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find 3 points on the plane...( 4 , 0 , 0) , ( 0 , 6 , 0 ) , & ( -1 , 0 , -1 )

now two vectors v1 = <-4,6,0 > and < 1 , 6 , 1 > = v2

then r = s v1 + t v2 is the plane , x = - 4 s + t , y = 6 s + 6t , z = t are parametric

Vectors U, V W are co-planar, ensure the Cartesian equation of the plane II containing the three vectors U,V,W passing via element (-2,3,7). U= 2i-j+ok V= i-j+3k w= 3i-j-ok This plane is overdescribed. It basically takes 2 non-collinear vectors lieing in a plane and one element to uniquely describe the plane. permit's pick vectors u and v. the traditional vector n, to the plane could properly be desperate with the aid of taking the circulate manufactured from u and v. n = u X v = <2, -a million, a million> X <a million, -a million, 3> = <-2, -5, -a million> Any non-0 different of n is likewise a classic vector to the plane. Multiply with the aid of -a million. n = <2, 5, a million> Now that we've n and the element P(-2, 3, 7) we are in a position to write the equation of the plane. 2(x + 2) + 5(y - 3) + a million(z - 7) = 0 2x + 4 + 5y - 15 + z - 7 = 0 2x + 5y + z - 18 = 0 ______ permit's examine to make beneficial the vector w is in the plane. whether that is it is going to be describable as a distinctive of u and v. And certainly 2u - v = w So w is in the plane as predicted.