What is the vector representation, parametric equations and rectangular equations for the line through the points P(3,2,1) and Q(-1,2,4)?

Question

1845 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-04T17:31:41

# vecr=<<3,2,1>> + lambda <<-4,0,3>>#, or, # vecr=((3),(2),(1)) + lamda ((-4),(0),(3)) #

Parametric Equation:

#{: (x=3-4lamda), (y=2), (z=1+3lamda) :} #

Cartessia Equation:

# (3-x)/4 = (z-1)/3 " "; y=2 #

We have:

# vec(OP)=<<3,2,1>> #
# vec(OQ)=<<-1,2,4>> #

So a line passing through #P# and #Q# will be in the direction;

# vec(PQ) = vec(OQ) - vec(OP) #
# " "= <<-1-3,2-2,4-1>> #
# " "= <<-4,0,3>> #

The vector equation of a straight line (using #vec(OP)#, but equally we could use #vec(OQ)#) is given by;

# \ \ \ \ \ vecr=veca + lambda vecd#
# :. vecr=<<3,2,1>> + lambda <<-4,0,3>>#

Or, in column notation

# vecr=((3),(2),(1)) + lamda ((-4),(0),(3)) #

For parametric equations we jus extract the #x#, #y# and #z# components, leaving lmda# is the parameter:

# {: (x=3-4lamda), (y=2+0lamda), (z=1+3lamda) :} => {: (x=3-4lamda), (y=2), (z=1+4lamda) :} #

For Cartesian equations we use the parameters form and eliminate the parameter:

# {: (x=3-4lamda), (y=2), (z=1+4lamda) :} => {: (lamda=(3-x)/4), (y=2), (lamda=(1-z)/3) :} #

# :. (3-x)/4 = (z-1)/3 " "; y=2 #