Question

(2.1) Find parametric equations of the line that passes through the point P = (2, 0, −1) and is

parallel to the vector ~n =< 2, 1, 3 >.

(2.2) Find parametric equations of the line that passes through the points A = (1, 2, −3) and B =

(7, 2, −4).

(2.3) Find parametric equations for the line of intersection of the planes −5x + y − 2z = 3 and

2x − 3y + 5z = −7.

Answer 1

Vector equation of a line is:

$r=r_0+tv$

where r₀ is position vector of any point on the line.

And v is a vector parallel to the line.

2.1

it is given that (2,0,-1) is a point on the line.

Its position vector is $r_0=2i+0j-k$

Therefore the vector equation of the line is

$r=(2i+oj-k)+t(2i+j+3k)\\r=(2+2t)i+tj+(-1+3t)k$

Therefore the parametric equations are

$x=2+2t, \space y=t, \space z=-1+3t$

2.2

it is given that (1,2,-3) and (7,2,-4) are points on the line.

The direction vector (a,b,c) will be$<7-1,2-2,-4-(-3)>\\<a,b,c>=<6,0,-1>$

The vector equation of the line is

$r=(x_0,y_0,z_0)+t(a,b,c)\\r=(1,2,-3)+t(6,0,-1)$

Therefore the parametric equations are

$x=1+6t, \space y=2, \space z=-3-t$

2.3

$-5x+y-2z=3\\2x-3y+5z=-7$

set z=0, solve for x and y

$-5x+y-0=3\\2x-3y+0=-7$

$y=5x+3\\2x-3(5x+3)=-7\\2x-15x-9=-7\\-13x=2\\x=-\frac{2}{13}$

$y=5(-\frac{2}{13})+3\\=-\frac{10}{13}+3\\=\frac{29}{13}$

point 1: $(x_0,y_0,z_0)=(-\frac{2}{13},\frac{29}{13},0)$

We can use the cross product to find a direction vector for the line

The cross product of the normals to the planes results in a vector that is orthogonal to both and would be parallel to the line of intersection.

$<a,b,c>\times<d,e,f>=<bf-ce, cd-af, ae-bd>\\<-5,1,-2>\times<2,-3,5>=1(5)--2(-3),-2(2)--5(5),-5(-3)-1(2)\\=<-1,21,13>$

plug the point into $(x_0,y_0,z_0)$ and the vector found into <a,b,c>

$x=-\frac{2}{13}-t$

$y=\frac{29}{13}+21t$

$z=13t$