Need a little bit of vectors help. Calc III

ohshiznit422

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Oct 4, 2012
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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)?
 
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.
Correct!

I don't know what to do afterwards

Since \(\displaystyle \vec n = \langle 1,2,1 \rangle \) is perpendicular to the plane it must be the direction vector of the line.
A line through P whose staionary vector is \(\displaystyle \vec p\) wit the direction vector \(\displaystyle \vec u\) has the equation:

\(\displaystyle \vec r = \vec p + t \cdot \vec u\)

2. Consider the vector-valued function r(t) = (e^(t-1),cos(((pi*t)/2))
...
 
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)?
Yes, find a vector normal to f'(1). There exist an infinite number of correct answers (that's why they said "a vector ..."), vectors of different lengths but the same or opposite directions.
 
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