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3-1.Vectors
hard
If a vector $\overrightarrow P $ making angles $\alpha, \beta\ and\ \gamma$ respectively with the $X, Y$ and $Z$ axes respectively. Then ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma = $
A$0$
B$1$
C$2$
D$3$
Solution
(c) ${\sin ^2}\alpha + {\sin ^2}\beta + {\sin^2} \gamma $
$ = 1 – {\cos ^2}\alpha + 1 – {\cos ^2}\beta + 1 – {\cos ^2}\gamma $
$ = 3 – ({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma )$
$ = 3 – 1 = 2$
$ = 1 – {\cos ^2}\alpha + 1 – {\cos ^2}\beta + 1 – {\cos ^2}\gamma $
$ = 3 – ({\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma )$
$ = 3 – 1 = 2$
Standard 11
Physics
