How can I answer vector on the triangle
$begingroup$
How can I get the vector OP to be this form :
I think that
- OP = b + t
- OP = a + s
I do not know how to do this problem, can anyone gives any idea and some steps to solve this ?
vector-spaces vectors triangles vector-fields
asked Dec 21 '18 at 2:31
Aster ZenAster Zen
438
$endgroup$
add a comment |
$begingroup$
How can I get the vector OP to be this form :
I think that
- OP = b + t
- OP = a + s
I do not know how to do this problem, can anyone gives any idea and some steps to solve this ?
vector-spaces vectors triangles vector-fields
asked Dec 21 '18 at 2:31
Aster ZenAster Zen
438
$endgroup$
add a comment |
$begingroup$
How can I get the vector OP to be this form :
I think that
- OP = b + t
- OP = a + s
I do not know how to do this problem, can anyone gives any idea and some steps to solve this ?
vector-spaces vectors triangles vector-fields
asked Dec 21 '18 at 2:31
Aster ZenAster Zen
438
$endgroup$
How can I get the vector OP to be this form :
I think that
- OP = b + t
- OP = a + s
I do not know how to do this problem, can anyone gives any idea and some steps to solve this ?
vector-spaces vectors triangles vector-fields
vector-spaces vectors triangles vector-fields
asked Dec 21 '18 at 2:31
Aster ZenAster Zen
438
asked Dec 21 '18 at 2:31
Aster ZenAster Zen
438
asked Dec 21 '18 at 2:31
Aster ZenAster Zen
438
asked Dec 21 '18 at 2:31
Aster ZenAster Zen
438
asked Dec 21 '18 at 2:31
Aster ZenAster Zen
438
438
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Using your definition of s and t, from triangle OAN we have
$$vec{OP} = (1 - s)vec{OA} + svec{ON} = (1 - s)vec{a} + frac{1}{3}svec{b}$$
Hence A = 1, B = 1 and C = 3.
Also from triangle OBM we have
$$vec{OP} = tvec{OM} + (1 - t)vec{OB} = frac{3}{4}tvec{a} + (1 - t)vec{b}$$
Hence D = 3, E = 4 and F = 1.
By comparing these 2 equations, we have
$1 - s = frac{3}{4}t$ and $frac{s}{3} = 1 - t$
The solution is $s = frac{1}{3}$ and $t = frac{8}{9}$
Therefore $vec{OP} = frac{2}{3}vec{a} + frac{1}{9}vec{b}$
answered Dec 21 '18 at 19:33
KY TangKY Tang
50436
$endgroup$
$begingroup$
Oh I get more insights, thxu, but I am confused bcs the right answer is : s= 1/3 and t = 8/9
$endgroup$
– Aster Zen
Dec 22 '18 at 0:54
$begingroup$
And I wonder how is the PN. OA looks like and how can you get them ?
$endgroup$
– Aster Zen
Dec 22 '18 at 1:36
$begingroup$
Sorry to have made a mistake by missing an 's' in the first equation. Now is fixed as shown above.
$endgroup$
– KY Tang
Dec 23 '18 at 15:22
$begingroup$
Oh alright thanks, but can you give more explanation about what PN x OA looks like ?
$endgroup$
– Aster Zen
Dec 24 '18 at 4:17
$begingroup$
PN = (1-s)AN =(1-s)(AO+ON) = (1-s)(-a+b/3)=-2a/3+2b/9 Also OA=a
$endgroup$
– KY Tang
Dec 24 '18 at 19:08
|
show 1 more comment
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using your definition of s and t, from triangle OAN we have
$$vec{OP} = (1 - s)vec{OA} + svec{ON} = (1 - s)vec{a} + frac{1}{3}svec{b}$$
Hence A = 1, B = 1 and C = 3.
Also from triangle OBM we have
$$vec{OP} = tvec{OM} + (1 - t)vec{OB} = frac{3}{4}tvec{a} + (1 - t)vec{b}$$
Hence D = 3, E = 4 and F = 1.
By comparing these 2 equations, we have
$1 - s = frac{3}{4}t$ and $frac{s}{3} = 1 - t$
The solution is $s = frac{1}{3}$ and $t = frac{8}{9}$
Therefore $vec{OP} = frac{2}{3}vec{a} + frac{1}{9}vec{b}$
answered Dec 21 '18 at 19:33
KY TangKY Tang
50436
$endgroup$
$begingroup$
Oh I get more insights, thxu, but I am confused bcs the right answer is : s= 1/3 and t = 8/9
$endgroup$
– Aster Zen
Dec 22 '18 at 0:54
$begingroup$
And I wonder how is the PN. OA looks like and how can you get them ?
