Finding Distances and Area???

Question

Q1a) Find the distance of the point A(1,0,1) from the line through P(2,-3,1) and parallel to (-1,2,3)

b) Find the distance from the point B(3,6,1) to the line x = (2-t, 1+4t, 2+t)

 

Q2a) Given points A,B and C with coordinates (-2,0,-3), (3,2,2) and (4,3,1) calculate the area of the triangle ABC

 

Answer 1

1(b).

The point B = (1, 0, 1) and the line x = (2 - t, 1 + 4t, 2 + t) ⇒ x = (2, 1, 2) + t( - 1, 4, 1).

Let the points P = (2, 1, 2) and Q = ( - 1, 4, 1).

Distance d = { || BP cross Q || } / || Q ||.

BP = P - B = <2 - 1, 1 - 0, 2 - 1> = <1, 1, 1>

Answer 2

2(a).

The triangle points are A = (- 2, 0, - 3), B = (3, 2, 2) and C = (4, 3, 1).

Let a, b and c are lengths of the sides BC, AC and AB respectively.

BC = <C - B> = <4 - 3, 3 - 2, 1 - 2> = <1, 1, - 1>

a = || BC || = √[ (1)² + (1)² + (- 1)² ] = √3 ≅ 1.732.

AC = <C - A> = <4 - (- 2), 3 - 0, 1 - (- 3)> = <6, 3, 4>

b = || AC || = √[ (6)² + (3)² + (4)² ] = √[ 36 + 9 + 16 ] = √61 ≅ 7.81.

AB = <B - A> = <3 - (- 2), 2 - 0, 2 - (- 3)> = <5, 2, 5>

c = || AB || = √[ (5)² + (2)² + (5)² ] = √[ 25 + 4 + 25 ] = √54 ≅ 7.348.

Calculate area of triangle by using Heron's formula A = √[s·(s-a)·(s-b)·(s-c).

where s = (a+b+c)/2 = (1.732 + 7.81 + 7.348) / 2 = 8.445.

A = √[ 8.445·(8.445 - 1.732)·(8.445 - 7.81)·(8.445 - 7.348) ]

A = √[ 8.445·(6.713)·(0.635)·(1.097) ]

A = √[ 39.49 ]

A ≅ 6.284 square units.

Answer 3

1(a).

The points are A = (1, 0, 1) and P = (2, - 3, 1).

Let the parallel point Q = (- 1, 2, 3).

Distance d = { || AP cross Q || } / || Q ||.

AP = P - A = <2 - 1, - 3 - 0, 1 - 1> = <1, - 3, 0>