Linear Algebra and planes in Cartesian space
$begingroup$
I was asked this question from the course Linear Algebra and I need to show all working.
The question is in 5 parts:
Consider the xyz-space R3 with the origin O. Let l be the line given by the Cartesian equation $$x = frac{z - 1}2, y = 1 $$ Let p be the plane given by the Cartesian equation $$2 x + y - z = 1$$
a) Find two unit vectors parallel to the line l.
b) Find the point Q which is the intersection of the plane p and z-axis.
c) Take n = 2 i + j - k as a normal vector of the plane p. Decompose the vector QO into the sum of two vectors: one of them is parallel to n and the other one is orthogonal to n.
d) The plane p divides R3 into two parts. Find the unit vector perpendicular to p and pointing into the part containing the origin O.
e) Let P(x, y, z) be a point on the line l. Letting x = t for some constant t, find the y and z coordinates of P. Calculate the distance from P to the plane p.
I would like to thank everyone who takes time in helping me with this problem and I really appreciate the help.
Thanks again.
linear-algebra vectors
asked Apr 27 '14 at 8:03
CheyChey
12
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add a comment |
$begingroup$
I was asked this question from the course Linear Algebra and I need to show all working.
The question is in 5 parts:
Consider the xyz-space R3 with the origin O. Let l be the line given by the Cartesian equation $$x = frac{z - 1}2, y = 1 $$ Let p be the plane given by the Cartesian equation $$2 x + y - z = 1$$
a) Find two unit vectors parallel to the line l.
b) Find the point Q which is the intersection of the plane p and z-axis.
c) Take n = 2 i + j - k as a normal vector of the plane p. Decompose the vector QO into the sum of two vectors: one of them is parallel to n and the other one is orthogonal to n.
d) The plane p divides R3 into two parts. Find the unit vector perpendicular to p and pointing into the part containing the origin O.
e) Let P(x, y, z) be a point on the line l. Letting x = t for some constant t, find the y and z coordinates of P. Calculate the distance from P to the plane p.
I would like to thank everyone who takes time in helping me with this problem and I really appreciate the help.
Thanks again.
linear-algebra vectors
asked Apr 27 '14 at 8:03
CheyChey
12
$endgroup$
2
$begingroup$
What have you already tried? Where are you facing difficulties?
$endgroup$
– user60306
Apr 27 '14 at 8:39
$begingroup$
I found b) as Q(0,0,-1) by subbing in x,y=0. Then I did c) using the vector projection and dot products. I got the answers parallel: -1/6*(2i+j-k); orthogonal 1/6*(2i+j+5k). For a) I manged to get out the answer +/-1/root(2)*(j+k) (+/- because they point at opposite directions. Not sure if right or not). I'm just not sure how to go about d). What are the two parts? For e) I subbed in x=t into the line equation and got the point P=(t,1,2t+1). I used P with Q(0,1,0), which is a point on the plane (got by inspection). I got the distance 1/root(6). Thanks for answering my question.
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– Chey
Apr 27 '14 at 10:33
$begingroup$
for c) how did you get the 1/6 part. I can get the vectors but where did the 1/6 come from. is it sqrt(2^2+1^2+1^2+2^2+1^2+5^2)? but what's the reason for that
$endgroup$
– user146138
Apr 27 '14 at 13:47
add a comment |
$begingroup$
I was asked this question from the course Linear Algebra and I need to show all working.
The question is in 5 parts:
Consider the xyz-space R3 with the origin O. Let l be the line given by the Cartesian equation $$x = frac{z - 1}2, y = 1 $$ Let p be the plane given by the Cartesian equation $$2 x + y - z = 1$$
a) Find two unit vectors parallel to the line l.
b) Find the point Q which is the intersection of the plane p and z-axis.
c) Take n = 2 i + j - k as a normal vector of the plane p. Decompose the vector QO into the sum of two vectors: one of them is parallel to n and the other one is orthogonal to n.
d) The plane p divides R3 into two parts. Find the unit vector perpendicular to p and pointing into the part containing the origin O.
e) Let P(x, y, z) be a point on the line l. Letting x = t for some constant t, find the y and z coordinates of P. Calculate the distance from P to the plane p.
I would like to thank everyone who takes time in helping me with this problem and I really appreciate the help.
Thanks again.
linear-algebra vectors
asked Apr 27 '14 at 8:03
CheyChey
12
$endgroup$
I was asked this question from the course Linear Algebra and I need to show all working.
