Question

If \(\vec{a} = \vec{i} + p \vec{j} - 3 \vec{k}, \vec{b} = p \vec{i} - 3 \vec{j} + \vec{k}, \vec{c} = -3 \vec{i} + \vec{j} + 2 \vec{k}\) are three vectors such that \[ |\vec{a} \times \vec{b}| = |\vec{a} \times \vec{c}|, \] then \(p =\)

• A. \(-2\)
• B. \(-1\)
• C. \(1\)
• D. \(2\)

Hint:

Calculate cross products and use magnitudes to find unknown vector components.

Answer 1

The Correct Option is D

Step 1: Calculate \(|\vec{a} \times \vec{b}|\)
\[ \vec{a} = (1, p, -3), \quad \vec{b} = (p, -3, 1) \] \[ \vec{a} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & p & -3 \\ p & -3 & 1 \end{vmatrix} = \vec{i} (p \times 1 - (-3) \times (-3)) - \vec{j} (1 \times 1 - (-3) \times p) + \vec{k} (1 \times (-3) - p \times p) \] \[ = \vec{i} (p - 9) - \vec{j} (1 + 3p) + \vec{k} (-3 - p^2) \] \[ |\vec{a} \times \vec{b}|^2 = (p - 9)^2 + (1 + 3p)^2 + (-3 - p^2)^2 \] Step 2: Calculate \(|\vec{a} \times \vec{c}|\)
\[ \vec{c} = (-3, 1, 2) \] \[ \vec{a} \times \vec{c} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 1 & p & -3 \\ -3 & 1 & 2 \end{vmatrix} = \vec{i} (p \times 2 - (-3) \times 1) - \vec{j} (1 \times 2 - (-3) \times (-3)) + \vec{k} (1 \times 1 - p \times (-3)) \] \[ = \vec{i} (2p + 3) - \vec{j} (2 - 9) + \vec{k} (1 + 3p) \] \[ |\vec{a} \times \vec{c}|^2 = (2p + 3)^2 + (-7)^2 + (1 + 3p)^2 \] Step 3: Set \(|\vec{a} \times \vec{b}| = |\vec{a} \times \vec{c}|\)
\[ (p - 9)^2 + (1 + 3p)^2 + (-3 - p^2)^2 = (2p + 3)^2 + 49 + (1 + 3p)^2 \] Step 4: Simplify and solve for \(p\)
\[ (p - 9)^2 + (-3 - p^2)^2 = (2p + 3)^2 + 49 \] Expanding and simplifying yields \(p = 2\).