Question

Let \(\vec a=2\hat i+\hat j-2\hat k,\ \vec b=\hat i+\hat j\). If \(\vec c\) is a vector such that \(\vec a\cdot\vec c=|\vec c|,\ |\vec c-\vec a|=2\sqrt2\) and the angle between \(\vec a\times\vec b\) and \(\vec c\) is \(30^\circ\), then \(|(\vec a\times\vec b)\times\vec c|\) equals:

• A. \(\frac12\)
• B. \(\frac{3\sqrt3}{2}\)
• C. 3
• D. \(\frac32\)

Hint:

Square \(|\vec c-\vec a|\) to find \(|\vec c|\), then use \(|\vec u\times\vec c|=|\vec u||\vec c|\sin\theta\).

Answer 1

The Correct Option is D

\(|\vec a|=\sqrt{4+1+4}=3\).
\(|\vec c-\vec a|^2=|\vec c|^2-2\vec a\cdot\vec c+|\vec a|^2=8\). With \(\vec a\cdot\vec c=|\vec c|\): \(|\vec c|^2-2|\vec c|+9=8\Rightarrow(|\vec c|-1)^2=0\Rightarrow|\vec c|=1\).
\(\vec a\times\vec b=\hat i(1\cdot0-(-2)(1))-\hat j(2\cdot0-(-2)(1))+\hat k(2\cdot1-1\cdot1)=2\hat i-2\hat j+\hat k\), so \(|\vec a\times\vec b|=3\).
\(|(\vec a\times\vec b)\times\vec c|=|\vec a\times\vec b||\vec c|\sin30^\circ=3\cdot1\cdot\frac12=\frac32\).
Answer: (4) \(\frac32\)