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
Jan 28, 2017

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 #((7),(3),(3)) #

The new line, (whose equation we seek) passes through
#P(-1,1,-4)#, and has the same direction vector, so it has the equation:

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

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So now we need to find the coordinates where this new line, given by #vec(s)# meets the coordinate planes,

For the #x#-axis, #-1+7lamda = 0=> lamda=1/7#, giving

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

For the #y#-axis, #1+3lamda = 0=> lamda=-1/3#, giving

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

For the #z#-axis, #-4+3lamda = 0=> lamda=4/3#, giving

# 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))#