Question

Let \(\vec{a}=2\hat{i}+\hat{j}-2\hat{k}\) and \(\vec{b}=\hat{i}+\hat{j}\). Let \(\vec{c}\) be a vector such that \(|\vec{c}-\vec{a}|=3\), \(|(\vec{a}\times\vec{b})\times\vec{c}|=3\) and the angle between \(\vec{c}\) and \(\vec{a}\times\vec{b}\) is \(30^\circ\). Then \(\vec{a}\cdot\vec{c}\) is equal to :

• A. 2
• B. 5
• C. \(\frac{1}{8}\)
• D. \(\frac{25}{8}\)

Hint:

Use \(|\vec{n}\times\vec{c}|=|\vec{n}||\vec{c}|\sin30^\circ\) to get \(|\vec{c}|\), then expand \(|\vec{c}-\vec{a}|^2\).

Answer 1

The Correct Option is A

\(\vec{a}\times\vec{b}=\begin{vmatrix}\hat i&\hat j&\hat k\\2&1&-2\\1&1&0\end{vmatrix}=2\hat i-2\hat j+\hat k\), so \(|\vec{a}\times\vec{b}|=3\); also \(|\vec{a}|=3\).
\(|(\vec{a}\times\vec{b})\times\vec{c}|=|\vec{a}\times\vec{b}||\vec{c}|\sin30^\circ\Rightarrow 3=3\cdot|\vec{c}|\cdot\frac{1}{2}\Rightarrow|\vec{c}|=2\).
\(|\vec{c}-\vec{a}|^2=|\vec{c}|^2-2\vec{a}\cdot\vec{c}+|\vec{a}|^2\Rightarrow 9=4-2\vec{a}\cdot\vec{c}+9\).
So \(\vec{a}\cdot\vec{c}=2\).
Answer: (1) 2