Finding point of intersection using cartesian equation?
$begingroup$
L1 is defined by $ dfrac{x+1}{2} = dfrac{y-3}{3} = 1-z$
So we can write this as $((2t-1), (3t+3), (1-t)) = r $
L2 passes through $(5,4,2)$ and intersects with L1 at right angles.
I am asked to determine the point of intersections between those two lines.
I have tried to find the dot product of the directional vector of L1 and another directional vector (x,y,z)
So I got this $2x+3y-z = 0$ and said that $ x = 3, y=2$ and $ z=12$
So now L2 has the following equation
$ r = ((5+3t), (4+2t), (2+12t))$
However when i equate L1 and L2 to find $t$. The equations aren't satisfied, meaning that $t$ isn't common.
vectors
asked Jan 9 at 18:06
user154844user154844
33
$endgroup$
add a comment |
$begingroup$
L1 is defined by $ dfrac{x+1}{2} = dfrac{y-3}{3} = 1-z$
So we can write this as $((2t-1), (3t+3), (1-t)) = r $
L2 passes through $(5,4,2)$ and intersects with L1 at right angles.
I am asked to determine the point of intersections between those two lines.
I have tried to find the dot product of the directional vector of L1 and another directional vector (x,y,z)
So I got this $2x+3y-z = 0$ and said that $ x = 3, y=2$ and $ z=12$
So now L2 has the following equation
$ r = ((5+3t), (4+2t), (2+12t))$
However when i equate L1 and L2 to find $t$. The equations aren't satisfied, meaning that $t$ isn't common.
vectors
asked Jan 9 at 18:06
user154844user154844
33
$endgroup$
$begingroup$
You can't just assume $x=3,y=2,z=12$, because there are infinitely many vectors perpendicular to $(2,3,-1)$.
$endgroup$
– Shubham Johri
Jan 9 at 18:23
add a comment |
$begingroup$
L1 is defined by $ dfrac{x+1}{2} = dfrac{y-3}{3} = 1-z$
So we can write this as $((2t-1), (3t+3), (1-t)) = r $
L2 passes through $(5,4,2)$ and intersects with L1 at right angles.
I am asked to determine the point of intersections between those two lines.
I have tried to find the dot product of the directional vector of L1 and another directional vector (x,y,z)
So I got this $2x+3y-z = 0$ and said that $ x = 3, y=2$ and $ z=12$
So now L2 has the following equation
$ r = ((5+3t), (4+2t), (2+12t))$
However when i equate L1 and L2 to find $t$. The equations aren't satisfied, meaning that $t$ isn't common.
vectors
asked Jan 9 at 18:06
user154844user154844
33
$endgroup$
L1 is defined by $ dfrac{x+1}{2} = dfrac{y-3}{3} = 1-z$
So we can write this as $((2t-1), (3t+3), (1-t)) = r $
L2 passes through $(5,4,2)$ and intersects with L1 at right angles.
I am asked to determine the point of intersections between those two lines.
I have tried to find the dot product of the directional vector of L1 and another directional vector (x,y,z)
So I got this $2x+3y-z = 0$ and said that $ x = 3, y=2$ and $ z=12$
So now L2 has the following equation
$ r = ((5+3t), (4+2t), (2+12t))$
However when i equate L1 and L2 to find $t$. The equations aren't satisfied, meaning that $t$ isn't common.
vectors
vectors
asked Jan 9 at 18:06
user154844user154844
33
asked Jan 9 at 18:06
user154844user154844
33
asked Jan 9 at 18:06
user154844user154844
33
asked Jan 9 at 18:06
user154844user154844
33
asked Jan 9 at 18:06
user154844user154844
33
33
$begingroup$
You can't just assume $x=3,y=2,z=12$, because there are infinitely many vectors perpendicular to $(2,3,-1)$.
$endgroup$
– Shubham Johri
Jan 9 at 18:23
add a comment |
$begingroup$
You can't just assume $x=3,y=2,z=12$, because there are infinitely many vectors perpendicular to $(2,3,-1)$.
$endgroup$
– Shubham Johri
Jan 9 at 18:23
$begingroup$
You can't just assume $x=3,y=2,z=12$, because there are infinitely many vectors perpendicular to $(2,3,-1)$.
$endgroup$
– Shubham Johri
Jan 9 at 18:23
$begingroup$
You can't just assume $x=3,y=2,z=12$, because there are infinitely many vectors perpendicular to $(2,3,-1)$.
