Question

Suppose we have the triangle with vertices P(1, 3, 1), Q(-4, 9, -3), and R(6, -4, 1). Answer the following questions. 1. Find a non-zero vector orthogonal to the plane through the points P, Q, and R. Answer: 2. Find the area of the triangle PQR. Area: 

          Suppose we have the triangle with vertices P(1, 3, 1), Q(-4, 9, -3), and R(6, -4, 1). Answer the following questions.
1. Find a non-zero vector orthogonal to the plane through the points P, Q, and R.
Answer:
2. Find the area of the triangle PQR.
Area:
Show more…
Suppose we have the triangle with vertices P(1, 3, 1), Q(-4, 9, -3), and R(6, -4, 1). Answer the following questions. 1. Find a non-zero vector orthogonal to the plane through the points P, Q, and R. Answer: 2. Find the area of the triangle PQR. Area:

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Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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Suppose we have the triangle with vertices P(1,3,1), Q(-4,9,-3), and R(6,-4,1). Answer the following questions. 1. Find a non-zero vector orthogonal to the plane through the points P, Q, and R. Answer: 2. Find the area of the triangle PQR. Area:
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Suppose we have the triangle with vertices P(1,3,1), Q(-4,9,-3), and R(6,-4,1). Answer the following questions. 1. Find a non-zero vector orthogonal to the plane through the points P, Q, and R. Answer: 2. Find the area of the triangle PQR. Area:

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Transcript

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00:01 Hi, given a triangle with vertices p as 131, q as minus 4, 9 and minus 3 and r as 6 minus 4 1.
00:13 Now in first we have to find a non -zero vector orthogonal to the plane through the points pqr.
00:21 That is we have to find the normal vector to the plane passing through pqr.
00:26 So equation of plane passing through pqr.
00:30 Q and r is given by the formula x minus x1 x2 minus x1 x3 minus x1 y minus y 1 y 2 minus y 1 y 2 minus y 1 y 3 minus y 1 z z 1 z 2 minus z 1 and z 3 minus z 1 equal to 0 so now putting this is x 1 y 1 1 x2 y 2 x 3 y 3 so this comes out to be equal to x minus 1 minus 5 and 5 y minus 3, 6 and minus 7, z minus 1 minus 4 and 0 equal to 0.
01:06 So from here we get the equation of the plane as 28x plus 20y minus 5 z minus 83 is equal to 0.
01:15 So we had to find the normal vector.
01:18 So therefore normal vector is equal to 28i plus 20 j minus 5k...
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