Q.If a unit vector $\vec{a}$ makes angles $\frac{\pi}{3}$ with $\hat{i}$, $\frac{\pi}{4}$ with $\hat{j}$ and an acute angle $\theta$ with $\hat{k}$, then find $\theta$ and hence, the components of $\vec{a}$.

Concept: Direction Vectors — the cosines of the angles a unit vector makes with the coordinate axes are its components, and their squares sum to 1.

Let $\vec{a} = a_1 \hat{i} + a_2 \hat{j} + a_3 \hat{k}$. Since $\vec{a}$ is a unit vector, $a_1^2 + a_2^2 + a_3^2 = 1$.

The angle with $\hat{i}$ is $\frac{\pi}{3}$, so $a_1 = \cos\frac{\pi}{3} = \frac12$.

The angle with $\hat{j}$ is $\frac{\pi}{4}$, so $a_2 = \cos\frac{\pi}{4} = \frac{1}{\sqrt{2}}$.

Substitute into the unit vector condition:

Since $\theta$ is acute, $a_3 = \cos\theta > 0$, so $a_3 = \frac12$. Hence $\theta = \cos^{-1}\!\left(\frac12\right) = \frac{\pi}{3}$.

The key idea here is direction cosines. For any unit vector in 3D space, the cosines of the angles it makes with the $x$, $y$, and $z$ axes are exactly its components. That is, if a unit vector $\vec{a}$ makes angles $\alpha, \beta, \gamma$ with $\hat{i}, \hat{j}, \hat{k}$ respectively, then:

And because it's a unit vector, the sum of squares of these cosines must equal 1:

This is the fundamental relation we'll use.

1. Write what's given.

   $\alpha = \frac{\pi}{3}$, so $\cos\alpha = \cos\frac{\pi}{3} = \frac12$.

   $\beta = \frac{\pi}{4}$, so $\cos\beta = \cos\frac{\pi}{4} = \frac{1}{\sqrt2}$.

   $\gamma = \theta$, which is acute (so $\cos\theta > 0$).

2. Apply the direction cosine relation.

3. Solve for $\cos^2\theta$.

4. Determine $\theta$. Since $\theta$ is acute, $\cos\theta > 0$, so $\cos\theta = \frac12$. Therefore $\theta = \frac{\pi}{3}$.

5. Write the components of $\vec{a}$. The components are just the direction cosines:

Method: The Direction-Cosine Identity $l^2+m^2+n^2=1$

Use this whenever a unit vector's angles with the coordinate axes are (partly) known.

Steps

Step 1: Write each component as the cosine of its axis angle.

For a unit vector, the components ARE the direction cosines: $a_1=\cos\alpha$, $a_2=\cos\beta$, $a_3=\cos\gamma$.

Step 2: Apply the identity.

Substitute the known cosines and solve for the unknown $\cos^2$ term.

Step 3: Choose the sign from the stated angle condition.

The square root gives two signs; use the problem's condition (e.g. 'acute angle' $\Rightarrow$ positive cosine) to pick the correct one, then read off the components.

Common Mistakes

Mistake 1: Using the angles themselves instead of their cosines.

Why it's wrong: the identity is $\cos^2\alpha+\cos^2\beta+\cos^2\gamma=1$, not a sum of the angles. Correct approach: convert each angle to its cosine first.

Mistake 2: Taking the negative cosine despite an acute angle.

Why it's wrong: $\cos^2\theta=\frac14$ gives $\cos\theta=\pm\frac12$, but 'acute' forces the positive root, so $\theta=\frac{\pi}{3}$, not $\frac{2\pi}{3}$. Correct approach: use the stated acute condition to pick $+\frac12$.

Mistake 3: Writing $l+m+n=1$.

Why it's wrong: only the sum of squares equals $1$. Correct approach: apply $l^2+m^2+n^2=1$.