Solving Vector Questions: Find Relationship, Point C & Value of k

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In summary: exactly one solution.(ii) infinitely many solutions.should i solve this using a standard matrix? what happens when i get down to x = something, y = something, z = something/k. i don't know really what to do differently for i) and ii), what would each part entail?
  • #1
masterofthewave124
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1. Consider two lines L1: r = (2,0,0) + t(1,2,−1) and L2: r =(3,2,3) + s(p,q,1)
Find a relationship between p and q [independent of s and t] that ensures that L1 and L2 intersect.

i proceeded like any other interesection of line question but got stuck when i got single equations with 3 unknowns (including p and q)


2. The position vectors of the points A and B, with respect to the origin O, are 2i−3j+3k and 5i + j+ck respectively, where c is a constant. The point C is such that OABC is a rectangle.
(a) Find the value of c
(b) Find the point C

how do i find c? if i had the exact coordinates of B, I am pretty sure developing an equation like OA = CB would give be the point C (which is part b))


3. In the following system of equations, k is a real number.
x−2y+z=4
x−y−z=3
x+y+kz=1

(a) Determine the value(s) of k for which the system of equations has
(i) exactly one solution
(ii) infinitely many solutions

should i solve this using a standard matrix? what happens when i get down to x = something, y = something, z = something/k. i don't know really what to do differently for i) and ii), what would each part entail?


i know i haven't shown a tremendous amount of work but I'm not the greatest thinker and i really have a deficiency of knowledge in this area. thanks for any help given!
 
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  • #2
1. for this one, note that 2 lines will intersect iff their direction vectors are not parallel, i.e. not scalar multiples of each other. so do you know of a condition you can put on (p,q,1) such that it is not a multiple of (1,2,-1)?

3. when you put it into matrix form and reduce it, which values of k will give you a row of zeros? for those values, there will be infinitely many solutions. if there are no rows of zeros there will be one solution.
 
  • #3
hmm ill start working with what you said, thanks by the way. any further advice is welcome too!

i also have a question about the distance between two skew lines. geometrically, what is the relationship between the line that represents the distance and the skew lines. the distance line is perpendicular to both no?
 
  • #4
For Q2, use the fact that adjacent sides of a rectangle are perpendicular to each other.
 
  • #5
eok20 said:
1. for this one, note that 2 lines will intersect iff their direction vectors are not parallel, i.e. not scalar multiples of each other. so do you know of a condition you can put on (p,q,1) such that it is not a multiple of (1,2,-1)?
No, that's not true. In 3 dimensions two lines may be "skew", neither parallel nor intersecting. A simple example is the x axis, r= t(1, 0, 0) and the line r= (1, 1, 0)+ t(0, 0, 1), through (1, 1, 0) parallel to the z-axis.

masterofthewave124, you started correctly. in order for the lines, r = (2,0,0) + t(1,2,−1) and L2: r =(3,2,3) + s(p,q,1) to intesect, you must have x= 2+ t= 3+ ps, y= 2t= 2+ qs, z= -t= 3+ s for some t and s. Multiply the last, z, equation by 2 and add to the second, y, equation to eliminate t: 0= 8+ (2+ q)s so s= -8/(2+q). Now use the second equation to determine t: -t= 3+ s= 3- 8/(2+q)= (3q- 2)/(2+ q) so t= (2- 3q)/(2+ q). Finally, those two values must satisfy the first, x, equation:
2+ t= 3+ ps or 2+ (2-3q)/(2+q)= 3- 8p/(2+q).
(6- q)/(2+q)= (6+ 3q- 8p)/(2+q). In order for that to be true, and the two lines intersect, we must have 6- q= 6+3q- 8p or q= 2p (and, of course, q not equal to -2).

2. The position vectors of the points A and B, with respect to the origin O, are 2i−3j+3k and 5i + j+ck respectively, where c is a constant. The point C is such that OABC is a rectangle.
(a) Find the value of c
(b) Find the point C
As pizzasky said, to be a rectangle, two sides, here the vectors 2i- 3j+ 3k and 5i+ j+ ck, must be perpendicular. Set the dot product of those vectors equal to 0 and solve for c.
I'm not sure why you say "if i had the exact coordinates of B"- you do. Being told that the "position vector of B, with respect ot the origin O, is 5i+ j+ ck" tells you that the "exact coordinates" of B are (5, 1, c) and you just found c.


3. In the following system of equations, k is a real number.
x−2y+z=4
x−y−z=3
x+y+kz=1

(a) Determine the value(s) of k for which the system of equations has
(i) exactly one solution
(ii) infinitely many solutions
If you are familiar with matrix methods, sure, go ahead and row-reduce the augmented matrix. For what value of k is the last row (excluding the "augmented part") not 0? What does that tell you? For what value of k are all the terms of the last row, (including the "augmented part") equal to 0? What does that tell you? There is, of course, a possible (iii) NO solution. For what value of k does that happen?

You don't have to use matrix methods of course. I you use you old algebraic methods to solve for x, y, and z, you will eventually have to divide by something involving k. For what values of k is that non-zero so you can solve the equation? For what values of k does that give a fraction in which both numerator and denominator are 0 (and what does that mean)? For what values of k (if any) does that give a fraction in which the numerator is non-zero but the denominator is zero?
 
