Vectors Help

two equations of lines:

L1 (Point A)

r = (9,5,2) + \lambda (-9,8,-2)

L2 (Point B)

r = (31,-3,20) + \mu (2,a,7)

the first part was to find a which i did as the dot product of the direction vectors is zero ,because the lines are perpendicular so:

(-9,8,-2) * (2,a,7) = a - 32

therefore a = 32

so we can re write eq 2 as

r = (31,-3,20) + \mu (2,32,7)

So this is the part i don't understand:

It says find the coordinates of the point C where L1 and L2 intersect?

How do I do this?

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Reply 1

Farhan.Hanif93

Original post by a10

two equations of lines:

L1 (Point A)

r = (9,5,2) + \lambda (-9,8,-2)

L2 (Point B)

r = (31,-3,20) + \mu (2,a,7)

the first part was to find a which i did as the dot product of the direction vectors is zero ,because the lines are perpendicular so:

(-9,8,-2) * (2,a,7) = a - 32

therefore a = 32

so we can re write eq 2 as

r = (31,-3,20) + \mu (2,32,7)

So this is the part i don't understand:

It says find the coordinates of the point C where L1 and L2 intersect?

How do I do this?

Your value of

a

is incorrect, check the dot product again.

To find the intersection (after you've found

a

correctly), you seek

\lambda

and

\mu

such that the following equation holds:

\begin{pmatrix} 9 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -9 \\ 8 \\ -2 \end{pmatrix} = \begin{pmatrix} 31 \\ -3 \\ 20 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ a \\ 7 \end{pmatrix}

Can you see how this gives 3 equations in terms of

\lambda

and

\mu

? Use two of them to find values that work and then verify them in the 3rd equation.

Reply 2

a10

Original post by Farhan.Hanif93

Your value of

a

is incorrect, check the dot product again.

To find the intersection (after you've found

a

correctly), you seek

\lambda

and

\mu

such that the following equation holds:

\begin{pmatrix} 9 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -9 \\ 8 \\ -2 \end{pmatrix} = \begin{pmatrix} 31 \\ -3 \\ 20 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ a \\ 7 \end{pmatrix}

Can you see how this gives 3 equations in terms of

\lambda

and

\mu

? Use two of them to find values that work and then verify them in the 3rd equation.

Okay thanks, i did this with the previous question and worked out lambda and mew. But when i have the values what do i do with them to get the C co-ordinates.

EDIT: I worked out a to be 4?

\begin{pmatrix} 9 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -9 \\ 8 \\ -2 \end{pmatrix} = \begin{pmatrix} 31 \\ -3 \\ 20 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix}

(edited 11 years ago)

Reply 3

ghostwalker

Original post by a10

Okay thanks, i did this with the previous question and worked out lambda and mew. But when i have the values what do i do with them to get the C co-ordinates.

EDIT: I worked out a to be 4?

Yep.

\begin{pmatrix} 9 \\ 5 \\ 2 \end{pmatrix} + \lambda \begin{pmatrix} -9 \\ 8 \\ -2 \end{pmatrix} = \begin{pmatrix} 31 \\ -3 \\ 20 \end{pmatrix} + \mu \begin{pmatrix} 2 \\ 4 \\ 7 \end{pmatrix}

Just substitute the value of lambda into L1, or mu into L2, they should give the same result - the coordinates of C.

In fact it's a useful check to do both.

Reply 4

a10

Original post by ghostwalker

Yep.

Just substitute the value of lambda into L1, or mu into L2, they should give the same result - the coordinates of C.

In fact it's a useful check to do both.

i think i did something wrong on the last part the answers seem a bit obsurd,

lambda = -54/10

mew = -179/5

EDIT: i just checked it in the third equation it seems iv done something wrong :/

I'll try again:

9-9\lambda = 31 + 2\mu

(multiply this eq by 2)

18-18\lambda = 62 + 4\mu

5 + 8\lambda = -3+4\mu

subtracting the two:

13 - 26\lambda = 65[br][br]-26\lambda = 52[br][br]\lambda =-2

(edited 11 years ago)

Reply 5

ghostwalker

Original post by a10

I had:

9-9\lambda = 32 + 2\mu

(multiply this eq by 2)

Where did the 32 come from? It's not correct.

Reply 6

a10

Original post by ghostwalker

Where did the 32 come from? It's not correct.

OMG

hahaha thanks!! it's still early in the morning :cry2:

Reply 7

c471

Original post by a10

9-9\lambda = 32 + 2\mu

(multiply this eq by 2)

18-18\lambda = 64 + 4\mu

5 + 8\lambda = -3+4\mu

subtracting the two:

13 - 26\mu = 67[br][br]-26\mu = 54[br][br]\mu =-27/13

this is full of arithmetic mistakes.

firstly, its 31, not 32, and -18 - 8 isnt 10. [EDIT: you've already corrected that while I was typing!]
your answer is an integer. and of course you switched from lambda to mu halfway through so make sure you get that right or it will mess you up!

