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

2023 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-04T17:31:41

$\to r = ⟨ 3, 2, 1 ⟩ + λ ⟨ - 4, 0, 3 ⟩$ $vecr=<<3,2,1>> + lambda <<-4,0,3>>$ , 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:

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

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