LinAlg4.2.3

From Exampleproblems

If  are mutually perpendicular vectors of equal magnitude,show that  is equally inclined to

Given that the vectors a,b,c are mutually perpendicular to each other.

Hence $\bar{a}\cdot\bar{b}=\bar{b}\cdot\bar{c}=\bar{c}\cdot\bar{a}=0\,$

Also given the three of equal magnitude $|a|=|b|=|c|\,$

Now $|a+b+c|^2=(\bar{a}+\bar{b}+\bar{c})^2=(\bar{a})^2+(\bar{b})^2+(\bar{c})^2+2(\bar{a}\cdot\bar{b}+\bar{b}\cdot\bar{c}+\bar{c}\cdot\bar{a})=|a|^2+|b|^2+|c|^2+2(0)=3|a|^2\,$

Hence $|\bar{a}+\bar{b}+\bar{c}|=3\sqrt{\bar{a}}\,$

Let $\bar{a}+\bar{b}+\bar{c}\,$ make angles $\alpha,\beta,\gamma\,$ respectively.

Therefore $\cos\alpha=\frac{(\bar{a}+\bar{b}+\bar{c})\cdot\bar{a}}{|a+b+c||a|}=\frac{|a|^2}{\sqrt{3}|a|^2}=\frac{1}{\sqrt{3}}\,$

Similarly $\cos\beta=\frac{1}{\sqrt{3}},\cos\gamma=\frac{1}{\sqrt{3}}\,$

Therefore $\cos\alpha=\cos\beta=\cos\gamma\,$ implies that $\alpha=\beta=\gamma\,$

Hence the sum of the three vectors is equally inclined to the three individual vectors.

Main Page:Linear Algebra

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