Linear Algebra and planes in Cartesian space












0












$begingroup$


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







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  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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.
    $endgroup$
    – 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














0












0








0


1



$begingroup$


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







share|cite|improve this question











$endgroup$




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






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share|cite|improve this question













share|cite|improve this question




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edited Apr 27 '14 at 8:17







Chey

















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














  • 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










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! :)






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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






    share|cite|improve this answer











    $endgroup$









    • 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











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    2 Answers
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    2 Answers
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    oldest

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    0












    $begingroup$

    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! :)






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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! :)






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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! :)






        share|cite|improve this answer









        $endgroup$



        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! :)







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Apr 27 '14 at 11:50









        user146118user146118

        11




        11























            0












            $begingroup$

            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






            share|cite|improve this answer











            $endgroup$









            • 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
















            0












            $begingroup$

            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






            share|cite|improve this answer











            $endgroup$









            • 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














            0












            0








            0





            $begingroup$

            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






            share|cite|improve this answer











            $endgroup$



            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







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 28 '14 at 11:00

























            answered Apr 27 '14 at 12:42









            EnyaEnya

            13




            13








            • 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














            • 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








            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


















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