How to find a point which lies at distance d on 3D line, given a position vector and direction vector? Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to find line parallel to direction vector and passing through a specific point?What is the modulus in $mathbbC^3$?Calculate velocity vector given position on sphere, heading, and pitchFind the Distance Point to Line with Point on Line and Direction VectorGiven a line and a point in 3D, how to find the closest point on the line?Finding shortest distance from a point to line through direction vectorMovement of a 3D point along its direction vectorAdding direction vector to a positionHow to calculate a straight with a position vector (x,y) and a direction vector (x,y)Finding the position vector of a point on a line
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How to find a point which lies at distance d on 3D line, given a position vector and direction vector?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to find line parallel to direction vector and passing through a specific point?What is the modulus in $mathbbC^3$?Calculate velocity vector given position on sphere, heading, and pitchFind the Distance Point to Line with Point on Line and Direction VectorGiven a line and a point in 3D, how to find the closest point on the line?Finding shortest distance from a point to line through direction vectorMovement of a 3D point along its direction vectorAdding direction vector to a positionHow to calculate a straight with a position vector (x,y) and a direction vector (x,y)Finding the position vector of a point on a line
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I have a position vector $(p_x, p_y, p_z)$ and direction vector $(v_x, v_y, v_z)$. I need to find a point on along the direction vector which is at distance $d$ from $(p_x, p_y, p_z)$.
vectors 3d
edited Feb 11 '16 at 18:51
Frentos
2,5151622
asked Feb 11 '16 at 18:03
user2454092user2454092
1
$endgroup$
add a comment |
$begingroup$
I have a position vector $(p_x, p_y, p_z)$ and direction vector $(v_x, v_y, v_z)$. I need to find a point on along the direction vector which is at distance $d$ from $(p_x, p_y, p_z)$.
vectors 3d
edited Feb 11 '16 at 18:51
Frentos
2,5151622
asked Feb 11 '16 at 18:03
user2454092user2454092
1
$endgroup$
$begingroup$
What about $frac d (v_x, v_y, v_z)+(p_x, p_y, p_z)$?
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– Gyro Gearloose
Feb 11 '16 at 18:15
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can you please explain ? @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:18
$begingroup$
Got it, Thank you :) @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:22
add a comment |
$begingroup$
I have a position vector $(p_x, p_y, p_z)$ and direction vector $(v_x, v_y, v_z)$. I need to find a point on along the direction vector which is at distance $d$ from $(p_x, p_y, p_z)$.
vectors 3d
edited Feb 11 '16 at 18:51
Frentos
2,5151622
asked Feb 11 '16 at 18:03
user2454092user2454092
1
$endgroup$
I have a position vector $(p_x, p_y, p_z)$ and direction vector $(v_x, v_y, v_z)$. I need to find a point on along the direction vector which is at distance $d$ from $(p_x, p_y, p_z)$.
vectors 3d
vectors 3d
edited Feb 11 '16 at 18:51
Frentos
2,5151622
asked Feb 11 '16 at 18:03
user2454092user2454092
1
edited Feb 11 '16 at 18:51
Frentos
2,5151622
asked Feb 11 '16 at 18:03
user2454092user2454092
1
edited Feb 11 '16 at 18:51
Frentos
2,5151622
edited Feb 11 '16 at 18:51
Frentos
2,5151622
edited Feb 11 '16 at 18:51
Frentos
2,5151622
2,5151622
asked Feb 11 '16 at 18:03
user2454092user2454092
1
asked Feb 11 '16 at 18:03
user2454092user2454092
1
asked Feb 11 '16 at 18:03
user2454092user2454092
1
1
$begingroup$
What about $frac d (v_x, v_y, v_z)+(p_x, p_y, p_z)$?
$endgroup$
– Gyro Gearloose
Feb 11 '16 at 18:15
$begingroup$
can you please explain ? @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:18
$begingroup$
Got it, Thank you :) @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:22
add a comment |
$begingroup$
What about $frac d (v_x, v_y, v_z)+(p_x, p_y, p_z)$?
