### Video Transcript

In this video, we will learn how to
recognize parallel and perpendicular vectors in space. We will begin by looking at the
conditions that must be true for two vectors to be parallel or perpendicular.

Two vectors π and π are parallel
if and only if they are scalar multiples of each other. Vector π must be equal to π
multiplied by vector π, where π is a constant not equal to zero. If vector π has components π sub
π₯, π sub π¦, and π sub π§ and vector π has components π sub π₯, π sub π¦, π
sub π§, then the vectors are parallel if the ratio of their components are
equal. π sub π₯ over π sub π₯ must be
equal to π sub π¦ over π sub π¦, which is also equal to π sub π§ over π sub
π§. To prove that two vectors are
parallel, we must show that these conditions are true.

Letβs now consider the conditions
for two vectors to be perpendicular. Two vectors π and π are
perpendicular if and only if their scalar product, sometimes known as the dot
product, is equal to zero. If vector π once again has
components π sub π₯, π sub π¦, π sub π§ and vector π has components π sub π₯,
π sub π¦, π sub π§, then the vectors are perpendicular if π sub π₯, π sub π₯
plus π sub π¦, π sub π¦ plus π sub π§, π sub π§ is equal to zero. We find the sum of the product of
the π₯-, π¦-, and π§-components, and if this is equal to zero, the vectors are
perpendicular.

Once again, to prove that two
vectors are perpendicular, we must show that the above conditions are true. We will now look at some questions
where we need to determine whether vectors are perpendicular or parallel.

Determine whether the following is
true or false: If the component of a vector in the direction of another vector is
zero, then the two are parallel.

Letβs begin by considering two
vectors π and π as shown. The component of vector π on
vector π is shown in the diagram. We can see that this creates a
right angle triangle. And we can let the angle at the
origin be π. In right angle trigonometry, the
cosine ratio is equal to the adjacent over the hypotenuse. We know that the hypotenuse is
equal to the magnitude of vector π. And if we let the component of
vector π on vector π be equal to π₯, then cos π is equal to π₯ over the magnitude
of vector π. Rearranging this equation, we get
that π₯ is equal to the magnitude of vector π multiplied by cos π.

Weβre told in the question that π₯
is equal to zero. Therefore, zero is equal to the
magnitude of vector π multiplied by cos π. The magnitude of any of vector must
be positive; it could not equal zero. This means that cos π must be
equal to zero. π is therefore equal to 90
degrees. This means that the two vectors are
actually perpendicular. We can therefore conclude that the
statement is false. The vectors are not parallel but
are perpendicular.

In our next question, we will find
the missing value using a pair of parallel vectors.

Find the values of π and π so
that vector two π’ plus seven π£ plus ππ€ is parallel to vector six π’ plus ππ£
minus 21π€.

We know that two vectors are
parallel if they are scalar multiples of each other. Vector π must be equal to π
multiplied by vector π, where π is a constant not equal to zero. In this question, two π’ plus seven
π£ plus ππ€ must be equal to the constant π multiplied by six π’ plus π π£ minus
21π€. By distributing the parentheses or
expanding the brackets on the right-hand side and then equating components, we see
that two is equal to six π. Equating the π£-components, we have
seven is equal to ππ. Equating the π€-components, we have
π is equal to negative 21 multiplied by the constant π.

We can solve the first equation by
dividing both sides by six. Therefore, π is equal to two over
six or two-sixths. This fraction can be simplified so
that π is equal to one-third. We can then substitute this value
into our other two equations. Seven is equal to π multiplied by
one-third. Multiplying both sides of this
equation by three gives us π is equal to 21. In our third equation, π is equal
to negative 21 multiplied by one-third. One-third of 21 is equal to
seven. Therefore, one-third of negative 21
is equal to negative seven. If our two vectors are parallel,
then π is equal to negative seven and π is equal to 21.

An alternative method here would be
to write each of the components as a ratio. We could do this either way
round. By putting the larger π’-component
on top, we have six over two is equal to π over seven which is equal to negative 21
over π. Six divided by two is equal to
three. So we can therefore solve three is
equal to π over seven and three is equal to negative 21 over π to calculate the
values of π and π. This would once again give us
answers of π equals 21 and π equals negative seven.

In our next question, we will
identify the vector that is not perpendicular to a given vector.

Which of the following vectors is
not perpendicular to the line whose direction vector π« is six, negative five? Is it (A) π« is equal to negative
five, negative six? (B) π« is equal to five, six. (C) π« is equal to 10, 12. Or (D) π« is equal to 12, negative
10.

We recall that two vectors are
perpendicular if their scalar or dot product is equal to zero. To calculate the scalar product, we
find the sum of the product of the individual components. For option (A), we need to multiply
six by negative five and then add negative five multiplied by negative six. Multiplying a positive number by a
negative number gives a negative answer. And multiplying two negative
numbers together gives a positive answer. This leaves us with negative 30
plus 30. As this is equal to zero, the two
vectors are perpendicular. Option (A) is therefore not the
correct answer.

