Let ` vec a , vec b , vec c` be three non-zero vectors such that ` vec c` is a unit vector perpendicular to both ` vec aa n d vec b` . If the between ` vec aa n d vec b` is `pi//6` , prove that `[ vec a vec b vec c]^2=1/4| vec a|^2| vec b|^2dot`

Explanation

To prove that \([ \vec{a}, \vec{b}, \vec{c} ]^2 = \frac{1}{4} |\vec{a}|^2 |\vec{b}|^2\), given that \(\vec{c}\) is a unit vector perpendicular to both \(\vec{a}\) and \(\vec{b}\) and the angle between \(\vec{a}\) and \(\vec{b}\) is \(\frac{\pi}{6}\), we can follow these steps:

### Step 1: Understand the Scalar Triple Product
The scalar triple product \([ \vec{a}, \vec{b}, \vec{c} ]\) can be expressed as:
\[ [ \vec{a}, \vec{b}, \vec{c} ] = \vec{c} \cdot (\vec{a} \times \vec{b}) \]
Thus, the square of the scalar triple product is:
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