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anonymous
 one year ago
The question 1B6?
anonymous
 one year ago
The question 1B6?

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phi
 one year ago
Best ResponseYou've already chosen the best response.1can you post the question?

phi
 one year ago
Best ResponseYou've already chosen the best response.1is it what is the locus of points for the equation \[ \vec{OP} \cdot \textbf{u}= c  \vec{OP} \] ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yes, 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

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01B6 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 halfline) c) empty

phi
 one year ago
Best ResponseYou've already chosen the best response.1if 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
 one year ago
Best ResponseYou've already chosen the best response.1now consider c=0 any vector OP perpendicular to u , when dotted with u will give 0 dw:1439222905180:dw

phi
 one year ago
Best ResponseYou've already chosen the best response.1any 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
 one year ago
Best ResponseYou've already chosen the best response.1if 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
 one year ago
Best ResponseYou've already chosen the best response.0If 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
 one year ago
Best ResponseYou've already chosen the best response.0points 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
 one year ago
Best ResponseYou've already chosen the best response.1for 0 < c < 1 we get a cone

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ok Thanks Phi, I will see it carefully, thanks by help me!

phi
 one year ago
Best ResponseYou've already chosen the best response.1what part of the answer is not clear?

phi
 one year ago
Best ResponseYou've already chosen the best response.1***"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
 one year ago
Best ResponseYou've already chosen the best response.1let 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
 one year ago
Best ResponseYou've already chosen the best response.0Ok, Itś clear now! Thaks... Now I will remake this exercise.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Unfortunately 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
 one year ago
Best ResponseYou've already chosen the best response.0it is just because cos or sen modules can't be larger than 1. It is  sen(x)  <=1 and cos(x)<=1 ?

phi
 one year ago
Best ResponseYou've already chosen the best response.1yes, 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 \[ <0,0,0> \cdot <a,b,c> = 0 \] where I am assuming we are in 3D space.

phi
 one year ago
Best ResponseYou've already chosen the best response.1and <0,0,0> dot u gives 0 therefore the zero vector \( \vec{\textbf{0}} \) is a solution for c>1

phi
 one year ago
Best ResponseYou've already chosen the best response.1v 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
 one year ago
Best ResponseYou've already chosen the best response.0Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.1we can do it algebraically say we have v= <a,b,c> 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
 one year ago
Best ResponseYou've already chosen the best response.1if 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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