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rotate 3d unit vector a, on plane of a and the j (up) axis



The 2019 Stack Overflow Developer Survey Results Are InComponents of a 3d vector given specific anglesRotate Existing Vectortesting parallelity/perpendicularity of two 3D vectors with lengths close to zero using dot productFinding the exact value of the sine of the angle between a line and a planeFinding the angle between two 3 dimensional vectorsDerivative in Vector vs Index NotationReflect a ray off a circle so it hits another pointRotating an N-dimensional vector in a planeThe simplest billiards problemUnit vector of a planeRotate vector by angles between a unit vector and the positive x axis










1












$begingroup$


vector $tilde a = ctilde i+dtilde j+etilde k$ is our input, and vector $tilde b = ftilde i+gtilde j+htilde k$ is our output, and $theta$ is the angle to rotate by. Essentially $tilde b = f(tilde a, theta)$



This is similar to Components of a 3d vector given specific angles, but I need to rotate the vector on the plane created by $tilde a$ and $tilde j$.



I attempted to create a function by converting $tilde a$ to a 2d point on this plane:
$(x, y) = (sqrt1-d^2, d)$

converting to a polar coordinate, adding alpha, converting back to cartesian, and taking the $y$ component as the value for $g$:



$$g = sqrtd^2+sqrt1-d^2 times sinbigg(tan^-1Big(fracdsqrt1-d^2Big)bigg)$$
$$f = frac c timessqrt1-d^2 sqrtc^2+e^2$$
$$h = frac etimes fc$$



This formula produces unexpected results when given a value close to $pmtilde j$, eg. $f(0.174126tilde i+0.984723tilde j, frac -pi16)$ returns $g = 0.996584$ which is higher the input value of $d=0.984723$. The expected value is aproximately $g=0.779171$ (I think). Continuing to tilt the resulting $tilde b$ by $frac-pi16$, cases $g$ to aproach $0.996$ ish.



Can anyone spot my error, or find a better way of doing this?










share|cite|improve this question









$endgroup$











  • $begingroup$
    Find the angle of rotation by dot multiplication. Why moving to 2D?
    $endgroup$
    – Moti
    Mar 24 at 6:32










  • $begingroup$
    @Moti because f and h can be found from a and g, and 2d allowed me to use radial coordinates which (I think) did exactly what I want, just in 2d
    $endgroup$
    – Alice Jacka
    Mar 24 at 7:05










  • $begingroup$
    @Moti could you elaborate on the dot multiplication?
    $endgroup$
    – Alice Jacka
    Mar 24 at 7:26















1












$begingroup$


vector $tilde a = ctilde i+dtilde j+etilde k$ is our input, and vector $tilde b = ftilde i+gtilde j+htilde k$ is our output, and $theta$ is the angle to rotate by. Essentially $tilde b = f(tilde a, theta)$



This is similar to Components of a 3d vector given specific angles, but I need to rotate the vector on the plane created by $tilde a$ and $tilde j$.



I attempted to create a function by converting $tilde a$ to a 2d point on this plane:
$(x, y) = (sqrt1-d^2, d)$

converting to a polar coordinate, adding alpha, converting back to cartesian, and taking the $y$ component as the value for $g$:



$$g = sqrtd^2+sqrt1-d^2 times sinbigg(tan^-1Big(fracdsqrt1-d^2Big)bigg)$$
$$f = frac c timessqrt1-d^2 sqrtc^2+e^2$$
$$h = frac etimes fc$$



This formula produces unexpected results when given a value close to $pmtilde j$, eg. $f(0.174126tilde i+0.984723tilde j, frac -pi16)$ returns $g = 0.996584$ which is higher the input value of $d=0.984723$. The expected value is aproximately $g=0.779171$ (I think). Continuing to tilt the resulting $tilde b$ by $frac-pi16$, cases $g$ to aproach $0.996$ ish.



Can anyone spot my error, or find a better way of doing this?










share|cite|improve this question









$endgroup$











  • $begingroup$
    Find the angle of rotation by dot multiplication. Why moving to 2D?
    $endgroup$
    – Moti
    Mar 24 at 6:32










  • $begingroup$
    @Moti because f and h can be found from a and g, and 2d allowed me to use radial coordinates which (I think) did exactly what I want, just in 2d
    $endgroup$
    – Alice Jacka
    Mar 24 at 7:05










  • $begingroup$
    @Moti could you elaborate on the dot multiplication?
    $endgroup$
    – Alice Jacka
    Mar 24 at 7:26













1












1








1





$begingroup$


vector $tilde a = ctilde i+dtilde j+etilde k$ is our input, and vector $tilde b = ftilde i+gtilde j+htilde k$ is our output, and $theta$ is the angle to rotate by. Essentially $tilde b = f(tilde a, theta)$



This is similar to Components of a 3d vector given specific angles, but I need to rotate the vector on the plane created by $tilde a$ and $tilde j$.



