Q.Show that the vector $\hat{i} + \hat{j} + \hat{k}$ is equally inclined to the axes OX, OY and OZ.

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

For a vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, the direction cosines are:

Here $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, so $a_1 = a_2 = a_3 = 1$. Its magnitude is:

Thus the direction cosines are:

$l = m = n = \frac{1}{\sqrt{3}}$ …

The idea is simple: a vector is "equally inclined" to the three axes if its direction cosines — the cosines of the angles it makes with the positive x, y, and z axes — are all equal. For any vector, these cosines are just the components divided by the magnitude. So the question reduces to checking whether the three components of $\hat{i} + \hat{j} + \hat{k}$ are equal in magnitude relative to the vector's length.

Let's walk through it.

1. Write the vector in component form.

   The given vector is $\vec{v} = \hat{i} + \hat{j} + \hat{k}$. In component notation, that's $(1, 1, 1)$. Each component is 1.

2. Find the magnitude.

   The magnitude (or length) of $\vec{v}$ is

3. Recall what direction cosines are. If a vector $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$ makes angles $\alpha$, $\beta$, $\gamma$ with the OX, OY, OZ axes respectively, then

These three numbers are the direction cosines.

4. Apply to our vector. Here $a = b = c = 1$ and $|\vec{v}| = \sqrt{3}$. So

All three direction cosines are identical. That means $\alpha = \beta = \gamma$ (since the cosine function is one-to-one on $[0, \pi]$, the range of angles between a vector and an axis).

5. State the common angle. The angle each axis makes with the vector is $\theta = \cos^{-1}\left(\frac{1}{\sqrt{3}}\right).$ …

Method: Proving a Vector is Equally Inclined to the Axes

This is a reasoning task: "equally inclined to OX, OY, OZ" means the vector makes the same angle with each axis. The strategy is to show its three direction cosines are equal.

Steps

Step 1: Translate the condition into direction cosines.

The angles $\alpha,\beta,\gamma$ with the axes are equal $\iff \cos\alpha=\cos\beta=\cos\gamma$, i.e. $l=m=n$ (cosine is one-to-one on $[0,\pi]$, so equal cosines force equal angles).

Step 2: Compute the direction cosines.

Step 3: Show they are equal. …

Common Mistakes

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

Why it's wrong: only the sum of squares of direction cosines equals $1$; the plain sum does not. Correct approach: verify with $l^2+m^2+n^2=1$.

Mistake 2: Reporting the inclination as $\tfrac{1}{\sqrt3}$.

Why it's wrong: $\tfrac{1}{\sqrt3}$ is the cosine of the angle, not the angle. Correct approach: the angle is $\cos^{-1}\!\left(\tfrac{1}{\sqrt3}\right)\approx54.7^\circ$. …