In this explainer, we will learn how to use vector operations and vector properties to solve problems involving geometrical shapes.
Before we begin discussing the applications of vectors to geometric problems, let’s start by reviewing some of the important properties that vectors have.
Theorem: Properties of Vectors
For any points , , and
- , in other words, .
For any vectors and
- and are parallel when they are scalar multiples of each other,
- two vectors are equal if they have the same magnitude and direction.
Finally, we can work geometrically with our vectors, or we can work with their components algebraically. Sometimes one method will be easier than the other, so we should consider both options for each problem.
Let’s see some examples of geometric problems that we can solve using the properties of vectors.
Example 1: Using Vectors to Find the Coordinates of a Vertex in a Rectangle
is a rectangle in which the coordinates of the points , , and are , , and respectively. Use vectors to find the value of and the coordinates of point .
Since this question specifies that we should use vectors, we will begin by converting this problem into one involving vectors. Whenever we do this, it is generally a good idea to sketch the given information. We start with points , , , and in a rectangle.
This gives us rectangle , where is one unit below and is adjacent to . To use vectors to answer this question, let’s start by replacing the sides of the rectangle with the vectors between the adjacent vertices of our rectangle.
We see that and . We can also see that the opposing sides have the same magnitude since they are opposite sides of a rectangle. Since they have the same magnitude and direction, we must have
Now, we use the given coordinates to find these vectors:
To find the value of , we need to use the fact that all the internal angles of a rectangle are and the fact that is vertical. Since is perpendicular to a vertical vector, it must be horizontal, in other words its vertical component must be 0. Setting the vertical component equal to 0 gives us
Therefore, has coordinates . We can use the fact that to find the coordinates of . Substituting into the vector gives
This is equal to , so
Substituting this into gives us
Rearranging, we get
So, has the coordinates .
Therefore, we have shown that and .
In our previous example, we used the properties of vectors along with our knowledge of rectangles to solve the problem. Let’s now see an example that involves more complicated geometric rules alongside the properties of vectors.
Example 2: Finding the Scalar That Satisfies a Given Operation on Vectors Represented in a Figure
Given the information in the diagram below, find the value of such that .
Let’s start by examining the vector equation
We can simplify this equation by noticing that .
This means we want to find the value of such that
We know we can find such a value of because . To find this value of , we observe that . We can then apply a useful geometric property. We have the following corresponding angles in our diagram:
This gives us that triangle and triangle have the same internal angles. In other words, these two triangles are similar. In particular, we can use the fact that the ratio of the lengths of corresponding sides is the same in similar triangles.
This means that
We can rearrange this equation to see which means that
We chose these particular sides because we know the magnitude of three of these values from the diagram:
Substituting these into the equation, we get
Therefore, the value of is .
When we are given a vector equation in the context of a geometric problem, we should sketch the given information to help us solve the problem. Let’s see a few examples to become more familiar with the context.
Example 3: Finding a Missing Value Using Vectors
In the triangle , , where . Given that , find the value of .
We begin by sketching the given information, starting with triangle .
We add a point, , to the side , so that .
As pictured above, we can say that and for some value .
We want to find the value of such that . This means that we want to write an equation involving each of these vectors. We will start by finding an expression for by considering the following figure.
We notice that , and this expression for is similar to the expression on the left-hand side of our equation. Substituting this expression for into our equation gives us
We can then simplify
We now need to determine which scalar multiples of these vectors add together to give us , and we can do this using the diagram.
We can then write in terms of by noticing that and have the same direction with and .
Remember that two vectors are equal if they have the same magnitude and direction. The size of is times the size of . If we multiply by , the resulting vector will have the same size as . Therefore,
We have just shown that
Then, equating these two vector expressions for gives us
Finally, we can then equate the scalar coefficients of the vectors
This system has one solution, which is .
Example 4: Using Vectors to Find the Coordinates of a Vertex in a Square and Its Area
is a square in which the coordinates of the points , , and are , , and . Use vectors to determine the coordinates of the point and the area of the square.
We want to use vectors to find the coordinates of the missing point in a square. Let’s start by sketching the given points and point .
Since this is a square, we know that the opposite sides are parallel and equal in length. Since we have the coordinates of , , and , we can find the vectors and . Then, we represent our square as shown.
There are then two methods of finding the coordinates of . We can use the fact that , and we write this in terms of the initial and terminal points:
Then we substitute in the position vectors and solve
Therefore, the coordinates of are .
We could also see from the diagram that . This gives us
Therefore, the coordinates of are .
Finally, to find the area of our square, we need to square the length of one side. We can find the side length by calculating the magnitude of vector :
Then, the area of the square is given by
Hence, the coordinates of are , and the area of square is 8 square units.
Let’s now see an example involving the area of a trapezoid using vectors.
Example 5: Using Vectors to Find the Area of a Right Trapezoid
Trapezoid has vertices , , , and . Given that and , find the area of that trapezoid.
We want to find the area of a trapezoid and we are given the coordinates of the vertices, the parallel sides, and the perpendicular side. We start by sketching this information.
To find the area of this trapezoid, we remember the formula for the area of a trapezoid. If and are the lengths of the parallel sides in a trapezoid, and is the perpendicular height, then:
For our trapezoid, the lengths of the parallel sides are and , and the perpendicular height is equal to .
This means that the area of our trapezoid is given by
We can find these vectors by using the coordinates of the points given to us:
Then, the lengths of the sides are
Finally, we substitute these lengths into our formula for the area of the trapezoid:
Hence, the area of trapezoid is 248 area units.
In the previous examples, we have shown many geometric properties of given shapes using vectors. It is also possible to show geometric properties of shapes in general using vectors.
For example, suppose that we have a parallelogram with diagonals, as shown.
We can show, by using vectors, that these diagonals bisect each other. To do this, we will call the midpoint of , . Note that and will have the same magnitude and direction, so
We can then sketch the vectors and . For to be the midpoint of , we need to show these two vectors are equal. We add vectors to the diagram, as shown below.
We can see from the diagram that
We recall that opposite sides in a parallelogram have the same length and are parallel, so
Substituting the expressions from equations (1) and (2) into the above, we get
and are equal, so we can remove the equal vectors from each side of the equation, leading to
In particular, this means their magnitudes and directions are equal, and hence is the midpoint of .
Another example of a geometric property we can prove using vectors is the statement that the line segment connecting the midpoints of two sides of a triangle is parallel to the third side. Consider the triangle with midpoints of two sides labeled and , as shown.
We want to show that . From the diagram, we have
Since and have the same magnitude and direction, we have
We can substitute these into our expression for to get
In our diagram, we can also see
Applying this identity to our expression for , we get
Since is a scalar multiple of , they must be parallel. Infact, by taking the magnitude of both sides of this equation, we see that
We can also see that is twice as long as .
Let us finish by recapping some of the most important things to take away from this explainer.
- We can convert problems involving geometric shapes into ones involving vectors.
- For any points , , and
- If and are parallel, then there exists a scalar , such that In particular, for any nonzero parallel vectors and , where the sign is positive if the directions of and are equal and the sign is negative if their directions are opposite.
- We can prove many geometric relationships from the properties of vectors.