Question

If ` vec a , vec b , vec c` are mutually perpenedicular vectors of equal magnitudes, show that the vector ` vec a+ vec b+ vec c` is equally inclined to ` vec a , vec b ,a n d vec cdot`

Answer 1

Let `|vec(a)|=|vec(b)| = |vec(c )|=a`
Given that, `vec(a).vec(b) = vec(b).vec(c) = vec(c).vec(a) =0`
Now `(vec(a) + vec(b) + vec(c ))^(2)`
`=|vec(a)|^(2) + |vec(b)|^(2) + |vec(c )|^(2) + 2(vec(a).vec(b) + vec(b).vec(c ) + vec( c) .vec(a))`
`=a^(2) + a^(2) +a ^(2) +0=3a^(2)`
`implies |vec(a)+vec(b)+vec(c)|=sqrt(3)a`
Let `theta` be the angle between `vec(a)` and `(vec(a) + vec(b) + vec(c ))`
`therefore vec(a). (vec(a) + vec(b) + vec(c )) = |vec(a)||vec(a)+vec(b)+vec(c )|costheta_(1)`
`implies vec(a) .vec(a)+vec(a).vec(b) + vec(a).vec(c ) = a(asqrt(3))costheta_(1)`
`implies a^(2) + 0 + 0 = a^(2) sqrt(3) cos theta_(1)`
`implies cos theta_(1) = (a^(2))/(a^(2)sqrt(3)) = (1)/(sqrt(3))`
`implies theta_(1) = cos^(-1)((1)/(sqrt(3)))`
Similarly,
angle between `vec(b)` and `vec(a) + vec(b) + vec(c) = cos^(-1) .(1)/(sqrt(3))`
and angle between `vec(c ) and vec(a) + vec(b) + vec(c ) = cos^(-1).(1)/(sqrt(3))`.
Therefore, `vec(a) + vec(b) + vec(c )` makes equal angles with the vectors `vec(a),vec(b)` and `vec(c )`.
Hence Proved.