The question is in 5 parts:
Consider the xyz-space R3 with the origin O. Let l be the line given by the Cartesian equation $$x = frac{z - 1}2, y = 1 $$ Let p be the plane given by the Cartesian equation $$2 x + y - z = 1$$
a) Find two unit vectors parallel to the line l.
b) Find the point Q which is the intersection of the plane p and z-axis.
c) Take n = 2 i + j - k as a normal vector of the plane p. Decompose the vector QO into the sum of two vectors: one of them is parallel to n and the other one is orthogonal to n.
d) The plane p divides R3 into two parts. Find the unit vector perpendicular to p and pointing into the part containing the origin O.
e) Let P(x, y, z) be a point on the line l. Letting x = t for some constant t, find the y and z coordinates of P. Calculate the distance from P to the plane p.
I would like to thank everyone who takes time in helping me with this problem and I really appreciate the help.
Thanks again.
linear-algebra vectors
linear-algebra vectors
asked Apr 27 '14 at 8:03
CheyChey
12
asked Apr 27 '14 at 8:03
CheyChey
12
edited Apr 27 '14 at 8:17
Chey
asked Apr 27 '14 at 8:03
CheyChey
12
asked Apr 27 '14 at 8:03
CheyChey
12
asked Apr 27 '14 at 8:03
CheyChey
12
12
2
$begingroup$
What have you already tried? Where are you facing difficulties?
$endgroup$
– user60306
Apr 27 '14 at 8:39
$begingroup$
I found b) as Q(0,0,-1) by subbing in x,y=0. Then I did c) using the vector projection and dot products. I got the answers parallel: -1/6*(2i+j-k); orthogonal 1/6*(2i+j+5k). For a) I manged to get out the answer +/-1/root(2)*(j+k) (+/- because they point at opposite directions. Not sure if right or not). I'm just not sure how to go about d). What are the two parts? For e) I subbed in x=t into the line equation and got the point P=(t,1,2t+1). I used P with Q(0,1,0), which is a point on the plane (got by inspection). I got the distance 1/root(6). Thanks for answering my question.
$endgroup$
– Chey
Apr 27 '14 at 10:33
$begingroup$
for c) how did you get the 1/6 part. I can get the vectors but where did the 1/6 come from. is it sqrt(2^2+1^2+1^2+2^2+1^2+5^2)? but what's the reason for that
$endgroup$
– user146138
Apr 27 '14 at 13:47
add a comment |
2
$begingroup$
What have you already tried? Where are you facing difficulties?
$endgroup$
– user60306
Apr 27 '14 at 8:39
$begingroup$
I found b) as Q(0,0,-1) by subbing in x,y=0. Then I did c) using the vector projection and dot products. I got the answers parallel: -1/6*(2i+j-k); orthogonal 1/6*(2i+j+5k). For a) I manged to get out the answer +/-1/root(2)*(j+k) (+/- because they point at opposite directions. Not sure if right or not). I'm just not sure how to go about d). What are the two parts? For e) I subbed in x=t into the line equation and got the point P=(t,1,2t+1). I used P with Q(0,1,0), which is a point on the plane (got by inspection). I got the distance 1/root(6). Thanks for answering my question.
$endgroup$
– Chey
Apr 27 '14 at 10:33
$begingroup$
for c) how did you get the 1/6 part. I can get the vectors but where did the 1/6 come from. is it sqrt(2^2+1^2+1^2+2^2+1^2+5^2)? but what's the reason for that
$endgroup$
– user146138
Apr 27 '14 at 13:47
2
2
$begingroup$
What have you already tried? Where are you facing difficulties?
$endgroup$
– user60306
Apr 27 '14 at 8:39
$begingroup$
What have you already tried? Where are you facing difficulties?
$endgroup$
– user60306
Apr 27 '14 at 8:39
$begingroup$
I found b) as Q(0,0,-1) by subbing in x,y=0. Then I did c) using the vector projection and dot products. I got the answers parallel: -1/6*(2i+j-k); orthogonal 1/6*(2i+j+5k). For a) I manged to get out the answer +/-1/root(2)*(j+k) (+/- because they point at opposite directions. Not sure if right or not). I'm just not sure how to go about d). What are the two parts? For e) I subbed in x=t into the line equation and got the point P=(t,1,2t+1). I used P with Q(0,1,0), which is a point on the plane (got by inspection). I got the distance 1/root(6). Thanks for answering my question.
