Question

Two planes are orthogonal if their normal vectors are orthogonal. Find the equation of a plane that is orthogonal to the plane 3x − y + 2z = 1 and that contains the points (1, 1, 2) and (2, 2, 1).

Answer 1

Answer:

X-5Y-4Z+12=0

Step-by-step explanation:

Plan P1 = 3x-y+2z=1. with A=(1,1,2); B=(2,2,1); we must will find V1 from P1,

V1 =(3,-1,2) and AB = (2-1,2-1,1-2) = (1,1,-1) now we find V2 by vector product.

V2 =V1.X,AB =$\left[\begin{array}{ccc}i&j&k\\3&-1&2\\1&1&-1\end{array}\right] = i+2j+3k-(2i-3j-k)=V2=(-i+5j+4k)$; (aplying sarrus method)

according to general plan formula a(X-Xo)+b(Y-Yo)+c(Z-Zo); V=(a,b,c)

V2=(-1,5,4) ; A=(1,1,2);  so -(X-1)+5(Y-1)+4(z-2)=0  → -X+1+5Y-5+4Z-8=0  →

-X+5Y+4Z+1-5-8=0 → -X+5Y+4Z-12=0 (-1) → Plan equation X-5Y-4Z+12=0