Find an equation of a plane π that passes through point P = (0, 1, 0) and is parallel to vectors ~a = [−1,...
-1
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I need some help with the second one, can someone help me to solve this?enter image description here
linear-algebra vectors analytic-geometry
edited Jan 13 at 19:23
KReiser
10.2k21435
asked Jan 13 at 17:59
Aliaksei KlimovichAliaksei Klimovich
536
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closed as off-topic by KReiser, Namaste, mrtaurho, Shailesh, zz20s Jan 14 at 4:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – KReiser, Namaste, mrtaurho, Shailesh, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
-1
$begingroup$
I need some help with the second one, can someone help me to solve this?enter image description here
linear-algebra vectors analytic-geometry
edited Jan 13 at 19:23
KReiser
10.2k21435
asked Jan 13 at 17:59
Aliaksei KlimovichAliaksei Klimovich
536
$endgroup$
closed as off-topic by KReiser, Namaste, mrtaurho, Shailesh, zz20s Jan 14 at 4:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – KReiser, Namaste, mrtaurho, Shailesh, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
How did you solve the first one?
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– John Douma
Jan 13 at 18:15
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@JohnDouma ibb.co/PDjTJGC
$endgroup$
– Aliaksei Klimovich
Jan 13 at 19:05
$begingroup$
This is the same. A plane parallel to $a$ and $b$ will have its normal vector parallel to the cross product of $a$ and $b$.
$endgroup$
– John Douma
Jan 13 at 20:59
add a comment |
-1
-1
-1
$begingroup$
I need some help with the second one, can someone help me to solve this?enter image description here
linear-algebra vectors analytic-geometry
edited Jan 13 at 19:23
KReiser
10.2k21435
asked Jan 13 at 17:59
Aliaksei KlimovichAliaksei Klimovich
536
$endgroup$
I need some help with the second one, can someone help me to solve this?enter image description here
linear-algebra vectors analytic-geometry
linear-algebra vectors analytic-geometry
edited Jan 13 at 19:23
KReiser
10.2k21435
asked Jan 13 at 17:59
Aliaksei KlimovichAliaksei Klimovich
536
edited Jan 13 at 19:23
KReiser
10.2k21435
asked Jan 13 at 17:59
Aliaksei KlimovichAliaksei Klimovich
536
edited Jan 13 at 19:23
KReiser
10.2k21435
edited Jan 13 at 19:23
KReiser
10.2k21435
edited Jan 13 at 19:23
KReiser
10.2k21435
10.2k21435
asked Jan 13 at 17:59
Aliaksei KlimovichAliaksei Klimovich
536
asked Jan 13 at 17:59
Aliaksei KlimovichAliaksei Klimovich
536
asked Jan 13 at 17:59
Aliaksei KlimovichAliaksei Klimovich
536
536
closed as off-topic by KReiser, Namaste, mrtaurho, Shailesh, zz20s Jan 14 at 4:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – KReiser, Namaste, mrtaurho, Shailesh, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by KReiser, Namaste, mrtaurho, Shailesh, zz20s Jan 14 at 4:32
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – KReiser, Namaste, mrtaurho, Shailesh, zz20s
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
How did you solve the first one?
$endgroup$
– John Douma
Jan 13 at 18:15
$begingroup$
@JohnDouma ibb.co/PDjTJGC
$endgroup$
– Aliaksei Klimovich
Jan 13 at 19:05
$begingroup$
This is the same. A plane parallel to $a$ and $b$ will have its normal vector parallel to the cross product of $a$ and $b$.
$endgroup$
– John Douma
Jan 13 at 20:59
add a comment |
$begingroup$
How did you solve the first one?
$endgroup$
– John Douma
Jan 13 at 18:15
$begingroup$
@JohnDouma ibb.co/PDjTJGC
$endgroup$
– Aliaksei Klimovich
Jan 13 at 19:05
$begingroup$
This is the same. A plane parallel to $a$ and $b$ will have its normal vector parallel to the cross product of $a$ and $b$.
