# Consider the line which passes through the point P(-1, 1, -4), and which is parallel to the line x=1+7t, y=2+3t, z=3+3t How do you find the point of intersection of this new line with each of the coordinate planes?

##### 1 Answer

#### Answer:

#((0),(10/7),(-25/7))# ,#((-10/3),(0),(-5))# and#((4/3),(5),(0))#

#### Explanation:

The vector equation of the line given by:

# x=1+7t; y=2+3t; z=3+3t #

is:

# vec(r) = ((1),(2),(3)) + t((7),(3),(3)) #

So the line is in the direction of the vector

The new line, (whose equation we seek) passes through

# vec(s) = ((-1),(1),(-4)) +lamda((7),(3),(3)) #

So now we need to find the coordinates where this new line, given by

For the

# vec(s) = ((-1),(1),(-4)) +1/7((7),(3),(3)) = ((0),(10/7),(-25/7))#

For the

# vec(s) = ((-1),(1),(-4)) -1/3((7),(3),(3)) = ((-10/3),(0),(-5))#

For the

# vec(s) = ((-1),(1),(-4)) +4/3((7),(3),(3)) = ((4/3),(5),(0))#

Hence our coordinates are:

#((0),(10/7),(-25/7))# ,#((-10/3),(0),(-5))# and#((4/3),(5),(0))#