Finding the intersection of two planes, given their normal vectors
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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 Vastine
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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 Vastine
11
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 Vastine
11
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 Vastine
11
edited Oct 17 '15 at 12:43
asked Oct 17 '15 at 3:37
Gwen Vastine
11
asked Oct 17 '15 at 3:37
Gwen Vastine
11
asked Oct 17 '15 at 3:37
Gwen Vastine
11
11
add a comment |Â
add a comment |Â
2 Answers
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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
mesel
10.1k21644
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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
Lorenzo
11.5k31537
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.
– Gwen Vastine
Oct 17 '15 at 13:03
@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 $b$.)
– Lorenzo
Oct 17 '15 at 18:14
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
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
mesel
10.1k21644
add a comment |Â
up vote
0
down vote
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
mesel
10.1k21644
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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
mesel
10.1k21644
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
mesel
10.1k21644
answered Oct 17 '15 at 3:43
mesel
10.1k21644
answered Oct 17 '15 at 3:43
mesel
10.1k21644
answered Oct 17 '15 at 3:43
mesel
10.1k21644
10.1k21644
add a comment |Â
add a comment |Â
up vote
0
down vote
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
Lorenzo
11.5k31537
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.
– Gwen Vastine
Oct 17 '15 at 13:03
@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 $b$.)
– Lorenzo
Oct 17 '15 at 18:14
add a comment |Â
up vote
0
down vote
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
Lorenzo
11.5k31537
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.
– Gwen Vastine
Oct 17 '15 at 13:03
@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 $b$.)
– Lorenzo
Oct 17 '15 at 18:14
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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
Lorenzo
11.5k31537
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
Lorenzo
11.5k31537
answered Oct 17 '15 at 3:44
Lorenzo
11.5k31537
answered Oct 17 '15 at 3:44
Lorenzo
11.5k31537
answered Oct 17 '15 at 3:44
Lorenzo
11.5k31537
11.5k31537
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.
– Gwen Vastine
Oct 17 '15 at 13:03
@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 $b$.)
– Lorenzo
Oct 17 '15 at 18:14
add a comment |Â
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.
– Gwen Vastine
Oct 17 '15 at 13:03
@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 $b$.)
– Lorenzo
Oct 17 '15 at 18:14
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.
– Gwen Vastine
Oct 17 '15 at 13:03
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.
– Gwen Vastine
Oct 17 '15 at 13:03
@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 $b$.)
– Lorenzo
Oct 17 '15 at 18:14
@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 $b$.)
– Lorenzo
Oct 17 '15 at 18:14
add a comment |Â
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