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Question
If `veca, vecb, vecc` are mutually perpendicular vectors of equal magnitudes, show that the vector `vecc* vecd = 15` is equally inclined to `veca, vecb "and" vecc.`
Solution
`veca, vecb, vecc` are of equal magnitude
`|veca| = |vecb| = |vecc|`
`veca, vecb, vecc` are mutedly ⊥ to each other `veca * vecb = vecb * vecc = vecc * veca = 0`
`(veca + vecb + vecc)* veca = |veca + vecb + vecc| cos alpha`
α angle of `(veca + vecb + vecc)* veca`
⇒ `veca * vecd + vecb * veca + vecc * veca`
`= |veca + vecb + vecc| |veca| cos alpha`
`|veca|^2 = |veca + vecb + vecc| |veca|cos alpha`
`cos alpha = |veca|/ |veca + vecb + vecc|`
`(veca + vecb + vecc) * vecb = |veca + vecb + vecc| |vecb| cos beta`
`beta ` angle of `(veca + vecb + vecc)* vecb`
⇒ `veca* vecb + vecb * vecb + vecc * vecb = |veca + vecb + vecc| |vecb| cos beta`
`|vecb|^2 = |veca + vecb + vecc| |vecc| cos gamma`
`gamma "angle of" (veca + vecb + vecc)*vecc`
`veca *vecc + vecb*vecc + vecc* vecc = |veca + vecb + vecc| |vecc| cos gamma`
`cos gamma = |vecc|/|veca + vecb + vecc|`
But `|veca| = |vecb| = |vecc|`
`cos alpha = cos beta = cos gamma`
`alpha = beta = gamma`
`veca + vecb + vecc` are inclined to `veca, vecb, vecc`
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