Lesson Explainer: Geometric Applications of Vectors Mathematics

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 𝐴, 𝐡, and 𝐢

  • 𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄,
  • 𝐴𝐡+οƒŸπ΅πΆ=𝐴𝐢,
  • 𝐴𝐡=βˆ’οƒ π΅π΄, in other words, 𝐴𝐡+𝐡𝐴=0.

For any vectors ⃑𝑒 and ⃑𝑣

  • ⃑𝑒 and ⃑𝑣 are parallel when they are scalar multiples of each other, ⃑𝑒=π‘˜βƒ‘π‘£,
  • 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

𝐴𝐡𝐢𝐷 is a rectangle in which the coordinates of the points 𝐴, 𝐡, and 𝐢 are (βˆ’18,βˆ’2), (βˆ’18,βˆ’3), and (βˆ’8,π‘˜) respectively. Use vectors to find the value of π‘˜ and the coordinates of point 𝐷.

Answer

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 𝐴(βˆ’18,βˆ’2), 𝐡(βˆ’18,βˆ’3), 𝐢(βˆ’8,π‘˜), and 𝐷 in a rectangle.

This gives us rectangle 𝐴𝐡𝐢𝐷, where 𝐡 is one unit below 𝐴 and 𝐢 is adjacent to 𝐡. 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 οƒŸπ΅πΆβ«½οƒ π΄π· and 𝐴𝐡⫽𝐷𝐢. 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 οƒŸπ΅πΆ=𝐴𝐷,𝐴𝐡=𝐷𝐢.

Now, we use the given coordinates to find these vectors: 𝐴𝐡=οƒŸπ‘‚π΅βˆ’οƒ π‘‚π΄=(βˆ’18,βˆ’3)βˆ’(βˆ’18,βˆ’2)=(βˆ’18βˆ’(βˆ’18),βˆ’3βˆ’(βˆ’2))=(0,βˆ’1).

Similarly, οƒŸπ΅πΆ=οƒ π‘‚πΆβˆ’οƒŸπ‘‚π΅=(βˆ’8,π‘˜)βˆ’(βˆ’18,βˆ’3)=(βˆ’8βˆ’(βˆ’18),π‘˜βˆ’(βˆ’3))=(10,π‘˜+3).

To find the value of π‘˜, we need to use the fact that all the internal angles of a rectangle are 90∘ and the fact that 𝐴𝐡 is vertical. Since οƒŸπ΅πΆ 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 π‘˜+3=0,π‘˜=βˆ’3.

Therefore, 𝐢 has coordinates (βˆ’8,βˆ’3). We can use the fact that οƒŸπ΅πΆ=𝐴𝐷 to find the coordinates of 𝐷. Substituting π‘˜=βˆ’3 into the vector οƒŸπ΅πΆ gives οƒŸπ΅πΆ=(10,π‘˜+3)=(10,0).

This is equal to 𝐴𝐷, so 𝐴𝐷=(10,0).

Substituting this into 𝐴𝐷=οƒ π‘‚π·βˆ’οƒ π‘‚π΄ gives us (10,0)=𝐴𝐷=οƒ π‘‚π·βˆ’οƒ π‘‚π΄=οƒ π‘‚π·βˆ’(βˆ’18,βˆ’2).

Rearranging, we get 𝑂𝐷=(10,0)+(βˆ’18,βˆ’2)=(10+(βˆ’18),0+(βˆ’2))=(βˆ’8,βˆ’2).

So, 𝐷 has the coordinates (βˆ’8,βˆ’2).

Therefore, we have shown that π‘˜=βˆ’3 and 𝐷(βˆ’8,βˆ’2).

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 𝑛 such that 𝐴𝐷+𝐷𝐸=𝑛𝐴𝐢.

Answer

Let’s start by examining the vector equation 𝐴𝐷+𝐷𝐸=𝑛𝐴𝐢.

We can simplify this equation by noticing that 𝐴𝐷+𝐷𝐸=𝐴𝐸.

This means we want to find the value of 𝑛 such that 𝐴𝐸=𝑛𝐴𝐢.