$endgroup$
– Shubham Johri
Jan 9 at 18:23
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The point of intersection lies on $L_1$, so we can express it as $P(2t-1,3t+3,1-t)$ for some $tinBbb R$. We require that the vector from $Pto(5,4,2)$ be orthogonal to the vector $(2,3,-1)$, which is colinear with the line. That is,$$(2t-1,3t+3,1-t)to(5,4,2)perp(2,3,-1)$$This implies,$$(2t-6,3t-1,-1-t)cdot(2,3,-1)=0$$giving the value of $t=1$ and the required point as $(1,6,0)$.
answered Jan 9 at 18:18
Shubham JohriShubham Johri
4,760717
$endgroup$
add a comment |
$begingroup$
Let $L_2$ $$vec{x}=[5,4,2]+s[a_x,a_y,a_z]$$
Then we have the condition $$2a_x+3a_y+4a_z=0$$ (by the dot-product)
and the system
$$2t-sa_x=6$$
$$3t-sa_y=1$$
$$-t-sa_z=1$$
Can you solve this?
answered Jan 9 at 18:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
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votes
$begingroup$
The point of intersection lies on $L_1$, so we can express it as $P(2t-1,3t+3,1-t)$ for some $tinBbb R$. We require that the vector from $Pto(5,4,2)$ be orthogonal to the vector $(2,3,-1)$, which is colinear with the line. That is,$$(2t-1,3t+3,1-t)to(5,4,2)perp(2,3,-1)$$This implies,$$(2t-6,3t-1,-1-t)cdot(2,3,-1)=0$$giving the value of $t=1$ and the required point as $(1,6,0)$.
answered Jan 9 at 18:18
Shubham JohriShubham Johri
4,760717
$endgroup$
add a comment |
$begingroup$
The point of intersection lies on $L_1$, so we can express it as $P(2t-1,3t+3,1-t)$ for some $tinBbb R$. We require that the vector from $Pto(5,4,2)$ be orthogonal to the vector $(2,3,-1)$, which is colinear with the line. That is,$$(2t-1,3t+3,1-t)to(5,4,2)perp(2,3,-1)$$This implies,$$(2t-6,3t-1,-1-t)cdot(2,3,-1)=0$$giving the value of $t=1$ and the required point as $(1,6,0)$.
answered Jan 9 at 18:18
Shubham JohriShubham Johri
4,760717
$endgroup$
add a comment |
$begingroup$
The point of intersection lies on $L_1$, so we can express it as $P(2t-1,3t+3,1-t)$ for some $tinBbb R$. We require that the vector from $Pto(5,4,2)$ be orthogonal to the vector $(2,3,-1)$, which is colinear with the line. That is,$$(2t-1,3t+3,1-t)to(5,4,2)perp(2,3,-1)$$This implies,$$(2t-6,3t-1,-1-t)cdot(2,3,-1)=0$$giving the value of $t=1$ and the required point as $(1,6,0)$.
answered Jan 9 at 18:18
Shubham JohriShubham Johri
4,760717
$endgroup$
The point of intersection lies on $L_1$, so we can express it as $P(2t-1,3t+3,1-t)$ for some $tinBbb R$. We require that the vector from $Pto(5,4,2)$ be orthogonal to the vector $(2,3,-1)$, which is colinear with the line. That is,$$(2t-1,3t+3,1-t)to(5,4,2)perp(2,3,-1)$$This implies,$$(2t-6,3t-1,-1-t)cdot(2,3,-1)=0$$giving the value of $t=1$ and the required point as $(1,6,0)$.
answered Jan 9 at 18:18
Shubham JohriShubham Johri
4,760717
answered Jan 9 at 18:18
Shubham JohriShubham Johri
4,760717
answered Jan 9 at 18:18
Shubham JohriShubham Johri
4,760717
answered Jan 9 at 18:18
Shubham JohriShubham Johri
4,760717
4,760717
add a comment |
add a comment |
$begingroup$
Let $L_2$ $$vec{x}=[5,4,2]+s[a_x,a_y,a_z]$$
Then we have the condition $$2a_x+3a_y+4a_z=0$$ (by the dot-product)
and the system
$$2t-sa_x=6$$
$$3t-sa_y=1$$
$$-t-sa_z=1$$
Can you solve this?
answered Jan 9 at 18:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
$endgroup$
add a comment |
$begingroup$
Let $L_2$ $$vec{x}=[5,4,2]+s[a_x,a_y,a_z]$$
Then we have the condition $$2a_x+3a_y+4a_z=0$$ (by the dot-product)
and the system
$$2t-sa_x=6$$
$$3t-sa_y=1$$
$$-t-sa_z=1$$
Can you solve this?
answered Jan 9 at 18:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
$endgroup$
add a comment |
$begingroup$
Let $L_2$ $$vec{x}=[5,4,2]+s[a_x,a_y,a_z]$$
Then we have the condition $$2a_x+3a_y+4a_z=0$$ (by the dot-product)
and the system
$$2t-sa_x=6$$
$$3t-sa_y=1$$
$$-t-sa_z=1$$
Can you solve this?
answered Jan 9 at 18:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
$endgroup$
Let $L_2$ $$vec{x}=[5,4,2]+s[a_x,a_y,a_z]$$
Then we have the condition $$2a_x+3a_y+4a_z=0$$ (by the dot-product)
and the system
$$2t-sa_x=6$$
$$3t-sa_y=1$$
$$-t-sa_z=1$$
Can you solve this?
answered Jan 9 at 18:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
answered Jan 9 at 18:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
answered Jan 9 at 18:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
answered Jan 9 at 18:18
Dr. Sonnhard GraubnerDr. Sonnhard Graubner
74k42865
74k42865
add a comment |
add a comment |
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$begingroup$
You can't just assume $x=3,y=2,z=12$, because there are infinitely many vectors perpendicular to $(2,3,-1)$.
$endgroup$
– Shubham Johri
Jan 9 at 18:23