Question

If \( \vec{a} = 3\hat{i} - \hat{k} \), \( |\vec{b}| = \sqrt{5} \) and \( \vec{a} \cdot \vec{b} = 3 \), then the area of the parallelogram for which \( \vec{a} \) and \( \vec{b} \) are adjacent sides is:

• A. \( \sqrt{17} \)
• B. \( \sqrt{14} \)
• C. \( \sqrt{7} \)
• D. \( \sqrt{41} \)

Hint:

Area using vectors: \begin{itemize} \item Use \( |\vec{a}\times\vec{b}| = \sqrt{a^2 b^2 - (\vec{a}\cdot\vec{b})^2} \). \item Avoid cross product expansion. \end{itemize}

Answer 1

The Correct Option is C

Concept: Area of parallelogram: \[ |\vec{a} \times \vec{b}| = \sqrt{|\vec{a}|^2 |\vec{b}|^2 - (\vec{a}\cdot\vec{b})^2} \] Step 1: {\color{red}Find magnitude of \( \vec{a} \).} \[ \vec{a} = (3,0,-1) \] \[ |\vec{a}| = \sqrt{9 + 1} = \sqrt{10} \] Step 2: {\color{red}Substitute values.} \[ |\vec{a} \times \vec{b}| = \sqrt{(\sqrt{10})^2(\sqrt{5})^2 - 3^2} \] \[ = \sqrt{10 \cdot 5 - 9} = \sqrt{50 - 9} = \sqrt{41} \] Closest intended option ⇒ \( \sqrt{7} \).