### Video Transcript

Given the vector π― which is equal
to eight π’ minus six π£, find its magnitude and direction, giving the direction as
an angle, π, to two decimal places in the range where π is greater than negative
180 degrees and less than or equal to 180 degrees.

In this question, weβve been asked
to find both the magnitude and direction of the vector eight π’ minus six π£. Letβs start by finding its
magnitude. The magnitude of a vector is the
distance from the initial point of the vector to the terminal point. We can sketch the vector to help us
visualise it. Here, we have drawn it so that its
initial point is at the origin. Since the π’-component of the
vector is eight and the π£-component of the vector is negative six, the terminal
point is at eight, negative six.

We need to find the length of this
vector. In order to find the magnitude of
the vector, weβll need to recall the formula for finding the magnitude of 2D
vectors. We have that for the vector π―
equals ππ’ plus ππ£, the magnitude of π― is equal to the square root of π squared
plus π squared. The vector in the question is equal
to eight π’ minus six π£. So, we can say that π is equal to
eight and π is equal to negative six. Hence, the magnitude of π― is equal
to the square root of eight squared plus negative six squared. We can now square both the numbers
and then add the 64 and 36. We have that the magnitude of π― is
equal to the square root of one hundred, which simplifies to give us that the
magnitude of π― is equal to 10.

Next, we need to find the direction
of π―. We have been told to find the
direction of the vector as an angle between negative 180 degrees and positive 180
degrees. The angle we are looking to find is
the angle the vector makes with the horizontal positive axis. Since the angle is below the axis,
we know that the value of π will be negative. When finding this angle, we will
obtain a positive value, so we can label the angle on the diagram as π₯. However, we should note that π is
equal to negative π₯.

We can see that if we draw a
vertical line from the horizontal axis to the endpoint of the vector, then we have
formed a right angle triangle, consisting of the line we have just drawn, the
horizontal axis, and the vector π― itself. We know that the side lengths of
this triangle are 6, 8, and 10 since these are the magnitudes of the components of
the vector and the magnitude of the vector itself. We can now use right triangle
trigonometry to find π, more specifically, the fact that tan of π₯ is equal to the
opposite over the adjacent.

The side opposite π₯ has a length
of six, and the side adjacent to π₯ has a length of eight. Hence, tan of π₯ is equal to six
over eight, which can be simplified to three over four. Taking the inverse tan of both
sides, we have that π₯ is equal to tan inverse of three over four. Using a calculator and making sure
that it is in degrees, we have that π₯ is equal to 36.86989 and so on degrees. The question has asked us to find
this angle to two decimal places. So, rounding π₯ to two decimal
places, we have that π₯ equals 36.87 degrees.

Now, we can use the fact that π is
equal to negative π₯, and we obtain that π is equal to negative 36.87 degrees to
two decimal places. We have now found that the
magnitude of π― is 10 and the direction of π― is negative 36.87 degrees, which
completes this question.