Mixing
Calculate angle
$begingroup$
What angle is between vector a and vector b if the angle between vector p and vector q is 90°, where vector p=5a-2b and vector q=-3a-6b?
(Yes this is homework)
vectors angle
asked Jan 26 at 18:37
InviteMeInviteMe
1
$endgroup$
add a comment |
$begingroup$
What angle is between vector a and vector b if the angle between vector p and vector q is 90°, where vector p=5a-2b and vector q=-3a-6b?
(Yes this is homework)
vectors angle
asked Jan 26 at 18:37
InviteMeInviteMe
1
$endgroup$
$begingroup$
I think you need more information.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:41
$begingroup$
Can you add an image? Are $vec a$ and $vec b$ at $90^circ$ to one another?
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 18:41
$begingroup$
Sadly i don't have an image. The correct answer is supposed to be 97.18°. This is all the information that is given
$endgroup$
– InviteMe
Jan 26 at 18:48
$begingroup$
The inner product of $p$ and $q$ is 0. Did you try using it?
$endgroup$
– Damien
Jan 26 at 19:03
add a comment |
$begingroup$
What angle is between vector a and vector b if the angle between vector p and vector q is 90°, where vector p=5a-2b and vector q=-3a-6b?
(Yes this is homework)
vectors angle
asked Jan 26 at 18:37
InviteMeInviteMe
1
$endgroup$
What angle is between vector a and vector b if the angle between vector p and vector q is 90°, where vector p=5a-2b and vector q=-3a-6b?
(Yes this is homework)
vectors angle
vectors angle
asked Jan 26 at 18:37
InviteMeInviteMe
1
asked Jan 26 at 18:37
InviteMeInviteMe
1
asked Jan 26 at 18:37
InviteMeInviteMe
1
asked Jan 26 at 18:37
InviteMeInviteMe
1
asked Jan 26 at 18:37
InviteMeInviteMe
1
1
$begingroup$
I think you need more information.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:41
$begingroup$
Can you add an image? Are $vec a$ and $vec b$ at $90^circ$ to one another?
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 18:41
$begingroup$
Sadly i don't have an image. The correct answer is supposed to be 97.18°. This is all the information that is given
$endgroup$
– InviteMe
Jan 26 at 18:48
$begingroup$
The inner product of $p$ and $q$ is 0. Did you try using it?
$endgroup$
– Damien
Jan 26 at 19:03
add a comment |
$begingroup$
I think you need more information.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:41
$begingroup$
Can you add an image? Are $vec a$ and $vec b$ at $90^circ$ to one another?
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 18:41
$begingroup$
Sadly i don't have an image. The correct answer is supposed to be 97.18°. This is all the information that is given
$endgroup$
– InviteMe
Jan 26 at 18:48
$begingroup$
The inner product of $p$ and $q$ is 0. Did you try using it?
$endgroup$
– Damien
Jan 26 at 19:03
$begingroup$
I think you need more information.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:41
$begingroup$
I think you need more information.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:41
$begingroup$
Can you add an image? Are $vec a$ and $vec b$ at $90^circ$ to one another?
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 18:41
$begingroup$
Can you add an image? Are $vec a$ and $vec b$ at $90^circ$ to one another?
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 18:41
$begingroup$
Sadly i don't have an image. The correct answer is supposed to be 97.18°. This is all the information that is given
$endgroup$
– InviteMe
Jan 26 at 18:48
$begingroup$
Sadly i don't have an image. The correct answer is supposed to be 97.18°. This is all the information that is given
$endgroup$
– InviteMe
Jan 26 at 18:48
$begingroup$
The inner product of $p$ and $q$ is 0. Did you try using it?
$endgroup$
– Damien
Jan 26 at 19:03
$begingroup$
The inner product of $p$ and $q$ is 0. Did you try using it?
$endgroup$
– Damien
Jan 26 at 19:03
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Remember the scalar product of two vectors is
a•b == |a| |b| cos(theta) so
cos(theta) = (a•b)/(|a| |b|)
Solve your two equations for a and b in terms of
p and q. Make your life simpler by assuming p
And q are normalized, e.g. p•p = 1. Noting that
since the angle between p and q is 90,
p•q =0, you can calculate (a•b), |a|, and |b|
easily, if laboriously. Substitute those into the
scalar product equation above and you have
the cosine of the angle you’re looking for.
answered Jan 26 at 19:39
Nick BoshaftNick Boshaft
64
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Remember the scalar product of two vectors is
a•b == |a| |b| cos(theta) so
cos(theta) = (a•b)/(|a| |b|)
Solve your two equations for a and b in terms of
p and q. Make your life simpler by assuming p
And q are normalized, e.g. p•p = 1. Noting that
since the angle between p and q is 90,
p•q =0, you can calculate (a•b), |a|, and |b|
easily, if laboriously. Substitute those into the
scalar product equation above and you have
the cosine of the angle you’re looking for.
answered Jan 26 at 19:39
Nick BoshaftNick Boshaft
64
$endgroup$
add a comment |
$begingroup$
Remember the scalar product of two vectors is
a•b == |a| |b| cos(theta) so
cos(theta) = (a•b)/(|a| |b|)
Solve your two equations for a and b in terms of
p and q. Make your life simpler by assuming p
And q are normalized, e.g. p•p = 1. Noting that
since the angle between p and q is 90,
p•q =0, you can calculate (a•b), |a|, and |b|
easily, if laboriously. Substitute those into the
scalar product equation above and you have
the cosine of the angle you’re looking for.
answered Jan 26 at 19:39
Nick BoshaftNick Boshaft
64
$endgroup$
add a comment |
$begingroup$
Remember the scalar product of two vectors is
a•b == |a| |b| cos(theta) so
cos(theta) = (a•b)/(|a| |b|)
Solve your two equations for a and b in terms of
p and q. Make your life simpler by assuming p
And q are normalized, e.g. p•p = 1. Noting that
since the angle between p and q is 90,
p•q =0, you can calculate (a•b), |a|, and |b|
easily, if laboriously. Substitute those into the
scalar product equation above and you have
the cosine of the angle you’re looking for.
answered Jan 26 at 19:39
Nick BoshaftNick Boshaft
64
$endgroup$
Remember the scalar product of two vectors is
a•b == |a| |b| cos(theta) so
cos(theta) = (a•b)/(|a| |b|)
Solve your two equations for a and b in terms of
p and q. Make your life simpler by assuming p
And q are normalized, e.g. p•p = 1. Noting that
since the angle between p and q is 90,
p•q =0, you can calculate (a•b), |a|, and |b|
easily, if laboriously. Substitute those into the
scalar product equation above and you have
the cosine of the angle you’re looking for.
answered Jan 26 at 19:39
Nick BoshaftNick Boshaft
64
answered Jan 26 at 19:39
Nick BoshaftNick Boshaft
64
answered Jan 26 at 19:39
Nick BoshaftNick Boshaft
64
answered Jan 26 at 19:39
Nick BoshaftNick Boshaft
64
64
add a comment |
add a comment |
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$begingroup$
I think you need more information.
$endgroup$
– Lord Shark the Unknown
Jan 26 at 18:41
$begingroup$
Can you add an image? Are $vec a$ and $vec b$ at $90^circ$ to one another?
$endgroup$
– Mohammad Zuhair Khan
Jan 26 at 18:41
$begingroup$
Sadly i don't have an image. The correct answer is supposed to be 97.18°. This is all the information that is given
$endgroup$
– InviteMe
Jan 26 at 18:48
$begingroup$
The inner product of $p$ and $q$ is 0. Did you try using it?
$endgroup$
– Damien
Jan 26 at 19:03