Question

Let \(\overset{⃗}{a} = \hat{i}+2\hat{j}-2\hat{k}\) and \(\overset{⃗}{b} = \hat{i}-\hat{j}+\hat{k}\). If \(\overset{⃗}{c}\) is a vector such that \(\overset{⃗}{a}\cdot \overset{⃗}{c} = |\overset{⃗}{c}|\), \(|\overset{⃗}{c}-\overset{⃗}{a}| = 2\sqrt{2}\) and the angle between \(\overset{⃗}{a}\times \overset{⃗}{b}\) and \(\overset{⃗}{c}\) is \(60^{\circ}\), then \(|(\overset{⃗}{a}\times \overset{⃗}{b})\times \overset{⃗}{c}|\) is equal to

• A. \(\frac{3\sqrt{3}}{2}\)
• B. \(\sqrt{\frac{3}{2}}\)
• C. \(3\sqrt{\frac{3}{2}}\)
• D. \(\frac{9\sqrt{3}}{2}\)

Hint:

Find |c| from the two given conditions, then use |p x c| = |p||c| sin(angle).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
We need \(|\vec c|\). The magnitude of \((\vec a \times \vec b) \times \vec c\) equals \(|\vec a \times \vec b
\vec c|\sin 60^{\circ}\).

Step 2: Find |c|
\(|\vec a| = \sqrt{1 + 4 + 4} = 3\). From \(|\vec c - \vec a|^2 = 8\):
\[ |\vec c|^2 - 2\vec a\cdot\vec c + |\vec a|^2 = 8 \Rightarrow |\vec c|^2 - 2|\vec c| + 9 = 8 \Rightarrow (|\vec c| - 1)^2 = 0 \]
So \(|\vec c| = 1\).

Step 3: Find |a x b|
\[ \vec a \times \vec b = (2\cdot1 - (-2)(-1))\hat i - (1\cdot1 - (-2)(1))\hat j + (1\cdot(-1) - 2\cdot1)\hat k = 0\hat i - 3\hat j - 3\hat k \]
\[ |\vec a \times \vec b| = 3\sqrt2 \]

Step 4: Final value
\[ |(\vec a \times \vec b)\times\vec c| = 3\sqrt2 \times 1 \times \sin 60^{\circ} = 3\sqrt2\cdot\frac{\sqrt3}{2} = \frac{3\sqrt6}{2} = 3\sqrt{\frac32} \]
This is option (C).

Final Answer:
The value is \(3\sqrt{\frac32}\), option (C). \[ \boxed{3\sqrt{\frac{3}{2}}} \]