is the line through (4, 1, -1) and (2, 5, 3) perpendicular to the line through (-3, 2, 0) and (5, 1, 4)?

Answer 1

yes, if the vectors through those points are perpendicular

vector from (4, 1, -1) to (2, 5, 3): <-2,4,4>
vector from (-3, 2, 0) to (5, 1, 4): <8,-1,0>

to check if those vector are perpendicular, one takes the dot product:
<-2,4,4>.<8,-1,0>= -16-4=-20 which is not cero,
so
the vectors are not perpendicular,
therefore the lines are not perpendicular

Answer 2

vector through the first set of points:
(4, 1, -1) and (2, 5, 3) is v1: [ -2, 4, 4 ]
let us write the cartesian equation of a plane perpendicular to that first vector v1:
-2x+4y+4z = k,
where k is defined by a point on that plane
In fact for the two lines of your question to be perpendicular, the second two points must be on that plane.
let us find k so that (-3,2,0) is on it.
(-2)*(-3)+4*4+4*0 = -6+16 = 10 = k
let us verify whether the second point give the same k
(-2)*5+4*1+4*4 = -10+4+16 = 10 = the same .
The two lines are perpendicular

In addition and for fun, let us find out if the lines cross with one another ...

The plane equation -2x+4y+4z = 10 contains the second line and is perpendicular to the first line. Let us find the intersection of the first line with the plane and check if the intersecting point is also on the second line.

The parametric equation of the first line is (using v1 and the first point):
{ x = -2t+4 ; y = 4t+1 ; z = 4t-1 } where gives you all point on the line.
To find the intersection let us find t that will give you the point on the plane:
(-2)*(-2t+4) + 4*(4t+1) + 4*(4t-1) = 10
4t-8+16t+4+16t-4 = 10
36t-8 = 10
t = 18/36 = 1/2
so the point is (x =3;y=3;z=1)
This point is on the second line is two vectors of the three point are colinear:
v2: [8,-1,4] v3: [2,-2,3], which is clearly not the case .... one is not a multiplier of the other.

Answer 3

First you need to find equations for the lines. From the equations you need to determine if the lines cross at some point. If they don't cross, they aren't perpendicular. If the lines cross, you can find vectors that point in the same direction as the lines. Then you can use the dot product to find the angle between them. If the angle is 90 ( pi/2) then they are perpendicular.

Answer 4

Let A(4 ,1,-1) and B(2,5,3) therefore
vector AB={(2-4),(5-1),3-(-1)}
vector AB={(-2),(4),(4)}..................i

Let C(-3 ,2,0) and D(5,1,4) therefore
vector CD={[(5-(-3)],(1-2),4-(0)}
vector AB={(8),(-1),(4)}..................ii

Vector AB dot CD =(-2)(8) + (4)(-1) +(4)(4)
=-16 -4+16
=-4.................iii
Since dot product is not zero, therefore they are not perpendicular.