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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.
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Weβre given four points, which weβre told are coplanar; that is, they all lie in the same plane.
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And weβre asked to find the value of the unknown constant π.
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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.
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Before we do this, however, we need to check that our three known points are noncolinear.
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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.
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Letβs label our points π, π, π, and π.
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And we can take any two of the vectors formed by the points π, π, and π.
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Letβs choose ππ and ππ.
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We know that ππ is equal to ππ minus ππ.
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And subtracting like-for-like components, we have negative two, two, negative eight.
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Similarly, ππ is equal ππ minus ππ.
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And thatβs equal to negative four, one, negative 10.
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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.
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Making some room, this evaluates to negative 12, 12, six, which is not equal to the zero vector.
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So we know that our three points are not colinear.
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And we can move on to using the scalar triple product to find the value of π.
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Weβre told that all four points π, π, π, and π are coplanar.
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And we know that the scalar triple product of three coplanar vectors is equal to zero.
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We already have part of our scalar triple product.
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Thatβs with the cross product of ππ and ππ.
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Now if all four points lie in the plane, then the vector ππ must also lie in the plane.
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The vector ππ is given by ππ minus ππ.
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And that is negative five, negative 10, π plus two.
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So taking the scalar triple product of ππ with ππ and ππ, we simply have the scalar product of the two vectors shown.
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This evaluates to negative five times negative 12 plus negative 10 times 12 plus π plus two times six.
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That is 60 minus 120 plus six π plus 12, which simplifies to six π minus 48.
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If the vectors are coplanar, this must be equal to zero.
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Adding 48 to both sides, this gives us six π is equal to 48.
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And dividing both sides by six, this gives us π is equal to eight.
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Hence, the value of π for which the four given points all lie in a single plane is π is equal to eight.