a) u = (-1, 2, -1) , v = (2, -1, 0) , w = (1, 1, -1)
b) you want the cross-product of u and v, that is,
a vector which is perpendicular to both u and v
I don't know which method you have learned, but you should get (1, 2, 3)
c) the normal to the triangle is (1,2,3)
so the equation of the plane containing the triangle is
x + 2y + 3z = k , plug in A(-1,-2,0)
-1 -4 + 0 = k = -5
x + 2y + 3z = -5
(check by subbing in all 3 given points, it works!)
c) using A and the normal
x = -1 + k
y = -2 + 2k
z = 0 + 3k
d) a possible cartesian equation using A and the normal is
(x+1)/1 = (y+2)/2 = z/3
e)
let r be the vector, the line containing AC
r = (-1, -2, 0) + t(1, 2, 3)
Let the points A = (−1, −2, 0), B = (−2, 0, −1), C = (0, −1, −1) be the vertices of a triangle.
(a) Write down the vectors u=AB (vector), v=BC(vector) and w=AC(vector)
(b) Find a vector n that is orthogonal to both u and v.
(c) Find a parametric equation for the plane on which the triangle lies.
(d) Find a Cartesian equation for the plane on which the triangle lies.
(e) Find a parametric equation for the line containing the points A and C.
Thanks for help
1 answer