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# Can a vector have direction angles $30^{\circ},45^{\circ},60^{\circ}$?

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• If $l, m , n$ are the d.c.s of a vector, then $l^2+m^2+n^2=1$
$Cos^2 30^{\circ} +Cos^2 45^{\circ} +Cos^2 60^{\circ} = \large\frac{3}{4}+ \large\frac{1}{2}+ \large\frac{1}{4}= \large\frac{3}{2} \neq 1$
A vector cannot have direction angles $30^{\circ}, 45^{\circ}, 60^{\circ}$

edited Jun 23, 2013

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