$endgroup$
– John Douma
Jan 13 at 20:59
$begingroup$
How did you solve the first one?
$endgroup$
– John Douma
Jan 13 at 18:15
$begingroup$
How did you solve the first one?
$endgroup$
– John Douma
Jan 13 at 18:15
$begingroup$
@JohnDouma ibb.co/PDjTJGC
$endgroup$
– Aliaksei Klimovich
Jan 13 at 19:05
$begingroup$
@JohnDouma ibb.co/PDjTJGC
$endgroup$
– Aliaksei Klimovich
Jan 13 at 19:05
$begingroup$
This is the same. A plane parallel to $a$ and $b$ will have its normal vector parallel to the cross product of $a$ and $b$.
$endgroup$
– John Douma
Jan 13 at 20:59
$begingroup$
This is the same. A plane parallel to $a$ and $b$ will have its normal vector parallel to the cross product of $a$ and $b$.
$endgroup$
– John Douma
Jan 13 at 20:59
add a comment |
1 Answer
1
active
oldest
votes
1
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Hint: Write $$vec{x}=vec{OP}+svec{a}+tvec{b}$$ where $s,t$ are real numbers.
It is $$vec{x}=[0,1,0]+s[-1,3,0]+t[3,1,-5]$$
and from here we get the system
$$x=-s+3t$$
$$y=1+3s+t$$
$$z=-5t$$
It is better for you now? With the equation $$t=-frac{z}{5}$$ you can eliminate $t$ then with $$s=-x-frac{3}{5}z$$ you can eliminate $s$
answered Jan 13 at 18:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
$endgroup$
$begingroup$
I can't really understand your equation, can you show me on my example?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:03
$begingroup$
So, the answer is 3 equations, am i right?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:58
$begingroup$
You can also eliminate the parameters to obtain an equation of the form $$ax+by+cz+d=0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 19:33
$begingroup$
Sorry, what do you mean with the number $3$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:48
$begingroup$
I get $$3x+y+2z-1=0$$ as an equation of your plane.
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:53
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
1
$begingroup$
Hint: Write $$vec{x}=vec{OP}+svec{a}+tvec{b}$$ where $s,t$ are real numbers.
It is $$vec{x}=[0,1,0]+s[-1,3,0]+t[3,1,-5]$$
and from here we get the system
$$x=-s+3t$$
$$y=1+3s+t$$
$$z=-5t$$
It is better for you now? With the equation $$t=-frac{z}{5}$$ you can eliminate $t$ then with $$s=-x-frac{3}{5}z$$ you can eliminate $s$
answered Jan 13 at 18:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
$endgroup$
$begingroup$
I can't really understand your equation, can you show me on my example?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:03
$begingroup$
So, the answer is 3 equations, am i right?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:58
$begingroup$
You can also eliminate the parameters to obtain an equation of the form $$ax+by+cz+d=0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 19:33
$begingroup$
Sorry, what do you mean with the number $3$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:48
$begingroup$
I get $$3x+y+2z-1=0$$ as an equation of your plane.
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:53
add a comment |
1
$begingroup$
Hint: Write $$vec{x}=vec{OP}+svec{a}+tvec{b}$$ where $s,t$ are real numbers.
It is $$vec{x}=[0,1,0]+s[-1,3,0]+t[3,1,-5]$$
and from here we get the system
$$x=-s+3t$$
$$y=1+3s+t$$
$$z=-5t$$
It is better for you now? With the equation $$t=-frac{z}{5}$$ you can eliminate $t$ then with $$s=-x-frac{3}{5}z$$ you can eliminate $s$
answered Jan 13 at 18:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
$endgroup$
$begingroup$
I can't really understand your equation, can you show me on my example?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:03
$begingroup$
So, the answer is 3 equations, am i right?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:58
$begingroup$
You can also eliminate the parameters to obtain an equation of the form $$ax+by+cz+d=0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 19:33
$begingroup$
Sorry, what do you mean with the number $3$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:48
$begingroup$
I get $$3x+y+2z-1=0$$ as an equation of your plane.
