Vector Equation Practice Problems

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Hello guys, I've got two questions here which I'm really unsure about on how to actually tackle. As a result, maybe it would be better if I could possibly be linked to some resources where I could read up on how to actually solve them.

Homework Statement

1. Find the equation of the plane in 3 dimensions through the point (1, 0, 1) perpendicular to the vector i + j - k

2. What is the parametric form of the line joining the points (5, 0, 1) and (7, 4, 7)

(Sorry if I posted this in the wrong forum).

Thanks

(I think I might have worked out question 1.)
 
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You haven't posted in the wrong but you also haven't shown any work! We won't do your work for you. Both of those look to me like simple applications of the definitions. What formulas does your textbook give for equations of line and plane?
 
If you want us to check your answer to the first question, we will willingly do so.

As for the second one: What is the general form of the parametric form for a line? You will need a point on the line and a vector pointing in the direction of the line, won't you?
 
HallsofIvy said:
You haven't posted in the wrong but you also haven't shown any work! We won't do your work for you. Both of those look to me like simple applications of the definitions. What formulas does your textbook give for equations of line and plane?

I was hoping I could be linked to some resources to find the answer actually.

Pere Callahan said:
If you want us to check your answer to the first question, we will willingly do so.

Equation of a plane takes the form of Ax + By + Cz + D = 0

Point = (1, 0, 1)
Normal point = (i + j - k) = (1, 1, -1)

Point Normal Form
= (1, 1, -1) x ((x, y, z) - (1, 0, 1)
= 1(x -1) + 1(y-0) -1(z-1)
= x - 1 + y - z + 1

Equation of Plane
= x + y - z = 0

I don't really know if that's right though.

Pere Callahan said:
As for the second one: What is the general form of the parametric form for a line? You will need a point on the line and a vector pointing in the direction of the line, won't you?

From my notes, it's OP(t) = (1-t)OA + (t)OB

I think I should maybe take a look through my notes again, see if I can figure this out. A check on the first question would be great though.

Thanks for the responses guys.
 
Crosshash said:
Equation of Plane
= x + y - z = 0
Correct, well done.


Crosshash said:
From my notes, it's OP(t) = (1-t)OA + (t)OB
Yes, or equivalently

OP(t)=OA + t BA

In both equations, A,B are two points on the line.

Crosshash said:
I think I should maybe take a look through my notes again, see if I can figure this out.

Good idea. Just post your answer if you'd like to have it checked.
 
Pere Callahan said:
Correct, well done.

Thanks a lot

Pere Callahan said:
Yes, or equivalently

OP(t)=OA + t BA

In both equations, A,B are two points on the line.



Good idea. Just post your answer if you'd like to have it checked.

Yeah, just got back, I'll take a stab at this now and post my result :)
 
Ok, well, I'm really not sure what I'm doing but I did this:

OP(t) = OA + (t)(BA) = (1-t)OA + (t)OB

(1-t)(5,0,1) + (t)(7,4,7)
(-6t + 6) + (7t + 4t + 7t)
12t + 6

OP(t) = 12t + 6

Surely there's more to it then that?
 
Crosshash said:
Ok, well, I'm really not sure what I'm doing:
Never mind, to err is human:smile:
Crosshash said:
(1-t)(5,0,1) + (t)(7,4,7)
(-6t + 6) + (7t + 4t + 7t)
12t + 6

How did you get from the first line to the second?? In the first line you add two multiples of 3-vectors whule in the secind line you have just a singe real number!

Remember that OP(t) should be a three-component vector. You're essentially done after the first line you wrote. There's no more to the question than inserting the given points into the formula. (If you are only interested in the result, that is. From the mistakes you made I wouldn't be surprised if there is a lot more work to do of you want to get a thorough understanding of what you acutally do, not only how)
 
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Crosshash said:
Ok, well, I'm really not sure what I'm doing but I did this:

OP(t) = OA + (t)(BA) = (1-t)OA + (t)OB

(1-t)(5,0,1) + (t)(7,4,7)
(-6t + 6) + (7t + 4t + 7t)
12t + 6

OP(t) = 12t + 6

Surely there's more to it then that?
Yes, there is! OP(t) is a vector, not a number! (1-t)(5, 0, 1) does NOT mean 1-t times the sum 5+ 0+ 1, it is the vector (5(1-t), 0, 1-t) and tOB= (7t, 4t, 7t). Add those.
 
  • #10
ah, I see the error of my ways, ok:

OP(t) = OA + (t)(BA) = (1-t)OA + (t)OB

(1-t)OA = (5 -5t, 0, 1-t)
(t)OB = (7t, 4t, 7t)

OP(t) = (5 -5t, 0, 1-t) + (7t, 4t, 7t)
OP(t) = (5 + 2t, 4t, 1 + 6t)

So that's the answer? I think I'll have to do more research on this regardless?
 
  • #11
Crosshash said:
OP(t) = (5 + 2t, 4t, 1 + 6t)

So that's the answer?
Indeed.
Crosshash said:
I think I'll have to do more research on this regardless?

It's not so much about research as about practicing, studying your notes, textbook and whatever else you can find and not forgetting to think about what you're actually doing (for example what should be a vector, what should be a number, what does an equation mean geometrically and so on)
 
  • #12
Pere Callahan said:
Indeed.

Thankyou both very much for the help


Pere Callahan said:
It's not so much about research as about practicing, studying your notes, textbook and whatever else you can find and not forgetting to think about what you're actually doing (for example what should be a vector, what should be a number, what does an equation mean geometrically and so on)

Yeah, I completely understand where my problem is, I'll definitely do this. Thanks again.
 
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