### Video Transcript

Find the angle π between the
vectors π negative π’ plus two π£ plus π€ and π negative three π’ plus six π£ plus
three π€.

In this question, weβre given two
vectors π and π in terms of the unit directional vectors π’, π£, and π€. We need to use this to determine
the angle π between these two vectors. To answer this question, letβs
start by recalling how we find the angle between two vectors. We recall if π is the angle
between two vectors π and π, then the cos of π will be equal to the dot product
of vectors π and π divided by the magnitude of vector π multiplied by the
magnitude of vector π.

The same is also true in
reverse. If π satisfies this equation, then
π can be said to be an angle between vectors π and π. However, usually when weβre talking
about the angle between two vectors, we mean the smallest nonnegative angle between
these two vectors. We can find this by taking the
inverse cosine of both sides of this equation because, for example, in degrees, the
inverse cosine function has a range between zero and 180 degrees. Therefore, we can find the angle π
between the vectors π and π by finding the dot product between π and π and the
magnitudes of vector π and vector π.

Letβs start by finding the dot
product between vector π and vector π. Thatβs the dot product between the
vector negative π’ plus two π£ plus π€ and the vector negative three π’ plus six π£
plus three π€. Remember, to find the dot product
between two vectors, we need to find the sum of the products of the corresponding
components of these two vectors. We could write these vectors in
component form; however, itβs not necessary. The components of the vectors will
be the coefficients of the unit directional vectors π’, π£, and π€. Therefore, in the dot product
between vectors π and π, the sum of the products of the corresponding components
of each vector is negative one times negative three plus two times six plus one
multiplied by three. And if we calculate this
expression, we see itβs equal to 18.

We now need to determine the
magnitude of vector π and the magnitude of vector π. To do this, we recall the magnitude
of a vector is equal to the square root of the sums of the squares of its
components. In other words, the magnitude of
the vector ππ’ plus ππ£ plus ππ€ is equal to the square root of π squared plus
π squared plus π squared. We can use this to find the
magnitude of vector π and the magnitude of vector π. The magnitude of vector π will be
equal to the magnitude of negative π’ plus two π£ plus π€. We need to find the square root of
the sum of the squares of its components. Remember, the components of this
vector will be the coefficients of the unit directional vectors π’, π£, and π€. The magnitude of vector π will be
the square root of negative one all squared plus two squared plus one squared, which
we can calculate is equal to the square root of six.

We can do the same to find the
magnitude of π. The square root of the sum of the
squares of the components of π will be the square root of negative three all
squared plus six squared plus three squared, which we can calculate is equal to the
square root of 54. We can simplify this even further
by noticing that 54 is equal to three squared multiplied by six. We can then use this to simplify
root 54 to be the square root of three squared multiplied by root six, which is, of
course, three times root six.

Weβre now ready to use our formula
to find the value of π, the angle between the two vectors π and π. First, we know the cos of π will
be equal to the dot product between vectors π and π divided by the magnitude of
vector π times the magnitude of vector π. Next, we can substitute the values
we found for the dot product between π and π, the magnitude of vector π and the
magnitude of vector π, into this equation. This gives us the cos of π is
equal to 18 divided by root six multiplied by three root six. We can simplify the right-hand side
of this equation. In our denominator, we have root
six times three root six. Root six times root six is equal to
six, and three times six is equal to 18. So, the right- hand side of this
equation simplifies to give us 18 divided by 18, which is, of course, just equal to
one. So, our equation simplifies to give
us the cos of π is equal to one.

Finally, we can solve for our value
of π by taking the inverse cosine of both sides of this equation. Doing this gives us that π is
equal to the inverse cos of one, which we know is equal to zero. In this case, weβll use
degrees. So, π is equal to zero
degrees.

Now, we could stop here. Weβve shown that the angle π
between our two vectors π and π is equal to zero degrees. However, itβs worth pointing
something out. Because the angle between these two
vectors is equal to zero degrees, we can conclude that the two vectors must have the
same direction. In other words, these two vectors
must be parallel, and we can show this. If we were to multiply all of the
components of vector π by three, we would get back to π. In other words, three π is equal
to π. Itβs also worth pointing out if the
scalar value had instead been negative, then we know that π and π would point in
opposite directions. However, in this case, the angle
between the two vectors would have been 180 degrees.

Therefore, we were able to show the
angle between two vectors π negative π’ plus two π£ plus π€ and π negative three
π’ plus six π£ plus three π€ is equal to zero degrees.