$endgroup$
– Aster Zen
Dec 22 '18 at 1:36
$begingroup$
Sorry to have made a mistake by missing an 's' in the first equation. Now is fixed as shown above.
$endgroup$
– KY Tang
Dec 23 '18 at 15:22
$begingroup$
Oh alright thanks, but can you give more explanation about what PN x OA looks like ?
$endgroup$
– Aster Zen
Dec 24 '18 at 4:17
$begingroup$
PN = (1-s)AN =(1-s)(AO+ON) = (1-s)(-a+b/3)=-2a/3+2b/9 Also OA=a
$endgroup$
– KY Tang
Dec 24 '18 at 19:08
|
show 1 more comment
$begingroup$
Using your definition of s and t, from triangle OAN we have
$$vec{OP} = (1 - s)vec{OA} + svec{ON} = (1 - s)vec{a} + frac{1}{3}svec{b}$$
Hence A = 1, B = 1 and C = 3.
Also from triangle OBM we have
$$vec{OP} = tvec{OM} + (1 - t)vec{OB} = frac{3}{4}tvec{a} + (1 - t)vec{b}$$
Hence D = 3, E = 4 and F = 1.
By comparing these 2 equations, we have
$1 - s = frac{3}{4}t$ and $frac{s}{3} = 1 - t$
The solution is $s = frac{1}{3}$ and $t = frac{8}{9}$
Therefore $vec{OP} = frac{2}{3}vec{a} + frac{1}{9}vec{b}$
answered Dec 21 '18 at 19:33
KY TangKY Tang
50436
$endgroup$
$begingroup$
Oh I get more insights, thxu, but I am confused bcs the right answer is : s= 1/3 and t = 8/9
$endgroup$
– Aster Zen
Dec 22 '18 at 0:54
$begingroup$
And I wonder how is the PN. OA looks like and how can you get them ?
$endgroup$
– Aster Zen
Dec 22 '18 at 1:36
$begingroup$
Sorry to have made a mistake by missing an 's' in the first equation. Now is fixed as shown above.
$endgroup$
– KY Tang
Dec 23 '18 at 15:22
$begingroup$
Oh alright thanks, but can you give more explanation about what PN x OA looks like ?
$endgroup$
– Aster Zen
Dec 24 '18 at 4:17
$begingroup$
PN = (1-s)AN =(1-s)(AO+ON) = (1-s)(-a+b/3)=-2a/3+2b/9 Also OA=a
$endgroup$
– KY Tang
Dec 24 '18 at 19:08
|
show 1 more comment
$begingroup$
Using your definition of s and t, from triangle OAN we have
$$vec{OP} = (1 - s)vec{OA} + svec{ON} = (1 - s)vec{a} + frac{1}{3}svec{b}$$
Hence A = 1, B = 1 and C = 3.
Also from triangle OBM we have
$$vec{OP} = tvec{OM} + (1 - t)vec{OB} = frac{3}{4}tvec{a} + (1 - t)vec{b}$$
Hence D = 3, E = 4 and F = 1.
By comparing these 2 equations, we have
$1 - s = frac{3}{4}t$ and $frac{s}{3} = 1 - t$
The solution is $s = frac{1}{3}$ and $t = frac{8}{9}$
Therefore $vec{OP} = frac{2}{3}vec{a} + frac{1}{9}vec{b}$
answered Dec 21 '18 at 19:33
KY TangKY Tang
50436
$endgroup$
Using your definition of s and t, from triangle OAN we have
$$vec{OP} = (1 - s)vec{OA} + svec{ON} = (1 - s)vec{a} + frac{1}{3}svec{b}$$
Hence A = 1, B = 1 and C = 3.
Also from triangle OBM we have
$$vec{OP} = tvec{OM} + (1 - t)vec{OB} = frac{3}{4}tvec{a} + (1 - t)vec{b}$$
Hence D = 3, E = 4 and F = 1.
