Question

If \(\vec{a},\vec{b}\), and \(\vec{c}\) are unit vectors such that \(\vec{a}+2\vec{b}+2\vec{c}=\vec{0}\), then \(|\vec{a}\times\vec{c}|\) is equal to :

• A. \(\frac{\sqrt{15}}{4}\)
• B. \(\frac{1}{4}\)
• C. \(\frac{15}{16}\)
• D. \(\frac{\sqrt{15}}{16}\)

Hint:

Use \(\vec a+2\vec c=-2\vec b\) and take the modulus-square to find the angle between \(\vec a\) and \(\vec c\).

Answer 1

The Correct Option is A

From \(\vec a+2\vec c=-2\vec b\): \(|\vec a+2\vec c|^2=4|\vec b|^2\Rightarrow1+4+4\,\vec a\cdot\vec c=4\Rightarrow\vec a\cdot\vec c=-\frac{1}{4}\).
So \(\cos\theta=-\frac14\) and \(\sin\theta=\sqrt{1-\frac{1}{16}}=\frac{\sqrt{15}}{4}\).
\(|\vec a\times\vec c|=|\vec a||\vec c|\sin\theta=\frac{\sqrt{15}}{4}\).
Answer: (1) \(\frac{\sqrt{15}}{4}\)