$endgroup$
– Chey
Apr 27 '14 at 10:33
$begingroup$
I found b) as Q(0,0,-1) by subbing in x,y=0. Then I did c) using the vector projection and dot products. I got the answers parallel: -1/6*(2i+j-k); orthogonal 1/6*(2i+j+5k). For a) I manged to get out the answer +/-1/root(2)*(j+k) (+/- because they point at opposite directions. Not sure if right or not). I'm just not sure how to go about d). What are the two parts? For e) I subbed in x=t into the line equation and got the point P=(t,1,2t+1). I used P with Q(0,1,0), which is a point on the plane (got by inspection). I got the distance 1/root(6). Thanks for answering my question.
$endgroup$
– Chey
Apr 27 '14 at 10:33
$begingroup$
for c) how did you get the 1/6 part. I can get the vectors but where did the 1/6 come from. is it sqrt(2^2+1^2+1^2+2^2+1^2+5^2)? but what's the reason for that
$endgroup$
– user146138
Apr 27 '14 at 13:47
$begingroup$
for c) how did you get the 1/6 part. I can get the vectors but where did the 1/6 come from. is it sqrt(2^2+1^2+1^2+2^2+1^2+5^2)? but what's the reason for that
$endgroup$
– user146138
Apr 27 '14 at 13:47
add a comment |
2 Answers
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for part d), i'm not 100% sure but couldn't we say that it will either be the normal vector n, or the one in the opposite direction -n?
and then use the formula for a distance (d) from a point P (the origin) to a plane containing Q (found from part a) with normal vector (n) as d = |n.PQ|/|n|, but then take away the absolute values from the top part of the fraction to give us a "direction" to the distance between the plane and the origin and thus the unit vector will be the one with the same sign as that direction?
like i said this could be total rubbish, but just a thought! :)
answered Apr 27 '14 at 11:50
user146118user146118
11
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add a comment |
$begingroup$
Here is my solutions. FYI I can do only half d). Will maybe continue working on it someday later. Hope it helps.
a) use parametric form of 3D-line, and the coefficients of the parameter 't' will be your direction vector: i + 2k. Then simply work out it's +/-ve two unit vectors by dividing its modulus: sqrt(1^2 + 2^2), and eventually I got: +-(i + 2K)/sqrt5
b) same.
c) same.
d) However you don't know what the two parts are, you should be able to find the unit vector perpendicular/normal/orthogonal to a plane. I don't know how to show which vector is pointing into where the origin is, but I can show you from which two you can choose, i.e. the unit vectors pair:
The normal vector to a plane is simply of the coefficients of the plane equation. In this case, our plane is given: 2x + y - z = 1 (Don't worry about the 1 at last, it can be any constant), → normal vectors pair: +-(2, 1, -1), or say: n = +-(2i + j - k).
Therefore the unit vector of n: n_hat = n/|n| = +-(2i + j - k)/sqrt(2^2 + 1^2 + (-1)^2)
I.e. the unit vectors pair n_hat = +-(2i + j - k)/sqrt6
Then you may wish to find someone else to teach you which one is pointing in where the origin part is, either the +ve n_hat or the -ve.
e) I use your P and Q, which I agree with you:
QP.n = (t, 0, 2t+1).(2, 1, -1) = -1
where |n| = sqrt6
therefore distance d = |QP.n|/|n| = 1/sqrt6
answered Apr 27 '14 at 12:42
EnyaEnya
13
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1
$begingroup$
Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good.
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– Daniel R
Apr 27 '14 at 13:04
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@enya. I don't understand how you even got 3a.
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– user146374
Apr 28 '14 at 14:17
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Imagine that you are standing at point (0,0,0). Then, as 2*0+1*0-0*0=0<1, the origin O is under the plane. This means you have to look upside to see the plane. Thus, sign the vector hat{k} should be -. Therefore, (2i+j-k)/sqrt(6) is the vector contained in origin part.
