How do you find the unit vector that is orthogonal to the plane through the points P = (3, -3, 0), Q = (5, -1, 2), and R = (5, -1, 6)?

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Answer 1 · 2017-01-09T14:19:35

The reqd. unit normal $i/sqrt2-j/sqrt2$ .

The given points $P, Q, and, R$ are co-planer; therefore so are the

vectors $vec(PQ) and vec(QR).$

Knowing that $vec n=vec(PQ) xx vec(QR)$ is orthogonal to both of

these vectors, $vec n$ is the normal vector of the plane containing

the points $P,Q, &, R$ .

Now, $vec(PQ)=(5,-1,2)-(3,-3,0)=(2,2,2)=2(1,1,1),$

$vec(QR)=(0,0,4)=4(0,0,1)$

$vecn=8|(i,j,k),(1,1,1),(0,0,1)|=8(i-j) rArr ||vecn||=8sqrt2.$

Hence, the reqd. unit normal $hatn=vecn/||vecn||=i/sqrt2-j/sqrt2$ .

Enjoy Maths.!