Need a little bit of vectors help. Calc III

This isn't my hw, but I want to know how to do these questions

1. Consider the point P=(1,3,0).
(a) Find the equation for the plane containing P and lying perpendicular
to the vector <1,2,1>
(b) Find the vector equation for the line passing through P that is orthog-
onal to the plane found in part (a).

I found (a) to be (x-1)+2(y-3)+z=0. I don't know what to do afterwards

2. Consider the vector-valued function r(t) = (e^(t-1),cos(((pi*t)/2))
(a) Find r'(1).
(b) Find a vector orthogonal to the curve r(t) at the point (1,0).

I know how to do a, not (b). Do you just find the normal vector at point (1,0)?

Quote:

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Originally Posted by ohshiznit422

This isn't my hw, but I want to know how to do these questions

1. Consider the point P=(1,3,0).
(a) Find the equation for the plane containing P and lying perpendicular
to the vector <1,2,1>
(b) Find the vector equation for the line passing through P that is orthog-
onal to the plane found in part (a).

I found (a) to be (x-1)+2(y-3)+z=0. I don't know what to do afterwards

2. Consider the vector-valued function r(t) = (e^(t-1),cos(((pi*t)/2))
(a) Find r'(1).
(b) Find a vector orthogonal to the curve r(t) at the point (1,0).

I know how to do a, not (b). Do you just find the normal vector at point (1,0)?

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1. a) is correct. As for part b) lines are simply infinitely long vectors that have been positioned somewhere. What is the vector that goes in the direction perpendicular to that plane? How can you make it infinitely long? How can you position it so it goes through P?

As for 2. b), yes, you should find the normal vector to the curve at (1, 0).

Dude, I have absolutely no idea

Then you need to think about it instead of sitting there expecting us to do it for you... You've been given some instructions...

No, I dont know what step to take. I feel like this requires me to use the cross product. Would I form PQ into a vector and take that cross product with the <1,2,1>?

Wait, I need to find the parametric equations, dont I? x=1-t, y=3-2t, z=t?

Quote:

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Originally Posted by ohshiznit422

Wait, I need to find the parametric equations, dont I? x=1-t, y=2-3t, z=t?

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You COULD find the parametric equations and use these to get the vector form of the line. I however, always get the parametric equations using the vector form of the line. Also, your equations for y and z are incorrect.

I always find it easier to find the equation of a line by finding/using a vector you know goes in the same direction of the line, making it infinitely long by multiplying by a parameter, and finally adding one of the "points" you know lies on the line to this infinitely long vector (really you are adding a vector starting at the origin and ending up at that point).

My y is supposed to be y= 3-2t, right? What is wrong with the z?

I think my professor prefers if we use the parametric equations.

If you read the question you posted, you'll know you're looking for the vector form.

You know that the line goes in the direction of the vector <1, 2, 1>, so to make it infinitely long, multiply it by t, giving <t, 2t, t>, then add the point you know lies on the line, giving <t, 2t, t> + <1, 3, 0> = <t + 1, 2t + 3, t>. So the parametric equations could be x = t + 1, y = 2t + 3, z = t.

What you have done is used the vector, but swapped its direction (which is fine, the inclination is what's important, not which way it points), but to do that, you would have

<-t, -2t, -t> + <1, 3, 0> = <1 - t, 3 - 2t, -t>.

So what should your z actually be?

oh, z=-t lol. i thought it was some large mistake