Q.A vector $\vec{r}$ has magnitude 14 and direction ratios 2, 3, -6. Find the direction cosines and components of $\vec{r}$, given that $\vec{r}$ makes an acute angle with x-axis.

Concept: Direction Cosines Properties — direction cosines are the unit vector components, obtained by normalising direction ratios.

Step 1: Normalise the direction ratios.

The magnitude of the direction ratios is

$\sqrt{2^2 + 3^2 + (-6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.

Step 2: Write direction cosines.

Since $\vec{r}$ makes an acute angle with the x‑axis, the x‑direction cosine must be positive.

$l = \frac{2}{7},\quad m = \frac{3}{7},\quad n = \frac{-6}{7}$.

Step 3: Find components of $\vec{r}$. …

The key idea here is the relationship between direction ratios and direction cosines. Direction ratios are any three numbers proportional to the direction cosines. If a vector has direction ratios $a, b, c$, then its direction cosines are:

Why? Because direction cosines are the actual cosines of the angles the vector makes with the axes — they must satisfy $l^2+m^2+n^2=1$. Direction ratios are just a scaled version, so we normalise them.

Now, once we have the direction cosines, the components of $\vec{r}$ are simply:

The only twist here is the condition "makes an acute angle with the x-axis". That tells us the x-component is positive — which resolves any sign ambiguity.

Step-by-step solution

1. Find the magnitude of the direction ratios. The direction ratios are $2, 3, -6$. Their magnitude is:

2. Write the direction cosines. Divide each direction ratio by 7:

Check: $\left(\frac{2}{7}\right)^2 + \left(\frac{3}{7}\right)^2 + \left(-\frac{6}{7}\right)^2 = \frac{4+9+36}{49} = 1$. Good.

3. Interpret the acute-angle condition. The angle $\alpha$ with the x-axis satisfies $\cos\alpha = l$. An acute angle means $\cos\alpha > 0$, so $l > 0$. Here $l = \frac{2}{7} > 0$, so the sign is already correct. …

Method: From direction ratios to direction cosines to components

Use this when a vector's direction ratios (or any set of proportional numbers) and its magnitude are given, and you need the direction cosines and/or the actual components.

Steps

Step 1: Normalise the direction ratios.

Direction ratios $a,b,c$ are only proportional to the direction cosines, not equal to them. Divide each by their root-sum-of-squares:

Check $l^2+m^2+n^2=1$.

Step 2: Fix the overall sign using the stated angle condition. …

Common Mistakes

Mistake 1: Treating the direction ratios $2,3,-6$ as the direction cosines.

Why it's wrong: direction ratios are only proportional to direction cosines; you must divide by $\sqrt{2^2+3^2+(-6)^2}=7$ first. Correct approach: normalise to get $\left(\tfrac27,\tfrac37,-\tfrac67\right)$.

Mistake 2: Ignoring the "acute angle with the $x$-axis" condition.

Why it's wrong: a line has two opposite direction-cosine sets; the acute condition forces $l>0$ to pick the correct one. Correct approach: check $l=\tfrac27>0$ (already correct here) and flip all three signs together if it were negative. …