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  • If vector a b c are mutually perpendicular vectors of equal magnitudes, show that the vector is equally inclined to .

Q.14 If \vec a , \vec b , \vec c are mutually perpendicular vectors of equal magnitudes, show that the vector \vec a+\vec b +\vec c  is equally inclined to \vec a , \vec b \: \: and \: \: \vec c .

Answers (1)

best_answer

Given 

|\vec a|=|\vec b|=|\vec c| and 

\vec a.\vec b=\vec b.\vec c=\vec c.\vec a=0

Now, let vector \vec a+\vec b +\vec cis inclined to \vec a , \vec b \: \: and \: \: \vec c at \theta_1,\theta_2\:and\:\theta_3 respectively.

Now, 

cos\theta_1=\frac{(\vec a+\vec b+\vec c).\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a +\vec a.\vec b +\vec c.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{\vec a.\vec a}{|\vec a+\vec b+\vec c||\vec a|}=\frac{|\vec a|}{|\vec a+\vec b+\vec c|}

cos\theta_2=\frac{(\vec a+\vec b+\vec c).\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec a.\vec b +\vec b.\vec b +\vec c.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{\vec b.\vec b}{|\vec a+\vec b+\vec c||\vec b|}=\frac{|\vec b|}{|\vec a+\vec b+\vec c|}

cos\theta_3=\frac{(\vec a+\vec b+\vec c).\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec a.\vec c +\vec b.\vec c +\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{\vec c.\vec c}{|\vec a+\vec b+\vec c||\vec c|}=\frac{|\vec c|}{|\vec a+\vec b+\vec c|}

Now, Since,|\vec a|=|\vec b|=|\vec c| 

cos\theta_1=cos\theta_2=cos\theta_3

\theta_1=\theta_2=\theta_3   

Hence vector \vec a+\vec b +\vec c  is equally inclined to \vec a , \vec b \: \: and \: \: \vec c .

Posted by

Pankaj Sanodiya

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