### Video Transcript

In this video, we will learn how to
find direction angles and direction cosines for a given vector in space. We will begin by considering the
three-dimensional coordinate grid. We know that in three-dimensional
space, we have the π₯-, π¦-, and π§- or π§-axis. These are all perpendicular or at
right angles to one another.

Letβs assume we have the vector π―
in the direction shown, where vector π― has components π― sub π₯, π― sub π¦, and π―
sub π§. It is also important to recall that
the unit vectors π’, π£, and π€ act in the π₯-, π¦-, and π§-direction,
respectively. The first direction angle we need
to be aware of is πΌ. This is the angle between the unit
vector π’ and π―. Our second angle is the angle π½,
which is the angle between the unit vector π£ and vector π―. Finally, we have the angle πΎ,
which is the angle between the unit vector π€ and vector π―. These are the three direction
angles πΌ, π½, and πΎ.

Now letβs consider the direction
cosines. We know that in right-angle
trigonometry, the cos of any angle is equal to the adjacent over the hypotenuse. The direction cosines are simply
the cos of our three angles, cos of πΌ, cos of π½, and cos of πΎ. cos of πΌ is equal to the
π₯-component of the vector, in this case, π― sub π₯, divided by the magnitude of
vector π―. In the same way, the cos of angle
π½ is equal to π― sub π¦ divided by the magnitude of π―. And finally, the cos of angle πΎ is
equal to π― sub π§ divided by the magnitude of π―. The cos of πΌ, cos of π½, and cos
of πΎ are known as the direction cosines.

By taking the inverse cos of both
sides of these three equations, we get expressions for angle πΌ, π½, and πΎ. πΌ is equal to the inverse cos of
π― sub π₯ over the magnitude of π―. Likewise, angle π½ is equal to the
inverse cos of π― sub π¦ divided by the magnitude of π―. The angle πΎ is equal to the
inverse cos of π― sub π§ over the magnitude of π―. We recall that if vector π― has
components π― sub π₯, π― sub π¦, and π― sub π§, then the magnitude of vector π― is
equal to the square root of π― sub π₯ squared plus π― sub π¦ squared plus π― sub π§
squared. Note that this magnitude of a
vector is also sometimes called the norm. In our first question, we will find
a vector given its direction angles.

Find vector π whose norm is 41 and
whose direction angles are 135 degrees, 120 degrees, and 60 degrees.

We know that the norm of a vector
is its magnitude and that the direction vectors πΌ, π½, and πΎ are the angles
between the unit vectors π’, π£, and π€ and the vector π―. We also know that the direction
cosines are such that cos πΌ is equal to π― sub π₯ over the magnitude of π―, cos π½
is equal to π― sub π¦ over the magnitude of π―, and cos πΎ is equal to π― sub π§
over the magnitude of π―, where the vector π― has components π― sub π₯, π― sub π¦,
and π― sub π§. Substituting in our values for πΌ
and the magnitude, we see that the cos of 135 degrees is equal to π sub π₯ over
41. We can then multiply both sides of
this equation by 41. π sub π₯ is equal to negative 41
root two over two. This is the π₯-component of our
vector π.

In the same way, considering angle
π½, we have the cos of 120 degrees is equal to π sub π¦ over 41. Once again, we multiply both sides
of the equation by 41 such that π sub π¦ is equal to negative 41 over two. Finally, we have the cos of 60
degrees, angle πΎ, is equal to π sub π§ divided by 41. π sub π§ is therefore equal to 41
multiplied by the cos of 60 degrees, which in turn is equal to 41 over two. This means that vector π has
components negative 41 root two over two, negative 41 over two, and 41 over two. This is the vector whose magnitude
or norm is 41 and direction angles are 135, 120, 60 degrees.

Before looking at our next example,
we will consider a formula that links the direction cosines. cos squared πΌ plus cos squared π½
plus cos squared πΎ is equal to one. From our knowledge of the direction
cosines, letβs consider why this is true. We know that the cos of πΌ, π½, and
πΎ are as shown. Squaring both sides of this
equation, we see that cos squared πΌ is equal to π― sub π₯ squared divided by the
magnitude of π― squared. Doing the same to our other two
equations, we get cos squared π½ is equal to π― sub π¦ squared over the magnitude of
π― squared and cos squared πΎ is equal to π― sub π§ squared over the magnitude of π―
squared.

We can now add the three
expressions on the right-hand side as this would be equal to cos squared πΌ plus cos
squared π½ plus cos squared πΎ. As all three fractions have the
same denominator, this is equal to π― sub π₯ squared plus π― sub π¦ squared plus π―
sub π§ squared all divided by the magnitude of π― squared. We know that the magnitude of π― is
equal to the square root of π― sub π₯ squared plus π― sub π¦ squared plus π― sub π§
squared.

Squaring both sides of this
equation gives us the magnitude of π― squared is equal to π― sub π₯ squared plus π―
sub π¦ squared plus π― sub π§ squared. Substituting this into our
expression gives us the magnitude of π― squared over the magnitude of π―
squared. We know that dividing any scalar by
itself gives us an answer of one. This proves that cos squared πΌ
plus cos squared π½ plus cos squared πΎ is equal to one. We will now use this equation to
help us solve a problem.

Suppose that 31 degrees, 65
degrees, and π are the direction angles of a vector. Which of the following, to the
nearest hundredth, is π? Is it (A) 72.88 degrees, (B) 84.00
degrees, (C) 85.03 degrees, or (D) 264.00 degrees?

