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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,
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