# A) Show that the line containing the point (1, 7, 5) and (3, 2, -1) is parallel to the line...

## Question:

A) Show that the line containing the point {eq}(1, 7, 5) {/eq} and {eq}(3, 2, -1) {/eq} is parallel to the line containing {eq}(2, -2, 5) {/eq} and {eq}(-2, 8, 17) {/eq}.

B) Find the distance {eq}D {/eq} from the point {eq}(2, 1, 0) {/eq} to the line with equations {eq}\; x = -2, \; y + 1 = z {/eq}.

C) Find the distance {eq}D {/eq} between the lines {eq}x - 2y = 1 {/eq} and {eq}2x - 4y = 3 {/eq} in the {eq}xy {/eq}-plane.

## Vectors:

Suppose 2 points are (a1,b1,c1) and (a2,b2,c2), then the vector joining those 2 points is given by

(a2-a1)i + (b2-b1)j + (c2-c1)k

To get the distance between a point and a line, we take their dot product divided by the magnitude of the line

To find the distance between 2 lines ax+by+c1=0 and ax+by+c2=0, it is given by{eq}\frac{c2-c1}{\sqrt{a^2+b^2}} {/eq}

A) The vector containing the first two points becomes

2i -5j -6k

and the second vector containing the other two points becomes

4i -10j -12k

For these two to be parallel, their cross product should be 0

So

i(-5*-12 - (-6*-10)) -j(2*-12 -(-6*4)) + k(2*-10 -(-5*4)) = 0, So they are parallel

B) The distance of (2,1,0) from x + 2 = 0 is {eq}\frac{2+2}{\sqrt{1^2}} = 4 {/eq}

The distance of (2,1,0) from z-y-1=0 is {eq}\frac{0-1-1}{\sqrt{1^2+1^2}} = 1.41 {/eq}

C) Distance between the two parallel lines

x - 2y = 1 and 2x - 4y = 3

{eq}\frac{1.5-1}{\sqrt{1^2+2^2}} = 0.22 {/eq} 