We know we can find such a value of 𝑛 because 𝐴𝐸⫽𝐴𝐢. To find this value of 𝑛, we observe that οƒ πΈπ·β«½οƒŸπΆπ΅. We can then apply a useful geometric property. We have the following corresponding angles in our diagram:

π‘šβˆ π΄πΈπ·=π‘šβˆ π΄πΆπ΅,π‘šβˆ π΄π·πΈ=π‘šβˆ π΄π΅πΆ.

This gives us that triangle 𝐴𝐷𝐸 and triangle 𝐴𝐡𝐢 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 ‖‖𝐴𝐡‖‖‖‖𝐴𝐷‖‖=‖‖𝐴𝐢‖‖‖‖𝐴𝐸‖‖.

We can rearrange this equation to see ‖‖𝐴𝐸‖‖=‖‖𝐴𝐷‖‖‖‖𝐴𝐡‖‖‖‖𝐴𝐢‖‖, which means that 𝑛=‖‖𝐴𝐷‖‖‖‖𝐴𝐡‖‖.

We chose these particular sides because we know the magnitude of three of these values from the diagram: ‖‖𝐴𝐡‖‖=15,‖‖𝐴𝐷‖‖=7.5.cmcm

Substituting these into the equation, we get 𝑛=7.515=12.

Therefore, the value of 𝑛 is 12.

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 𝐴𝐡𝐢, 𝐷∈𝐡𝐢, where 𝐡𝐷∢𝐷𝐢=2∢3. Given that 3𝐴𝐡+2𝐴𝐢=π‘˜οƒ π΄π·, find the value of π‘˜.

Answer

We begin by sketching the given information, starting with triangle 𝐴𝐡𝐢.

We add a point, 𝐷, to the side 𝐡𝐢, so that 𝐡𝐷∢𝐷𝐢=2∢3.

As pictured above, we can say that ‖‖𝐡𝐷‖‖=2π‘Ÿ and ‖‖𝐷𝐢‖‖=3π‘Ÿ for some value π‘Ÿ>0.

We want to find the value of π‘˜ such that 3𝐴𝐡+2𝐴𝐢=π‘˜οƒ π΄π·. This means that we want to write an equation involving each of these vectors. We will start by finding an expression for 𝐴𝐷 by considering the following figure.

We notice that 𝐴𝐷=𝐴𝐡+𝐡𝐷, and this expression for 𝐴𝐷 is similar to the expression on the left-hand side of our equation. Substituting this expression for 𝐴𝐷 into our equation gives us 3𝐴𝐡+2𝐴𝐢=π‘˜ο€Ίοƒ π΄π΅+𝐡𝐷.

We can then simplify 3𝐴𝐡+2𝐴𝐢=π‘˜οƒ π΄π΅+π‘˜οƒ π΅π·2𝐴𝐢=(π‘˜βˆ’3)𝐴𝐡+π‘˜οƒ π΅π·οƒ π΄πΆ=(π‘˜βˆ’3)2𝐴𝐡+π‘˜2𝐡𝐷.

We now need to determine which scalar multiples of these vectors add together to give us 𝐴𝐢, and we can do this using the diagram.

First, 𝐴𝐢=𝐴𝐡+οƒŸπ΅πΆ.

We can then write οƒŸπ΅πΆ in terms of 𝐡𝐷 by noticing that οƒŸπ΅πΆ and 𝐡𝐷 have the same direction with β€–β€–οƒŸπ΅πΆβ€–β€–=5π‘Ÿ and ‖‖𝐡𝐷‖‖=2π‘Ÿ.

Remember that two vectors are equal if they have the same magnitude and direction. The size of οƒŸπ΅πΆ is 52 times the size of 𝐡𝐷. If we multiply 𝐡𝐷 by 52, the resulting vector will have the same size as οƒŸπ΅πΆ. Therefore, οƒŸπ΅πΆ=52𝐡𝐷.

We have just shown that 𝐴𝐢=𝐴𝐡+οƒŸπ΅πΆ=𝐴𝐡+52𝐡𝐷.

Then, equating these two vector expressions for 𝐴𝐢 gives us 𝐴𝐡+52𝐡𝐷=(π‘˜βˆ’3)2𝐴𝐡+π‘˜2𝐡𝐷.

