Lesson Explainer: Magnitude of a 2D Vector | Nagwa Lesson Explainer: Magnitude of a 2D Vector | Nagwa

Lesson Explainer: Magnitude of a 2D Vector Mathematics • First Year of Secondary School

In this explainer, we will learn how to determine the magnitude of two-dimensional vectors.

The absolute value of a real number tells us the size of the number, or the distance between zero and the number on a number line. The magnitude of a vector is an analogue of the absolute value for real numbers; hence, the notation for the magnitude inherits the use of the absolute value.

Definition: Magnitude of a Vector

The magnitude of a vector 𝑣, denoted 𝑣, is the length of the vector, or the distance between the initial and terminal points of a vector.

In particular, a unit vector is a vector whose magnitude is equal to 1. For instance, we recall the fundamental unit vectors in two dimensions 𝑖=(1,0),𝑗=(0,1).

These unit vectors have magnitude 1, because if we measure their lengths drawn on a 2D plane, they are equal to 1 unit.

Another special case is a vector whose magnitude is equal to zero. There is only one vector whose length is equal to zero, and we call this vector the zero vector: 0=(0,0).

Let us begin by considering a simple example for the magnitude of a vector.

Example 1: The Magnitude of a Vertical Vector

The vector 𝑣 is shown on the grid of unit squares below. Find the value of 𝑣.

Answer

Recall that the notation 𝑣 represents the magnitude of the vector, which is the length of the vector. From the given diagram, we can see that the given vector 𝑣 spans two vertical grid lengths. Since the grids are unit squares, one side of the smallest square has a length of 1 unit. This tells us that the length of 𝑣 is 2 length units.

Because vector 𝑣 points downward, it may be tempting to conclude that the sign of the magnitude is negative. However, we should remember that a length, hence the magnitude of a vector, cannot be negative. Thus, 𝑣=2.

In the first example, we found the magnitude of a vertical vector from its graph. The magnitude of a vertical or a horizontal vector is simple to find when we are given the graph. If a vector is not vertical or horizontal, we can find its magnitude by applying the Pythagorean theorem, as we will see in the next example.

Example 2: The Magnitude of a Vector

The vector 𝑣 is shown on the grid of unit squares below. Find the value of 𝑣.

Answer

Recall that the notation 𝑣 represents the magnitude of the vector, which is the length of the vector. We can see that vector 𝑣 lies diagonally over multiple squares. We can form a right triangle so that the vector forms the hypotenuse of the right triangle. Since the grids are unit squares, one side of the smallest square has a length of 1 unit, which tells us the lengths of the two sides of the right triangle, as shown below.

Let us apply the Pythagorean theorem to find the length of this hypotenuse, which will be the magnitude of vector 𝑣. The lengths of the two non-hypotenuse sides of the right triangle are 3 and 4, and the length of the hypotenuse is equal to the magnitude 𝑣. Hence, 3+4=𝑣.

This means 𝑣=25.

Taking the square root of both sides of this equation leads to 𝑣=±5. Since the magnitude 𝑣 is a length, it cannot be negative, so we can ignore the negative answer. Hence, 𝑣=5.

In the previous example, we computed the magnitude of a vector by applying the Pythagorean theorem. We can use the same method to find the magnitude of any two-dimensional vector, as long as it is not vertical or horizontal.

Consider a vector 𝑣=(𝑎,𝑏), where both 𝑎 and 𝑏 are nonzero. Similarly to the previous example, we can draw a right triangle where the magnitude of this vector gives the hypotenuse of this right triangle.

Hence, the application of the Pythagorean theorem gives us 𝑎+𝑏=𝑣.

Taking the square root of both sides of this equation and ignoring the negative solution leads to the following formula.

Definition: Magnitude of a 2D Vector

Let 𝑣=(𝑎,𝑏) be a vector in two dimensions. Then, the magnitude of this vector is given by 𝑣=𝑎+𝑏.

While we have only shown this formula for vectors lying in the first quadrant, the formula holds for any 2D vector. For instance, we can see that if the vector 𝑣=(𝑎,𝑏) lies in the third quadrant, we have the following diagram:

In the figure, we can see that the lengths of the two non-hypotenuse sides of the right triangles are given by the absolute values |𝑎| and |𝑏|. Applying the Pythagorean theorem and taking the positive square root leads to 𝑣=|𝑎|+|𝑏|.

But we know that for any real number 𝑎, |𝑎|=𝑎. This means that the formula for the magnitude of this vector agrees with the formula given above for general 2D vectors.

We can also verify this formula for vertical and horizontal vectors that have the form (𝑎,0) or (0,𝑏). In this case, as we saw in the first example, the length of the 𝑥- or 𝑦-coordinate determines the magnitude of the vector. This means that the magnitudes of these vectors are given by |𝑎| or |𝑏|. Since either the 𝑦- or 𝑥-coordinate of these vectors is zero, this agrees with the formula for the magnitude of 2D vectors given above.

In the next example, we will apply this formula to find the magnitude of a vector in component form.

Example 3: The Magnitude of a Vector

What is the magnitude of the vector (4,5)?

Answer

In this example, we need to find the magnitude of a vector given in component form. Recall that the magnitude of a vector 𝑣=(𝑎,𝑏) is given by 𝑣=𝑎+𝑏.

