Finding the intersection of two planes, given their normal vectors
$begingroup$
Two vectors $v_1=(-1,1,0), v_2=(1,0,1)$ span a plane $P$ in $Bbb R^3$.
Two vectors $w_1=(0,1,0), w_2=(1,1,0)$ span a plane $Q$ in $Bbb R^3$.
I am asked to show that $P$ and $Q$ are different and to find $P ∩ Q$.
I have computed the normal vectors to each plane:
normal to $P$ is $n_3∗ (0,-1,1)^t$
normal to $Q$ is $n_3∗(0,0,1)^t$
If the normal vectors are different, then the planes must be different.
How do I find $P ∩ Q$?
Thank you for your help.
linear-algebra vectors
asked Oct 17 '15 at 3:37
Gwen VastineGwen Vastine
12
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add a comment |
$begingroup$
Two vectors $v_1=(-1,1,0), v_2=(1,0,1)$ span a plane $P$ in $Bbb R^3$.
Two vectors $w_1=(0,1,0), w_2=(1,1,0)$ span a plane $Q$ in $Bbb R^3$.
I am asked to show that $P$ and $Q$ are different and to find $P ∩ Q$.
I have computed the normal vectors to each plane:
normal to $P$ is $n_3∗ (0,-1,1)^t$
normal to $Q$ is $n_3∗(0,0,1)^t$
If the normal vectors are different, then the planes must be different.
How do I find $P ∩ Q$?
Thank you for your help.
linear-algebra vectors
asked Oct 17 '15 at 3:37
Gwen VastineGwen Vastine
12
$endgroup$
add a comment |
$begingroup$
Two vectors $v_1=(-1,1,0), v_2=(1,0,1)$ span a plane $P$ in $Bbb R^3$.
Two vectors $w_1=(0,1,0), w_2=(1,1,0)$ span a plane $Q$ in $Bbb R^3$.
I am asked to show that $P$ and $Q$ are different and to find $P ∩ Q$.
I have computed the normal vectors to each plane:
normal to $P$ is $n_3∗ (0,-1,1)^t$
normal to $Q$ is $n_3∗(0,0,1)^t$
If the normal vectors are different, then the planes must be different.
How do I find $P ∩ Q$?
Thank you for your help.
linear-algebra vectors
asked Oct 17 '15 at 3:37
Gwen VastineGwen Vastine
12
$endgroup$
Two vectors $v_1=(-1,1,0), v_2=(1,0,1)$ span a plane $P$ in $Bbb R^3$.
Two vectors $w_1=(0,1,0), w_2=(1,1,0)$ span a plane $Q$ in $Bbb R^3$.
I am asked to show that $P$ and $Q$ are different and to find $P ∩ Q$.
I have computed the normal vectors to each plane:
normal to $P$ is $n_3∗ (0,-1,1)^t$
normal to $Q$ is $n_3∗(0,0,1)^t$
If the normal vectors are different, then the planes must be different.
How do I find $P ∩ Q$?
Thank you for your help.
linear-algebra vectors
linear-algebra vectors
asked Oct 17 '15 at 3:37
Gwen VastineGwen Vastine
12
asked Oct 17 '15 at 3:37
Gwen VastineGwen Vastine
12
edited Oct 17 '15 at 12:43
Gwen Vastine
asked Oct 17 '15 at 3:37
Gwen VastineGwen Vastine
12
asked Oct 17 '15 at 3:37
Gwen VastineGwen Vastine
12
asked Oct 17 '15 at 3:37
Gwen VastineGwen Vastine
12
12
add a comment |
add a comment |
2 Answers
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$begingroup$
Hint: Let $n_1$ and $n_2$ be normal vectors. If $n_1neq cn_2$ then intersetion is a line with direction vector $n_1times n_2$.
answered Oct 17 '15 at 3:43
meselmesel
10.4k21746
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add a comment |
$begingroup$
Here is a trick, since you have already computed the normal vectors.
$Normal_P$ and $Normal_Q$ are orthogonal to the planes $P$ and $Q$, respectively. Hence they are both orthogonal to the intersection of the planes $P$ and $Q$. But the intersection of 2 planes is a line, and it is a line spanned by a vector that is orthogonal to $N_P$ and $N_Q$.
The cross product of the normals will give you a vector orthogonal to $N_P$ and $N_Q$, which is necessarily nonzero, and which lies on the intersection. This is enough to describe the line that is the intersection.
answered Oct 17 '15 at 3:44
LorenzoLorenzo
11.7k31638
$endgroup$
$begingroup$
Great! I found the cross product of the two vectors which is (-1,0,0). I checked by computing the dot product of this new vector with the normal vectors and they are indeed perpendicular.
