Find the line perpendicular to these two vectors












0












$begingroup$


A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.



What is the equation of the line?



The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.



I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.



I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?



Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
    $endgroup$
    – B. Goddard
    Jan 18 at 13:00










  • $begingroup$
    The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
    $endgroup$
    – kelvin hong 方
    Jan 18 at 13:05










  • $begingroup$
    I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
    $endgroup$
    – Madison
    Jan 18 at 13:15










  • $begingroup$
    I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
    $endgroup$
    – John Hughes
    Jan 18 at 13:20










  • $begingroup$
    Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
    $endgroup$
    – Madison
    Jan 18 at 13:27
















0












$begingroup$


A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.



What is the equation of the line?



The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.



I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.



I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?



Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?






linear-algebra







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
    $endgroup$
    – B. Goddard
    Jan 18 at 13:00










  • $begingroup$
    The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
    $endgroup$
    – kelvin hong 方
    Jan 18 at 13:05










  • $begingroup$
    I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
    $endgroup$
    – Madison
    Jan 18 at 13:15










  • $begingroup$
    I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
    $endgroup$
    – John Hughes
    Jan 18 at 13:20










  • $begingroup$
    Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
    $endgroup$
    – Madison
    Jan 18 at 13:27














0












0








0





$begingroup$


A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.



What is the equation of the line?



The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.



I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.



I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?



Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?






linear-algebra







share|cite|improve this question











$endgroup$




A line goes through point $A = (9, -7, 31)$ and the line is perpendicular to vectors $[7, 1, 2]$ and $[3, 0, 1]$.



What is the equation of the line?



The cross product of $[7, 1, 2]$ and $[3, 0, 1]$ is $[1, -1, -3]$.



I believe the line can be written as $x = [9, -7, 31] + t [1, -1, -3]$, where $t$ is a real number.



I'm somewhat skeptical of my own work. If you find the cross product (which will be perpendicular to the two given vectors); why does that mean it is automatically parallel with the line that is also perpendicular to these two vectors?



Given the vectors are in three-dimensional space; is it possible to produce a cross product of these two vectors and that vector not to be parallel with the line I found?






linear-algebra




linear-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 18 at 13:52









idriskameni

641319




641319










asked Jan 18 at 12:35









MadisonMadison

31




31








  • 1




    $begingroup$
    The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
    $endgroup$
    – B. Goddard
    Jan 18 at 13:00










  • $begingroup$
    The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
    $endgroup$
    – kelvin hong 方
    Jan 18 at 13:05










  • $begingroup$
    I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
    $endgroup$
    – Madison
    Jan 18 at 13:15










  • $begingroup$
    I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
    $endgroup$
    – John Hughes
    Jan 18 at 13:20










  • $begingroup$
    Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
    $endgroup$
    – Madison
    Jan 18 at 13:27














  • 1




    $begingroup$
    The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
    $endgroup$
    – B. Goddard
    Jan 18 at 13:00










  • $begingroup$
    The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
    $endgroup$
    – kelvin hong 方
    Jan 18 at 13:05










  • $begingroup$
    I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
    $endgroup$
    – Madison
    Jan 18 at 13:15










  • $begingroup$
    I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
    $endgroup$
    – John Hughes
    Jan 18 at 13:20










  • $begingroup$
    Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
    $endgroup$
    – Madison
    Jan 18 at 13:27








1




1




$begingroup$
The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
$endgroup$
– B. Goddard
Jan 18 at 13:00




$begingroup$
The answer to your last question is "no." Two vectors determine a plane. There is only one direction normal to that plane and the cross product gives it.
$endgroup$
– B. Goddard
Jan 18 at 13:00












$begingroup$
The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
$endgroup$
– kelvin hong 方
Jan 18 at 13:05




$begingroup$
The people first think how to formulate a vector to be perpendicular to other two vectors in 3-dimensional space, then they invented the cross product.
$endgroup$
– kelvin hong 方
Jan 18 at 13:05












$begingroup$
I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
$endgroup$
– Madison
Jan 18 at 13:15




$begingroup$
I believe two non-parallel vectors make a plane. That's true. That makes it easy to visualise. The cross product makes a vector that pierces the plane at right angles from above and below said plane. That's my understanding.
$endgroup$
– Madison
Jan 18 at 13:15












$begingroup$
I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
$endgroup$
– John Hughes
Jan 18 at 13:20




$begingroup$
I would slightly amend @B.Goddard's assertion to say that there are two opposite directions that are normal (perpendicular) to the plane, and if your vectors are $u$ and $v$, then these two directions are given by $u times v$ and $v times u$.
$endgroup$
– John Hughes
Jan 18 at 13:20












$begingroup$
Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
$endgroup$
– Madison
Jan 18 at 13:27




$begingroup$
Yes, above and below the plane. I believe. The opposite direction being the anti-commutative.
$endgroup$
– Madison
Jan 18 at 13:27










2 Answers
2






active

oldest

votes


















0












$begingroup$

When you find a cross product,



$$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$



you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).