The given vector is (4,5), so we can substitute 𝑎=4 and 𝑏=5 into this formula to obtain (4,5)=(4)+5=16+25=41.

Hence, the magnitude of the given vector is 41.

In the previous example, we computed the magnitude of a vector given in component form by using a formula. Recall that any 2D vector can be expressed in terms of the fundamental unit vectors 𝑖 and 𝑗 by (𝑎,𝑏)=𝑎𝑖+𝑏𝑗.

This means that we can apply the same formula to find the magnitude of the vector expressed in terms of fundamental unit vectors: 𝑎𝑖+𝑏𝑗=𝑎+𝑏.

In the next example, we will find the magnitude of a vector expressed in terms of fundamental unit vectors using this formula.

Example 4: Finding the Norm of a Given Vector

Given that 𝐴=5𝑖3𝑗, where 𝑖 and 𝑗 are perpendicular unit vectors, find 𝐴.

Answer

In this example, we need to find the magnitude 𝐴 of a vector that is given in terms of fundamental unit vectors. Recall that if 𝑣=𝑎𝑖+𝑏𝑗, its magnitude is given by 𝑣=𝑎+𝑏.

We are given 𝐴=5𝑖3𝑗, so we can substitute 𝑎=5 and 𝑏=3 above to obtain 𝐴=(5)+(3)=25+9=34.

Hence, 𝐴=34.

So far, we have discussed how to find the magnitude of a vector given either in component form or in terms of fundamental unit vectors. Another way a vector can be defined is by specifying two endpoints. Recall that a vector beginning at point 𝐴 and ending at point 𝐵 is denoted 𝐴𝐵. If we have the coordinates of points 𝐴=(𝑥,𝑦) and 𝐵=(𝑥,𝑦), then this vector is given by 𝐴𝐵=(𝑥𝑥,𝑦𝑦).

If we apply the formula for the magnitude of a 2D vector to this vector, we obtain 𝐴𝐵=(𝑥𝑥)+(𝑦𝑦).

In the next example, we will first find the magnitude of a vector specified by its endpoints.

Example 5: The Magnitude of a Vector

What is the magnitude of the vector 𝐴𝐵, where 𝐴=(5,9) and 𝐵=(9,1)?

Answer

In this example, we need to find the magnitude of a vector specified by its endpoints. Recall that a vector from point 𝐴(𝑥,𝑦) to point 𝐵(𝑥,𝑦) is defined by 𝐴𝐵=(𝑥𝑥,𝑦𝑦).

From the given coordinates of 𝐴 and 𝐵, we know 𝑥=5,𝑦=9,𝑥=9,𝑦=1.

Substituting these values above, we obtain 𝐴𝐵=(95,1(9))=(4,10).

Now, we need to find the magnitude of this vector. Recall that the magnitude of a vector 𝑣=(𝑎,𝑏) is given by 𝑣=𝑎+𝑏.

Since our vector is (4,10), we can substitute 𝑎=4 and 𝑏=10 into this formula to obtain (4,10)=4+10=16+100=116=229.

Hence, the magnitude of 𝐴𝐵 is 229.

In our final example, we will find an unknown constant using concepts from the magnitude of a vector.

Example 6: Finding an Unknown Constant Using the Magnitude of a Vector

If 𝐴=(3,4), 𝐵=(2,𝑚), and 𝐴𝐵=5 length units, then 𝑚=.

Answer

In this example, we need to find an unknown constant when the magnitude of a vector is provided. Let us begin by finding the vector 𝐴𝐵, which is specified by its endpoints. Recall that a vector from point 𝐴(𝑥,𝑦) to point 𝐵(𝑥,𝑦) is defined by 𝐴𝐵=(𝑥𝑥,𝑦𝑦).

From the given coordinates of 𝐴 and 𝐵, we know 𝑥=3,𝑦=4,𝑥=2,𝑦=𝑚.

Substituting these values above, we obtain 𝐴𝐵=(23,𝑚4)=(5,𝑚4).

Now, let us find the magnitude of this vector. Recall that the magnitude of a vector 𝑣=(𝑎,𝑏) is given by 𝑣=𝑎+𝑏.

Since our vector is (5,𝑚4), we can substitute 𝑎=5 and 𝑏=𝑚4 into this formula to obtain 𝐴𝐵=(5)+(𝑚4).

We are given that this magnitude is equal to 5 length units. Hence, (5)+(𝑚4)=5.

Squaring both sides of this equation and simplifying, (5)+(𝑚4)=525+(𝑚4)=25(𝑚4)=0𝑚4=0𝑚=4.

Hence, 𝑚=4.

Let us finish by recapping a few important concepts from this explainer.

Key Points

  • The magnitude of a vector 𝑣, denoted 𝑣, is the length of the vector, or the distance between the initial and terminal points of a vector.
  • A unit vector is a vector whose magnitude is equal to 1. A zero vector is a vector whose magnitude is equal to 0.
  • Let 𝑣=(𝑎,𝑏) be a vector in two dimension. Then, the magnitude of this vector is given by 𝑣=𝑎+𝑏.
  • If 𝐴=(𝑥,𝑦) and 𝐵=(𝑥,𝑦), 𝐴𝐵=(𝑥𝑥)+(𝑦𝑦).

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