Question

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]`

Answer 1

∵ \(\vec d\)makes equal angle with \(\vec a,\vec b\) and \(\vec c\).

\(\therefore \vec d = \frac{\mu(\vec a + \vec b + \vec c)}{3}\)    ...(i)

Again \([\vec a\; \vec b\;\vec c] = [\vec d\;\vec b\; \vec c]\vec a + [\vec d\;\vec c\;\vec a]\vec b + [\vec d\; \vec a\;\vec b]\vec c\)   ....(ii)

From (i) and (ii), we get

\( [\vec d\;\vec b\; \vec c] = [\vec d\;\vec c\;\vec a]= [\vec d\; \vec a\;\vec b]\)

Answer 2

Since `vecd` makes equalw angles with the vectors `veca1 , vecb and vecc`, we have,
`d= (mu(veca + vecb + vecc))/3`
(`vecd` passes through the centroid of the triangle with position vectors, `veca , vecb and vecc`)
Again `[veca vecb vecc]vecd = [ vecd vecb vecc] + [vecd vecc vecd] vecb`
`+ [vecd veca vecb]vecc`
From (i) and (ii) , we get `[veca vecb vecc] = [vecd vecc veca] = [ vecd veca vecb] `