Video Transcript
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.
