### Video Transcript

If vector π is equal to negative π’ minus two π£, vector π is equal to negative
four π’ minus four π£, the cross product of vector π and π is negative three π€,
and the cross product of vector π and vector π is four π€, find vector π.

In this question, weβre given vectors π and π which lie in the coordinate plane
with π’ and π£ as unit vectors. We are also given the cross product of vector π and π together with the cross
product of vector π and π. And we are asked to find vector π.

We will let this vector be π₯π’ plus π¦π£, where π₯ and π¦ are constants. We recall that if we have two vectors π¦ and π§ 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 equal to the determinant of the two-by-two matrix π, π, π, β multiplied by the
unit vector π€, where the unit vector π€ is perpendicular to the π₯π¦-plane.

Letβs begin by considering the cross product of vectors π and π. Using our general rule, this must be equal to the determinant of the two-by-two
matrix negative one, negative two, π₯, π¦ multiplied by the unit vector π€. The determinant of a matrix is equal to negative one multiplied by π¦ minus π₯
multiplied by negative two. This is equal to negative π¦ plus two π₯. So the cross product of vector π and vector π is negative π¦ plus two π₯ multiplied
by the unit vector π€. We already know that this is equal to negative three π€. We can therefore conclude that negative three is equal to negative π¦ plus two
π₯.

Adding π¦ and three to both sides of this equation, we have π¦ is equal to two π₯
plus three. We will call this equation one and now repeat the process for the cross product of
vector π and π. This is equal to the determinant of the two-by-two matrix π₯, π¦, negative four,
negative four multiplied by π€. The determinant of the matrix is π₯ multiplied by negative four minus negative four
multiplied by π¦. This simplifies to negative four π₯ plus four π¦ such that the cross product of π
and π is negative four π₯ plus four π¦ multiplied by the unit vector π€. We are told that this is equal to four π€. And negative four π₯ plus four π¦ is therefore equal to four.

We can divide both sides of this equation by four such that one is equal to negative
π₯ plus π¦. And adding π₯ to both sides, we have π¦ is equal to π₯ plus one. We now have a pair of simultaneous equations we can solve to calculate the values of
π₯ and π¦. Since π¦ is equal to two π₯ plus three and π₯ plus one, then these two expressions
must be equal to one another. We can then subtract π₯ and three from both sides of the equation. Two π₯ minus π₯ is π₯, and one minus three is negative two, giving us π₯ is equal to
negative two. We can then substitute this value of π₯ into equation one or equation two.

In Equation two, we have π¦ is equal to negative two plus one, which gives us π¦ is
equal to negative one. Vector π is therefore equal to negative two π’ minus π£. If π is equal to negative π’ minus two π£, π is equal to negative four π£ minus
four π£, the cross product of π and π is negative three π€, and the cross product
of π and π is four π€, then vector π is equal to negative two π’ minus π£.