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Consider the points below. P(1, 0, 1), Q(-2, 1, 4), R(6, 2, 7) (a) Find a nonzero vector orthogonal to the plane through the points P, Q, and R. Need Help? (b) Find the area of the triangle PQR. SCALC9 12.4.029.

User Agelbess
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1 Answer

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Answer:

(a) (0, 3, -1)

(b) (11/2)√10 ≈ 17.3925

Explanation:

Given the points P(1, 0, 1), Q(-2, 1, 4), R(6, 2, 7), you want a normal vector to the plane containing them, and the area of the triangle they form.

Cross product

The cross product of two vectors is orthogonal to both. Its magnitude is ...

|PQ × PR| = |PQ|·|PR|·sin(θ) . . . . where θ is the angle between PQ and PR

The area of triangle PQR can be found from the side lengths PQ and PR as ...

A = 1/2·PQ·PR·sin(θ)

where θ is the angle between the sides.

This means the area of the triangle is half the magnitude of the cross product of two vectors that are its sides.

(a) Orthogonal vector

The attachment shows the cross product of vectors PQ and PR is (0, 33, -11). The components of this vector have a common factor of 11, so we can reduce it to (0, 3, -1).

A normal vector plane PQR is (0, 3, -1).

(b) Area

The area of the triangle is ...

A = 1/2√(0² +33² +(-11)²) = 1/2(11√10)

The area of triangle PQR is (11/2)√10 ≈ 17.3925 square units.

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Additional comment

The area figure can be confirmed by finding the triangle side lengths using the distance formula, then Heron's formula for area from side lengths. The arithmetic is messy, but the result is the same.

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Consider the points below. P(1, 0, 1), Q(-2, 1, 4), R(6, 2, 7) (a) Find a nonzero-example-1
User Tomi Seus
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