Vector linear combination problem
$begingroup$
Point $A$ and $B$ have position vectors $vec a$ and $vec b$ respectively relative to an orgin $O$.
The point $D$ is such that $overrightarrow{OD} = koverrightarrow{OA}$ and the point $E$ is such that $overrightarrow{AE} = loverrightarrow{AB}$.
The line segments $BD$ and $OE$ intersect at $X$.
If $overrightarrow{OX} = frac{2}{5}overrightarrow{OE}$ and $overrightarrow{XB} = dfrac{4}{5}overrightarrow{DB}$.
Express $overrightarrow{OX}$ and $XB$ in terms of $vec a, vec b, k, l$ and hence evaluate $k$ and $l$.
I have worked out most of the problem but can't figure out how to evaluate $l$.
Using ratio theorem, I got $overrightarrow{OX}$ as,
$$
overrightarrow{OX} = dfrac{2}{5}Big[(1-l)a + lbBig]
$$
And similarly, $overrightarrow{XB}$
$$
overrightarrow{XB} = dfrac{4}{5}(b - ka)
$$
Then using,
$$
begin{align}
overrightarrow{OX} + overrightarrow{XB} &= overrightarrow{OB} \
dfrac{2}{5}Big[(1-l)a + lbBig] + dfrac{4}{5}(b - ka) &= b\
\
text{...}\
\
[2(1-l) - 4k]a &= (1-2l)b\
2(1-l)- 4k &= 1 - 2l & text{(a and b are non-zero and non-parallel)}\
-4k &= -1 \
k &= dfrac{1}{4}\
end{align}
$$
I can't seem to figure out how to get $l$. I tried $overrightarrow{DA} + overrightarrow{AE} + overrightarrow{E} = overrightarrow{DX}$ and $overrightarrow{OA} + overrightarrow{AE} = overrightarrow{OE}$, these just give an equality statement.
How do I evaluate $l$?
Thanks for your help.
algebra-precalculus
edited Dec 11 '18 at 22:24
Glorfindel
3,41981830
asked Jul 20 '11 at 14:37
mathguy80mathguy80
621720
$endgroup$
add a comment |
$begingroup$
Point $A$ and $B$ have position vectors $vec a$ and $vec b$ respectively relative to an orgin $O$.
The point $D$ is such that $overrightarrow{OD} = koverrightarrow{OA}$ and the point $E$ is such that $overrightarrow{AE} = loverrightarrow{AB}$.
The line segments $BD$ and $OE$ intersect at $X$.
If $overrightarrow{OX} = frac{2}{5}overrightarrow{OE}$ and $overrightarrow{XB} = dfrac{4}{5}overrightarrow{DB}$.
Express $overrightarrow{OX}$ and $XB$ in terms of $vec a, vec b, k, l$ and hence evaluate $k$ and $l$.
I have worked out most of the problem but can't figure out how to evaluate $l$.
Using ratio theorem, I got $overrightarrow{OX}$ as,
$$
overrightarrow{OX} = dfrac{2}{5}Big[(1-l)a + lbBig]
$$
And similarly, $overrightarrow{XB}$
$$
overrightarrow{XB} = dfrac{4}{5}(b - ka)
$$
Then using,
$$
begin{align}
overrightarrow{OX} + overrightarrow{XB} &= overrightarrow{OB} \
dfrac{2}{5}Big[(1-l)a + lbBig] + dfrac{4}{5}(b - ka) &= b\
\
text{...}\
\
[2(1-l) - 4k]a &= (1-2l)b\
2(1-l)- 4k &= 1 - 2l & text{(a and b are non-zero and non-parallel)}\
-4k &= -1 \
k &= dfrac{1}{4}\
end{align}
$$
I can't seem to figure out how to get $l$. I tried $overrightarrow{DA} + overrightarrow{AE} + overrightarrow{E} = overrightarrow{DX}$ and $overrightarrow{OA} + overrightarrow{AE} = overrightarrow{OE}$, these just give an equality statement.
How do I evaluate $l$?
Thanks for your help.
algebra-precalculus
edited Dec 11 '18 at 22:24
Glorfindel
3,41981830
asked Jul 20 '11 at 14:37
mathguy80mathguy80
621720
$endgroup$
add a comment |
$begingroup$
Point $A$ and $B$ have position vectors $vec a$ and $vec b$ respectively relative to an orgin $O$.