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– Ryoungwoo Jang
Mar 3 '16 at 6:19
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2 Answers
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active
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2 Answers
2
active
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active
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$begingroup$
for part d), i'm not 100% sure but couldn't we say that it will either be the normal vector n, or the one in the opposite direction -n?
and then use the formula for a distance (d) from a point P (the origin) to a plane containing Q (found from part a) with normal vector (n) as d = |n.PQ|/|n|, but then take away the absolute values from the top part of the fraction to give us a "direction" to the distance between the plane and the origin and thus the unit vector will be the one with the same sign as that direction?
like i said this could be total rubbish, but just a thought! :)
answered Apr 27 '14 at 11:50
user146118user146118
11
$endgroup$
add a comment |
$begingroup$
for part d), i'm not 100% sure but couldn't we say that it will either be the normal vector n, or the one in the opposite direction -n?
and then use the formula for a distance (d) from a point P (the origin) to a plane containing Q (found from part a) with normal vector (n) as d = |n.PQ|/|n|, but then take away the absolute values from the top part of the fraction to give us a "direction" to the distance between the plane and the origin and thus the unit vector will be the one with the same sign as that direction?
like i said this could be total rubbish, but just a thought! :)
answered Apr 27 '14 at 11:50
user146118user146118
11
$endgroup$
add a comment |
$begingroup$
for part d), i'm not 100% sure but couldn't we say that it will either be the normal vector n, or the one in the opposite direction -n?
and then use the formula for a distance (d) from a point P (the origin) to a plane containing Q (found from part a) with normal vector (n) as d = |n.PQ|/|n|, but then take away the absolute values from the top part of the fraction to give us a "direction" to the distance between the plane and the origin and thus the unit vector will be the one with the same sign as that direction?
like i said this could be total rubbish, but just a thought! :)
answered Apr 27 '14 at 11:50
user146118user146118
11
$endgroup$
for part d), i'm not 100% sure but couldn't we say that it will either be the normal vector n, or the one in the opposite direction -n?
and then use the formula for a distance (d) from a point P (the origin) to a plane containing Q (found from part a) with normal vector (n) as d = |n.PQ|/|n|, but then take away the absolute values from the top part of the fraction to give us a "direction" to the distance between the plane and the origin and thus the unit vector will be the one with the same sign as that direction?
like i said this could be total rubbish, but just a thought! :)
answered Apr 27 '14 at 11:50
user146118user146118
11
answered Apr 27 '14 at 11:50
user146118user146118
11
answered Apr 27 '14 at 11:50
user146118user146118
11
answered Apr 27 '14 at 11:50
user146118user146118
11
11
add a comment |
add a comment |
$begingroup$
Here is my solutions. FYI I can do only half d). Will maybe continue working on it someday later. Hope it helps.
a) use parametric form of 3D-line, and the coefficients of the parameter 't' will be your direction vector: i + 2k. Then simply work out it's +/-ve two unit vectors by dividing its modulus: sqrt(1^2 + 2^2), and eventually I got: +-(i + 2K)/sqrt5
b) same.
c) same.
d) However you don't know what the two parts are, you should be able to find the unit vector perpendicular/normal/orthogonal to a plane. I don't know how to show which vector is pointing into where the origin is, but I can show you from which two you can choose, i.e. the unit vectors pair:
The normal vector to a plane is simply of the coefficients of the plane equation. In this case, our plane is given: 2x + y - z = 1 (Don't worry about the 1 at last, it can be any constant), → normal vectors pair: +-(2, 1, -1), or say: n = +-(2i + j - k).
Therefore the unit vector of n: n_hat = n/|n| = +-(2i + j - k)/sqrt(2^2 + 1^2 + (-1)^2)
I.e. the unit vectors pair n_hat = +-(2i + j - k)/sqrt6
Then you may wish to find someone else to teach you which one is pointing in where the origin part is, either the +ve n_hat or the -ve.
e) I use your P and Q, which I agree with you:
QP.n = (t, 0, 2t+1).(2, 1, -1) = -1
where |n| = sqrt6
therefore distance d = |QP.n|/|n| = 1/sqrt6
answered Apr 27 '14 at 12:42
EnyaEnya
13
$endgroup$
1
$begingroup$
Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good.
$endgroup$
– Daniel R
Apr 27 '14 at 13:04
$begingroup$
@enya. I don't understand how you even got 3a.
$endgroup$
– user146374
Apr 28 '14 at 14:17
$begingroup$
Imagine that you are standing at point (0,0,0). Then, as 2*0+1*0-0*0=0<1, the origin O is under the plane. This means you have to look upside to see the plane. Thus, sign the vector hat{k} should be -. Therefore, (2i+j-k)/sqrt(6) is the vector contained in origin part.
