# Question Video: Using the Scalar Triple Product to Find an Unknown Mathematics

Find the value of π for which the four points (1, 7, β2), (3, 5, 6), (β1, 6, β4), and (β4, β3, π) all lie in a single plane.

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### 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.

Weβre given four points, which weβre told are coplanar; that is, they all lie in the same plane. And weβre asked to find the value of the unknown constant π. To do this, weβre 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 weβre not simply taking the scalar product with the zero vector in our scalar triple product.

Letβs 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. Letβs choose ππ and ππ. We know that ππ is equal to ππ minus ππ. And subtracting like-for-like components, we have negative two, two, negative eight. Similarly, ππ is equal ππ minus ππ. And thatβs equal to negative four, one, negative 10. We know that the cross product of ππ with ππ 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. Weβre 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. Thatβs with the cross product of ππ and ππ. Now if all four points lie in the plane, then the vector ππ must also lie in the plane. The vector ππ is given by ππ minus ππ. And that is negative five, negative 10, π plus two.

So taking the scalar triple product of ππ with ππ and ππ, 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.