Find the equation of a line in vector form that is parallel to 3x−2y+z= 4, passes through the point (2,0,1), and is perpendicular to r(t) =〈2−t,2t,1 +t〉.

Answer 1

Adam does a good job and his answer matches mine.

He uses cross product to find the direction vector for line

while I use scalar dot product.

the perpendicular line direction (-1,2,1)dot(a,b,c)=0

when vector (a,b,c) is perpendicular to (-1,2,1)

-a+2b+c=0

the normal to the plane direction (3,-2,1)dot(a,b,c)=0

when vector (a,b,c) is perpendicular to (3,-2,1)

3a-2b+c=0

subtracting find

a=b, and then c=-a.

choose a=1

line equation at point (2,0,1)

r(t)=(1*t+2,1*t+0,-1*t+1)

Answer 2

The normal vector of the plane is n = < 3, -2 ,1 > and the directing vector of the given line

is v = < -1 , 2 , 1 >.

Since the line in question is orthogonal to both n and v , then its directing vector u is any scalar multiple of

n x v = < 3, -2 ,1 > x < -1 , 2 , 1 > = < -4 ,-4, 4 >

Then a "nice" u = < 1, 1, -1 > and the parametric equations of the unknown line are

x = 2 + t

y = t

z = 1 - t