Question

If vectors \(\vec a,\vec b,\vec c\) are mutually perpendicular and having equal magnitude, then show that the vector \(\vec a+\vec b+\vec c\) is equally inclined to the vectors \(\vec a,\vec b,\vec c\).

Hint:

Use \(\vec s\cdot\vec a=\vec s\cdot\vec b=\vec s\cdot\vec c\) (only the matching term survives by perpendicularity) together with equal magnitudes.

Answer 1

Step 1: Coordinate-based cross-check:
Since \(\vec a,\vec b,\vec c\) are mutually perpendicular with equal magnitude \(k\), set up axes along them: \(\vec a=(k,0,0)\), \(\vec b=(0,k,0)\), \(\vec c=(0,0,k)\).

Step 2: Computing s in these coordinates:
\(\vec s=\vec a+\vec b+\vec c=(k,k,k)\), so \(|\vec s|=\sqrt{3k^2}=k\sqrt3\).

Step 3: Computing each angle by the standard cosine formula:
\(\cos\theta_a=\dfrac{\vec s\cdot\vec a}{|\vec s||\vec a|}=\dfrac{k^2}{k\sqrt3\cdot k}=\dfrac{1}{\sqrt3}\). By the coordinates' symmetry, \(\cos\theta_b=\cos\theta_c=\dfrac{1}{\sqrt3}\) too.

Final Answer:
All three angles have cosine \(\dfrac{1}{\sqrt3}\), so \(\boxed{\vec a+\vec b+\vec c\text{ is equally inclined to }\vec a,\vec b,\vec c}\), matching the coordinate-free proof.