The point $D$ is such that $overrightarrow{OD} = koverrightarrow{OA}$ and the point $E$ is such that $overrightarrow{AE} = loverrightarrow{AB}$.
The line segments $BD$ and $OE$ intersect at $X$.
If $overrightarrow{OX} = frac{2}{5}overrightarrow{OE}$ and $overrightarrow{XB} = dfrac{4}{5}overrightarrow{DB}$.
Express $overrightarrow{OX}$ and $XB$ in terms of $vec a, vec b, k, l$ and hence evaluate $k$ and $l$.
I have worked out most of the problem but can't figure out how to evaluate $l$.
Using ratio theorem, I got $overrightarrow{OX}$ as,
$$
overrightarrow{OX} = dfrac{2}{5}Big[(1-l)a + lbBig]
$$
And similarly, $overrightarrow{XB}$
$$
overrightarrow{XB} = dfrac{4}{5}(b - ka)
$$
Then using,
$$
begin{align}
overrightarrow{OX} + overrightarrow{XB} &= overrightarrow{OB} \
dfrac{2}{5}Big[(1-l)a + lbBig] + dfrac{4}{5}(b - ka) &= b\
\
text{...}\
\
[2(1-l) - 4k]a &= (1-2l)b\
2(1-l)- 4k &= 1 - 2l & text{(a and b are non-zero and non-parallel)}\
-4k &= -1 \
k &= dfrac{1}{4}\
end{align}
$$
I can't seem to figure out how to get $l$. I tried $overrightarrow{DA} + overrightarrow{AE} + overrightarrow{E} = overrightarrow{DX}$ and $overrightarrow{OA} + overrightarrow{AE} = overrightarrow{OE}$, these just give an equality statement.
How do I evaluate $l$?
Thanks for your help.
algebra-precalculus
edited Dec 11 '18 at 22:24
Glorfindel
3,41981830
asked Jul 20 '11 at 14:37
mathguy80mathguy80
621720
$endgroup$
Point $A$ and $B$ have position vectors $vec a$ and $vec b$ respectively relative to an orgin $O$.
The point $D$ is such that $overrightarrow{OD} = koverrightarrow{OA}$ and the point $E$ is such that $overrightarrow{AE} = loverrightarrow{AB}$.
The line segments $BD$ and $OE$ intersect at $X$.
If $overrightarrow{OX} = frac{2}{5}overrightarrow{OE}$ and $overrightarrow{XB} = dfrac{4}{5}overrightarrow{DB}$.
Express $overrightarrow{OX}$ and $XB$ in terms of $vec a, vec b, k, l$ and hence evaluate $k$ and $l$.
I have worked out most of the problem but can't figure out how to evaluate $l$.
Using ratio theorem, I got $overrightarrow{OX}$ as,
$$
overrightarrow{OX} = dfrac{2}{5}Big[(1-l)a + lbBig]
$$
And similarly, $overrightarrow{XB}$
$$
overrightarrow{XB} = dfrac{4}{5}(b - ka)
$$
Then using,
$$
begin{align}
overrightarrow{OX} + overrightarrow{XB} &= overrightarrow{OB} \
dfrac{2}{5}Big[(1-l)a + lbBig] + dfrac{4}{5}(b - ka) &= b\
\
text{...}\
\
[2(1-l) - 4k]a &= (1-2l)b\
2(1-l)- 4k &= 1 - 2l & text{(a and b are non-zero and non-parallel)}\
-4k &= -1 \
k &= dfrac{1}{4}\
end{align}
$$
I can't seem to figure out how to get $l$. I tried $overrightarrow{DA} + overrightarrow{AE} + overrightarrow{E} = overrightarrow{DX}$ and $overrightarrow{OA} + overrightarrow{AE} = overrightarrow{OE}$, these just give an equality statement.
How do I evaluate $l$?
Thanks for your help.
algebra-precalculus
algebra-precalculus
edited Dec 11 '18 at 22:24
Glorfindel
3,41981830
asked Jul 20 '11 at 14:37
mathguy80mathguy80
621720
edited Dec 11 '18 at 22:24
Glorfindel
3,41981830
asked Jul 20 '11 at 14:37
mathguy80mathguy80
621720
edited Dec 11 '18 at 22:24
Glorfindel
3,41981830
edited Dec 11 '18 at 22:24
Glorfindel
3,41981830
edited Dec 11 '18 at 22:24
Glorfindel
3,41981830
3,41981830
asked Jul 20 '11 at 14:37
mathguy80mathguy80
621720
asked Jul 20 '11 at 14:37
mathguy80mathguy80
621720
asked Jul 20 '11 at 14:37
mathguy80mathguy80
621720
621720
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
At some point you said
$$
2(1-ell)- 4k = 1-2 ell
$$
But you should've noticed that you could say that because
$$
2(1-ell) - 4k = 0 = 1-2 ell.
$$
I think that gives you $ell = 1/2$.
