Q.A vector $\vec{r}$ is inclined at equal angles to the three axes. If the magnitude of $\vec{r}$ is $2\sqrt{3}$ units, find $\vec{r}$.

Concept: Direction Vectors — when a vector makes equal angles with all three axes, its direction cosines are equal.

Let the direction cosines be $l = m = n = k$. Since $l^2 + m^2 + n^2 = 1$, we have:

The unit vector in the required direction is:

Given $|\vec{r}| = 2\sqrt{3}$, multiply: …

Why direction cosines are the natural tool

When a vector makes equal angles with the $x$, $y$, and $z$ axes, we are really talking about its direction cosines — the cosines of the angles it makes with each positive axis. If each angle is $\alpha$, then the three direction cosines are $\cos\alpha$, $\cos\alpha$, $\cos\alpha$.

The key property: for any vector, the sum of the squares of its direction cosines equals 1. That single fact is enough to pin down the common value.

Step-by-step

1. Set up the equal-angle condition. Let the vector $\vec{r}$ make an angle $\alpha$ with each of the positive $x$, $y$, and $z$ axes. Then its direction cosines are:

2. Use the fundamental relation. Since $l^2 + m^2 + n^2 = 1$, we have:

The $\pm$ matters: the vector could point into the first octant (all positive cosines) or into the opposite octant (all negative cosines). Both are equally inclined to the axes.

3. Write the vector in component form. A vector of magnitude $|\vec{r}|$ with direction cosines $l, m, n$ is:

Here $|\vec{r}| = 2\sqrt{3}$ and $l = m = n = \pm\frac{1}{\sqrt{3}}$. So:

4. Simplify. The $\sqrt{3}$ cancels: …

Method: Building a vector from equal direction cosines and a given magnitude

Use this when a vector makes equal angles with all three axes (or any specified direction cosines) and its magnitude is known.

Steps

Step 1: Convert "equal angles" into equal direction cosines.

Equal angles with the $x,y,z$ axes mean $l=m=n$. Apply the fundamental identity

Step 2: Write the unit vector, keeping the sign consistent.

$\hat{r}=\pm\frac{1}{\sqrt{3}}(\hat{i}+\hat{j}+\hat{k}).$ …

Common Mistakes

Mistake 1: Keeping only the positive sign, giving one answer.

Why it's wrong: "equally inclined" does not say "acute", so the vector pointing into the opposite octant (all negative cosines) is equally valid. Correct approach: report both $\vec{r}=2(\hat{i}+\hat{j}+\hat{k})$ and $-2(\hat{i}+\hat{j}+\hat{k})$.

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

Why it's wrong: it is the sum of squares of the direction cosines that equals $1$; the plain sum has no such property. Correct approach: set $3l^2=1$ to get $l=\pm\tfrac{1}{\sqrt3}$. …