If a, b, c Are Three Mutually Perpendicular Vectors of the Same Magnitude, Prove That a + b + c Is Equally Inclined With a, b and c

A standard vector algebra problem, not tied to a specific exam paper.

Problem

If $\vec{a},\vec{b},\vec{c}$ are three mutually perpendicular vectors of the same magnitude, prove that $\vec{a}+\vec{b}+\vec{c}$ is equally inclined with $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Key insight. “Equally inclined” means the cosine of the angle with each vector is the same. Since $\cos\theta = \dfrac{\vec{d}\cdot\vec{a}}{|\vec{d}||\vec{a}|}$ for $\vec{d}=\vec{a}+\vec{b}+\vec{c}$, and mutual perpendicularity makes every cross-dot-product (like $\vec{b}\cdot\vec{a}$) vanish, each of the three cosines collapses to the exact same simple expression.

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Approach

Let $\vec{d}=\vec{a}+\vec{b}+\vec{c}$, and let $k=|\vec{a}|=|\vec{b}|=|\vec{c}|$. Compute $\vec{d}\cdot\vec{a}$, $\vec{d}\cdot\vec{b}$, $\vec{d}\cdot\vec{c}$ individually, using the fact that mutually perpendicular vectors have zero dot product with each other. Then compute $|\vec{d}|$, and show that $\cos\theta$ (the angle between $\vec{d}$ and each of $\vec{a},\vec{b},\vec{c}$) comes out identical in all three cases.

Solution

Step 1 — Set up the given conditions

Let $k=|\vec{a}|=|\vec{b}|=|\vec{c}|$. Since $\vec{a},\vec{b},\vec{c}$ are mutually perpendicular:

$$\vec{a}\cdot\vec{b}=\vec{b}\cdot\vec{c}=\vec{c}\cdot\vec{a}=0$$

Step 2 — Compute d·a, where d = a+b+c

$$\vec{d}\cdot\vec{a} = (\vec{a}+\vec{b}+\vec{c})\cdot\vec{a} = \vec{a}\cdot\vec{a}+\vec{b}\cdot\vec{a}+\vec{c}\cdot\vec{a} = k^2+0+0 = k^2$$

Step 3 — Compute d·b and d·c similarly

$$\vec{d}\cdot\vec{b} = \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{b}+\vec{c}\cdot\vec{b} = 0+k^2+0 = k^2$$

$$\vec{d}\cdot\vec{c} = \vec{a}\cdot\vec{c}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{c} = 0+0+k^2 = k^2$$

Step 4 — Compute |d|

$$|\vec{d}|^2 = |\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a})$$

$$= k^2+k^2+k^2+2(0) = 3k^2 \implies |\vec{d}| = k\sqrt3$$

Step 5 — Compute the cosine of the angle between d and a

$$\cos\theta_a = \frac{\vec{d}\cdot\vec{a}}{|\vec{d}||\vec{a}|} = \frac{k^2}{(k\sqrt3)(k)} = \frac{1}{\sqrt3}$$

Step 6 — Compute the cosines for b and c

By the identical computations in Steps 2–5 (with $\vec{d}\cdot\vec{b}$ and $\vec{d}\cdot\vec{c}$ both equal to $k^2$, and the same $|\vec{d}|$ and $|\vec{b}|=|\vec{c}|=k$):

$$\cos\theta_b = \cos\theta_c = \frac{1}{\sqrt3}$$

Step 7 — Conclude

Since $\cos\theta_a=\cos\theta_b=\cos\theta_c=\dfrac{1}{\sqrt3}$, the vector $\vec{d}=\vec{a}+\vec{b}+\vec{c}$ makes the same angle with each of $\vec{a}$, $\vec{b}$, $\vec{c}$ — it is equally inclined to all three.

Answer

Common mistakes

• Forgetting to use “same magnitude” and instead leaving $|\vec{a}|,|\vec{b}|,|\vec{c}|$ as different unknowns. Without this condition, the three cosines would generally come out different (each divided by a different magnitude), and the “equally inclined” conclusion wouldn’t follow.
• Missing that mutually perpendicular vectors give zero dot products between different pairs, but not $\vec{a}\cdot\vec{a}$ itself. The self dot-product $\vec{a}\cdot\vec{a}=|\vec{a}|^2=k^2$ is what survives in each computation — it’s the cross-terms between different vectors that vanish.

Practise next

• If $\vec{p}$ and $\vec{q}$ are two mutually perpendicular vectors of the same magnitude, prove that $\vec{p}+\vec{q}$ is equally inclined to each of them, using the same dot-product approach.