Hope that helps,
answered Jul 20 '11 at 14:50
Patrick Da SilvaPatrick Da Silva
32k354106
$endgroup$
$begingroup$
Thanks. Can you please clarify why you equated it to 0?
$endgroup$
– mathguy80
Jul 20 '11 at 15:05
$begingroup$
Well, since you have a multiple of $a$ and a multiple of $b$ that are equal, and that those two vectors are not colinear, it must be because they are both the zero vectors, hence their coefficients are both zero. Isn't that why you said that $2(1-ell) - 4k = 1-2ell$ in the first place?
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 15:19
$begingroup$
The proof that I was taught is, $$ $$ Let $a$ and $b$ be two non-zero and non-parallel vectors. If $$alpha a + beta b = lambda a + mu b$$ for some $alpha, beta, lambda, mu in mathbb{R}$, then $$(alpha - lambda)a = (mu - beta)b$$ and hence, $$alpha = lambda text{ and } mu = beta$$. It was assumed that a and b are not zero vectors. Does that hold for zero vectors also?
$endgroup$
– mathguy80
Jul 20 '11 at 16:06
$begingroup$
It does not, indeed. Having $(1-2) vec{0} = vec 0$ does not imply $1=2$. =) But the idea in your proof was that since $a$ and $b$ cannot be parallel, it is impossible for them that a multiple of $a$ is equal to a multiple of $b$, unless they are both $0$, which leads to both coefficients being $0$.
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 16:15
$begingroup$
Ah! Thanks, that makes sense. I was beginning to think I had learned some basic ideas incorrectly!
$endgroup$
– mathguy80
Jul 21 '11 at 2:51
|
show 2 more comments
Your Answer
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1 Answer
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1 Answer
1
active
oldest
votes
active
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active
oldest
votes
$begingroup$
At some point you said
$$
2(1-ell)- 4k = 1-2 ell
$$
But you should've noticed that you could say that because
$$
2(1-ell) - 4k = 0 = 1-2 ell.
$$
I think that gives you $ell = 1/2$.
Hope that helps,
answered Jul 20 '11 at 14:50
Patrick Da SilvaPatrick Da Silva
32k354106
$endgroup$
$begingroup$
Thanks. Can you please clarify why you equated it to 0?
$endgroup$
– mathguy80
Jul 20 '11 at 15:05
$begingroup$
Well, since you have a multiple of $a$ and a multiple of $b$ that are equal, and that those two vectors are not colinear, it must be because they are both the zero vectors, hence their coefficients are both zero. Isn't that why you said that $2(1-ell) - 4k = 1-2ell$ in the first place?
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 15:19
$begingroup$
The proof that I was taught is, $$ $$ Let $a$ and $b$ be two non-zero and non-parallel vectors. If $$alpha a + beta b = lambda a + mu b$$ for some $alpha, beta, lambda, mu in mathbb{R}$, then $$(alpha - lambda)a = (mu - beta)b$$ and hence, $$alpha = lambda text{ and } mu = beta$$. It was assumed that a and b are not zero vectors. Does that hold for zero vectors also?
$endgroup$
– mathguy80
Jul 20 '11 at 16:06
$begingroup$
It does not, indeed. Having $(1-2) vec{0} = vec 0$ does not imply $1=2$. =) But the idea in your proof was that since $a$ and $b$ cannot be parallel, it is impossible for them that a multiple of $a$ is equal to a multiple of $b$, unless they are both $0$, which leads to both coefficients being $0$.
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 16:15
$begingroup$
Ah! Thanks, that makes sense. I was beginning to think I had learned some basic ideas incorrectly!
$endgroup$
– mathguy80
Jul 21 '11 at 2:51
|
show 2 more comments
$begingroup$
At some point you said
$$
2(1-ell)- 4k = 1-2 ell
$$
But you should've noticed that you could say that because
$$
2(1-ell) - 4k = 0 = 1-2 ell.
$$
I think that gives you $ell = 1/2$.
