Q.The direction cosines of the vector $(2\hat{i} + 2\hat{j} - \hat{k})$ are __________.

Concept: Direction Vectors — the direction cosines are the cosines of the angles a vector makes with the coordinate axes, found by dividing each component by the vector’s magnitude.

Step 1: Find the magnitude of $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$:

$|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = 3.$ …

Why Direction Cosines?

A vector in 3D space points in some direction. Direction cosines give us a precise, unitless way to describe that direction. They are simply the components of the unit vector in the direction of the original vector. If you know the direction cosines, you know exactly which way the vector points, regardless of its length.

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

• $l = \cos \alpha = \frac{a_1}{|\vec{a}|}$
• $m = \cos \beta = \frac{a_2}{|\vec{a}|}$
• $n = \cos \gamma = \frac{a_3}{|\vec{a}|}$

where $\alpha, \beta, \gamma$ are the angles the vector makes with the $x, y, z$ axes respectively.

Step-by-step solution

1. Identify the components.

The vector is $\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}$. So:

• $a_1 = 2$
• $a_2 = 2$
• $a_3 = -1$

2. Find the magnitude of the vector.

The magnitude (or length) is:

3. Compute each direction cosine.

Divide each component by the magnitude:

• $l = \frac{a_1}{|\vec{a}|} = \frac{2}{3}$ …

Method: Direction cosines of a vector

Use this to convert any vector into its direction cosines — the cosines of the angles it makes with the three axes.

Steps

Step 1: Find the magnitude.

For $\vec a=a_1\hat i+a_2\hat j+a_3\hat k$, compute $|\vec a|=\sqrt{a_1^2+a_2^2+a_3^2}$.

Step 2: Divide each component by the magnitude.

$l=\frac{a_1}{|\vec a|},\quad m=\frac{a_2}{|\vec a|},\quad n=\frac{a_3}{|\vec a|}.$ …

Common Mistakes

Mistake 1: Giving the components $(2,2,-1)$ as the direction cosines.

Why it's wrong: those are direction ratios; direction cosines need the unit vector. Correct approach: divide by $|\vec a|=3$ to get $\left(\tfrac23,\tfrac23,-\tfrac13\right)$.

Mistake 2: Dropping the negative sign on the $\hat k$ component. …