they give you x
x=-12
so solve for t in the x equation by plugging in the x value
-12=-4-6t
-8=-6t
4/3=t
now plug in the t value for the y and z equation to find the coordinates
The parametric equations for a line L1 are as follows:
x = −4−6t
y = 2+2t
z = 1−4t
Let L2 be the line parallel to L1 and passing through the point (−3, −3, −3). Find the point P on L2 whose x-coordinate is −12.
P=P(-12, _ , _ ) ?
2 answers
It is easy to find the equation of the L2, since it is parallel to L1 and you have a point on it
direction vector of L1 is (-6,2,-4) or reduced:
(3, -1, 2)
a possible line equation for L2 is
x = -3 + 3k
y = -3 - k
z = -3 + 2k
you want x = -12, so -12 = -3 + 3k
3k = -9
k = -3
then y = -3 + 3 = 0
z = -3 - 6 = -9
P on L2 is (-12, 0, -9)
check my arithmetic.
Mia found the point P on L1 , not L2, which has x = -12
direction vector of L1 is (-6,2,-4) or reduced:
(3, -1, 2)
a possible line equation for L2 is
x = -3 + 3k
y = -3 - k
z = -3 + 2k
you want x = -12, so -12 = -3 + 3k
3k = -9
k = -3
then y = -3 + 3 = 0
z = -3 - 6 = -9
P on L2 is (-12, 0, -9)
check my arithmetic.
Mia found the point P on L1 , not L2, which has x = -12