I attempted to create a function by converting $tilde a$ to a 2d point on this plane:
$(x, y) = (sqrt1-d^2, d)$

converting to a polar coordinate, adding alpha, converting back to cartesian, and taking the $y$ component as the value for $g$:



$$g = sqrtd^2+sqrt1-d^2 times sinbigg(tan^-1Big(fracdsqrt1-d^2Big)bigg)$$
$$f = frac c timessqrt1-d^2 sqrtc^2+e^2$$
$$h = frac etimes fc$$



This formula produces unexpected results when given a value close to $pmtilde j$, eg. $f(0.174126tilde i+0.984723tilde j, frac -pi16)$ returns $g = 0.996584$ which is higher the input value of $d=0.984723$. The expected value is aproximately $g=0.779171$ (I think). Continuing to tilt the resulting $tilde b$ by $frac-pi16$, cases $g$ to aproach $0.996$ ish.



Can anyone spot my error, or find a better way of doing this?










share|cite|improve this question









$endgroup$




vector $tilde a = ctilde i+dtilde j+etilde k$ is our input, and vector $tilde b = ftilde i+gtilde j+htilde k$ is our output, and $theta$ is the angle to rotate by. Essentially $tilde b = f(tilde a, theta)$



This is similar to Components of a 3d vector given specific angles, but I need to rotate the vector on the plane created by $tilde a$ and $tilde j$.



I attempted to create a function by converting $tilde a$ to a 2d point on this plane:
$(x, y) = (sqrt1-d^2, d)$

converting to a polar coordinate, adding alpha, converting back to cartesian, and taking the $y$ component as the value for $g$:



$$g = sqrtd^2+sqrt1-d^2 times sinbigg(tan^-1Big(fracdsqrt1-d^2Big)bigg)$$
$$f = frac c timessqrt1-d^2 sqrtc^2+e^2$$
$$h = frac etimes fc$$



This formula produces unexpected results when given a value close to $pmtilde j$, eg. $f(0.174126tilde i+0.984723tilde j, frac -pi16)$ returns $g = 0.996584$ which is higher the input value of $d=0.984723$. The expected value is aproximately $g=0.779171$ (I think). Continuing to tilt the resulting $tilde b$ by $frac-pi16$, cases $g$ to aproach $0.996$ ish.



Can anyone spot my error, or find a better way of doing this?







vectors






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Mar 24 at 4:59









Alice JackaAlice Jacka

61




61











  • $begingroup$
    Find the angle of rotation by dot multiplication. Why moving to 2D?
    $endgroup$
    – Moti
    Mar 24 at 6:32










  • $begingroup$
    @Moti because f and h can be found from a and g, and 2d allowed me to use radial coordinates which (I think) did exactly what I want, just in 2d
    $endgroup$
    – Alice Jacka
    Mar 24 at 7:05










  • $begingroup$
    @Moti could you elaborate on the dot multiplication?
    $endgroup$
    – Alice Jacka
    Mar 24 at 7:26
















  • $begingroup$
    Find the angle of rotation by dot multiplication. Why moving to 2D?
    $endgroup$
    – Moti
    Mar 24 at 6:32










  • $begingroup$
    @Moti because f and h can be found from a and g, and 2d allowed me to use radial coordinates which (I think) did exactly what I want, just in 2d
    $endgroup$
    – Alice Jacka
    Mar 24 at 7:05










  • $begingroup$
    @Moti could you elaborate on the dot multiplication?
    $endgroup$
    – Alice Jacka
    Mar 24 at 7:26















$begingroup$
Find the angle of rotation by dot multiplication. Why moving to 2D?
$endgroup$
– Moti
Mar 24 at 6:32




$begingroup$
Find the angle of rotation by dot multiplication. Why moving to 2D?
$endgroup$
– Moti
Mar 24 at 6:32












$begingroup$
@Moti because f and h can be found from a and g, and 2d allowed me to use radial coordinates which (I think) did exactly what I want, just in 2d
$endgroup$
– Alice Jacka
Mar 24 at 7:05




$begingroup$
@Moti because f and h can be found from a and g, and 2d allowed me to use radial coordinates which (I think) did exactly what I want, just in 2d
$endgroup$
– Alice Jacka
Mar 24 at 7:05












$begingroup$
@Moti could you elaborate on the dot multiplication?
$endgroup$
– Alice Jacka
Mar 24 at 7:26




$begingroup$
@Moti could you elaborate on the dot multiplication?
$endgroup$
– Alice Jacka
Mar 24 at 7:26










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