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Suppose we have the triangle with vertices P(9, 0, 0), Q(0, 18, 0), and R(0, 0, 27). Answer the following questions. Find a non-zero vector orthogonal to the plane through the points P, Q, and R. Answer: Find the area of the triangle delta PQR. Area:

Sagot :

The non-zero orthogonal vector to the plane is 486i + 243j + 162j and the area of the triangle is 4374 unit squares.

The vertices of the triangle are P(9, 0, 0), Q(0, 18, 0), and R(0, 0, 27).

So, firstly we should find the vector PQ and PR,

Now,

PR = (0)i + (-18)j + (27)k

PR = -18j + 27j

And PQ,

PQ = (-9)i + (18)j + (0)k

PQ = -9i + 18j

Now, the normal vector which is also a non-zero orthogonal vector,

n = PQ x PR

n = 486i + 243j + 162j

Now, the area of the ΔPQR will be,

Ar(PQR) = [tex]\left[\begin{array}{ccc}9&0&0\\0&18&0\\0&0&27\end{array}\right][/tex]

Solving further,

Ar(PQR) = 4374 unit square.

So, the area of the triangle is 4374 unit square.

To know more about Orthogonal vector, visit,

https://brainly.com/question/2283992

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