If `veca , vecb , vecc and vecd` are four non-coplanar unit vectors such that `vecd` makes equal angles with all the three vectors `veca, vecb, vecc` then prove that `[vecd vecavecb]=[vecd veccvecb]=[vecd veccveca]`

Explanation

To prove that \([ \vec{d}, \vec{a} \times \vec{b} ] = [ \vec{d}, \vec{c} \times \vec{b} ] = [ \vec{d}, \vec{c} \times \vec{a} ]\) given that \(\vec{d}\) makes equal angles with the non-coplanar unit vectors \(\vec{a}, \vec{b}, \vec{c}\), we can follow these steps:

### Step 1: Understand the condition of equal angles
Since \(\vec{d}\) makes equal angles with \(\vec{a}, \vec{b}, \vec{c}\), we can express this condition mathematically. Let the angle between \(\vec{d}\) and \(\vec{a}\), \(\vec{b}\), and \(\vec{c}\) be \(\theta\). Therefore, we have:
\[ \vec{d} \cdot \vec{a} = \vec{d} \cdot \vec{b} = \vec{d} \cdot \vec{c} = \cos(\theta) \]

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