Hope that helps,
answered Jul 20 '11 at 14:50
Patrick Da SilvaPatrick Da Silva
32k354106
$endgroup$
$begingroup$
Thanks. Can you please clarify why you equated it to 0?
$endgroup$
– mathguy80
Jul 20 '11 at 15:05
$begingroup$
Well, since you have a multiple of $a$ and a multiple of $b$ that are equal, and that those two vectors are not colinear, it must be because they are both the zero vectors, hence their coefficients are both zero. Isn't that why you said that $2(1-ell) - 4k = 1-2ell$ in the first place?
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 15:19
$begingroup$
The proof that I was taught is, $$ $$ Let $a$ and $b$ be two non-zero and non-parallel vectors. If $$alpha a + beta b = lambda a + mu b$$ for some $alpha, beta, lambda, mu in mathbb{R}$, then $$(alpha - lambda)a = (mu - beta)b$$ and hence, $$alpha = lambda text{ and } mu = beta$$. It was assumed that a and b are not zero vectors. Does that hold for zero vectors also?
$endgroup$
– mathguy80
Jul 20 '11 at 16:06
$begingroup$
It does not, indeed. Having $(1-2) vec{0} = vec 0$ does not imply $1=2$. =) But the idea in your proof was that since $a$ and $b$ cannot be parallel, it is impossible for them that a multiple of $a$ is equal to a multiple of $b$, unless they are both $0$, which leads to both coefficients being $0$.
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 16:15
$begingroup$
Ah! Thanks, that makes sense. I was beginning to think I had learned some basic ideas incorrectly!
$endgroup$
– mathguy80
Jul 21 '11 at 2:51
|
show 2 more comments
$begingroup$
At some point you said
$$
2(1-ell)- 4k = 1-2 ell
$$
But you should've noticed that you could say that because
$$
2(1-ell) - 4k = 0 = 1-2 ell.
$$
I think that gives you $ell = 1/2$.
Hope that helps,
answered Jul 20 '11 at 14:50
Patrick Da SilvaPatrick Da Silva
32k354106
$endgroup$
At some point you said
$$
2(1-ell)- 4k = 1-2 ell
$$
But you should've noticed that you could say that because
$$
2(1-ell) - 4k = 0 = 1-2 ell.
$$
I think that gives you $ell = 1/2$.
Hope that helps,
answered Jul 20 '11 at 14:50
Patrick Da SilvaPatrick Da Silva
32k354106
answered Jul 20 '11 at 14:50
Patrick Da SilvaPatrick Da Silva
32k354106
answered Jul 20 '11 at 14:50
Patrick Da SilvaPatrick Da Silva
32k354106
answered Jul 20 '11 at 14:50
Patrick Da SilvaPatrick Da Silva
32k354106
32k354106
$begingroup$
Thanks. Can you please clarify why you equated it to 0?
$endgroup$
– mathguy80
Jul 20 '11 at 15:05
$begingroup$
Well, since you have a multiple of $a$ and a multiple of $b$ that are equal, and that those two vectors are not colinear, it must be because they are both the zero vectors, hence their coefficients are both zero. Isn't that why you said that $2(1-ell) - 4k = 1-2ell$ in the first place?
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 15:19
$begingroup$
The proof that I was taught is, $$ $$ Let $a$ and $b$ be two non-zero and non-parallel vectors. If $$alpha a + beta b = lambda a + mu b$$ for some $alpha, beta, lambda, mu in mathbb{R}$, then $$(alpha - lambda)a = (mu - beta)b$$ and hence, $$alpha = lambda text{ and } mu = beta$$. It was assumed that a and b are not zero vectors. Does that hold for zero vectors also?
$endgroup$
– mathguy80
Jul 20 '11 at 16:06
$begingroup$
It does not, indeed. Having $(1-2) vec{0} = vec 0$ does not imply $1=2$. =) But the idea in your proof was that since $a$ and $b$ cannot be parallel, it is impossible for them that a multiple of $a$ is equal to a multiple of $b$, unless they are both $0$, which leads to both coefficients being $0$.
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 16:15
$begingroup$
Ah! Thanks, that makes sense. I was beginning to think I had learned some basic ideas incorrectly!
$endgroup$
– mathguy80
Jul 21 '11 at 2:51
|
show 2 more comments
$begingroup$
Thanks. Can you please clarify why you equated it to 0?