$endgroup$
– Gwen Vastine
Oct 17 '15 at 13:03
$begingroup$
@GwenVastine The cross product of vectors $a$ and $b$ will always be perpendicular to $a$ and $b$, which makes it a convenient tool for producing perpendicular vectors. You should think of this as what the cross product does - it produces a vector perpendicular to two other vectors (moreover, this is done in a consistent way, i.e. the components of the new vector are a continuous function of the components of the original vectors). (The other important feature is that the cross product has length equal to the area of the parallelogram spanned by $a$ and $b$.)
$endgroup$
– Lorenzo
Oct 17 '15 at 18:14
add a comment |
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2 Answers
2
active
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2 Answers
2
active
oldest
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active
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votes
$begingroup$
Hint: Let $n_1$ and $n_2$ be normal vectors. If $n_1neq cn_2$ then intersetion is a line with direction vector $n_1times n_2$.
answered Oct 17 '15 at 3:43
meselmesel
10.4k21746
$endgroup$
add a comment |
$begingroup$
Hint: Let $n_1$ and $n_2$ be normal vectors. If $n_1neq cn_2$ then intersetion is a line with direction vector $n_1times n_2$.
answered Oct 17 '15 at 3:43
meselmesel
10.4k21746
$endgroup$
add a comment |
$begingroup$
Hint: Let $n_1$ and $n_2$ be normal vectors. If $n_1neq cn_2$ then intersetion is a line with direction vector $n_1times n_2$.
answered Oct 17 '15 at 3:43
meselmesel
10.4k21746
$endgroup$
Hint: Let $n_1$ and $n_2$ be normal vectors. If $n_1neq cn_2$ then intersetion is a line with direction vector $n_1times n_2$.
answered Oct 17 '15 at 3:43
meselmesel
10.4k21746
answered Oct 17 '15 at 3:43
meselmesel
10.4k21746
answered Oct 17 '15 at 3:43
meselmesel
10.4k21746
answered Oct 17 '15 at 3:43
meselmesel
10.4k21746
10.4k21746
add a comment |
add a comment |
$begingroup$
Here is a trick, since you have already computed the normal vectors.
$Normal_P$ and $Normal_Q$ are orthogonal to the planes $P$ and $Q$, respectively. Hence they are both orthogonal to the intersection of the planes $P$ and $Q$. But the intersection of 2 planes is a line, and it is a line spanned by a vector that is orthogonal to $N_P$ and $N_Q$.
The cross product of the normals will give you a vector orthogonal to $N_P$ and $N_Q$, which is necessarily nonzero, and which lies on the intersection. This is enough to describe the line that is the intersection.
answered Oct 17 '15 at 3:44
LorenzoLorenzo
11.7k31638
$endgroup$
$begingroup$
Great! I found the cross product of the two vectors which is (-1,0,0). I checked by computing the dot product of this new vector with the normal vectors and they are indeed perpendicular.
$endgroup$
– Gwen Vastine
Oct 17 '15 at 13:03
$begingroup$
@GwenVastine The cross product of vectors $a$ and $b$ will always be perpendicular to $a$ and $b$, which makes it a convenient tool for producing perpendicular vectors. You should think of this as what the cross product does - it produces a vector perpendicular to two other vectors (moreover, this is done in a consistent way, i.e. the components of the new vector are a continuous function of the components of the original vectors). (The other important feature is that the cross product has length equal to the area of the parallelogram spanned by $a$ and $b$.)
$endgroup$
– Lorenzo
Oct 17 '15 at 18:14
add a comment |
$begingroup$
Here is a trick, since you have already computed the normal vectors.
$Normal_P$ and $Normal_Q$ are orthogonal to the planes $P$ and $Q$, respectively. Hence they are both orthogonal to the intersection of the planes $P$ and $Q$. But the intersection of 2 planes is a line, and it is a line spanned by a vector that is orthogonal to $N_P$ and $N_Q$.
The cross product of the normals will give you a vector orthogonal to $N_P$ and $N_Q$, which is necessarily nonzero, and which lies on the intersection. This is enough to describe the line that is the intersection.
answered Oct 17 '15 at 3:44
LorenzoLorenzo
11.7k31638
$endgroup$
$begingroup$
Great! I found the cross product of the two vectors which is (-1,0,0). I checked by computing the dot product of this new vector with the normal vectors and they are indeed perpendicular.
$endgroup$
– Gwen Vastine
Oct 17 '15 at 13:03
$begingroup$
@GwenVastine The cross product of vectors $a$ and $b$ will always be perpendicular to $a$ and $b$, which makes it a convenient tool for producing perpendicular vectors. You should think of this as what the cross product does - it produces a vector perpendicular to two other vectors (moreover, this is done in a consistent way, i.e. the components of the new vector are a continuous function of the components of the original vectors). (The other important feature is that the cross product has length equal to the area of the parallelogram spanned by $a$ and $b$.)
