Question

The area of parallelogram formed by vectors \(\overset{⃗}{P} = 2\hat{i}-\hat{j}+5\hat{k}\) and \(\overset{⃗}{Q} = 3\hat{i}-2\hat{j}+4\hat{k}\) is

• A. \(\sqrt{72}\)
• B. \(\sqrt{86}\)
• C. \(\sqrt{104}\)
• D. \(\sqrt{240}\)

Hint:

Area equals the magnitude of the cross product P x Q.

Answer 1

The Correct Option is B

Step 1: Use the sine formula in a different way:
Area $= \sqrt{|P|^2|Q|^2 - (\vec P\cdot\vec Q)^2}$.

Step 2: Compute the pieces:
$|P|^2 = 4 + 1 + 25 = 30$. $|Q|^2 = 9 + 4 + 16 = 29$. $\vec P\cdot\vec Q = 6 + 2 + 20 = 28$.

Step 3: Combine:
$30 \times 29 - 28^2 = 870 - 784 = 86$. So the area is $\sqrt{86}$.

Final Answer:
The area is root 86, option (B). \[ \boxed{\sqrt{86}} \]