Finally, we can then equate the scalar coefficients of the vectors 1=(π‘˜βˆ’3)252=π‘˜2.and

This system has one solution, which is π‘˜=5.

Example 4: Using Vectors to Find the Coordinates of a Vertex in a Square and Its Area

𝐴𝐡𝐢𝐷 is a square in which the coordinates of the points 𝐴, 𝐡, and 𝐢 are (1,βˆ’8), (3,βˆ’10), and (5,βˆ’8). Use vectors to determine the coordinates of the point 𝐷 and the area of the square.

Answer

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 𝐷.

Since this is a square, we know that the opposite sides are parallel and equal in length. Since we have the coordinates of 𝐴, 𝐡, and 𝐢, we can find the vectors 𝐡𝐴 and οƒŸπ΅πΆ. Then, we represent our square as shown.

We have 𝐡𝐴=(1,βˆ’8)βˆ’(3,βˆ’10)=(βˆ’2,2),οƒŸπ΅πΆ=(5,βˆ’8)βˆ’(3,βˆ’10)=(2,2).

There are then two methods of finding the coordinates of 𝐷. We can use the fact that 𝐡𝐴=𝐢𝐷, and we write this in terms of the initial and terminal points: 𝐡𝐴=𝐢𝐷,οƒ π‘‚π΄βˆ’οƒŸπ‘‚π΅=οƒ π‘‚π·βˆ’οƒ π‘‚πΆ.

Then we substitute in the position vectors and solve (1,βˆ’8)βˆ’(3,βˆ’10)=οƒ π‘‚π·βˆ’(5,βˆ’8),(1βˆ’3,βˆ’8+10)=οƒ π‘‚π·βˆ’(5,βˆ’8),(βˆ’2,2)+(5,βˆ’8)=𝑂𝐷,(βˆ’2+5,2βˆ’8)=𝑂𝐷𝑂𝐷=(3,βˆ’6)

Therefore, the coordinates of 𝐷 are (3,βˆ’6).

We could also see from the diagram that 𝑂𝐷=𝑂𝐴+οƒŸπ΅πΆ. This gives us 𝑂𝐷=𝑂𝐴+οƒŸπ΅πΆ=(1,βˆ’8)+(2,2)=(3,βˆ’6).

Therefore, the coordinates of 𝐷 are (3,βˆ’6).

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 𝐡𝐴: ‖‖𝐡𝐴‖‖=β€–(βˆ’2,2)β€–=(βˆ’2)+2=√8.

Then, the area of the square is given by area=‖‖𝐡𝐴‖‖=ο€»βˆš8=8.

Hence, the coordinates of 𝐷 are (3,βˆ’6), and the area of square 𝐴𝐡𝐢𝐷 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 𝐴𝐡𝐢𝐷 has vertices 𝐴(4,14), 𝐡(4,βˆ’4), 𝐢(βˆ’12,βˆ’4), and 𝐷(βˆ’12,9). Given that 𝐴𝐡⫽𝐷𝐢 and οƒ π΄π΅βŸ‚οƒŸπΆπ΅, find the area of that trapezoid.

Answer

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 π‘Ž and 𝑏 are the lengths of the parallel sides in a trapezoid, and β„Ž is the perpendicular height, then: area=ο€½π‘Ž+𝑏2ο‰β„Ž.

For our trapezoid, the lengths of the parallel sides are ‖‖𝐡𝐴‖‖ and ‖‖𝐢𝐷‖‖, and the perpendicular height is equal to β€–β€–οƒŸπΆπ΅β€–β€–.

This means that the area of our trapezoid is given by area(𝐴𝐡𝐢𝐷)=‖‖𝐡𝐴‖‖+‖‖𝐢𝐷‖‖2οβ€–β€–οƒŸπΆπ΅β€–β€–.

We can find these vectors by using the coordinates of the points given to us: 𝐡𝐴=(4,14)βˆ’(4,βˆ’4)=(0,18),𝐢𝐷=(βˆ’12,9)βˆ’(βˆ’12,βˆ’4)=(0,13),οƒŸπΆπ΅=(4,βˆ’4)βˆ’(βˆ’12,βˆ’4)=(16,0).

