### Video Transcript

In this video, we will learn how to
represent a vector in space using a three-dimensional coordinate system. We will begin by looking at unit
vectors in the direction of the 𝑥-, 𝑦-, and 𝑧-axis. We will then move on to find the
components of a vector which connects two points in 3D space. We will do this both algebraically
and graphically.

Find the unit vector in the
direction of the 𝑦-axis.

We recall that a unit vector has a
magnitude equal to one. Let’s consider the
three-dimensional coordinate system or grid with center or origin 𝑂. We are told that the vector is
moving in the direction of the 𝑦-axis. This means that its 𝑥- and
𝑧-components must be equal to zero. In order for the vector to have a
magnitude of one, the 𝑦-component must also be one. The unit vector in the direction of
the 𝑦-axis is equal to zero, one, zero.

We can check this as a magnitude
equal to one by finding the square root of the sum of the squares of the individual
components. This is the square root of zero
squared plus one squared plus zero squared. This is equal to the square root of
one. And as the magnitude must be
positive, this is equal to one. We can use this information to find
the unit vector in the direction of the 𝑥- and 𝑧-axis.

The unit vector in the direction of
the 𝑥-axis is one, zero, zero. It has an 𝑥-component of one and a
𝑦- and 𝑧-component of zero. We have just seen that the unit
vector in the direction of the 𝑦-axis is zero, one, zero. Finally, the unit vector in the
direction of the 𝑧-axis is zero, zero, one. This time, we have a 𝑧-component
equal to one and an 𝑥- and 𝑦-component equal to zero.

We will now look at some questions
where we need to find the vector between two given points.

Which of the following is equal to
the vector 𝐀𝐁? Is it (A) 𝐀 plus 𝐁, (B) 𝐀 minus
𝐁, (C) 𝐁 minus 𝐀, or (D) 𝐀 times 𝐁?

Let’s begin by considering the two
points 𝐴 and 𝐵 on a two-dimensional coordinate grid. The vector 𝐀 will take us from the
origin 𝑂 to the point 𝐴. Likewise, vector 𝐁 will take us
from the origin to the point 𝐵. We need to work out how we can get
from point 𝐴 to point 𝐵. One way of doing this is going via
the origin 𝑂. This means that vector 𝐀𝐁 is
equal to vector 𝐀𝐎 plus vector 𝐎𝐁. As the vector 𝐎𝐀 equals 𝐀, then
the vector 𝐀𝐎 will be negative 𝐀, as it is going in the opposite direction but
has the same magnitude. The vector 𝐎𝐁 is equal to 𝐁. Negative 𝐀 plus 𝐁 can be
rewritten as 𝐁 minus 𝐀. This means that the correct answer
is option (C). The vector 𝐀𝐁 is equal to vector
𝐁 minus vector 𝐀.

We will now use this rule to find a
vector between two given points.

Given vector 𝐀 is equal to six,
one, four and vector 𝐁 is equal to three, one, two, find vector 𝐀𝐁.

We recall the general rule when
finding vectors between two points states that vector 𝐀𝐁 is equal to vector 𝐁
minus vector 𝐀. In this question, we want to
subtract the vector six, one, four from the vector three, one, two. When subtracting vectors, we
subtract the corresponding components. In this case, we need to subtract
six from three. This gives us negative three. One minus one is equal to zero. Finally, two minus four is equal to
negative two. The 𝑥-component is negative three,
the 𝑦-component is zero, and the 𝑧-component is negative two. The vector 𝐀𝐁 is equal to
negative three, zero, negative two.

In our next question, we will need
to find the position vector of a point given the vector joining it to another
point.

Given 𝐀𝐁 is equal to negative
one, negative three, zero and vector 𝐀 is equal to negative four, negative five,
negative five, express vector 𝐁 in terms of the fundamental unit vectors.

