To find the vector equation for the line through point P=(-3, 2, -1) and Q=(0, -1, 1), we can use the formula:
r = P + t(Q - P)
where r is the position vector of a point on the line, t is a parameter, and P and Q are the given points.
Substituting the values, we have:
r = <-3, 2, -1> + t(<0, -1, 1> - <-3, 2, -1>)
Simplifying:
r = <-3, 2, -1> + t(<0 - (-3), -1 - 2, 1 - (-1)>)
= <-3, 2, -1> + t(<3, -3, 2>)
= <-3 + 3t, 2 - 3t, -1 + 2t>
Therefore, the vector form of the line is:
r = <-3 + 3t, 2 - 3t, -1 + 2t>
For the parametric equations, we can express x(t), y(t), and z(t) in terms of the parameter t:
x(t) = -3 + 3t
y(t) = 2 - 3t
z(t) = -1 + 2t
Therefore, the parametric equations for the line passing through P when t=0 are:
x = -3 + 3t
y = 2 - 3t
z = -1 + 2t
Find the vector and parametric equations for the line through the point P=(−3,2,−1)
and the point Q=(0,−1,1)
.
Vector Form: r=⟨
,
, -1 ⟩+t⟨
,
, 2 ⟩
Parametric form (parameter t
, and passing through P
when t=0
):
x=x(t)=
y=y(t)=
z=z(t)=
1 answer