$endgroup$
– Ryoungwoo Jang
Mar 3 '16 at 6:19
add a comment |
$begingroup$
Here is my solutions. FYI I can do only half d). Will maybe continue working on it someday later. Hope it helps.
a) use parametric form of 3D-line, and the coefficients of the parameter 't' will be your direction vector: i + 2k. Then simply work out it's +/-ve two unit vectors by dividing its modulus: sqrt(1^2 + 2^2), and eventually I got: +-(i + 2K)/sqrt5
b) same.
c) same.
d) However you don't know what the two parts are, you should be able to find the unit vector perpendicular/normal/orthogonal to a plane. I don't know how to show which vector is pointing into where the origin is, but I can show you from which two you can choose, i.e. the unit vectors pair:
The normal vector to a plane is simply of the coefficients of the plane equation. In this case, our plane is given: 2x + y - z = 1 (Don't worry about the 1 at last, it can be any constant), → normal vectors pair: +-(2, 1, -1), or say: n = +-(2i + j - k).
Therefore the unit vector of n: n_hat = n/|n| = +-(2i + j - k)/sqrt(2^2 + 1^2 + (-1)^2)
I.e. the unit vectors pair n_hat = +-(2i + j - k)/sqrt6
Then you may wish to find someone else to teach you which one is pointing in where the origin part is, either the +ve n_hat or the -ve.
e) I use your P and Q, which I agree with you:
QP.n = (t, 0, 2t+1).(2, 1, -1) = -1
where |n| = sqrt6
therefore distance d = |QP.n|/|n| = 1/sqrt6
answered Apr 27 '14 at 12:42
EnyaEnya
13
$endgroup$
1
$begingroup$
Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good.
$endgroup$
– Daniel R
Apr 27 '14 at 13:04
$begingroup$
@enya. I don't understand how you even got 3a.
$endgroup$
– user146374
Apr 28 '14 at 14:17
$begingroup$
Imagine that you are standing at point (0,0,0). Then, as 2*0+1*0-0*0=0<1, the origin O is under the plane. This means you have to look upside to see the plane. Thus, sign the vector hat{k} should be -. Therefore, (2i+j-k)/sqrt(6) is the vector contained in origin part.
$endgroup$
– Ryoungwoo Jang
Mar 3 '16 at 6:19
add a comment |
$begingroup$
Here is my solutions. FYI I can do only half d). Will maybe continue working on it someday later. Hope it helps.
a) use parametric form of 3D-line, and the coefficients of the parameter 't' will be your direction vector: i + 2k. Then simply work out it's +/-ve two unit vectors by dividing its modulus: sqrt(1^2 + 2^2), and eventually I got: +-(i + 2K)/sqrt5
b) same.
c) same.
d) However you don't know what the two parts are, you should be able to find the unit vector perpendicular/normal/orthogonal to a plane. I don't know how to show which vector is pointing into where the origin is, but I can show you from which two you can choose, i.e. the unit vectors pair:
The normal vector to a plane is simply of the coefficients of the plane equation. In this case, our plane is given: 2x + y - z = 1 (Don't worry about the 1 at last, it can be any constant), → normal vectors pair: +-(2, 1, -1), or say: n = +-(2i + j - k).
Therefore the unit vector of n: n_hat = n/|n| = +-(2i + j - k)/sqrt(2^2 + 1^2 + (-1)^2)
I.e. the unit vectors pair n_hat = +-(2i + j - k)/sqrt6
Then you may wish to find someone else to teach you which one is pointing in where the origin part is, either the +ve n_hat or the -ve.
e) I use your P and Q, which I agree with you:
QP.n = (t, 0, 2t+1).(2, 1, -1) = -1
where |n| = sqrt6
therefore distance d = |QP.n|/|n| = 1/sqrt6
answered Apr 27 '14 at 12:42
EnyaEnya
13
$endgroup$
Here is my solutions. FYI I can do only half d). Will maybe continue working on it someday later. Hope it helps.
a) use parametric form of 3D-line, and the coefficients of the parameter 't' will be your direction vector: i + 2k. Then simply work out it's +/-ve two unit vectors by dividing its modulus: sqrt(1^2 + 2^2), and eventually I got: +-(i + 2K)/sqrt5
b) same.
c) same.
d) However you don't know what the two parts are, you should be able to find the unit vector perpendicular/normal/orthogonal to a plane. I don't know how to show which vector is pointing into where the origin is, but I can show you from which two you can choose, i.e. the unit vectors pair:
The normal vector to a plane is simply of the coefficients of the plane equation. In this case, our plane is given: 2x + y - z = 1 (Don't worry about the 1 at last, it can be any constant), → normal vectors pair: +-(2, 1, -1), or say: n = +-(2i + j - k).