That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.



Hence, what you have done is OK.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
    $endgroup$
    – Madison
    Jan 18 at 13:20



















1












$begingroup$

Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.



Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.



The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
$ell(t) = [9+t, -7-t, 31-3t]$ is that line:



$$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$



and $$3x + z = 3(9+t)+(31-3t) = 58$$






share|cite|improve this answer









$endgroup$













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    When you find a cross product,



    $$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$



    you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).



    That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.



    Hence, what you have done is OK.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
      $endgroup$
      – Madison
      Jan 18 at 13:20
















    0












    $begingroup$

    When you find a cross product,



    $$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$



    you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).



    That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.



    Hence, what you have done is OK.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
      $endgroup$
      – Madison
      Jan 18 at 13:20














    0












    0








    0





    $begingroup$

    When you find a cross product,



    $$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$



    you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).



    That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.



    Hence, what you have done is OK.






    share|cite|improve this answer











    $endgroup$



    When you find a cross product,



    $$A=begin{bmatrix}a&b&c\7&1&2\ 3& 0& 1end{bmatrix}$$



    you are finding a linearly independent vector, namely $vec{v}=(a,b,c)$, to the other two vectors, namely $vec{u}=(7,1,2),vec{w}=(3,0,1)$ (and it is like that because you are finding a vector which gives you $det(A)neq 0$, hence $dim(A)=3$, hence $vec{u},vec{v},vec{w}$ generate $mathbb{R}^3$).



    That means that you will have a basis of the $3$-dimensional space (since you have $3$ linearly independent vectors). An that is why if you have a fourth vector, namely $vec{alpha}$ which is perpendicular to the first two vectors, it has to be proportional to the one you have find with the cross product.



    Hence, what you have done is OK.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 18 at 13:20









    John Hughes

    64k24191




    64k24191










    answered Jan 18 at 13:08









    idriskameniidriskameni

    641319




    641319








    • 1




      $begingroup$
      Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
      $endgroup$
      – Madison
      Jan 18 at 13:20














    • 1




      $begingroup$
      Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
      $endgroup$
      – Madison
      Jan 18 at 13:20








    1




    1




    $begingroup$
    Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
    $endgroup$
    – Madison
    Jan 18 at 13:20




    $begingroup$
    Okay, this is good. I was basically taught to just follow that it's true; but, it was lacking intuition. Soon to study linear independence.
    $endgroup$
    – Madison
    Jan 18 at 13:20











    1












    $begingroup$

    Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.



    Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.



    The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
    $ell(t) = [9+t, -7-t, 31-3t]$ is that line:



    $$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$



    and $$3x + z = 3(9+t)+(31-3t) = 58$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.



      Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.



      The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
      $ell(t) = [9+t, -7-t, 31-3t]$ is that line:



      $$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$



      and $$3x + z = 3(9+t)+(31-3t) = 58$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.



        Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.



        The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
        $ell(t) = [9+t, -7-t, 31-3t]$ is that line:



        $$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$



        and $$3x + z = 3(9+t)+(31-3t) = 58$$






        share|cite|improve this answer









        $endgroup$



        Consider the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[7,1,2]$. Its equation is $7x + 1y + 2z = 7(9) + 1(-7) + 2(31)$, which simplifies to $7x + y + 2z = 118$.



        Similarly the plane containing the point $A=(9,-7,31)$ and perpendicular to the vector $[3,0,1]$ has equation $3x+z=58$.



        The intersection of these two planes will be the line that contains the point $A$ and is perpendicular to both vectors. It is not hard to check that the line
        $ell(t) = [9+t, -7-t, 31-3t]$ is that line:



        $$7x + y + 2z = 7(9+t)+(-7-t)+2(31-3t) = 118$$



        and $$3x + z = 3(9+t)+(31-3t) = 58$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 18 at 13:44









        steven gregorysteven gregory

        18.1k32258




        18.1k32258






























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            What does “Dominus providebit” mean?

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            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1 ...
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