Expert: Paul Klarreich - 1/26/2006

Question
I worked it as i was taught but i didnt get the answer. the ans. to the foll. question is:

r . (2,-6,3)= 4
cosine of the acute angle is: 5/63*root 21

THE QUES.
the position vectors a,b,c,d of the points A,B C,D respestively are given by: a = 2i+j+2k, b = -7i-j+4k       c = 7i+2j-2k , d = -8i-2k.

find in the form (r.n=p) (THE "N" IS A UNIT NORMAL) the equation of the plane containing A and B and parallel to the line CD.
find the cosine of the acute angle between the directions of the lines AC and BD.

can u plz give a detail workin or explanation to how u arrived at the ans.
thnx.

Answer
Hi, Jason,

Your question:
I worked it as i was taught but i didnt get the answer. the ans. to the foll. question is:

r . (2,-6,3)= 4
cosine of the acute angle is: 5/63*root 21

THE QUES.

the position vectors a,b,c,d of the points A,B C,D respestively are given by:
a =   2i +  j + 2k
b = - 7i -  j + 4k
c =   7i + 2j - 2k
d = - 8i      - 2k

find in the form (r.n=p) (THE "N" IS A UNIT NORMAL) the equation of the plane containing A and B and parallel to the line CD.

find the cosine of the acute angle between the directions of the lines AC and BD.

can u plz give a detail workin or explanation to how u arrived at the ans.
thnx.

-----------------------------------
I hate these questions.  No offense, but although the mathematics may be quite elegant and interesting, the examples are just messy and a little dropped sign here screws up the whole thing.

Good news.  (Bad news?)  I didn't get those answers, either.  Here is my arithmetic:  

STANDARD WARNING: THE MATERIAL BELOW MAY CONTAIN FRACTIONS AND OTHER MATERIAL INAPPROPRIATE FOR CERTAIN COMPUTING SYSTEMS.  BE SURE TO VIEW IT IN A FIXED-SIZE FONT, SUCH AS COURIER.

The points are: I don't always use the i,j,k's, but I don't think you'll have trouble with the notation.
a = (2,  1, 2)
b = (-7, -1, 4)
c = (7,  2, -2)
d = (-8, 0, -2)

Now the vector cd = (-15,-2, 0)  and the vector ab = (-9,-2,2) are both vectors in the desired plane.  So we need a vector normal to both.  One way to find it is to solve the equations:

-15x - 2y = 0
-9x - 2y + 2z = 0

Set x arbitrarily, since there are three variables.  I'll pick x = 2

-30 - 2y = 0
y = -15
-9(2) -2(-15) + 2z = 0
-18 + 30 + 2z = 0
12 + 2z = 0
z = -6

So a 'normal' vector is  (2, -15, -6)

Another way is to take the cross product: AB X CD, which I learned as the following determinant:

|  i   j   k  |
| -9  -2   2  |
|-15  -2   0  |

= i(0 + 4) + j(-30 - 0) + k(18 - 30) = 4i - 30j - 12k  which is a scalar multiple of  2i - 15j - 6k, exactly as we found before.

And the equation is in the form:

2x - 15y - 6z = K, where K is found by substituting A (or B)

Using a = (2,  1, 2)

2(2) - 15(1) - 6(2) = 4 - 15 - 12 = - 23

Using b = (-7, -1, 4)

2(-7) - 15(-1) - 6(4) = - 14 + 15 - 24 = - 23  (same thing)

So the equation is:

2x - 15y - 6z = - 23

Or, in your form:  r . (2,-15,-6) = -23

Actually, we are being a bit lazy here.  We are supposed to divide the vector (2,-15,-6) by its length, sqrt(265), to get a unit normal vector, but it's getting late.  I hope you understand.
-----------------------------------------------

Now about that cosine problem, which is unrelated:

find the cosine of the acute angle between the directions of the lines AC and BD.
(Repeating)
a = (2,  1, 2)
b = (-7, -1, 4)
c = (7,  2, -2)
d = (-8, 0, -2)

ac = (5,1,-4)        bd = (-1,1,-6)

The cosine of the angle between the lines is the same as the cosine of the angle between the two vectors, which is: (t = theta)
       ac (dot) bd
cos t = -----------
        |ac| |bd|

  (5,1,-4).(-1,1,-6)           -5 + 1 + 24
= ------------------------- = ----------------
 sqrt(25+1+16)sqrt(1+1+36)   sqrt(42)sqrt(38)

       20
= -------------------
 2 sqrt(21) sqrt(19)


      10
= -----------------
 sqrt(21) sqrt(19)

Well, the sqrt(21) came up, at least.