  • #6
HallsofIvy said:
As pizzasky said, to be a rectangle, two sides, here the vectors 2i- 3j+ 3k and 5i+ j+ ck, must be perpendicular. Set the dot product of those vectors equal to 0 and solve for c.
I'm not sure why you say "if i had the exact coordinates of B"- you do. Being told that the "position vector of B, with respect ot the origin O, is 5i+ j+ ck" tells you that the "exact coordinates" of B are (5, 1, c) and you just found c.

what i meant about the exact coordinates of B is i don't have c. i was suggesting that you needed the answer from a) to do part b).

i have a question about your method, being that the rectangle OABC is formed by taking the vertices in order, shouldn't the dot product of A and C be 0 and not A and B? simply because isn't B a diagonal of the rectangle?
 
  • #7
HallsofIvy said:
You don't have to use matrix methods of course. I you use you old algebraic methods to solve for x, y, and z, you will eventually have to divide by something involving k. For what values of k is that non-zero so you can solve the equation? For what values of k does that give a fraction in which both numerator and denominator are 0 (and what does that mean)? For what values of k (if any) does that give a fraction in which the numerator is non-zero but the denominator is zero?

ok, i tried this method and i need some help here:

so x-2y+z = 4 (1)
x-y-z = 3 (2)
x+y+kz = 1 (3)

(1) - (2) = -y + 2z = 1 (4)
(1) - (3) = -3y + z(1-k) = 3 (5)

Then
3x(4) - (5) = 6z - z(1-k) = 0
z(5+k) = 0

it appears that for any value k, z will be zero. if k is -5, z would = 0/0? does that mean for any value k that's not -5, the lines intersect at a point and when k is -5, they intersect at a line?

evidently, i still don't understand the implications of the values of k. (i.e. k is nonzero, the numerator and denominator are zero, the denominator is zero). this is all so confusing.
 
  • #8
Okay, you reduced it to z(5+k)= 0. No, it is not the case that "for any value k, z will be zero"- if k= -5, the equation reduces to 0*z= 0 and z can be anything. You have now answered your question: If z is any number except, -5, z= 0. Putting that into the previous equations
12x- 5= 0 so x= 5/12, and y= -1. That's a single solution. If k= -5, then z can be anything so there are an infinite number of solutions.

"does that mean for any value k that's not -5, the lines intersect at a point and when k is -5, they intersect at a line?"
Yes, that's exactly what it means!
 
  • #9
masterofthewave124 said:
what i meant about the exact coordinates of B is i don't have c. i was suggesting that you needed the answer from a) to do part b).
My experience is, that when a single problem has "part a" and "part b", that's often the case!:rolleyes:

i have a question about your method, being that the rectangle OABC is formed by taking the vertices in order, shouldn't the dot product of A and C be 0 and not A and B? simply because isn't B a diagonal of the rectangle?
Oh, I see your point. I misread the problem. I assumed that OA and OB were two adjacent sides of the rectangle but since it specifically says "2. The position vectors of the points A and B, with respect to the origin O, are 2i−3j+3k and 5i + j+ck respectively, where c is a constant. The point C is such that OABC is a rectangle." It clearly intends that the A and B be adjacent so that OB is a diagonal, not a side of the rectangle.
(Notice that A, B, C, O are points, not vectors. It's better to say that OB is a diagonal rather than "B is a diagonal".)
Of course, OA+ OC= OB so OC= OB- OA= 5i + j+ck - (2i−3j+3k)= 3i+ 4j+ (c-3)k. Put in the value of c you got in (a) and take the dot product of OA and OC to see if those vectors are pependicular.
 
  • #10
But at this point, we still haven't determined the value of c!

edit: nvm, your wording was just a bit confusing. thanks for all your help though!

for part b), should i just sub in the value for c and solve for OC using OB - OA?

has anyone given any consideration to the question: "geometrically, what is the relationship between the line that represents the distance and the skew lines. the distance line is perpendicular to both no?"
 
Last edited:

1. How do I find the relationship between two vectors?

The relationship between two vectors can be determined by calculating their dot product. If the dot product is equal to zero, the vectors are perpendicular. If the dot product is greater than zero, the vectors are acute. If the dot product is less than zero, the vectors are obtuse.

2. How do I find point C on a vector?

To find point C on a vector, you will need to have the coordinates of two other points on the vector. Use the distance formula to calculate the distance between the two points, then use this distance and the direction of the vector to find point C.

3. How do I solve for the value of k in a vector equation?

To solve for the value of k in a vector equation, you will need to use the properties of vectors and algebraic manipulation. First, isolate the vector with the unknown scalar (k) on one side of the equation. Then, use the properties of vectors to simplify the equation and solve for k.

4. What are some common mistakes to avoid when solving vector questions?

Some common mistakes to avoid when solving vector questions include not properly understanding vector properties, not using the correct formulas or equations, and not carefully checking your work for errors. It is also important to pay attention to the direction and orientation of vectors when performing calculations.

5. How can I apply vector concepts to real-world situations?

Vectors are commonly used in physics, engineering, and other scientific fields to represent physical quantities such as force, velocity, and acceleration. They can also be applied in navigation and mapping, computer graphics, and even in sports such as golf and baseball. By understanding vector concepts and properties, you can better understand and analyze real-world situations and problems.

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