How can I do this using vectors or parametric equations?

Question

580.Let A = (4,0,0), B = (0,3,0), and C = (0,0,5).

(a) Draw a diagram, then find coordinates for the point closest to C on segment AB. (b) Find the area of triangle ABC.

(c) Find the length of the altitude drawn from A to BC.

Answer 1

Line AB lies only in x-y plane and C lies only in y-z plane. line AB and line BC intersect and can define a plane. The shortest length of BC will be normal to AB and a line normal to AB intersecting C will lie in the y-z plane, thus B is closest to C.

Area of ABC = 1/2*Mag(CBXCA)=9.6

Height = ABXBC/mag(BC)= 13.9

Answer 2

just a quick rough calculation, but the point on AB closest to C will have z coordinate =0 and be closer to the x axis than to the y axis

and it's probably (2.4, .8, 0) = (2 2/5, 4/5, 0)

a) AB is a line segment of length = 5 = sqr(4^2+3^2) = sqr(16+9) = sqr25

C is on the z axis. You need a 3D graph. A is on the y axis, B on the x axis

closest point on AB to C is probably (2.4,.8,0)

sides of the triangle are: sqr41, sqr34, and 5

use Heron's formula to find the area

A = sqr((s)(s-5)(s-sqr41)(s-sqr34)) where s= the semi perimeter, half of 5+sqr41+sqr34

c) altitude from A to BC= length of the perpendicular line from A to BC

which also = A/BC = Area found in a) above divided by length of BC, length of BC = sqr(3^2 +5^2)=sqr34