(edited 11 years ago)

Reply 8

a10

Original post by c471

haha yeahh thanks I've fixed it now xD :biggrin:

Reply 9

ghostwalker

Original post by a10

OMG

hahaha thanks!! it's still early in the morning :cry2:

Reply 10

a10

Original post by ghostwalker

okay so i got

\mu = -2

and

\lambda = -2

i subbed them into the L1 and L2 equations and i have

On the LHS i have (9,5,2)(18,-16,4) do i multiply these now to get C?

(edited 11 years ago)

Reply 11

ghostwalker

Original post by a10

okay so i got

\mu = -2

and

\lambda = -2

i subbed them into the L1 and L2 equations and i have

On the LHS i have (9,5,2)(18,-16,4) do i multiply these now to get C?

No,

The definition of L1 is r=(9,5,2)+lambda(-9,8,-2)

You just sub the value of lambda in, and then r will be the point of intersection.

I.e. =(9,5,2)-2(-9,8,-2)

Reply 12

a10

Original post by ghostwalker

No,

The definition of L1 is r=(9,5,2)+lambda(-9,8,-2)

You just sub the value of lambda in, and then r will be the point of intersection.

I.e. =(9,5,2)-2(-9,8,-2)

i did that but then i don't understand how we then get the co-ordinates of C because we subbed both values of lambda and mew into either side?

Reply 13

ghostwalker

Original post by a10

i did that but then i don't understand how we then get the co-ordinates of C because we subbed both values of lambda and mew into either side?

Not sure what you're subbing the values of the parameters into.

You only need to sub lambda into L1, and this give you r (the position of C)

Reply 14

a10

Original post by Farhan.Hanif93

Your value of

a

is incorrect, check the dot product again.

Can you see how this gives 3 equations in terms of

\lambda

and

\mu

? Use two of them to find values that work and then verify them in the 3rd equation.

hi,

EDIT: Could you help on the question below? Please :biggrin:

(edited 11 years ago)

Reply 15

aznkid66

Example:

\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 1 \end{pmatrix}

In this case

\lambda = 2

and

\mu = 1

.

Given that those parameters take those values, both equations refer to the point

\begin{pmatrix} 1 \\ 0 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + 1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}

So it doesn't really matter which one you use, as they will both reach the same value.

Reply 16

a10

Original post by aznkid66

Example:

\begin{pmatrix} 1 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + \mu \begin{pmatrix} 0 \\ 1 \end{pmatrix}

In this case

\lambda = 2

and

\mu = 1

.

Given that those parameters take those values, both equations refer to the point

\begin{pmatrix} 1 \\ 0 \end{pmatrix} + 2 \begin{pmatrix} 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 1 \end{pmatrix} + 1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} 3 \\ 2 \end{pmatrix}

So it doesn't really matter which one you use, as they will both reach the same value.

thanks

how do you solve this question?

Point X is the point on L1 such that the vector CX(with the arrow at the top) is perpendicular to the vector AB (again with the arrow at the top, don't know how to do this in latex haha)

a) Find the co-ordinates of X
b) Find the shortest distance of C from L1

A= \begin{pmatrix} 4 \\ 2 \\ -5 \end{pmatrix}

B= \begin{pmatrix} 10 \\ 2 \\ -7 \end{pmatrix}

C= \begin{pmatrix} 12 \\ 5 \\ 9 \end{pmatrix}

I worked out AB and subbed that in L1 to get the equation of the line it came out to be:

r= \begin{pmatrix} 4 \\ 2 \\ -5 \end{pmatrix} + \lambda \begin{pmatrix} 6 \\ 0 \\ -2 \end{pmatrix}

From here i tried to work out vector CX which will be x-c so:

X= \begin{pmatrix} x \\ y \\ z \end{pmatrix}

C= \begin{pmatrix} 12 \\ 5 \\ 9 \end{pmatrix}

Vector CX:

\begin{pmatrix} x \\ y \\ z \end{pmatrix} - \begin{pmatrix} 12 \\ 5 \\ 9 \end{pmatrix} = \begin{pmatrix} x-12 \\ y-5 \\ z-9 \end{pmatrix}

Is this correct?? What do i do next im stuck :/

After trying the dot product i get:

6x -2z= 54?

(edited 11 years ago)

Reply 17

aznkid66

Wait, so are A and B also on the line?

Well, to find the coordinates of X...

Let the coordinates of X be (x,y,z). What is vector CX? What is vector AB? Knowing what CX dot AB is, can you make an equation of x, y, and z?

I don't know how you found the equation of the line, but knowing that...

(x,y,z) is a point on the line, so you can write the x, y, and z components in terms of lambda. Now, can you evaluate your previous equation (hint: substitution)?

There is also a method pertaining to the cross product, but I'm not to fond of that one. You can post for more information if you're more familiar with that method, or just curious.

Reply 18

a10

Original post by aznkid66

Could you please check...I've edited it but don't know how to continue? I tried the dot product being zero since it said perpendicular but then u get two unknowns x and z???

Reply 19

aznkid66

(x,y,z) is a point on the line, which means that there exists a lambda where r=(x,y,z)

For that lambda, in terms of lambda according to the equation of the line x=...?
Similarly, in terms of lambda z=...?