How to solve an equation with vector coefficients?

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1
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I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$ where

$$ d(t) = frac12at^2 + vt + r$$

Where a, v, and r are all 3-dimensional vectors in cartesian coordinates. How can I do this?

edited 12 mins ago

Carl Woll

59.9k279154

asked 41 mins ago

Sean McAllister

1084

New contributor

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
37 mins ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
36 mins ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
26 mins ago

add a commentÂ |Â

up vote
1
down vote

favorite

I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$ where

$$ d(t) = frac12at^2 + vt + r$$

Where a, v, and r are all 3-dimensional vectors in cartesian coordinates. How can I do this?

edited 12 mins ago

Carl Woll

59.9k279154

asked 41 mins ago

Sean McAllister

1084

New contributor

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
37 mins ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
36 mins ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
26 mins ago

add a commentÂ |Â

up vote
1
down vote

favorite

I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$ where

$$ d(t) = frac12at^2 + vt + r$$

Where a, v, and r are all 3-dimensional vectors in cartesian coordinates. How can I do this?

edited 12 mins ago

Carl Woll

59.9k279154

asked 41 mins ago

Sean McAllister

1084

New contributor

I want to solve $(ct)^2 = d(t)cdot d(t)$ for $t$ where

$$ d(t) = frac12at^2 + vt + r$$

Where a, v, and r are all 3-dimensional vectors in cartesian coordinates. How can I do this?

linear-algebra algebraic-manipulation tensors algebra

edited 12 mins ago

Carl Woll

59.9k279154

asked 41 mins ago

Sean McAllister

1084

New contributor

edited 12 mins ago

Carl Woll

59.9k279154

asked 41 mins ago

Sean McAllister

1084

New contributor

edited 12 mins ago

Carl Woll

59.9k279154

edited 12 mins ago

Carl Woll

59.9k279154

edited 12 mins ago

Carl Woll

59.9k279154

asked 41 mins ago

Sean McAllister

1084

New contributor

asked 41 mins ago

Sean McAllister

1084

asked 41 mins ago

Sean McAllister

1084

New contributor

Sean McAllister is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
37 mins ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
36 mins ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
26 mins ago

add a commentÂ |Â

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
37 mins ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
36 mins ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
26 mins ago

What is the variable that you want to solve for?
â€“Â Henrik Schumacher
37 mins ago

(ct)^2 is a constant?
â€“Â Ulrich Neumann
36 mins ago

t is the variable to solve for, c is the speed of light so a constant yes.
â€“Â Sean McAllister
26 mins ago

add a commentÂ |Â

2 Answers
2

active

oldest

votes

up vote
3
down vote

accepted

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

edited 4 mins ago

answered 17 mins ago

Ulrich Neumann

5,070413

add a commentÂ |Â

up vote
2
down vote

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 13 mins ago

Carl Woll

59.9k279154

add a commentÂ |Â

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

up vote
3
down vote

accepted

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

edited 4 mins ago

answered 17 mins ago

Ulrich Neumann

5,070413

add a commentÂ |Â

up vote
3
down vote

accepted

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

edited 4 mins ago

answered 17 mins ago

Ulrich Neumann

5,070413

add a commentÂ |Â

up vote
3
down vote

accepted

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

edited 4 mins ago

answered 17 mins ago

Ulrich Neumann

5,070413

If you define the three vectors explicitly as

A = a1, a2, a3;
V = v1, v2, v3;
R = r1, r2, r3;

your equation evaluates to a polynom in t of order 4

eq = (c t)^2 == #.# &[1/2 A t^2 + V t + R] // Collect[#, t] &

which you might solve using MMA Solve[eq, t]

edited 4 mins ago

answered 17 mins ago

Ulrich Neumann

5,070413

edited 4 mins ago

answered 17 mins ago

Ulrich Neumann

5,070413

answered 17 mins ago

Ulrich Neumann

5,070413

answered 17 mins ago

Ulrich Neumann

5,070413

add a commentÂ |Â

up vote
2
down vote

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 13 mins ago

Carl Woll

59.9k279154

add a commentÂ |Â

up vote
2
down vote

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 13 mins ago

Carl Woll

59.9k279154

add a commentÂ |Â

up vote
2
down vote

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 13 mins ago

Carl Woll

59.9k279154

You can use TensorExpand. First, some assumptions, and your distance function:

$Assumptions = (a|v|r) Ã¢ÂˆÂˆ Vectors[3];

d[t_] := 1/2 a t^2 + v t + r

Then, use Solve on the tensor expanded equation:

Solve[
 TensorExpand[d[t] . d[t] == (c t)^2],
 t,
 Quartics->False
]

t -> Root[
4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 1], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 2], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 3], t ->
Root[4 r.r + 8 r.v #1 + (-4 c^2 + 4 a.r + 4 v.v) #1^2 + 4 a.v #1^3 +
a.a #1^4 &, 4]

Without the Quartics option, you get a mess of hard to understand radicals.

answered 13 mins ago

Carl Woll

59.9k279154

answered 13 mins ago

Carl Woll

59.9k279154

answered 13 mins ago

Carl Woll

59.9k279154

answered 13 mins ago

Carl Woll

59.9k279154

add a commentÂ |Â

Sean McAllister is a new contributor. Be nice, and check out our Code of Conduct.

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