$endgroup$
– Lorenzo
Oct 17 '15 at 18:14
add a comment |
$begingroup$
Here is a trick, since you have already computed the normal vectors.
$Normal_P$ and $Normal_Q$ are orthogonal to the planes $P$ and $Q$, respectively. Hence they are both orthogonal to the intersection of the planes $P$ and $Q$. But the intersection of 2 planes is a line, and it is a line spanned by a vector that is orthogonal to $N_P$ and $N_Q$.
The cross product of the normals will give you a vector orthogonal to $N_P$ and $N_Q$, which is necessarily nonzero, and which lies on the intersection. This is enough to describe the line that is the intersection.
answered Oct 17 '15 at 3:44
LorenzoLorenzo
11.7k31638
$endgroup$
Here is a trick, since you have already computed the normal vectors.
$Normal_P$ and $Normal_Q$ are orthogonal to the planes $P$ and $Q$, respectively. Hence they are both orthogonal to the intersection of the planes $P$ and $Q$. But the intersection of 2 planes is a line, and it is a line spanned by a vector that is orthogonal to $N_P$ and $N_Q$.
The cross product of the normals will give you a vector orthogonal to $N_P$ and $N_Q$, which is necessarily nonzero, and which lies on the intersection. This is enough to describe the line that is the intersection.
answered Oct 17 '15 at 3:44
LorenzoLorenzo
11.7k31638
answered Oct 17 '15 at 3:44
LorenzoLorenzo
11.7k31638
answered Oct 17 '15 at 3:44
LorenzoLorenzo
11.7k31638
answered Oct 17 '15 at 3:44
LorenzoLorenzo
11.7k31638
11.7k31638
$begingroup$
Great! I found the cross product of the two vectors which is (-1,0,0). I checked by computing the dot product of this new vector with the normal vectors and they are indeed perpendicular.
$endgroup$
– Gwen Vastine
Oct 17 '15 at 13:03
$begingroup$
@GwenVastine The cross product of vectors $a$ and $b$ will always be perpendicular to $a$ and $b$, which makes it a convenient tool for producing perpendicular vectors. You should think of this as what the cross product does - it produces a vector perpendicular to two other vectors (moreover, this is done in a consistent way, i.e. the components of the new vector are a continuous function of the components of the original vectors). (The other important feature is that the cross product has length equal to the area of the parallelogram spanned by $a$ and $b$.)
$endgroup$
– Lorenzo
Oct 17 '15 at 18:14
add a comment |
$begingroup$
Great! I found the cross product of the two vectors which is (-1,0,0). I checked by computing the dot product of this new vector with the normal vectors and they are indeed perpendicular.
$endgroup$
– Gwen Vastine
Oct 17 '15 at 13:03
$begingroup$
@GwenVastine The cross product of vectors $a$ and $b$ will always be perpendicular to $a$ and $b$, which makes it a convenient tool for producing perpendicular vectors. You should think of this as what the cross product does - it produces a vector perpendicular to two other vectors (moreover, this is done in a consistent way, i.e. the components of the new vector are a continuous function of the components of the original vectors). (The other important feature is that the cross product has length equal to the area of the parallelogram spanned by $a$ and $b$.)
$endgroup$
– Lorenzo
Oct 17 '15 at 18:14
$begingroup$
Great! I found the cross product of the two vectors which is (-1,0,0). I checked by computing the dot product of this new vector with the normal vectors and they are indeed perpendicular.
$endgroup$
– Gwen Vastine
Oct 17 '15 at 13:03
$begingroup$
Great! I found the cross product of the two vectors which is (-1,0,0). I checked by computing the dot product of this new vector with the normal vectors and they are indeed perpendicular.
$endgroup$
– Gwen Vastine
Oct 17 '15 at 13:03
$begingroup$
@GwenVastine The cross product of vectors $a$ and $b$ will always be perpendicular to $a$ and $b$, which makes it a convenient tool for producing perpendicular vectors. You should think of this as what the cross product does - it produces a vector perpendicular to two other vectors (moreover, this is done in a consistent way, i.e. the components of the new vector are a continuous function of the components of the original vectors). (The other important feature is that the cross product has length equal to the area of the parallelogram spanned by $a$ and $b$.)
$endgroup$
– Lorenzo
Oct 17 '15 at 18:14
$begingroup$
@GwenVastine The cross product of vectors $a$ and $b$ will always be perpendicular to $a$ and $b$, which makes it a convenient tool for producing perpendicular vectors. You should think of this as what the cross product does - it produces a vector perpendicular to two other vectors (moreover, this is done in a consistent way, i.e. the components of the new vector are a continuous function of the components of the original vectors). (The other important feature is that the cross product has length equal to the area of the parallelogram spanned by $a$ and $b$.)
$endgroup$
– Lorenzo
Oct 17 '15 at 18:14
add a comment |
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