We recall that if the three
direction angles of a vector are πΌ, π½, and πΎ, then cos squared πΌ plus cos
squared π½ plus cos squared πΎ is equal to one. In this question, we will
substitute πΌ equal to 31 degrees, π½ equal to 65 degrees, and πΎ equal to π. This gives us the equation
shown. We recall that cos squared of 31
degrees is the same as cos of 31 degrees all squared. This is how we type this into a
scientific calculator. cos squared 31 degrees plus cos
squared 65 degrees is equal to 0.91334 and so on. We can then subtract this from both
sides of our equation, giving us cos squared π is equal to 0.08665 and so on.

We then square root both sides of
this equation. cos π is equal to 0.29437 and so
on. Finally, we take the inverse cos of
both sides, such that π is equal to 72.8797 and so on. We want the answer to the nearest
hundredth. This means that the nine in the
thousandths column is the deciding number. Rounding up gives us 72.88
degrees. Therefore, the correct answer is
option (A). The three direction angles of the
vector are 61 degrees, 65 degrees, and 72.88 degrees.

In our next question, we need to
work out the direction cosines of a vector.

Find the direction cosines of the
vector π with components five, two, and eight.

We recall that if vector π― has
components π― sub π₯, π― sub π¦, and π― sub π§ and direction angles πΌ, π½, and πΎ,
then the direction cosines cos of πΌ, cos of π½, and cos of πΎ are equal to π― sub
π₯ over the magnitude of π―, π― sub π¦ over the magnitude of π―, and π― sub π§ over
the magnitude of π―, respectively, where the magnitude of vector π― is equal to the
square root of π― sub π₯ squared plus π― sub π¦ squared plus π― sub π§ squared.

In this question, we are told that
vector π has components five, two, and eight. The magnitude of vector π is
therefore equal to the square roots of five squared plus two squared plus eight
squared. Five squared is equal to 25, two
squared is equal to four, and eight squared is equal to 64. Summing these three values gives us
an answer of 93. Therefore, the magnitude of vector
π is the square root of 93. This means that the cos of angle πΌ
is equal to five over the square root of 93, the cos of angle π½ is equal to two
over the square root of 93, and, finally, the cos of angle πΎ is equal to eight over
the square root of 93. The direction cosines of vector π
are five over the square root of 93, two over the square root of 93, and eight over
the square root of 93.

In our final question, we will need
to calculate the direction angles from a geometric problem.

Find the measure of the direction
angles of the vector π
represented by the given figure, correct to one decimal
place.

We will begin by writing the vector
π
in terms of its three components. In the π₯-direction, we travel
eight centimeters. Therefore, the π₯-component of the
vector is eight. In the π¦-direction, we travel 19
centimeters. Therefore, the π¦-component of
vector π
is 19. We travel nine centimeters in the
π§- or π§-direction. Therefore, the π§-component is
nine. Vector π
is equal to eight, 19,
nine. The magnitude of vector π
is equal
to the square root of eight squared plus 19 squared plus nine squared as we know
that the magnitude of any vector is equal to the square root of the sum of the
squares of the individual components.

Eight squared plus 19 squared plus
nine squared is equal to 506. Therefore, the magnitude of vector
π
is the square root of 506. We are asked to calculate the
direction angles. These are usually labeled πΌ, π½,
and πΎ, where πΌ is the angle between the π₯-axis and the vector π
, π½ is the angle
between the π¦-axis and the vector π
, and, finally, πΎ is the angle between the
π§-axis and the vector π
.

From our knowledge of the direction
cosines, we know that πΌ is equal to the inverse cos of π
sub π₯ over the magnitude
of π
, π½ is equal to the inverse cos of π
sub π¦ over the magnitude of π
, and πΎ
is equal to the inverse cos of π
sub π§ over the magnitude of π
, where π
sub π₯,
π
sub π¦, and π
sub π§ are the three components of vector π
. We will now clear some space to
calculate these values.

Angle πΌ is equal to the inverse
cos of eight over the square root of 506. This is equal to 69.1671 and so
on. Rounding to one decimal place,
angle πΌ is equal to 69.2 degrees. π½ is equal to the inverse cos of
19 over the square root of 506. This is equal to 32.3652 and so
on. Rounding this to one decimal place
gives us 32.4 degrees. Angle πΎ is equal to the inverse
cos of nine over the square root of 506. This is equal to 66.4156 and so
on. To one decimal place, angle πΎ is
66.4 degrees. The angles πΌ, π½, and πΎ can also
be written as π sub π₯, π sub π¦, and π sub π§. In this question, theyβre equal to
69.2 degrees, 32.4 degrees, and 66.4 degrees, respectively.

We will know summarize the key
points from this video. The direction angles β often
denoted πΌ, π½, and πΎ β are the angles between a vector and the π₯-, π¦-, and π§-
or π§-axes, respectively. The direction cosines of any vector
are such that the cos of angle πΌ is equal to π― sub π₯ divided by the magnitude of
vector π―, the cos of angle π½ is equal to π― sub π¦ over the magnitude of vector
π―, and the cos of angle πΎ is equal to π― sub π§ over the magnitude of vector π―,
where π― sub π₯, π― sub π¦, and π― sub π§ are the components of the vector in the
π₯-, π¦-, and π§-directions.

The magnitude of a vector is equal
to the square root of the sum of the squares of the individual components. We also proved in this video that
the sum of the squares of the direction cosines is equal to one. cos squared πΌ plus cos squared π½
plus cos squared πΎ is equal to one. We can use these equations to
calculate the direction angles, direction cosines, or any other unknowns.