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

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)

(a) To find the vectors u=AB, v=BC, and w=AC, we can use the coordinates of the given points. The vector from A to B can be found by subtracting the coordinates of A from the coordinates of B:

u = B - A = (-2, 0, -1) - (-1, -2, 0) = (-2 + 1, 0 - (-2), -1 - 0) = (-3, 2, -1)

Similarly, the vector from B to C can be found by subtracting the coordinates of B from the coordinates of C:

v = C - B = (0, -1, -1) - (-2, 0, -1) = (0 - (-2), -1 - 0, -1 - (-1)) = (2, -1, 0)

Finally, the vector from A to C can be found by subtracting the coordinates of A from the coordinates of C:

w = C - A = (0, -1, -1) - (-1, -2, 0) = (0 - (-1), -1 - (-2), -1 - 0) = (1, 1, -1)

(b) To find a vector n that is orthogonal to both u and v, we can take their cross product. We find the cross product by computing the determinant of the matrix formed by u, v, and n:

n = u x v = |i j k |
|-3 2 -1|
|2 -1 0 |

Expanding the determinant, we have:

n = (2 * 0 - (-1) * (-1), -1 * 0 - 2 * (-1), (-3) * (-1) - 2 * 2)
= (1, 2, -5)

Therefore, a vector n orthogonal to both u and v is n = (1, 2, -5).

(c) To find a parametric equation for the plane on which the triangle lies, we need to find a point on the plane and a set of two vectors that lie on the plane. We can choose one of the vertices, say A = (-1, -2, 0), as the point on the plane. Then, we can choose two vectors u = AB = (-3, 2, -1) and w = AC = (1, 1, -1) as the vectors on the plane.

The parametric equation for the plane is given as:

P = A + m * u + n * w

where m and n are parameters that vary over real numbers.

(d) To find a Cartesian equation for the plane on which the triangle lies, we can use the normal vector we found in part (b). The equation of a plane can be written as:

n · (P - A) = 0

where · represents the dot product. Substituting the values, we have:

(1, 2, -5) · (P - (-1, -2, 0)) = 0

Simplifying, we get:

(1, 2, -5) · (x + 1, y + 2, z - 0) = 0

Expanding the dot product, we have:

1 * (x + 1) + 2 * (y + 2) - 5 * (z - 0) = 0

Simplifying further, we obtain the Cartesian equation of the plane:

x + 2y - 5z + 1 - 4 - 0 = 0
x + 2y - 5z - 3 = 0

Therefore, the Cartesian equation for the plane on which the triangle lies is x + 2y - 5z - 3 = 0.

(e) To find a parametric equation for the line containing the points A and C, we can use the point-slope form of a line. Given two points P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂), the parametric equation of the line passing through them is:

x = x₁ + t(x₂ - x₁)
y = y₁ + t(y₂ - y₁)
z = z₁ + t(z₂ - z₁)

Using the points A = (-1, -2, 0) and C = (0, -1, -1), we can plug in the values and obtain:

x = -1 + t(0 - (-1)) = -1 + t(1) = -1 + t
y = -2 + t(-1 - (-2)) = -2 + t(1) = -2 + t
z = 0 + t(-1 - 0) = t(-1)

Therefore, the parametric equation for the line containing the points A and C is:

x = -1 + t
y = -2 + t
z = -t