Mathematics

Mathematics

Problem

Statement I Distance of point D(1,0,−1)D(1, 0, -1) from the plane of points A(1,−2,0)A(1, -2, 0) , B(3,1,2)B(3, 1, 2) and C(−1,1,−1)C(-1, 1, -1) is 8229\frac{8}{\sqrt{229}} . Statement II Volume of tetrahedron formed by the points AA , BB , CC and DD is 2292\frac{\sqrt{229}}{2} .
Intuitive explanation

The key is recognizing that both statements hinge on the same geometric object: the plane containing AA, BB, CC. Find the plane's equation using the normal vector AB⃗×AC⃗\vec{AB} \times \vec{AC}, then use the point-to-plane distance formula for Statement I. For Statement II, the tetrahedron volume equals 16∣AD⃗⋅(AB⃗×AC⃗)∣\frac{1}{6}|\vec{AD} \cdot (\vec{AB} \times \vec{AC})|—notice this shares the cross product, and the distance formula connects directly to volume through the base area. Check whether both statements are consistent with each other.

Intuitive explanation

The key is recognizing that both statements hinge on the same geometric object: the plane containing AA, BB, CC. Find the plane's equation using the normal vector AB⃗×AC⃗\vec{AB} \times \vec{AC}, then use the point-to-plane distance formula for Statement I. For Statement II, the tetrahedron volume equals 16∣AD⃗⋅(AB⃗×AC⃗)∣\frac{1}{6}|\vec{AD} \cdot (\vec{AB} \times \vec{AC})|—notice this shares the cross product, and the distance formula connects directly to volume through the base area. Check whether both statements are consistent with each other.

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