Q.If $\vec{a}$, $\vec{b}$, $\vec{c}$ are mutually perpendicular vectors of equal magnitudes, show that the vector $\vec{a} + \vec{b} + \vec{c}$ is equally inclined to $\vec{a}$, $\vec{b}$ and $\vec{c}$.

Concept: Vector Magnitude Properties — When vectors are mutually perpendicular and equal in magnitude, dot products simplify drastically.

Let $|\vec{a}| = |\vec{b}| = |\vec{c}| = k$.

Since they are mutually perpendicular:

Step 1: Compute the dot product of $\vec{a} + \vec{b} + \vec{c}$ with $\vec{a}$:

Step 2: By symmetry, the same result $k^2$ holds for dot products with $\vec{b}$ and $\vec{c}$.

Step 3: Find the magnitude of $\vec{a} + \vec{b} + \vec{c}$:

So $|\vec{a} + \vec{b} + \vec{c}| = k\sqrt{3}$. …

We start with the core idea: the angle between two vectors is determined by their dot product. If we can show that the dot product of $\vec{a}+\vec{b}+\vec{c}$ with $\vec{a}$ is the same as with $\vec{b}$ and with $\vec{c}$, then the cosines of those angles are equal — and since all angles lie between $0$ and $\pi$, equal cosine means equal angle.

The problem gives us two powerful conditions:

1. Mutually perpendicular: $\vec{a}\cdot\vec{b}=0$, $\vec{b}\cdot\vec{c}=0$, $\vec{c}\cdot\vec{a}=0$.
2. Equal magnitudes: $|\vec{a}|=|\vec{b}|=|\vec{c}|=k$ (say).

These are the only facts we need. No coordinates, no components — just vector algebra.

1. Find the dot product of the sum with one of the vectors.

   Take $\vec{a}$ first:

Because $\vec{b}\perp\vec{a}$ and $\vec{c}\perp\vec{a}$, the last two terms are zero. So:

By symmetry, the same calculation with $\vec{b}$ or $\vec{c}$ gives:

So the dot product of the sum with each original vector is identical.

2. Find the magnitude of the sum vector.

Expand:

All cross terms vanish (perpendicularity). So:

Hence:

3. Compute the cosine of the angle between the sum and each vector.

   Let $\theta_a$ be the angle between $\vec{a}+\vec{b}+\vec{c}$ and $\vec{a}$. Then:

Exactly the same calculation for $\theta_b$ and $\theta_c$ gives: …

Method: Proving a Sum Vector Is Equally Inclined to Given Vectors

Use this when vectors are mutually perpendicular with equal magnitude and you must show their sum makes equal angles with each.

Steps

Step 1: Encode the conditions as dot products

Mutually perpendicular: $\vec a\cdot\vec b=\vec b\cdot\vec c=\vec c\cdot\vec a=0$. Equal magnitudes: $|\vec a|=|\vec b|=|\vec c|=k$, so $\vec a\cdot\vec a=k^2$, etc.

Step 2: Dot the sum with each vector

Compute $(\vec a+\vec b+\vec c)\cdot\vec a$. The cross terms vanish, leaving $k^2$. By symmetry the dot with $\vec b$ and with $\vec c$ each also equal $k^2$. …

Common Mistakes

Mistake 1: Leaving the cross terms in $|\vec a+\vec b+\vec c|^2$

Why it's wrong: the expansion has $2(\vec a\cdot\vec b+\vec b\cdot\vec c+\vec c\cdot\vec a)$, all zero here; keeping them gives the wrong magnitude. Correct approach: use perpendicularity to drop them, leaving $3k^2$.

Mistake 2: Comparing only the dot products and forgetting the magnitude

Why it's wrong: equal dot products alone do not give equal angles — you must divide by $|\vec a+\vec b+\vec c|\,|\vec a|$. Correct approach: form $\cos\theta$ fully for each vector. …