \( 6 \)

Hint:

For collinear vectors \( \mathbf{a} \) and \( \mathbf{b} \), use the property \( \mathbf{a} = \lambda \mathbf{b} \). For scalar triple product, remember \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \begin{vmatrix} a_1 & a_2 & a_3
b_1 & b_2 & b_3
c_1 & c_2 & c_3 \end{vmatrix} \). Use the given conditions to form equations involving \( p \) and \( q \) and solve for them. Finally, find the value of \( 3p + q \).

Answer 1

The Correct Option is A

First, calculate the cross product \( \mathbf{a} \times \mathbf{b} \): $$ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}
1 & -2 & 0
0 & 2 & 3 \end{vmatrix} = \mathbf{i}(-6 - 0) - \mathbf{j}(3 - 0) + \mathbf{k}(2 - 0) = -6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} $$ Given \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = 3 \): $$ (-6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) \cdot (p\mathbf{i} + q\mathbf{j} + 0\mathbf{k}) = 3 $$ $$ -6p - 3q = 3 $$ Dividing by \( -3 \): $$ 2p + q = -1 \quad \cdots (1) $$ Given \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d} = 0 \): $$ (-6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) \cdot (0\mathbf{i} + p\mathbf{j} - q\mathbf{k}) = 0 $$ $$ -3p - 2q = 0 \quad \cdots (2) $$ Multiply equation (1) by 2: \( 4p + 2q = -2 \quad \cdots (3) \) Add equation (2) and (3): $$ (-3p - 2q) + (4p + 2q) = 0 + (-2) $$ $$ p = -2 $$ Substitute \( p = -2 \) into equation (1): $$ 2(-2) + q = -1 $$ $$ -4 + q = -1 $$ $$ q = 3 $$ Now, calculate \( 3p + q \): $$ 3p + q = 3(-2) + 3 = -6 + 3 = -3 $$ There seems to be a discrepancy with the correct answer provided.
Let's recheck the calculations.
\( \mathbf{a} \times \mathbf{b} = -6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k} \) \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = -6p - 3q = 3 \implies 2p + q = -1 \) \( (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{d} = -3p - 2q = 0 \) From \( 2p + q = -1 \), \( q = -1 - 2p \).
Substitute into \( -3p - 2q = 0 \): \( -3p - 2(-1 - 2p) = 0 \implies -3p + 2 + 4p = 0 \implies p = -2 \).
\( q = -1 - 2(-2) = -1 + 4 = 3 \).
\( 3p + q = 3(-2) + 3 = -6 + 3 = -3 \).
Let's check if there was a typo in the question or options.
If the correct answer is 0, let's see if that's possible.
If \( 3p + q = 0 \implies q = -3p \).
\( 2p + q = -1 \implies 2p - 3p = -1 \implies -p = -1 \implies p = 1 \).
\( q = -3(1) = -3 \).
Check \( -3p - 2q = -3(1) - 2(-3) = -3 + 6 = 3 \neq 0 \).
There appears to be an inconsistency.
Assuming the provided correct answer is indeed (A) 0, there might be an error in my derivation or the question statement.