In this explainer, we will learn how to find the scalar projection of a vector onto another vector.
Vectors are quantities that have both a magnitude and a direction. In this explainer, we will explore the concept of projecting a vector in the direction of another vector.
In order to mathematically describe our projections, we’ll need to use the dot product. This is somewhat of a vector analog to multiplication. Let us recap its definition for two vectors and :
Definition: The Dot Product
Given that the angle between and is , the dot product is defined as
An equivalent definition of the dot product is
On its own, the dot product does not have a particularly useful geometric representation; however, it becomes very useful when dealing with scalar projection.
As the name suggests, performing a scalar projection of a vector in the direction of will result in a scalar. This scalar describes the component of vector in the direction of .
Performing a vector projection has a very similar interpretation. The result of this process is itself a vector, which maintains the property of directionality parallel to vector .
For the purpose of this explainer, we will be solely dealing with: scalar projection, which is is defined below.
Definition: Scalar Projection
Given that the angle between and is , the scalar projection of in the direction of is given by
An equivalent statement can be formed using the dot product
The scalar projection tells us the component of a vector that points in the direction of another vector, .
We may have already seen this in action, since the component form of a vector can be viewed as an application of scalar projection. We can see this by recalling that the unit vector in the direction of is defined by
This means that an alternative definition of our scalar projection is
Since , , and are unit vectors in the -, -, and -directions, respectively, we could find the component of in one of cardinal directions by taking the dot product with these unit vectors. For example, the component of in the -direction can be found by taking .
To get an intuitive understanding of scalar projection, it may be helpful to remember right triangle geometry.
Consider the case where angle between two vectors is less than . In the diagram above, can be thought of as the length of the hypotenuse in a right triangle and our scalar projection as the length of the adjacent side to angle .
We will see an example later that scalar projection extends beyond right triangle geometry, since angle can be larger than , yet the relationship still holds.
A few specific cases worth noting are when our two vectors are parallel or perpendicular. If the vectors are parallel and point in the same direction, then the angle between them will be . If they are parallel and point in opposite directions, then the angle between them will be . Finally, if our vectors are perpendicular, the angle between them will be . These cases are given below:
It should make sense that the component of any vector pointing in the same parallel direction is simply equal to the magnitude of the vector. For vectors pointing in the opposite parallel direction, we see a negative magnitude that we’ll return to later in the explainer. Similarly, there is no component of a vector that points in the perpendicular direction, hence the scalar projection in this direction is equal to 0.
To improve our understanding, let’s take a look at some examples of scalar multiplication.
Example 1: Finding the Scalar Projection of a Vector given the Vector Magnitudes and the Angle between Them
If , , and the measure of the angle between them is , find the algebraic projection of in the direction of .
To solve this question, we should first recognize that the algebraic projection of one vector in some given direction is another way of describing a scalar projection.
The question has helpfully given us the angle between the two vectors, and , which we can represent as .
An important point to note is that we are not finding the projection of vector in the direction of , but rather we are projecting in the direction of . We should therefore use the formula
To solve this question, we can substitute in the values and :
Using the fact that is one of the exact trigonometric ratios gives us a direct route to our answer, which we can leave in surd form.
We might observe that is an unnecessary piece of information in this question. Since the question gives us angle , we do not need to use and it has only been included to trip up an unwary student!
It should also be noted that, by using the dot product, we can perform scalar multiplication without having to find the angle, , between the vectors:
This can be useful in cases where the vectors are simply given in component form.
Example 2: Determining the Magnitude of the Component of a Vector in a Given Direction
Determine, to the nearest hundredth, the component of vector along , given that and the coordinates of and are and respectively.
For this question, we are asked to find the projection of vector in the direction of , which can be denoted as .
The question has not given us vector but rather the coordinates of its start point and endpoint. Our first step will be to find the vector as follows:
We’ll also find the magnitude of our vector as this will be needed in a moment:
Now that we have our two vectors and the magnitude of , we can use the dot product definition of scalar projection to move forward:
Remember that we can find the dot product of two vectors using the components of the vectors:
Substituting in our values to the equation for our vector projection gives
As required by the question, our answer has been simplified and expressed in decimal form to the nearest hundredth (two decimal places).
Scalar projection can also be used in examples related to geometric systems. Let’s take a look at one such example.
Example 3: Finding the Scalar Projection Where Both Vectors are Represented as the Diagonals of a Cube
The cube shown has sides lengths of . Find the scalar projection of onto , giving your answer correct to two decimal places.
For this question, we are asked to find the scalar projection of one vector onto another. These two vectors are the diagonals of a cube.
Our intuition may lead us to believe that the diagonals of a cube are perpendicular to each other, and hence the result should be 0. This would be a mistake! In fact, the angle between the diagonals is not equal to and hence the scalar projection is nonzero. Let’s verify this with some calculations.
The question does not give us either of the two vectors, and instead we have been given the side length of the cube. We can use this information to find the vectors.
Our starting point will be to find the coordinates of point . Since is the origin, this will immediately give us vector .
Given that point lies at the origin and point is the opposite corner of the cube, it makes sense that the distances between these two points along the -, -, and -directions are all equal in magnitude. From our diagram we can also see that the magnitude of the separation is equal to the side length of our cube, and so we conclude the following:
Using similar logic, we might observe that point lies vertically above the origin, separated only in the -direction by the length of one side of the cube, while point is separated from the origin in the - and -directions, but not the -direction:
This gives us a way to find vector :
To find the required scalar projection, we can use the following formula:
To proceed, we’ll need the magnitude of vector :
We can substitute this value into our formula and expand the dot product as follows:
After eliminating the common factors of 44 and 17 on the top and bottom of our fraction, we are left with an irrational denominator. Although we could rationalize the denominator, instead we recall that our question asks for an answer to two decimal places:
It should be noted that in geometric examples such as the above, it is a perfectly valid (and sometimes more efficient) approach to find the angle between the vectors.
For our cube, however, since the angle between the diagonals is not commonly known, we chose to use a different approach with the components of the vectors. Finding the angle between these vectors would involve many of the same tools (such as the dot product).
As a point of interest, had we chosen the angle method, we would have found that rather than the that some might assume.
As a final note in this explainer, observe that our answer for the previous example is negative. We can see why by considering the following diagrams.
With the first few examples in this explainer, we observed that the angle between our vectors, , was less than . In these cases, vector will have a positive component in the direction of .
Had we used the supplementary angle (marked in the diagrams), this would have resulted in a sign error in our answer since .
The reason for this is that we would have actually been calculating the scalar projection of onto a separate vector , which points in the opposite direction to .
The geometric interpretation of a negative result to a scalar projection is that our original vector, , has a component pointing in the opposite direction to vector .
It therefore follows that scalar projection gives us positive answers when the angle between the vectors is less than and negative answers when the angle between the vectors is greater than .
- We can find the component of some vector in the direction of another vector using scalar projection.
- Given that is the angle between and , the scalar projection of in the direction of can be expressed as
- If and are parallel and point in the same direction, then and
- If and are parallel and point in opposite directions, then and
- If and are perpendicular, then and
- If the angle between the vectors is less than , then vector has a component in the same direction as and hence
- If the angle between the vectors is greater than , then vector has a component in the opposite direction to and hence