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: Understanding the Concept:
\"Equally inclined\" means the angle between \(\vec s=\vec a+\vec b+\vec c\) and each of \(\vec a,\vec b,\vec c\) is the same; since the cosine of the angle is \(\dfrac{\vec s\cdot(\cdot)}{|\vec s||\cdot|}\) and all three vectors share the same magnitude, it suffices to show \(\vec s\cdot\vec a=\vec s\cdot\vec b=\vec s\cdot\vec c\).

Step 2: Setting up notation:
Let \(|\vec a|=|\vec b|=|\vec c|=k\), with \(\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0\) (mutual perpendicularity).

Step 3: Computing s dot a:
\(\vec s\cdot\vec a=(\vec a+\vec b+\vec c)\cdot\vec a=\vec a\cdot\vec a+\vec b\cdot\vec a+\vec c\cdot\vec a=k^2+0+0=k^2\).

Step 4: Computing s dot b and s dot c similarly:
By the same reasoning (only the matching term survives, the other two dot products vanish by perpendicularity), \(\vec s\cdot\vec b=k^2\) and \(\vec s\cdot\vec c=k^2\).

Step 5: Comparing the angles:
\(\cos\theta_a=\dfrac{\vec s\cdot\vec a}{|\vec s|\,k}=\dfrac{k^2}{|\vec s|k}=\dfrac{k}{|\vec s|}\), and identically \(\cos\theta_b=\cos\theta_c=\dfrac{k}{|\vec s|}\).

Final Answer:
Since \(\cos\theta_a=\cos\theta_b=\cos\theta_c\), \(\boxed{\vec a+\vec b+\vec c\text{ is equally inclined to }\vec a,\vec b,\vec c}\).