 Lesson Video: Vector Operations in 2D | Nagwa Lesson Video: Vector Operations in 2D | Nagwa

# Lesson Video: Vector Operations in 2D Mathematics

In this video, we will learn how to perform operations on vectors algebraically such as vector addition, vector subtraction, and scalar multiplication in two dimensions.

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### Video Transcript

In this video, we will learn how to perform operations on vectors algebraically, such as vector addition, vector subtraction, scalar multiplication, and finding the magnitude of a vector in two dimensions. Specifically, we will look at combining two or more of these operations. Let’s begin by recalling how we perform the individual operations.

A two-dimensional vector has a horizontal and vertical component, which describe its size, or magnitude, and direction. These vectors can be represented graphically on the 𝑥𝑦-coordinate plane, where the unit vector 𝐢 is one unit in the positive 𝑥- direction and the unit vector 𝐣 is one unit in the positive 𝑦-direction. The vector 𝐮 equal to four 𝐢 plus three 𝐣 will move four units in the positive 𝑥-direction and three units in the positive 𝑦-direction. Likewise, vector 𝐯, which is equal to negative two 𝐢 plus five 𝐣, will move two units in the negative 𝑥-direction and five units in the positive 𝑦-direction. These vectors can also be written in terms of their components as shown.

When adding or subtracting two vectors, we treat the horizontal and vertical components separately. Adding vectors 𝐮 and 𝐯, we have four, three plus negative two, five. As four plus negative two is equal to two and three plus five is eight, this is equal to the vector two, eight. 𝐮 minus 𝐯 is equal to four, three minus negative two, five. Subtracting the corresponding components gives us the vector six, negative two. When multiplying a vector by a scalar, we multiply each of the components by that scalar. For example, four 𝐮 is equal to four multiplied by the vector four, three. We multiply four by four and then four by three, giving us the vector 16, 12.

Finally, we can calculate the magnitude of any vector by recalling that if vector 𝐮 is equal to 𝑥, 𝑦, then the magnitude of vector 𝐮 is equal to the square root of 𝑥 squared plus 𝑦 squared. In our example, the magnitude of vector 𝐮 is equal to the square root of four squared plus three squared. And since four squared is equal to 16 and three squared is equal to nine, this is equal to the square root of 25, which equals five. In the three examples that follow, we will see how we can use different combinations of the four operations to solve a variety of problems involving vectors.

Given that vector 𝐀 is equal to negative four, negative one and vector 𝐁 is equal to negative two, negative one, express vector 𝐂, which is equal to negative eight, negative one, in terms of vector 𝐀 and vector 𝐁.

As we want to express vector 𝐂 in terms of vectors 𝐀 and 𝐁, we will let 𝐂 equal some constant 𝑝 multiplied by 𝐀 plus a constant 𝑞 multiplied by 𝐁. Substituting in the values of vectors 𝐀, 𝐁, and 𝐂, we have negative eight, negative one is equal to 𝑝 multiplied by negative four, negative one plus 𝑞 multiplied by negative two, negative one. We recall that when multiplying any vector by a scalar or constant, we multiply each of the components by the scalar. This means that 𝑝 multiplied by negative four, negative one is equal to negative four 𝑝, negative 𝑝. Likewise, 𝑞 multiplied by negative two, negative one is equal to negative two 𝑞, negative 𝑞.

We now have the sum of two vectors, and we know this is equal to the vector negative eight, negative one. When adding two vectors, we simply add the corresponding components separately. This means that the right-hand side of our equation is equal to the vector negative four 𝑝 minus two 𝑞, negative 𝑝 minus 𝑞. As this must be equal to the vector negative eight, negative one, we can simply equate the components on either side of our equation. We have two equations: negative eight is equal to negative four 𝑝 minus two 𝑞, and negative one is equal to negative 𝑝 minus 𝑞.

We can simplify both of these equations and eliminate the negative signs by multiplying the top equation by negative one-half and the second equation by negative one. This gives us two simultaneous equations: four is equal to two 𝑝 plus 𝑞, and one is equal to 𝑝 plus 𝑞. One way of solving these equations to calculate the values of 𝑝 and 𝑞 is by elimination. Subtracting equation two from equation one, the 𝑞’s cancel, and we are left with 𝑝 is equal to three. We can then substitute this value of 𝑝 into equation two, giving us one is equal to three plus 𝑞. Subtracting three from both sides of this equation, we have 𝑞 is equal to negative two. We now have values for the constants 𝑝 and 𝑞. And we can therefore conclude that vector 𝐂 is equal to three 𝐀 minus two 𝐁.

Our next example is a more complicated problem also involving scalar multiplication and addition of vectors.

On a lattice, where vector 𝐀𝐂 is equal to three, three; vector 𝐁𝐂 is equal to 13, negative seven; and two 𝐂 plus two 𝐀𝐁 is equal to negative four, negative four, find the coordinates of the point 𝐶.

