Answer 1

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 = cosθ_{2 =} cosθ₃

∴ θ₁ = θ₂ = θ₃ 

Hence, the vector `(veca + vecb + vecc)` is equally inclined to `veca, vecb " and " vecc`