### 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.