By comparing these 2 equations, we have
$1 - s = frac{3}{4}t$ and $frac{s}{3} = 1 - t$
The solution is $s = frac{1}{3}$ and $t = frac{8}{9}$
Therefore $vec{OP} = frac{2}{3}vec{a} + frac{1}{9}vec{b}$
answered Dec 21 '18 at 19:33
KY TangKY Tang
50436
edited Dec 23 '18 at 15:21
answered Dec 21 '18 at 19:33
KY TangKY Tang
50436
answered Dec 21 '18 at 19:33
KY TangKY Tang
50436
answered Dec 21 '18 at 19:33
KY TangKY Tang
50436
50436
$begingroup$
Oh I get more insights, thxu, but I am confused bcs the right answer is : s= 1/3 and t = 8/9
$endgroup$
– Aster Zen
Dec 22 '18 at 0:54
$begingroup$
And I wonder how is the PN. OA looks like and how can you get them ?
$endgroup$
– Aster Zen
Dec 22 '18 at 1:36
$begingroup$
Sorry to have made a mistake by missing an 's' in the first equation. Now is fixed as shown above.
$endgroup$
– KY Tang
Dec 23 '18 at 15:22
$begingroup$
Oh alright thanks, but can you give more explanation about what PN x OA looks like ?
$endgroup$
– Aster Zen
Dec 24 '18 at 4:17
$begingroup$
PN = (1-s)AN =(1-s)(AO+ON) = (1-s)(-a+b/3)=-2a/3+2b/9 Also OA=a
$endgroup$
– KY Tang
Dec 24 '18 at 19:08
|
show 1 more comment
$begingroup$
Oh I get more insights, thxu, but I am confused bcs the right answer is : s= 1/3 and t = 8/9
$endgroup$
– Aster Zen
Dec 22 '18 at 0:54
$begingroup$
And I wonder how is the PN. OA looks like and how can you get them ?
$endgroup$
– Aster Zen
Dec 22 '18 at 1:36
$begingroup$
Sorry to have made a mistake by missing an 's' in the first equation. Now is fixed as shown above.
$endgroup$
– KY Tang
Dec 23 '18 at 15:22
$begingroup$
Oh alright thanks, but can you give more explanation about what PN x OA looks like ?
$endgroup$
– Aster Zen
Dec 24 '18 at 4:17
$begingroup$
PN = (1-s)AN =(1-s)(AO+ON) = (1-s)(-a+b/3)=-2a/3+2b/9 Also OA=a
$endgroup$
– KY Tang
Dec 24 '18 at 19:08
$begingroup$
Oh I get more insights, thxu, but I am confused bcs the right answer is : s= 1/3 and t = 8/9
$endgroup$
– Aster Zen
Dec 22 '18 at 0:54
$begingroup$
Oh I get more insights, thxu, but I am confused bcs the right answer is : s= 1/3 and t = 8/9
$endgroup$
– Aster Zen
Dec 22 '18 at 0:54
$begingroup$
And I wonder how is the PN. OA looks like and how can you get them ?
$endgroup$
– Aster Zen
Dec 22 '18 at 1:36
$begingroup$
And I wonder how is the PN. OA looks like and how can you get them ?
$endgroup$
– Aster Zen
Dec 22 '18 at 1:36
$begingroup$
Sorry to have made a mistake by missing an 's' in the first equation. Now is fixed as shown above.
$endgroup$
– KY Tang
Dec 23 '18 at 15:22
$begingroup$
Sorry to have made a mistake by missing an 's' in the first equation. Now is fixed as shown above.
$endgroup$
– KY Tang
Dec 23 '18 at 15:22
$begingroup$
Oh alright thanks, but can you give more explanation about what PN x OA looks like ?
$endgroup$
– Aster Zen
Dec 24 '18 at 4:17
$begingroup$
Oh alright thanks, but can you give more explanation about what PN x OA looks like ?
$endgroup$
– Aster Zen
Dec 24 '18 at 4:17
$begingroup$
PN = (1-s)AN =(1-s)(AO+ON) = (1-s)(-a+b/3)=-2a/3+2b/9 Also OA=a
$endgroup$
– KY Tang
Dec 24 '18 at 19:08
$begingroup$
PN = (1-s)AN =(1-s)(AO+ON) = (1-s)(-a+b/3)=-2a/3+2b/9 Also OA=a
$endgroup$
– KY Tang
Dec 24 '18 at 19:08
|
show 1 more comment
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