Q.Show that the direction cosines of a vector equally inclined to the axes OX, OY and OZ are $\pm \left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$.

Concept: Direction Cosines — the cosines of the angles a vector makes with the coordinate axes.

Let the direction cosines be $l, m, n$. For a vector equally inclined to OX, OY, OZ, we have $l = m = n$.

The fundamental relation for any vector is:

Substituting $l = m = n$ gives:

$3l^2 = 1 \quad\Rightarrow\quad l^2 = \frac{1}{3} \quad\Rightarrow\quad l = \pm \frac{1}{\sqrt{3}}$ …

The key idea is simple: if a vector makes the same angle with the x‑axis, y‑axis, and z‑axis, then its direction cosines — which are the cosines of those angles — must all be equal. Let’s call that common value $k$.

Direction cosines are not independent; they always satisfy $l^2 + m^2 + n^2 = 1$, because they are the components of a unit vector along the given direction. So if $l = m = n = k$, we get:

That’s the entire derivation. But let’s walk through it carefully, step by step.

1. Define direction cosines.

   For a vector $\vec{r}$, the direction cosines are $l = \cos \alpha$, $m = \cos \beta$, $n = \cos \gamma$, where $\alpha, \beta, \gamma$ are the angles $\vec{r}$ makes with OX, OY, OZ respectively.

2. Apply the condition “equally inclined”.

   “Equally inclined” means $\alpha = \beta = \gamma$. Therefore $\cos \alpha = \cos \beta = \cos \gamma$, so $l = m = n$. Let this common value be $k$.

3. Use the fundamental identity.

   For any vector, $l^2 + m^2 + n^2 = 1$. Substituting $l = m = n = k$ gives $3k^2 = 1$.

4. Solve for $k$.

   $k^2 = \frac{1}{3}$, so $k = \pm \frac{1}{\sqrt{3}}$.

5. Write the direction cosines. …

Method: Direction Cosines of a Vector Equally Inclined to the Axes

Use this reasoning whenever a direction makes equal angles with all three coordinate axes.

Steps

Step 1: Translate 'equally inclined' into equal direction cosines

Equal angles $\alpha=\beta=\gamma$ mean equal cosines, so $l=m=n$. Call the common value $k$.

Step 2: Apply the fundamental identity

Direction cosines always satisfy $l^2+m^2+n^2=1$. Substituting $l=m=n=k$ gives $3k^2=1$.

Step 3: Solve, keeping both signs together …

Common Mistakes

Mistake 1: Using $l+m+n=1$ instead of $l^2+m^2+n^2=1$

Why it's wrong: the identity is on the squares of the direction cosines, not their plain sum. Correct approach: $3k^2=1$, giving $k=\pm\tfrac1{\sqrt3}$.

Mistake 2: Mixing the signs across the three cosines …