Question

(a) Find a vector perpendicular to the plane through the points A(1, 0, 0), B(2, 0, -1), and C(1, 4, 3).
(b) Find the area of triangle ABC.

Answer 1

a. To find a vector perpendicular to the plane via the factors A, B, and C, we are able to use the cross products of two vectors in the plane. Let's first find vectors that lie in the plane.

We can begin through finding the vector AB, through subtracting the coordinates of A from the coordinates of B:

AB = (2, 0, -1) - (1, 0, 0) = (1, 0, -1)

We also can find the vector AC, through subtracting the coordinates of A from the coordinates of C:

AC = (1, 4, 3) - (1, 0, 0) = (0, 4, 3)

Now we are able to use the cross product to find a vector perpendicular to the plane. The pass obtained from vectors is described as:

cross(u, v) = (u2v3 - u3v2, u3v1 - u1v3, u1v2 - u2v1)

So, we find the cross product of  AB and AC as follows:

cross(AB, AC) = (0 × 3 - (-1) ×4, -1 ×0 - 1 ×3, 1 × 4 - 0 ×0) = (4, -3, 4)

This vector is perpendicular to the plane via the points A, B, and C.

The vector will be (4, -3, 4)

b. To find the area of triangle ABC, we use the cross product to find its area as follows:

area= 1/2 × |cross(AB, AC)|

Substituting  the values we observed above:

area = 1/2 × |(4, -3, 4)|

= 1/2 × √(4^2 + (-3)^2 + 4^2)

= 1/2 × √41  

=6.4/2

So, the area of triangle ABC is 3.2

Read more about vectors at:

brainly.com/question/15074216

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