$endgroup$
– Gyro Gearloose
Feb 11 '16 at 18:15
$begingroup$
can you please explain ? @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:18
$begingroup$
Got it, Thank you :) @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:22
$begingroup$
What about $frac d (v_x, v_y, v_z)+(p_x, p_y, p_z)$?
$endgroup$
– Gyro Gearloose
Feb 11 '16 at 18:15
$begingroup$
What about $frac d (v_x, v_y, v_z)+(p_x, p_y, p_z)$?
$endgroup$
– Gyro Gearloose
Feb 11 '16 at 18:15
$begingroup$
can you please explain ? @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:18
$begingroup$
can you please explain ? @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:18
$begingroup$
Got it, Thank you :) @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:22
$begingroup$
Got it, Thank you :) @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:22
add a comment |
1 Answer
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$begingroup$
There are two points on the line at distance $d$ from $mathbfp$. They are $mathbfppm d hatmathbfv$ where $hatmathbfv$ is a unit vector parallel to $mathbfv$. In coordinates, that's $$(p_x,p_y,p_z)pmdfracdsqrtv_x^2+v_y^2+v_z^2(v_x,v_y,v_z)$$
answered Feb 11 '16 at 18:51
FrentosFrentos
2,5151622
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
There are two points on the line at distance $d$ from $mathbfp$. They are $mathbfppm d hatmathbfv$ where $hatmathbfv$ is a unit vector parallel to $mathbfv$. In coordinates, that's $$(p_x,p_y,p_z)pmdfracdsqrtv_x^2+v_y^2+v_z^2(v_x,v_y,v_z)$$
answered Feb 11 '16 at 18:51
FrentosFrentos
2,5151622
$endgroup$
add a comment |
$begingroup$
There are two points on the line at distance $d$ from $mathbfp$. They are $mathbfppm d hatmathbfv$ where $hatmathbfv$ is a unit vector parallel to $mathbfv$. In coordinates, that's $$(p_x,p_y,p_z)pmdfracdsqrtv_x^2+v_y^2+v_z^2(v_x,v_y,v_z)$$
answered Feb 11 '16 at 18:51
FrentosFrentos
2,5151622
$endgroup$
add a comment |
$begingroup$
There are two points on the line at distance $d$ from $mathbfp$. They are $mathbfppm d hatmathbfv$ where $hatmathbfv$ is a unit vector parallel to $mathbfv$. In coordinates, that's $$(p_x,p_y,p_z)pmdfracdsqrtv_x^2+v_y^2+v_z^2(v_x,v_y,v_z)$$
answered Feb 11 '16 at 18:51
FrentosFrentos
2,5151622
$endgroup$
There are two points on the line at distance $d$ from $mathbfp$. They are $mathbfppm d hatmathbfv$ where $hatmathbfv$ is a unit vector parallel to $mathbfv$. In coordinates, that's $$(p_x,p_y,p_z)pmdfracdsqrtv_x^2+v_y^2+v_z^2(v_x,v_y,v_z)$$
answered Feb 11 '16 at 18:51
FrentosFrentos
2,5151622
answered Feb 11 '16 at 18:51
FrentosFrentos
2,5151622
answered Feb 11 '16 at 18:51
FrentosFrentos
2,5151622
answered Feb 11 '16 at 18:51
FrentosFrentos
2,5151622
2,5151622
add a comment |
add a comment |
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$begingroup$
What about $frac d (v_x, v_y, v_z)+(p_x, p_y, p_z)$?
$endgroup$
– Gyro Gearloose
Feb 11 '16 at 18:15
$begingroup$
can you please explain ? @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:18
$begingroup$
Got it, Thank you :) @Gyro
$endgroup$
– user2454092
Feb 11 '16 at 18:22