How do I find an equation from vectors of a certain form?












-3












$begingroup$


Consider the plane $P$ in $mathbb{R}^3$ containing all the vectors of the form:
$r begin{bmatrix} 1 \ 2 \ 3 end{bmatrix} + s begin{bmatrix} 3 \ 2 \ 1 end{bmatrix}
$



Find an equation which describes the plane and show the point $(5,6,5)$ is not on the plane $P$.






linear-algebra







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$endgroup$








  • 2




    $begingroup$
    What have you tried so far?
    $endgroup$
    – jwc845
    Jan 15 at 22:11
















-3












$begingroup$


Consider the plane $P$ in $mathbb{R}^3$ containing all the vectors of the form:
$r begin{bmatrix} 1 \ 2 \ 3 end{bmatrix} + s begin{bmatrix} 3 \ 2 \ 1 end{bmatrix}
$



Find an equation which describes the plane and show the point $(5,6,5)$ is not on the plane $P$.






linear-algebra







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    What have you tried so far?
    $endgroup$
    – jwc845
    Jan 15 at 22:11














-3












-3








-3





$begingroup$


Consider the plane $P$ in $mathbb{R}^3$ containing all the vectors of the form:
$r begin{bmatrix} 1 \ 2 \ 3 end{bmatrix} + s begin{bmatrix} 3 \ 2 \ 1 end{bmatrix}
$



Find an equation which describes the plane and show the point $(5,6,5)$ is not on the plane $P$.






linear-algebra







share|cite|improve this question











$endgroup$




Consider the plane $P$ in $mathbb{R}^3$ containing all the vectors of the form:
$r begin{bmatrix} 1 \ 2 \ 3 end{bmatrix} + s begin{bmatrix} 3 \ 2 \ 1 end{bmatrix}
$



Find an equation which describes the plane and show the point $(5,6,5)$ is not on the plane $P$.






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 15 at 23:04









jwc845

382215




382215










asked Jan 15 at 22:01









Brooklyn Van HellBrooklyn Van Hell

31




31








  • 2




    $begingroup$
    What have you tried so far?
    $endgroup$
    – jwc845
    Jan 15 at 22:11














  • 2




    $begingroup$
    What have you tried so far?
    $endgroup$
    – jwc845
    Jan 15 at 22:11








2




2




$begingroup$
What have you tried so far?
$endgroup$
– jwc845
Jan 15 at 22:11




$begingroup$
What have you tried so far?
$endgroup$
– jwc845
Jan 15 at 22:11










1 Answer
1






active

oldest

votes


















-1












$begingroup$

You can take the cross-product to get the normal: $(1,2,3)×(3,2,1)=begin{vmatrix}i&j&k\1&2&3\3&2&1end{vmatrix}=-4i+8j-4k$.



Then use the point $(1,2,3)$: $4x-8y+4z=dimplies 4(1)-8(2)+3(4)=0=d$. $d$ should be zero, because the plane goes through the origin.



So, $4x-8y+4z=0$ would be an equation of the plane.



Lastly, $4(5)-8(6)+4(5)=-8neq0$. So $(5,6,5)$ is not on the plane.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This helped a lot! Thanks so much!
    $endgroup$
    – Brooklyn Van Hell
    Jan 15 at 22:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Chris Custer
    Jan 15 at 23:08












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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









-1












$begingroup$

You can take the cross-product to get the normal: $(1,2,3)×(3,2,1)=begin{vmatrix}i&j&k\1&2&3\3&2&1end{vmatrix}=-4i+8j-4k$.



Then use the point $(1,2,3)$: $4x-8y+4z=dimplies 4(1)-8(2)+3(4)=0=d$. $d$ should be zero, because the plane goes through the origin.



So, $4x-8y+4z=0$ would be an equation of the plane.



Lastly, $4(5)-8(6)+4(5)=-8neq0$. So $(5,6,5)$ is not on the plane.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This helped a lot! Thanks so much!
    $endgroup$
    – Brooklyn Van Hell
    Jan 15 at 22:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Chris Custer
    Jan 15 at 23:08
















-1












$begingroup$

You can take the cross-product to get the normal: $(1,2,3)×(3,2,1)=begin{vmatrix}i&j&k\1&2&3\3&2&1end{vmatrix}=-4i+8j-4k$.



Then use the point $(1,2,3)$: $4x-8y+4z=dimplies 4(1)-8(2)+3(4)=0=d$. $d$ should be zero, because the plane goes through the origin.



So, $4x-8y+4z=0$ would be an equation of the plane.



Lastly, $4(5)-8(6)+4(5)=-8neq0$. So $(5,6,5)$ is not on the plane.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    This helped a lot! Thanks so much!
    $endgroup$
    – Brooklyn Van Hell
    Jan 15 at 22:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Chris Custer
    Jan 15 at 23:08














-1












-1








-1





$begingroup$

You can take the cross-product to get the normal: $(1,2,3)×(3,2,1)=begin{vmatrix}i&j&k\1&2&3\3&2&1end{vmatrix}=-4i+8j-4k$.



Then use the point $(1,2,3)$: $4x-8y+4z=dimplies 4(1)-8(2)+3(4)=0=d$. $d$ should be zero, because the plane goes through the origin.



So, $4x-8y+4z=0$ would be an equation of the plane.



Lastly, $4(5)-8(6)+4(5)=-8neq0$. So $(5,6,5)$ is not on the plane.






share|cite|improve this answer











$endgroup$



You can take the cross-product to get the normal: $(1,2,3)×(3,2,1)=begin{vmatrix}i&j&k\1&2&3\3&2&1end{vmatrix}=-4i+8j-4k$.



Then use the point $(1,2,3)$: $4x-8y+4z=dimplies 4(1)-8(2)+3(4)=0=d$. $d$ should be zero, because the plane goes through the origin.



So, $4x-8y+4z=0$ would be an equation of the plane.



Lastly, $4(5)-8(6)+4(5)=-8neq0$. So $(5,6,5)$ is not on the plane.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 15 at 22:37

























answered Jan 15 at 22:13









Chris CusterChris Custer

14.6k3827




14.6k3827












  • $begingroup$
    This helped a lot! Thanks so much!
    $endgroup$
    – Brooklyn Van Hell
    Jan 15 at 22:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Chris Custer
    Jan 15 at 23:08


















  • $begingroup$
    This helped a lot! Thanks so much!
    $endgroup$
    – Brooklyn Van Hell
    Jan 15 at 22:25










  • $begingroup$
    You're welcome.
    $endgroup$
    – Chris Custer
    Jan 15 at 23:08
















$begingroup$
This helped a lot! Thanks so much!
$endgroup$
– Brooklyn Van Hell
Jan 15 at 22:25




$begingroup$
This helped a lot! Thanks so much!
$endgroup$
– Brooklyn Van Hell
Jan 15 at 22:25












$begingroup$
You're welcome.
$endgroup$
– Chris Custer
Jan 15 at 23:08




$begingroup$
You're welcome.
$endgroup$
– Chris Custer
Jan 15 at 23:08


















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