$endgroup$
– mathguy80
Jul 20 '11 at 15:05
$begingroup$
Well, since you have a multiple of $a$ and a multiple of $b$ that are equal, and that those two vectors are not colinear, it must be because they are both the zero vectors, hence their coefficients are both zero. Isn't that why you said that $2(1-ell) - 4k = 1-2ell$ in the first place?
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 15:19
$begingroup$
The proof that I was taught is, $$ $$ Let $a$ and $b$ be two non-zero and non-parallel vectors. If $$alpha a + beta b = lambda a + mu b$$ for some $alpha, beta, lambda, mu in mathbb{R}$, then $$(alpha - lambda)a = (mu - beta)b$$ and hence, $$alpha = lambda text{ and } mu = beta$$. It was assumed that a and b are not zero vectors. Does that hold for zero vectors also?
$endgroup$
– mathguy80
Jul 20 '11 at 16:06
$begingroup$
It does not, indeed. Having $(1-2) vec{0} = vec 0$ does not imply $1=2$. =) But the idea in your proof was that since $a$ and $b$ cannot be parallel, it is impossible for them that a multiple of $a$ is equal to a multiple of $b$, unless they are both $0$, which leads to both coefficients being $0$.
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 16:15
$begingroup$
Ah! Thanks, that makes sense. I was beginning to think I had learned some basic ideas incorrectly!
$endgroup$
– mathguy80
Jul 21 '11 at 2:51
$begingroup$
Thanks. Can you please clarify why you equated it to 0?
$endgroup$
– mathguy80
Jul 20 '11 at 15:05
$begingroup$
Thanks. Can you please clarify why you equated it to 0?
$endgroup$
– mathguy80
Jul 20 '11 at 15:05
$begingroup$
Well, since you have a multiple of $a$ and a multiple of $b$ that are equal, and that those two vectors are not colinear, it must be because they are both the zero vectors, hence their coefficients are both zero. Isn't that why you said that $2(1-ell) - 4k = 1-2ell$ in the first place?
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 15:19
$begingroup$
Well, since you have a multiple of $a$ and a multiple of $b$ that are equal, and that those two vectors are not colinear, it must be because they are both the zero vectors, hence their coefficients are both zero. Isn't that why you said that $2(1-ell) - 4k = 1-2ell$ in the first place?
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 15:19
$begingroup$
The proof that I was taught is, $$ $$ Let $a$ and $b$ be two non-zero and non-parallel vectors. If $$alpha a + beta b = lambda a + mu b$$ for some $alpha, beta, lambda, mu in mathbb{R}$, then $$(alpha - lambda)a = (mu - beta)b$$ and hence, $$alpha = lambda text{ and } mu = beta$$. It was assumed that a and b are not zero vectors. Does that hold for zero vectors also?
$endgroup$
– mathguy80
Jul 20 '11 at 16:06
$begingroup$
The proof that I was taught is, $$ $$ Let $a$ and $b$ be two non-zero and non-parallel vectors. If $$alpha a + beta b = lambda a + mu b$$ for some $alpha, beta, lambda, mu in mathbb{R}$, then $$(alpha - lambda)a = (mu - beta)b$$ and hence, $$alpha = lambda text{ and } mu = beta$$. It was assumed that a and b are not zero vectors. Does that hold for zero vectors also?
$endgroup$
– mathguy80
Jul 20 '11 at 16:06
$begingroup$
It does not, indeed. Having $(1-2) vec{0} = vec 0$ does not imply $1=2$. =) But the idea in your proof was that since $a$ and $b$ cannot be parallel, it is impossible for them that a multiple of $a$ is equal to a multiple of $b$, unless they are both $0$, which leads to both coefficients being $0$.
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 16:15
$begingroup$
It does not, indeed. Having $(1-2) vec{0} = vec 0$ does not imply $1=2$. =) But the idea in your proof was that since $a$ and $b$ cannot be parallel, it is impossible for them that a multiple of $a$ is equal to a multiple of $b$, unless they are both $0$, which leads to both coefficients being $0$.
$endgroup$
– Patrick Da Silva
Jul 20 '11 at 16:15
$begingroup$
Ah! Thanks, that makes sense. I was beginning to think I had learned some basic ideas incorrectly!
$endgroup$
– mathguy80
Jul 21 '11 at 2:51
$begingroup$
Ah! Thanks, that makes sense. I was beginning to think I had learned some basic ideas incorrectly!
$endgroup$
– mathguy80
Jul 21 '11 at 2:51
|
show 2 more comments
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