Question

If \(\vec a\),\(\vec b\),\(\vec c\)are mutually perpendicular to equal magnitudes, showing that the vector \(\vec a\)+\(\vec b\)+\(\vec c\)is equally inclined to \(\vec a\),\(\vec b\),and \(\vec c\).

Answer 1

Since,\(\vec a\),\(\vec b\),and \(\vec c\) are mutually perpendicular vectors,we have
\(\vec a\).\(\vec b\)=\(\vec b\).\(\vec c\)=\(\vec c\).\(\vec a\)=0.
It is given that:
|\(\vec a\)|=|\(\vec b\)|=|\(\vec c\)|
Let vector \(\vec a\)+\(\vec b\)+\(\vec c\)be inclined to \(\vec a\),\(\vec b\)and \(\vec c\) at angles θ₁,θ₂,and θ₃ respectively.
Then,we have:
cosθ=(\(\vec a+\vec b+\vec c\)).\(\vec a\)/|\(\vec a+\vec b+\vec c\)||\(\vec a\)|=\(\vec a\).\(\vec a+\vec b\).\(\vec a+\vec c\).\(\vec a\)/|\(\vec a+\vec b+\vec c\)||\(\vec a\)|
=\(\frac{|\vec a|}{|\vec a+\vec b+\vec c|}\)|\(\vec a\)| [\(\vec a.\vec b\)=\(\vec a.\vec c\)=0]
=\(\frac{|\vec a|}{|\vec a+\vec b+\vec c|}\)
cosθ2=(\(\vec a+\vec b+\vec c\)).\(\vec b\)/|\(\vec a+\vec b+\vec c\)||\(\vec b\)|=\(\frac{\vec a.\vec b+\vec b.\vec b+\vec c.\vec b}{|\vec a+\vec b+\vec c|.|\vec b|}\)
=\(\frac{|\vec b|^2}{\vec a+\vec b+\vec c}.|\vec b|[\vec a.\vec b=\vec c.\vec b=0]\)
=\(\frac{|\vec b|}{|\vec a+\vec b+\vec c|}\)
cosθ3=(\(\vec a+\vec b+\vec c\)).\(\vec c\)/|\(\vec a+\vec b+\vec c\)||\(\vec c\)|=\(\vec a.\vec c\)+\(\vec b.\vec c\)+\(\vec c.\vec c\)/|\(\vec a+\vec b+\vec c\)||\(\vec c\)|
=|\(\vec c\)|²/|\(\vec a+\vec b+\vec c\)||\(\vec c\)| [\(\vec a\).\(\vec c\)=\(\vec b\).\(\vec c\)=0]
=\(\frac{|\vec c|}{|\vec a+\vec b+\vec c|}\)
Now,as |\(\vec a\)||\(\vec b\)|=|\(\vec c\)|,cosθ1=cosθ2=cosθ3.
∴θ1=θ2=θ3
Hence,(\(\vec a+\vec b+\vec c\))is equally inclined to \(\vec a\),\(\vec b\),and \(\vec c\).