Repeating this process for option
(B), we have six multiplied by five plus negative five multiplied by six. This simplifies to 30 plus negative
30. Once again, this is equal to
zero. So option (B) is not correct. In option (C), we have six
multiplied by 10 plus negative five multiplied by 12. Six multiplied by 10 is equal to
60, and negative five multiplied by 12 is equal to negative 60. Option (C) is not the correct
answer as the scalar product equals zero.

In option (D), our calculation is
six multiplied by 12 plus negative five multiplied by negative 10. Six multiplied by 12 is 72, and
negative five multiplied by negative 10 is positive 50. This means that the scalar product
is equal to 122. This is not equal to zero. Option (D) is therefore the correct
answer. The vector 12, negative 10 is not
perpendicular to the line whose direction vector is six, negative five. This is because their scalar
product is not equal to zero.

In our next question, we need to
identify whether two vectors are parallel, perpendicular, or neither.

Given the two vectors π is equal
to eight π’ minus seven π£ plus π€ and π is equal to 64π’ minus 56π£ plus eight π€,
determine whether these two vectors are parallel, perpendicular, or otherwise.

We know that two vectors are
parallel if they are scalar multiples of each other. Vector π must be equal to π
multiplied by vector π. Two vectors π and π are
perpendicular on the other hand, if their scalar or dot product is equal to
zero. Letβs firstly consider whether our
two vectors π and π are parallel. If one vector is a scalar multiple
of another vector, then the ratio of their individual components must be equal. In this case, 64 over eight must be
equal to negative 56 over negative seven, which must be equal to eight over one. 64 divided by eight is equal to
eight, and eight divided by one is equal to eight.

Dividing a negative number by a
negative number gives a positive answer. Therefore, negative 56 divided by
negative seven is also equal to eight. We can therefore conclude that
vector π is equal to eight multiplied by vector π or vector π is equal to
one-eighth of vector π. The two vectors π and π are
therefore parallel. Whilst they cannot be parallel and
perpendicular, letβs just check that the scalar product is not equal to zero. The π’-components of our vector are
eight and 64. The π£-components are negative
seven and negative 56. The π€-components are one and
eight. Eight multiplied by 64 is 512. Negative seven multiplied by
negative 56 is 392. One multiplied by eight is equal to
eight. The scalar product of vectors π
and π is therefore equal to 912. As this is not equal to zero, the
vectors π and π are not perpendicular.

In our final question, we will
solve a problem involving a pair of perpendicular straight lines.

If the straight line π₯ plus eight
over negative 10 is equal to π¦ plus eight over π which is equal to π§ plus 10 over
negative eight is perpendicular to π₯ plus five over negative four which is equal to
π¦ plus eight over 10 and π§ equals eight, find π.

We notice that both our straight
lines in this question are given in symmetric form. π₯ minus π₯ sub zero over π is
equal to π¦ minus π¦ sub zero over π which is equal to π§ minus π§ sub zero over
π. Any straight line in this form can
be rewritten in vector form π₯ sub zero, π¦ sub zero, π§ sub zero plus π‘ multiplied
by π, π, π. π₯ sub zero, π¦ sub zero, π§ sub
zero is the initial position and π, π, π is the direction of the straight
line.

The vector equation of our first
line is therefore equal to negative eight, negative eight, negative 10 plus π‘
multiplied by negative 10, π, negative eight. The second equation doesnβt appear
to have a π§-component; however, we are told that π§ equals eight. Therefore, the initial position is
negative five, negative eight, eight. As the π§-component is remaining
constant, the direction of the straight line is negative four, 10, zero. We know that when two straight
lines are perpendicular, the scalar product of the directions equals zero. To calculate the scalar products,
we find the sum of the products of the individual components.

This gives us negative 10
multiplied by negative four plus π multiplied by 10 plus negative eight multiplied
by zero. This must sum to zero. Negative 10 multiplied by negative
four is equal to 40, π multiplied by 10 is 10π, and negative eight multiplied by
zero is zero. Subtracting 40 from both sides of
this equation gives us 10π is equal to negative 40. We can then divide both sides of
this equation by 10, giving us π is equal to negative four. If the two straight lines are
perpendicular, the value of π is equal to negative four.

We will now summarize the key
points from this video. Two vectors are parallel if vector
π is equal to π multiplied by vector π, where π is a constant not equal to
zero. Any two vectors are only parallel
if they are scalar multiples of each other. This also means that the ratio of
their components are equal. Two vectors are perpendicular if
the scalar product or dot product is equal to zero. To calculate the scalar product, we
find the sum of the products of the individual components. By considering these two
properties, we can identify whether two vectors are parallel, perpendicular, or
otherwise.

When dealing with straight lines in
vector form, we consider the direction of the vector to identify whether two
straight lines are parallel or perpendicular. If a straight line has vector
equation π« equal to π₯ sub zero, π¦ sub zero, π§ sub zero plus π‘ multiplied by π,
π, π, it is the direction part π, π, π that is key.