The question 1B-6?

- anonymous

The question 1B-6?

- Stacey Warren - Expert brainly.com

Hey! We 've verified this expert answer for you, click below to unlock the details :)

- chestercat

I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!

- phi

can you post the question?

- phi

is it
what is the locus of points for the equation
\[ \vec{OP} \cdot \textbf{u}= c | \vec{OP}| \]
?

- 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

Looking for something else?

Not the answer you are looking for? Search for more explanations.

## More answers

- 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

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

now consider c=0
any vector OP perpendicular to u , when dotted with u will give 0
|dw:1439222905180:dw|

- 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

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

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

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

for 0 < |c| < 1 we get a cone

- anonymous

Ok Thanks Phi, I will see it carefully, thanks by help me!

- phi

what part of the answer is not clear?

- 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

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

Ok, Itś clear now! Thaks... Now I will remake this exercise.

- 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

it is just because cos or sen modules can't be larger than 1.
It is | sen(x) | <=1 and |cos(x)|<=1 ?

- 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

and <0,0,0> dot u gives 0
therefore the zero vector \( \vec{\textbf{0}} \) is a solution for |c|>1

- 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

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

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

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

Looking for something else?

Not the answer you are looking for? Search for more explanations.