If `veca, vecb, vecc` are mutually perpendicular vectors of equal magnitudes, show that the vector `veca + vecb+ vecc` is equally inclined to `veca, vecb` and `vecc`.
Solution
Since `veca , vecb "and" vec c` are mutually perpendicular vectors, we have
`veca . vecb = vecb . vecc = 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 `veca , vecb "and" vec c` at angles `theta_1,theta_2` and `theta_3` respectively.
Then, we have:
cos θ1 = `((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.veca)/(|vec a + vec b + vec c||vec a|)`
`= |vec a|^2/(|vec a + vec b + vec c||vec a|)` .....`[vecb. vec a = vec c. vec a = 0]`
`=|veca|/(|veca + vecb + vecc|)`
cosθ2 = `((vec a+vec b +vec c).vec b)/(|vec a + vec b + vec c|.|vec b|) =(vec a .vec b + vec b. vec b + vec c.vec b)/(|vec a + vec b + vec c||vec b|)`
`= |vec b|^2/(|vec a + vec b + vec c||vec b|)` .....`[vec a. vec b = vec c. vec b = 0]`
`=|vecb|/(|veca + vecb + vecc|)`
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]`
`=|vecc|/(|veca + vecb + vecc|)`
Now, as `|vec a| = |vec b| = |vec c|` , cos θ1 = cosθ2 = cosθ3
∴ θ1 = θ2 = θ3
Hence, the vector `(veca + vecb + vecc)` is equally inclined to `veca, vecb " and " vecc`