Consider the line which passes through the point P(−1,1,−2), and which is parallel to the line x=1+5t,y=2+2t,z=3+3t Find the point of intersection of this new line with each of the coordinate planes:

The intersection with xy -plane is P(7/3, 7/3, 0), the intersection with xz -plane is P(-7/2, 0, -7/2) and the intersection with yz-plane is P(0, 7/5, 13/5).

Explanation:

The director vector of the line x=1+5t,y=2+2t,z=3+3t is the vector (5,2,3). Since the new line is parallel to the above line then it has the same director vector.

So the equation of the new line is x=-1+5t, y=1+2t, z= -2+3t.

The equation of the xy - plane is z=0. Then

0=-2+3t

t= 2/3

Then x(2/3)= -1+5(2/3)=7/3, y(2/3)=1+2(2/3)=7/3. Then the point of intersection with the xy-plane is P(7/3, 7/3, 0).

The equation of the xz - plane is y=0. Then

0=1+2t

t=-1/2

Then x(1/2)= -1+5(-1/2)=7/2, z(-1/2)=-2+3(-1/2)=-7/2. Then the point of intersection with the xz-plane is P(-7/2, 0, -7/2).

The equation of the yz - plane is x=0. Then

0=-1+5t

t=1/5

Then y(1/5)= 1+2(1/5)=7/5, z(1/5)=-2+3(1/5)=13/5. Then the point of intersection with the yz-plane is P(0, 7/5, 13/5).