Pls Help (1 Viewer)

[Francis006]

if u,v and w are non-zero vectors such that u-v-w=0. It is also given that |u| = √2 |w| and |v| = √3 |w| . Let θ be the angle between u and v, what is the value of sinθ

 

[Lith_30]

The first equation means that all the vectors form a triangle, and from the given lengths we can say that this triangle is scalene. From there you can just use triangle rules to deduce what would be.

 

[ISAM77]

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[askit]

.

|u| = √2 |w| and |v| = √3 |w|

|u|^2 = 2|w|^2 and |v|^2 = 3|w|^2

hence: 2|w|^2 + 3|w|^2 - |w|^2 = 4|w|^2

not sure how you got 1 as your numerator shouldn't it be 2?

 

[liamkk112]

heres my algebraic method, since no one has posted one yet:
(i'm going to use x instead of theta)
first from the dotproduct we know that cosx = (u.v)/|u||v| = (u.v)/(sqrt(6)|w|^2)

then we also know that u-v = w
hence |u-v| = |w| -> |u-v|^2 = |w|^2, now because for any vector a, |a|^2 = a.a:

-> (u-v).(u-v) = |w|^2
u.u + v.v -2u.v = |w|^2
-> |u|^2 + |v|^2 - 2u.v = |w|^2, by the same property that for any vector a, |a|^2 = a.a
-> 2|w|^2 + 3|w|^2 - 2u.v = |w|^2, since we know the magnitudes of u and v
hence u.v = 2|w|^2

so now cosx = 2/sqrt(6)

then u can draw a right angled triangle and u get that sinx = 1/sqrt(3)

 

[liamkk112]

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i think your angle is in the wrong place, leading to the wrong result, as the angle should be here in the dot product, not the angle between the vectors tip to tail:

 

[ISAM77]

|u| = √2 |w| and |v| = √3 |w|

|u|^2 = 2|w|^2 and |v|^2 = 3|w|^2

hence: 2|w|^2 + 3|w|^2 - |w|^2 = 4|w|^2

not sure how you got 1 as your numerator shouldn't it be 2?

(|u|^2) / (2) = (2) / (2 |u| |v| ) = 1 / (|u| |v|)

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i think your angle is in the wrong place.

Oops... That means my pi - theta is actually theta.

So cos theta = root 6 / 6
And by constructing a right triangle: sin theta = root (30) / 6 which is still the same final answer (somehow).
New working is attached as Q00001i.jpg

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heres my algebraic method, since no one has posted one yet:

It's given in the question that |w| = root 3. Unless I'm misinterpreting what OP wrote. The second attached file is the solution using your method - giving the same answer as the cosine rule solution.

Pls correct me if I've done anything wrong. I need to know where my misconception is, if there is one.

 

[liamkk112]

(|u|^2) / (2) = (2) / (2 |u| |v| ) = 1 / (|u| |v|)

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Oops... That means my pi - theta is actually theta.

So cos theta = root 6 / 6
And by constructing a right triangle: sin theta = root (30) / 6 which is still the same final answer (somehow).
New working is attached as Q00001i.jpg

---


It's given in the question that |w| = root 3. Unless I'm misinterpreting what OP wrote. The second attached file is the solution using your method - giving the same answer as the cosine rule solution.

Pls correct me if I've done anything wrong. I need to know where my misconception is, if there is one.

i cant see that |w| = root3 was given in the question, only that |u| = √2 |w| and |v| = √3 |w| and that u-v-w = 0. this is probably why our answers differ, but even if |w| = root3, then |u| = sqrt(6) and |v| = 3, so then u would get that 6 + 9 - 2u.v = 3 which then means that u.v = 6, which is again a different answer to what u got. but as far as i can see |w| = root3 was not given in the question

 

[ISAM77]

i cant see that |w| = root3 was given in the question, only that |u| = √2 |w| and |v| = √3 |w| and that u-v-w = 0. this is probably why our answers differ, but even if |w| = root3, then |u| = sqrt(6) and |v| = 3, so then u would get that 6 + 9 - 2u.v = 3 which then means that u.v = 6, which is again a different answer to what u got. but as far as i can see |w| = root3 was not given in the question

If |v| = √3, how can |v| = 3 ? (bolded)

Also bolded: |w| and |v| = √3. Doesn't this mean |w| = √3 and |v| = √3 ? Otherwise, what does it mean?

 

[liamkk112]

If |v| = √3, how can |v| = 3 ? (bolded)

Also bolded: |w| and |v| = √3. Doesn't this mean |w| = √3 and |v| = √3 ? Otherwise, what does it mean?

if u are assuming that |w| = root3, then because |v| = sqrt(3) |w|, we then get that |v| = sqrt(3) sqrt(3) = 3

i think you are missing that the |w| is there

 

[ISAM77]

if u are assuming that |w| = root3, then because |v| = sqrt(3) |w|, we then get that |v| = sqrt(3) sqrt(3) = 3

i think you are missing that the |w| is there

where does the questions say |v| = root (3) |w|

The questions says |w| AND |v| = root(3) which means they are both equal to root (3)

 

[ISAM77]

oh wait, i see how you are interpretting it. You are reading it as:

|v| = root(2) |w|

Where as, I'm reading it as
|u| = root 2
|v| = root 3
|w| = root 3

It's ambiguous from the OP. (or im dumb. yeah, maybe im dumb haha)

 

[liamkk112]

in latex this is the question:

 

[liamkk112]

oh wait, i see how you are interpretting it. You are reading it as:

|v| = root(2) |w|

Where as, I'm reading it as
|u| = root 2
|v| = root 3
|w| = root 3

It's ambiguous from the OP. (or im dumb. yeah, maybe im dumb haha)

nah bc then why is there another |w| at the end (|w| and |v| = √3 |w| ) i bolded it

 

[ISAM77]

yeah, i completely misread it. My bad

 

[liamkk112]

yeah, i completely misread it. My bad

allg ez mistake

 

[ISAM77]

heres my algebraic method, since no one has posted one yet:
(i'm going to use x instead of theta)
first from the dotproduct we know that cosx = (u.v)/|u||v| = (u.v)/(sqrt(6)|w|^2)

then we also know that u-v = w
hence |u-v| = |w| -> |u-v|^2 = |w|^2, now because for any vector a, |a|^2 = a.a:

-> (u-v).(u-v) = |w|^2
u.u + v.v -2u.v = |w|^2
-> |u|^2 + |v|^2 - 2u.v = |w|^2, by the same property that for any vector a, |a|^2 = a.a
-> 2|w|^2 + 3|w|^2 - 2u.v = |w|^2, since we know the magnitudes of u and v
hence u.v = 2|w|^2

so now cosx = 2/sqrt(6)

then u can draw a right angled triangle and u get that sinx = 1/sqrt(3)

I re-did it and went the exact same route as you. I agree with you 100% now.

Thx for putting up with me

 

[liamkk112]

I re-did it and went the exact same route as you. I agree with you 100% now.

Thx for putting up with me

lmao