### Video Transcript

Find the value of π for which the
four points one, seven, negative two; three, five, six; negative one, six, negative
four; and negative four, negative three, π all lie in a single plane.

Weβre given four points, which
weβre told are coplanar; that is, they all lie in the same plane. And weβre asked to find the value
of the unknown constant π. To do this, weβre going to use the
scalar triple product since we know that if the scalar triple product of three
vectors is zero, then the three vectors must be coplanar. Before we do this, however, we need
to check that our three known points are noncolinear. We do this because we need to be
sure that weβre not simply taking the scalar product with the zero vector in our
scalar triple product.

Letβs label our points π, π, π,
and π. And we can take any two of the
vectors formed by the points π, π, and π. Letβs choose ππ and ππ. We know that ππ is equal to ππ
minus ππ. And subtracting like-for-like
components, we have negative two, two, negative eight. Similarly, ππ is equal ππ minus
ππ. And thatβs equal to negative four,
one, negative 10. We know that the cross product of
ππ with ππ is given by the determinant shown, where π’, π£, and π€ are the unit
vectors in the π₯-, π¦-, and π§-direction so that our determinant is π’ times two
times negative 10 minus negative eight times one, which is π’ times the determinant
of the two-by-two matrix in the bottom corner, minus π£ times the determinant of the
matrix with elements negative two, negative eight, negative four, and negative 10
plus π€ times the determinant of the matrix with elements negative two, two,
negative four, and one.

Making some room, this evaluates to
negative 12, 12, six, which is not equal to the zero vector. So we know that our three points
are not colinear. And we can move on to using the
scalar triple product to find the value of π. Weβre told that all four points π,
π, π, and π are coplanar. And we know that the scalar triple
product of three coplanar vectors is equal to zero. We already have part of our scalar
triple product. Thatβs with the cross product of
ππ and ππ. Now if all four points lie in the
plane, then the vector ππ must also lie in the plane. The vector ππ is given by ππ
minus ππ. And that is negative five, negative
10, π plus two.

So taking the scalar triple product
of ππ with ππ and ππ, we simply have the scalar product of the two vectors
shown. This evaluates to negative five
times negative 12 plus negative 10 times 12 plus π plus two times six. That is 60 minus 120 plus six π
plus 12, which simplifies to six π minus 48. If the vectors are coplanar, this
must be equal to zero. Adding 48 to both sides, this gives
us six π is equal to 48. And dividing both sides by six,
this gives us π is equal to eight. Hence, the value of π for which
the four given points all lie in a single plane is π is equal to eight.