Direction of a Vector Formula with Examples

Key Points

• The direction of a vector formula is $\theta = \arctan\left(\frac{y}{x}\right)$, giving the angle the vector makes with the positive $x$-axis.

• The arctangent returns values in only two quadrants, so vectors pointing left require adding 180 degrees to the raw result.

• The order is $y$ over $x$, the vertical component divided by the horizontal one, and reversing it gives the complementary angle instead.

• In three dimensions, direction is given by three direction cosines, each a component divided by the vector's magnitude.

• A vector's direction is a property of that single vector, which is a different question from the angle between two vectors.

What Is the Direction of a Vector Formula?

The direction of a vector formula gives the angle a vector makes with the positive $x$-axis, from its components:

$$\theta = \arctan\left(\frac{y}{x}\right)$$

The arctangent is the inverse tangent function. NIST's Digital Library of Mathematical Functions denotes its principal branch $\arctan z$ and tabulates its principal values, giving $\arctan(1) = \frac{\pi}{4}$ and $\arctan(\pm\infty) = \pm\frac{\pi}{2}$. Those principal values cover only the right half of the plane, which is why a quadrant check is part of the method.

Why the Quadrant Adjustment Is Needed

Dividing $y$ by $x$ loses information. The vector $(3, 4)$ and the vector $(-3, -4)$ give the same quotient, $\frac{4}{3}$, because two negatives cancel, yet they point in exactly opposite directions.

The arctangent cannot recover what the division discarded. Its principal values run only from $-90$ to $90$ degrees, which covers vectors pointing right but nothing pointing left.

So the rule is to compute the raw arctangent, then look at the signs of the components. If $x$ is negative the vector points left, and 180 degrees must be added to bring the angle into the correct quadrant.

Variable Key

Examples of the Direction of a Vector Formula

Example 1

Find the direction of the vector $(3, 4)$.

$$\theta = \arctan\left(\frac{4}{3}\right) \approx 53.13^\circ$$

Final answer: about $53.13^\circ$. Both components are positive, so the vector lies in the first quadrant and no adjustment is needed.

Example 2

Find the direction of the vector $(-3, -4)$.

The quotient is $\frac{-4}{-3} = \frac{4}{3}$, so the calculator returns

$$\arctan\left(\frac{4}{3}\right) \approx 53.13^\circ$$

But $53.13^\circ$ points up and to the right, into the first quadrant, while this vector has both components negative and so points down and to the left. The raw value cannot be the answer.

Both components being negative places the vector in the third quadrant, so add 180 degrees:

$$\theta \approx 53.13 + 180 = 233.13^\circ$$

Final answer: about $233.13^\circ$. Sketching the components before trusting a calculator catches this every time.

Example 3

Find the direction of the vector $(-5, 12)$.

$$\arctan\left(\frac{12}{-5}\right) \approx -67.38^\circ$$

Here $x$ is negative and $y$ positive, placing the vector in the second quadrant, so add 180:

$$\theta \approx -67.38 + 180 = 112.62^\circ$$

Final answer: about $112.62^\circ$

Example 4

Find the direction of the vector $(0, 7)$.

The formula would divide by zero, so read the geometry instead. The vector has no horizontal component and a positive vertical one, so it points straight up.

Final answer: exactly $90^\circ$. This matches the DLMF value $\arctan(\infty) = \frac{\pi}{2}$, approached as $x$ tends to zero from the positive side.

Example 5

Find the direction cosines of the three-dimensional vector $(2, 3, 6)$.

First the magnitude, then each component divided by it:

$$\lvert v \rvert = \sqrt{4+9+36} = \sqrt{49} = 7$$ $$\cos\alpha = \frac{2}{7}, \quad \cos\beta = \frac{3}{7}, \quad \cos\gamma = \frac{6}{7}$$

Final answer: $\frac{2}{7}$, $\frac{3}{7}$ and $\frac{6}{7}$.

A useful check: the squares of the three direction cosines sum to 1, since $\frac{4+9+36}{49} = 1$.

Don't Confuse Direction With These Three

The third row answers a different question entirely. A single vector has a direction; a pair of vectors has an angle between them, and that is found from the dot product rather than from an arctangent.

Where Do Students Trip Up on the Direction of a Vector?

Mistake 1: Trusting the raw arctangent

Where it slips in: A calculator returns a number and the number looks like an angle.

Don't do this: Report $\arctan\left(\frac{-4}{-3}\right) = 53.13^\circ$ for the vector $(-3,-4)$.

The correct way: Check the signs of $x$ and $y$, and add 180 degrees whenever $x$ is negative. The arctangent range covers only half the plane.

Students who sketch the two components on axes before computing anything get the quadrant right every time, because the sketch shows which way the vector actually points. The calculator cannot see the sketch, which is precisely the problem.

Mistake 2: Dividing $x$ by $y$

Where it slips in: The components are usually written $x$ first, so they get divided in that order.

Don't do this: $\theta = \arctan\left(\frac{x}{y}\right)$

The correct way: $\theta = \arctan\left(\frac{y}{x}\right)$. Reversing the fraction returns the complementary angle, so $53.13^\circ$ becomes $36.87^\circ$ and the two sum to 90.

Mistake 3: Confusing direction with the angle between two vectors

Where it slips in: Both are angles involving vectors.

Don't do this: Use an arctangent of components to find the angle separating two vectors.

The correct way: A direction angle belongs to one vector and is measured from the axis. The angle between two vectors comes from their dot product, which is a different calculation.

Bottom Line

• The direction of a vector formula is $\theta = \arctan\left(\frac{y}{x}\right)$, measured from the positive $x$-axis.

• The division loses the component signs, so the quadrant must be settled separately, adding 180 degrees when $x$ is negative.

• Reversing the fraction to $\frac{x}{y}$ returns the complementary angle rather than the direction.

• In three dimensions, direction becomes three direction cosines whose squares sum to 1.

• To work through vectors with a teacher, explore a trigonometry tutor or math classes online.

Where to Go From Here

Find the direction of $(5, -5)$ and of $(-2, 2)$, taking care with the quadrants. Then find the direction cosines of $(1, 2, 2)$ and check that their squares sum to 1.

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Read More

• Components of a Vector

• Angle Between Vectors

• Arctan

• Cartesian Coordinate System

• Vector Equation