Can you help me resolve this problem? Topic: vectors, lines & planes. ?

Question

I have a test the day after tomorrow, so I was going through some old exams and I found this exercise I cannot resolve.

I have a line L given by 2 equations : 3x + y + z = 16 and x - y + z = 6

then I have a plane Q : [(2,3,1),(0,-3,4)] + (1,1,3)

And I'm asked to find a line L' that is contained in the plane Q and that is perpendicular to L.

Thank you very much.

Answer 1

The line L is the line of intersection of the two given planes:

3x + y + z = 16 and x - y + z = 6

The directional vector u, of the line will be orthogonal to the normal vectors n1 and n2, of the given planes.  Take the cross product.

u = n1 X n2 = <3, 1, 1> X <1, -1, 1> = <2, -2, -4>

Any non-zero multiple of u is also a directional vector of the line.  Divide by 2.

u = <1, -1, -2>

Now we need to find a point on the line.  It will lie in both planes.  Let x = 0 and solve for y and z.

y + z = 16

-y + z = 6

Adding we get:

2z = 22

z = 11

Plug into the first equation and solve for y.

y + 11 = 16

y = 5

A point on the line is A(0, 5, 11).

The equation of the line is:

L = A + tu

L = <0, 5, 11> + t<1, -1, -2>

L = <t, 5 - t, 11 - 2t>

__________

Your notation for the plane Q is completely non-standard.  I can only guess at what it might mean.  I will assume that it means:

Q(s, t) = P + sv + tw

Q(s, t) = <1, 1, 3> + s<2, 3, 1> + t<0, -3, 4>

where s and t are parameters for the directional vectors v and w, respectively

The normal vector n, of Q is orthogonal to any vectors that lie in Q.  Take the cross product.

n = v X w = <2, 3, 1> X <0, -3, 4> = <15, -8, -6>

Let's write the equation of plane Q in Cartesian form.

With a point P(1, 1, 3), in the plane and a normal vector n, of the plane we can write the equation of the plane.  Remember, the normal vector of a plane is orthogonal to any vector that lies in the plane.  And the dot product of vectors is zero.  Define R(x,y,z) to be an arbitrary point in the plane.  Then vector PR lies in the plane.

n • PR = 0

n • <R - P> = 0

<15, -8, -6> • <x - 1, y - 1, z - 3> = 0

15(x - 1) - 8(y - 1) - 6(z - 3) = 0

15x - 15 - 8y + 8 - 6z + 18 = 0

15x - 8y - 6z + 11 = 0

Now we need to find where line L intersects with plane Q.  This will be the point at which perpendicular line L and L' meet.  Plug in the values for the variables x, y, and z in terms of t and solve for t.

15t - 8(5 - t) - 6(11 - 2t) + 11 = 0

15t - 40 + 8t - 66 + 12t + 11 = 0

35t - 95 = 0

t = 95/35 = 19/7

x = t = 19/7

y = 5 - t = 5 - 19/7 = 16/7

z = 11 - 2t = 11 - 2(19/7) = 39/7

The point of intersection is B(19/7, 16/7, 39/7).

Line L' is orthogonal to both line L and vector n, of plane Q.  Take the cross product to get its directional vector v2.

v2 = u X n = <1, -1, -2> X <15, -8, -6> = <-10, -24, 7>

The equation of line L' is:

L' = B + t(v2)

L' = <19/7, 16/7, 39/7> + t<-10, -24, 7>