Then, the lengths of the sides are ‖‖𝐡𝐴‖‖=β€–(0,18)β€–=√0+18=18,‖‖𝐢𝐷‖‖=β€–(0,13)β€–=√0+13=13,β€–β€–οƒŸπΆπ΅β€–β€–=β€–(16,0)β€–=√16+0=16.

Finally, we substitute these lengths into our formula for the area of the trapezoid: area(𝐴𝐡𝐢𝐷)=‖‖𝐡𝐴‖‖+‖‖𝐢𝐷‖‖2οβ€–β€–οƒŸπΆπ΅β€–β€–=ο€Ό18+13216=248.

Hence, the area of trapezoid 𝐴𝐡𝐢𝐷 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 𝐴𝐡𝐢𝐷 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 𝐴𝐢, 𝑀. Note that 𝐴𝑀 and 𝑀𝐢 will have the same magnitude and direction, so 𝐴𝑀=𝑀𝐢.

We can then sketch the vectors 𝐡𝑀 and 𝑀𝐷. For 𝑀 to be the midpoint of 𝐡𝐷, 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

𝐴𝐷=𝐴𝑀+𝑀𝐷(1)

and that

οƒŸπ΅πΆ=𝐡𝑀+𝑀𝐢.(2)

We recall that opposite sides in a parallelogram have the same length and are parallel, so 𝐴𝐷=οƒŸπ΅πΆ.

Substituting the expressions from equations (1) and (2) into the above, we get 𝐴𝑀+𝑀𝐷=𝐡𝑀+𝑀𝐢.

𝐴𝑀 and 𝑀𝐢 are equal, so we can remove the equal vectors from each side of the equation, leading to 𝑀𝐷=𝐡𝑀.

In particular, this means their magnitudes and directions are equal, and hence 𝑀 is the midpoint of 𝐡𝐷.

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 𝐴𝐡𝐢 with midpoints of two sides labeled 𝐷 and 𝐸, as shown.

We want to show that 𝐷𝐸⫽𝐴𝐡. From the diagram, we have 𝐴𝐡=𝐴𝐷+𝐷𝐸+οƒŸπΈπ΅.

Since 𝐴𝐷 and 𝐷𝐢 have the same magnitude and direction, we have 𝐴𝐷=𝐷𝐢.

Similarly, οƒŸπΈπ΅=οƒŸπΆπΈ.

We can substitute these into our expression for 𝐴𝐡 to get 𝐴𝐡=𝐷𝐢+𝐷𝐸+οƒŸπΆπΈ.

In our diagram, we can also see 𝐷𝐸=𝐷𝐢+οƒŸπΆπΈ.

Applying this identity to our expression for 𝐴𝐡, we get 𝐴𝐡=𝐷𝐢+𝐷𝐸+οƒŸπΆπΈ=𝐷𝐢+οƒŸπΆπΈο†+𝐷𝐸=𝐷𝐸+𝐷𝐸=2𝐷𝐸.

Since 𝐷𝐸 is a scalar multiple of 𝐴𝐡, they must be parallel. Infact, by taking the magnitude of both sides of this equation, we see that ‖‖𝐴𝐡‖‖=β€–β€–2𝐷𝐸‖‖=2‖‖𝐷𝐸‖‖.

We can also see that 𝐴𝐡 is twice as long as 𝐷𝐸.

Let us finish by recapping some of the most important things to take away from this explainer.

Key Points

  • We can convert problems involving geometric shapes into ones involving vectors.
  • For any points 𝐴, 𝐡, and 𝐢𝐴𝐡+οƒŸπ΅πΆ=𝐴𝐢.
  • If ⃑𝑒 and ⃑𝑣 are parallel, then there exists a scalar π‘˜, such that ⃑𝑒=π‘˜βƒ‘π‘£. In particular, for any nonzero parallel vectors ⃑𝑒 and ⃑𝑣, ⃑𝑒=±‖‖⃑𝑒‖‖‖‖⃑𝑣‖‖⃑𝑣, where the sign is positive if the directions of ⃑𝑒 and ⃑𝑣 are equal and the sign is negative if their directions are opposite.
  • We can prove many geometric relationships from the properties of vectors.

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