### Video Transcript

Given that vector π is equal to
seven π’ plus two π£, vector π is equal to negative π’ plus two π£, and vector π
is equal to six π’ plus six π£, determine the cross product of π plus π and
π.

In this question, weβre given three
vectors in two dimensions in terms of their π’- and π£-components. Firstly, we need to find the sum of
vector π and vector π. We then need to find the cross
product of this vector and vector π. Letβs begin by finding the sum of
vectors π and π. This is equal to six π’ plus six π£
plus seven π’ plus two π£. We can find the sum of the π’- and
π£-components separately. Six π’ plus seven π’ is equal to
13π’ and six π£ plus two π£ is eight π£, giving us the vector 13π’ plus eight
π£.

Next, we need to find the cross
product of this vector and vector π. This is the cross product of 13π’
plus eight π£ and negative π’ plus two π£. We recall that for two vectors π
and π in the coordinate plane with π’ and π£ as unit vectors such that π is equal
to ππ’ plus ππ£ and π is equal to ππ’ plus βπ£, then the cross product of
vectors π and π is the determinant of the two-by-two matrix π, π, π, β
multiplied by the unit vector π€. This is equal to πβ minus ππ
multiplied by the unit vector π€.

In this question, we need to find
the determinant of the two-by-two matrix 13, eight, negative one, two. This is equal to 13 multiplied by
two minus negative one multiplied by eight, giving us 26 plus eight. And multiplying by the unit vector
π€, this is equal to 34π€. If π is equal to seven π’ plus two
π£, π is equal to negative π’ plus two π£, and π is equal to six π’ plus six π£,
then the cross product of π plus π and π is 34π€.