If we begin by considering the three points 𝐴, 𝐵 and 𝐶 shown on the diagram, we know that vector 𝐀𝐂 is equal to three, three. This means that we move three units in the positive 𝑥-direction and three units in the positive 𝑦-direction. Vector 𝐁𝐂 is equal to 13, negative seven. To travel from point 𝐵 to point 𝐶, we move 13 units in the positive 𝑥-direction and seven units in the negative 𝑦-direction.

We can use this information to find the vector 𝐀𝐁. One way of traveling from point 𝐴 to point 𝐵 would be via point 𝐶. In order to do this, we would travel along the vectors 𝐀𝐂 and 𝐂𝐁. We know that vector 𝐀𝐂 is three, three. Vector 𝐂𝐁 would have the same magnitude as the vector 𝐁𝐂 but acts in the opposite direction. This means that vector 𝐂𝐁 is equal to negative 13, seven. Vector 𝐀𝐁 is therefore equal to three, three plus negative 13, seven.

We know that we can add two vectors by adding their corresponding components. Three plus negative 13 is equal to negative 10, and three plus seven is 10. Therefore, vector 𝐀𝐁 is equal to negative 10, 10. If we let the point 𝐶 have coordinates 𝑥, 𝑦, then the position vector of point 𝐶, also written 𝐎𝐂, is equal to the vector 𝑥, 𝑦. Substituting this together with the vector 𝐀𝐁 into the equation given, we have two multiplied by 𝑥, 𝑦 plus two multiplied by negative 10, 10 is equal to negative four, negative four.

We recall that we can multiply a vector by a scalar by multiplying each of the components by that scalar. Our equation simplifies to two 𝑥, two 𝑦 plus negative 20, 20 is equal to negative four, negative four. We can add the two vectors on the left-hand side such that two 𝑥 minus 20, two 𝑦 plus 20 is equal to negative four, negative four. And finally, to calculate the values of 𝑥 and 𝑦, we can equate the corresponding components. We have two equations: two 𝑥 minus 20 is equal to negative four, and two 𝑦 plus 20 is equal to negative four. Solving the first equation gives us 𝑥 is equal to eight. And solving the second equation, we have 𝑦 is equal to negative 12. We can therefore conclude that the coordinates of point 𝐶 are eight, negative 12.

In our final example, we will combine the skills of vector addition, scalar multiplication, and finding the magnitude of a vector.

If vector 𝐀 is equal to five, negative three and vector 𝐁 is equal to two, one, then the magnitude of 𝐀 plus three 𝐁 is equal to what length units.

In this question, we are given the two-dimensional vectors 𝐀 and 𝐁 in terms of their 𝑥- and 𝑦-components. We are asked to calculate the magnitude of vector 𝐀 plus three multiplied by vector 𝐁. We will do this in three steps, firstly by using scalar multiplication to calculate three 𝐁. When multiplying any vector by a scalar, we simply multiply each of the individual components by that scalar. This means that three multiplied by the vector two, one gives us the vector six, three. Our next step is to add this to vector 𝐀. And we will do this using the process of vector addition. We do this by adding the corresponding components separately, giving us the vector 11, zero.

Our final step is to find the magnitude of this vector. As the 𝑦-component of our vector is zero, there is a shortcut here. However, we will begin by looking at how we calculate the magnitude of any two-dimensional vector. If vector 𝐮 is equal to 𝑥, 𝑦, then the magnitude of vector 𝐮 is equal to the square root of 𝑥 squared plus 𝑦 squared. We find the sum of the squares of the individual components and then square root our answer. This means that the magnitude of the vector 11, zero is equal to the square root of 11 squared plus zero squared. This is equal to the square root of 121, which is equal to 11. If vector 𝐀 is equal to five, negative three, vector 𝐁 is equal to two, one, then the magnitude of 𝐀 plus three 𝐁 is equal to 11 length units.

As previously mentioned, there is a shortcut to calculate the magnitude when one of the components equals zero. In this question, the 𝑦-component was equal to zero. This means that the vector 11, zero moves 11 units in the positive 𝑥-direction. As the magnitude of a vector is its length, this confirms that the magnitude of the vector 11, zero is 11 length units. When a two-dimensional vector has one of its components equal to zero, then the magnitude of that vector will be equal to the nonzero component.

We will now summarize the key points from this video. We saw in this video that we can combine the skills of vector addition, vector subtraction, scalar multiplication, and finding the magnitude of a vector to solve problems involving two-dimensional vectors. We did this by recalling that if vector 𝐮 has components 𝑥 sub one, 𝑦 sub one and vector 𝐯 has components 𝑥 sub two, 𝑦 sub two, then we can add or subtract our two vectors by adding or subtracting the corresponding components. Multiplying a vector by a scalar or constant 𝑘 involves multiplying each of the components by that scalar. And the magnitude of a vector is the square root of the sum of the squares of the individual components.