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:53
add a comment |
1
1
1
$begingroup$
Hint: Write $$vec{x}=vec{OP}+svec{a}+tvec{b}$$ where $s,t$ are real numbers.
It is $$vec{x}=[0,1,0]+s[-1,3,0]+t[3,1,-5]$$
and from here we get the system
$$x=-s+3t$$
$$y=1+3s+t$$
$$z=-5t$$
It is better for you now? With the equation $$t=-frac{z}{5}$$ you can eliminate $t$ then with $$s=-x-frac{3}{5}z$$ you can eliminate $s$
answered Jan 13 at 18:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
$endgroup$
Hint: Write $$vec{x}=vec{OP}+svec{a}+tvec{b}$$ where $s,t$ are real numbers.
It is $$vec{x}=[0,1,0]+s[-1,3,0]+t[3,1,-5]$$
and from here we get the system
$$x=-s+3t$$
$$y=1+3s+t$$
$$z=-5t$$
It is better for you now? With the equation $$t=-frac{z}{5}$$ you can eliminate $t$ then with $$s=-x-frac{3}{5}z$$ you can eliminate $s$
answered Jan 13 at 18:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
edited Jan 14 at 18:51
answered Jan 13 at 18:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
answered Jan 13 at 18:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
answered Jan 13 at 18:02
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
79.1k42867
79.1k42867
$begingroup$
I can't really understand your equation, can you show me on my example?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:03
$begingroup$
So, the answer is 3 equations, am i right?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:58
$begingroup$
You can also eliminate the parameters to obtain an equation of the form $$ax+by+cz+d=0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 19:33
$begingroup$
Sorry, what do you mean with the number $3$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:48
$begingroup$
I get $$3x+y+2z-1=0$$ as an equation of your plane.
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:53
add a comment |
$begingroup$
I can't really understand your equation, can you show me on my example?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:03
$begingroup$
So, the answer is 3 equations, am i right?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:58
$begingroup$
You can also eliminate the parameters to obtain an equation of the form $$ax+by+cz+d=0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 19:33
$begingroup$
Sorry, what do you mean with the number $3$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:48
$begingroup$
I get $$3x+y+2z-1=0$$ as an equation of your plane.
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:53
$begingroup$
I can't really understand your equation, can you show me on my example?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:03
$begingroup$
I can't really understand your equation, can you show me on my example?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:03
$begingroup$
So, the answer is 3 equations, am i right?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:58
$begingroup$
So, the answer is 3 equations, am i right?
$endgroup$
– Aliaksei Klimovich
Jan 13 at 18:58
$begingroup$
You can also eliminate the parameters to obtain an equation of the form $$ax+by+cz+d=0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 19:33
$begingroup$
You can also eliminate the parameters to obtain an equation of the form $$ax+by+cz+d=0$$
$endgroup$
– Dr. Sonnhard Graubner
Jan 13 at 19:33
$begingroup$
Sorry, what do you mean with the number $3$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:48
$begingroup$
Sorry, what do you mean with the number $3$?
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:48
$begingroup$
I get $$3x+y+2z-1=0$$ as an equation of your plane.
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:53
$begingroup$
I get $$3x+y+2z-1=0$$ as an equation of your plane.
$endgroup$
– Dr. Sonnhard Graubner
Jan 14 at 18:53
add a comment |
$begingroup$
How did you solve the first one?
$endgroup$
– John Douma
Jan 13 at 18:15
$begingroup$
@JohnDouma ibb.co/PDjTJGC
$endgroup$
– Aliaksei Klimovich
Jan 13 at 19:05
$begingroup$
This is the same. A plane parallel to $a$ and $b$ will have its normal vector parallel to the cross product of $a$ and $b$.
$endgroup$
– John Douma
Jan 13 at 20:59