Therefore the unit vector of n: n_hat = n/|n| = +-(2i + j - k)/sqrt(2^2 + 1^2 + (-1)^2)
I.e. the unit vectors pair n_hat = +-(2i + j - k)/sqrt6
Then you may wish to find someone else to teach you which one is pointing in where the origin part is, either the +ve n_hat or the -ve.
e) I use your P and Q, which I agree with you:
QP.n = (t, 0, 2t+1).(2, 1, -1) = -1
where |n| = sqrt6
therefore distance d = |QP.n|/|n| = 1/sqrt6
answered Apr 27 '14 at 12:42
EnyaEnya
13
edited Apr 28 '14 at 11:00
answered Apr 27 '14 at 12:42
EnyaEnya
13
answered Apr 27 '14 at 12:42
EnyaEnya
13
answered Apr 27 '14 at 12:42
EnyaEnya
13
13
1
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Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good.
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– Daniel R
Apr 27 '14 at 13:04
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@enya. I don't understand how you even got 3a.
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– user146374
Apr 28 '14 at 14:17
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Imagine that you are standing at point (0,0,0). Then, as 2*0+1*0-0*0=0<1, the origin O is under the plane. This means you have to look upside to see the plane. Thus, sign the vector hat{k} should be -. Therefore, (2i+j-k)/sqrt(6) is the vector contained in origin part.
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– Ryoungwoo Jang
Mar 3 '16 at 6:19
add a comment |
1
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Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good.
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– Daniel R
Apr 27 '14 at 13:04
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@enya. I don't understand how you even got 3a.
$endgroup$
– user146374
Apr 28 '14 at 14:17
$begingroup$
Imagine that you are standing at point (0,0,0). Then, as 2*0+1*0-0*0=0<1, the origin O is under the plane. This means you have to look upside to see the plane. Thus, sign the vector hat{k} should be -. Therefore, (2i+j-k)/sqrt(6) is the vector contained in origin part.
$endgroup$
– Ryoungwoo Jang
Mar 3 '16 at 6:19
1
1
$begingroup$
Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good.
$endgroup$
– Daniel R
Apr 27 '14 at 13:04
$begingroup$
Welcome to math.SE! Please consider taking the time to read the faq to familiarise yourself with some of our common practices. In addition, this page should give you a start at learning how to typeset mathematics here so that your posts say what you want them to, and also look good.
$endgroup$
– Daniel R
Apr 27 '14 at 13:04
$begingroup$
@enya. I don't understand how you even got 3a.
$endgroup$
– user146374
Apr 28 '14 at 14:17
$begingroup$
@enya. I don't understand how you even got 3a.
$endgroup$
– user146374
Apr 28 '14 at 14:17
$begingroup$
Imagine that you are standing at point (0,0,0). Then, as 2*0+1*0-0*0=0<1, the origin O is under the plane. This means you have to look upside to see the plane. Thus, sign the vector hat{k} should be -. Therefore, (2i+j-k)/sqrt(6) is the vector contained in origin part.
$endgroup$
– Ryoungwoo Jang
Mar 3 '16 at 6:19
$begingroup$
Imagine that you are standing at point (0,0,0). Then, as 2*0+1*0-0*0=0<1, the origin O is under the plane. This means you have to look upside to see the plane. Thus, sign the vector hat{k} should be -. Therefore, (2i+j-k)/sqrt(6) is the vector contained in origin part.
$endgroup$
– Ryoungwoo Jang
Mar 3 '16 at 6:19
add a comment |
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What have you already tried? Where are you facing difficulties?
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– user60306
Apr 27 '14 at 8:39
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I found b) as Q(0,0,-1) by subbing in x,y=0. Then I did c) using the vector projection and dot products. I got the answers parallel: -1/6*(2i+j-k); orthogonal 1/6*(2i+j+5k). For a) I manged to get out the answer +/-1/root(2)*(j+k) (+/- because they point at opposite directions. Not sure if right or not). I'm just not sure how to go about d). What are the two parts? For e) I subbed in x=t into the line equation and got the point P=(t,1,2t+1). I used P with Q(0,1,0), which is a point on the plane (got by inspection). I got the distance 1/root(6). Thanks for answering my question.
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– Chey
Apr 27 '14 at 10:33
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for c) how did you get the 1/6 part. I can get the vectors but where did the 1/6 come from. is it sqrt(2^2+1^2+1^2+2^2+1^2+5^2)? but what's the reason for that
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– user146138
Apr 27 '14 at 13:47