We recall that when finding the
vector between two points, vector 𝐀𝐁 is equal to vector 𝐁 minus vector 𝐀. If we let vector 𝐁 have components
𝑥, 𝑦, 𝑧, then negative one, negative three, zero is equal to 𝑥, 𝑦, 𝑧 minus
negative four, negative five, negative five. Adding vector 𝐀 to both sides of
this equation gives us negative one, negative three, zero plus negative four,
negative five, negative five is equal to 𝑥, 𝑦, 𝑧.

When adding and subtracting
vectors, we can look at each component separately. This means that 𝑥 is equal to
negative one plus negative four. This is the same as negative one
minus four, which is equal to negative five. 𝑦 is equal to negative three plus
negative five. This equals negative eight. Finally, 𝑧 is equal to negative
five. Vector 𝐁 is, therefore, equal to
negative five, negative eight, negative five.

We were asked to write vector 𝐁 in
terms of the fundamental unit vectors. This means we need to write it in
the form 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤. Vector 𝐁 is, therefore, equal to
negative five 𝐢 minus eight 𝐣 minus five 𝐤.

In our next question, we will find
the components of a 3D position vector which is represented graphically.

Using the graph, write the vector
𝐀 in terms of its components.

As we have a three-dimensional
coordinate grid, vector 𝐀 will have three components, 𝑥, 𝑦, and 𝑧. Moving along the 𝑥-axis, we can
see that the 𝑥-component is two. The 𝑦-component is equal to
three. Finally, the 𝑧-component is equal
to four. This means that vector 𝐀 is equal
to two, three, four. The vector 𝐀 in terms of its
components is the displacement of point 𝐴 from the origin in the 𝑥-, 𝑦-, and
𝑧-directions.

In our final question, we will find
the components of a 3D vector which is again represented graphically.

Find the vector 𝐀𝐆 using the
graph.

One way of answering this question
would be to recall that we can find the vector 𝐀𝐁 by subtracting vector 𝐀 from
vector 𝐁. This means that in our question, we
need to subtract vector 𝐀 from vector 𝐆. Vector 𝐀 is the displacement of
point 𝐴 from the origin. This has an 𝑥-component of one, a
𝑦-component of one, and a 𝑧-component of zero. This means that vector 𝐀 is equal
to one, one, zero. Vector 𝐆 has an 𝑥-component of
four, a 𝑦-component of four, and a 𝑧-component of three. This means that vector 𝐆 is equal
to four, four, three.

To calculate vector 𝐀𝐆, we need
to subtract one, one, zero from four, four, three. When subtracting vectors, we
subtract each component separately. Four minus one is equal to
three. When subtracting the 𝑦-components,
we also get three. The same is true of the
𝑧-components as three minus zero is equal to three. Vector 𝐀𝐆 is, therefore, equal to
three, three, three.

An alternative method here would be
to recognize that we have a cube of side length three. Vertices 𝐴 and 𝐺 are opposite
corners of the cube. This means that we need to move
three units in the 𝑥-, 𝑦-, and 𝑧-direction to get from point 𝐴 to point 𝐺. This confirms that vector 𝐀𝐆 is
equal to three, three, three.

We will now summarize the key
points from this video. We saw in our first question that a
unit vector has a magnitude of one. A vector in three-dimensional space
can be written in terms of its three components, in triangular brackets 𝑥, 𝑦, 𝑧
or 𝑥𝐢 plus 𝑦𝐣 plus 𝑧𝐤. The vector connecting two points 𝐴
and 𝐵 in 3D space is written vector 𝐀𝐁. This is equal to vector 𝐁 minus
vector 𝐀. We can, therefore, calculate the
components of vector 𝐀𝐁 given the coordinates of points 𝐴 and 𝐵.

We also saw that we can find the
coordinates of an unknown point using the coordinates of a known point and the
components of a known vector. Finally, we saw in the last two
questions that we can find the components of a 3D vector which is represented
graphically.