anonymous
  • anonymous
The question 1B-6?
OCW Scholar - Multivariable Calculus
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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phi
  • phi
can you post the question?
phi
  • phi
is it what is the locus of points for the equation \[ \vec{OP} \cdot \textbf{u}= c | \vec{OP}| \] ?
anonymous
  • anonymous
yes, I can't understand: "Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ" How can I reach this conclusion

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anonymous
  • anonymous
1B-6 Let O be the origin, c a given number, and u a given direction (i.e., a unit vector). Describe geometrically the locus of all points P in space that satisfy the vector equation OP · u = c | OP | . In particular, tell for what value(s) of c the locus will be (Hint: divide through by | OP | ): a) a plane b) a ray (i.e., a half-line) c) empty
phi
  • phi
if we divide both sides by |OP| we get \[ \frac{OP}{|OP|} \cdot u = c \] that is two unit vectors. the maximum value of v dot u is 1 when v and u are unit vectors
phi
  • phi
now consider c=0 any vector OP perpendicular to u , when dotted with u will give 0 |dw:1439222905180:dw|
phi
  • phi
any point P in the plane defined by u (as the normal) will form vector OP, and after dividing by |OP| will still be a vector (of unit length) in that plane. and dotted with u, will give 0 thus c=0 results in a plane
phi
  • phi
if |c| = 1 (c=1 or c=-1) the locus will be a ray if c=1 then any point P where O to P is in the same direction as u will, after being divided by the length of OP will give us unit vector v = u and u dot v = 1
anonymous
  • anonymous
If we applied the equation of Dot Product: \[\left| A \right| \left| B \right| \cos \theta = A.B = \sum_{i }^{...} a _{i}.b_{i}\] W ell, u is a unit vecton, then \[\left| u \right| = 1\] then, \[ \frac{ OP.u }{ \left| OP \right|}=\cos \theta = c\] Off course, this is the angle between these two vector, The points P need stay
anonymous
  • anonymous
points P are at \[\theta \] of in relation to vectior u. But and when the corretion says: Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ 0 . I can understand.
phi
  • phi
for 0 < |c| < 1 we get a cone
anonymous
  • anonymous
Ok Thanks Phi, I will see it carefully, thanks by help me!
phi
  • phi
what part of the answer is not clear?
phi
  • phi
***"Then the locus is the nappe of a right circular cone with axis in the direction u and vertex angle 2θ" How can I reach this conclusion**** |dw:1439223719035:dw|
phi
  • phi
let c= 0.5 and OP/|OP| = v v dot u = 0.5 theta = 60 degrees any point P so that OP is in the direction 60 degrees away from u (in any direction) will be scaled to unit length. and then dotted with u = 0.5
anonymous
  • anonymous
Ok, Itś clear now! Thaks... Now I will remake this exercise.
anonymous
  • anonymous
Unfortunately I can´t get the concept on the last part of this question, c) empty response: Locus is the origin, if c > 1 or c < −1 (division by | OP | is illegal, notice).
anonymous
  • anonymous
it is just because cos or sen modules can't be larger than 1. It is | sen(x) | <=1 and |cos(x)|<=1 ?
phi
  • phi
yes, I was confused on that for a bit. if we want \[ \frac{OP}{|OP|} \cdot u = 2 \] for all P not at the origin, the vector OP divided by its length will give us a unit vector, call it v; and v dot u =2 will never be true. I originally thought this meant "no solution" but if P is the origin, then OP is the zero length vector and \[ \vec{OP} \cdot \textbf{u} = 2| OP| \] becomes \[ \vec{OP} \cdot \textbf{u} = 2\cdot 0= 0 \] or \[ <0,0,0> \cdot = 0 \] where I am assuming we are in 3D space.
phi
  • phi
and <0,0,0> dot u gives 0 therefore the zero vector \( \vec{\textbf{0}} \) is a solution for |c|>1
phi
  • phi
v dot u =2 to elaborate v dot u = |u| |v| cos A because |u| and |v| are both 1 (both are unit vectors) and the max value of cos A is 1 the max value for v dot u is 1 we can never get a value larger than 1
anonymous
  • anonymous
Ok, Let's a simplest way to show "a vector divided by its length" is a unit vector.... I used matlab... >> OP = [-4 -45 1] OP = -4 -45 1 >>mag=sqrt(4^2+45^2+1^2) mag = 45.1885 >>OP_unitVector=OP./mag OP_unitVector = -0.0885 -0.9958 0.0221 mag_unitVector=sqrt(0.0885^2 + 0.9958^2 + 0.0221^2 ) mag_unitVector = 1.0000
phi
  • phi
we can do it algebraically say we have v= then \( |v|= \sqrt{a^2+b^2 + c^2} \) (which follows from pythagoras) and \[ |v|^2 = a^2 + b^2 + c^2 \] \[ \frac{v}{|v|} \cdot \frac{v}{|v|}= \frac{v \cdot v}{|v|^2} \] \[ v \cdot v= a^2 + b^2 + c^2 \] and therefore \[ \frac{v \cdot v}{|v|^2} = \frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1 \]
phi
  • phi
if we let \[ u = \frac{v}{|v|} \] we have shown \[ u \cdot u = |u|^2= 1 \\ \sqrt{|u|^2} = \sqrt{1}\\ |u|= 1\] and therefore \[ \Bigg|\frac{v}{|v|}\Bigg|= 1\] ie. is unit length

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