### Video Transcript

Suppose vector π¨ is equal to
negative one, two, sevens, the magnitude of vector π© is equal to 13, and the angle
between the two vectors is 135 degrees. Find the dot product of vector π¨
and vector π© to the nearest hundredth.

We can calculate the dot product of
vector π¨ and vector π© by finding their magnitudes, multiplying these together, and
then multiplying this by the cos of angle π, where π is the angle between the two
vectors. In this question, we are told that
the magnitude of vector π© is equal to 13. And the angle between the two
vectors is 135 degrees. In order to calculate the dot
product of vector π¨ and vector π©, we firstly need to calculate the magnitude of
vector π¨.

Vector π¨ is written in terms of
its three components, which we will call π sub one, π sub two, and π sub
three. The magnitude of any vector is a
scalar quantity. It is equal to the square root of
π sub one squared plus π sub two squared plus π sub three squared. In this question, we begin by
squaring negative one, two, and seven. These are equal to one four and 49,
respectively. One plus four plus 49 is equal to
54. Therefore, the magnitude of vector
π¨ is equal to the square root of 54.

Using the laws of radicals or
surds, this can be rewritten as the square root of nine multiplied by the square
root of six, as nine multiplied by six is 54. The square root of nine is equal to
three. Therefore, the square root of 54
can be rewritten as three root six. This is the magnitude of vector
π¨.

We can now calculate the dot
product of vector π¨ and vector π©. It is equal to three root six
multiplied by 13 multiplied by the cos of 135 degrees. The cos of 135 degrees is equal to
negative root two over two. This means that the dot product is
equal to negative 39 root 12 over two. Root 12 can be rewritten as root
four multiplied by root three. As root four is equal to two, this
simplifies to negative 39 root three.

This isnβt the end of the question,
however, as we are asked to give our answer to the nearest hundredth. Negative 39 root three is equal to
negative 67.5499 and so on. Rounding this to two decimal places
or the nearest hundredth is negative 67.55. The dot product of vector π¨ and
vector π© to the nearest hundredth is negative 67.55.