Mixing
Sum of parallel and perpendicular vectors
$begingroup$
If $v_1 + v_2 = langle-5,5rangle$ where
$v_1$ is parallel to $langle-3,5rangle$ and
$v_2$ is perpendicular to $langle-3,5rangle$.
Then what are the two vectors. I’m not quite sure how I should approach this question
vectors
edited Jan 27 at 6:40
Martin Sleziak
44.9k10122275
asked Jan 27 at 4:21
Alex MaslachAlex Maslach
275
$endgroup$
add a comment |
$begingroup$
If $v_1 + v_2 = langle-5,5rangle$ where
$v_1$ is parallel to $langle-3,5rangle$ and
$v_2$ is perpendicular to $langle-3,5rangle$.
Then what are the two vectors. I’m not quite sure how I should approach this question
vectors
edited Jan 27 at 6:40
Martin Sleziak
44.9k10122275
asked Jan 27 at 4:21
Alex MaslachAlex Maslach
275
$endgroup$
$begingroup$
What do we know about two vectors if they are perpendicular?
$endgroup$
– Aniruddh Venkatesan
Jan 27 at 4:27
$begingroup$
I understand that their dot product is zero but I’m not quite sure what I should do with that fact
$endgroup$
– Alex Maslach
Jan 27 at 4:31
add a comment |
$begingroup$
If $v_1 + v_2 = langle-5,5rangle$ where
$v_1$ is parallel to $langle-3,5rangle$ and
$v_2$ is perpendicular to $langle-3,5rangle$.
Then what are the two vectors. I’m not quite sure how I should approach this question
vectors
edited Jan 27 at 6:40
Martin Sleziak
44.9k10122275
asked Jan 27 at 4:21
Alex MaslachAlex Maslach
275
$endgroup$
If $v_1 + v_2 = langle-5,5rangle$ where
$v_1$ is parallel to $langle-3,5rangle$ and
$v_2$ is perpendicular to $langle-3,5rangle$.
Then what are the two vectors. I’m not quite sure how I should approach this question
vectors
vectors
edited Jan 27 at 6:40
Martin Sleziak
44.9k10122275
asked Jan 27 at 4:21
Alex MaslachAlex Maslach
275
edited Jan 27 at 6:40
Martin Sleziak
44.9k10122275
asked Jan 27 at 4:21
Alex MaslachAlex Maslach
275
edited Jan 27 at 6:40
Martin Sleziak
44.9k10122275
edited Jan 27 at 6:40
Martin Sleziak
44.9k10122275
edited Jan 27 at 6:40
Martin Sleziak
44.9k10122275
44.9k10122275
asked Jan 27 at 4:21
Alex MaslachAlex Maslach
275
asked Jan 27 at 4:21
Alex MaslachAlex Maslach
275
asked Jan 27 at 4:21
Alex MaslachAlex Maslach
275
275
$begingroup$
What do we know about two vectors if they are perpendicular?
$endgroup$
– Aniruddh Venkatesan
Jan 27 at 4:27
$begingroup$
I understand that their dot product is zero but I’m not quite sure what I should do with that fact
$endgroup$
– Alex Maslach
Jan 27 at 4:31
add a comment |
$begingroup$
What do we know about two vectors if they are perpendicular?
$endgroup$
– Aniruddh Venkatesan
Jan 27 at 4:27
$begingroup$
I understand that their dot product is zero but I’m not quite sure what I should do with that fact
$endgroup$
– Alex Maslach
Jan 27 at 4:31
$begingroup$
What do we know about two vectors if they are perpendicular?
$endgroup$
– Aniruddh Venkatesan
Jan 27 at 4:27
$begingroup$
What do we know about two vectors if they are perpendicular?
$endgroup$
– Aniruddh Venkatesan
Jan 27 at 4:27
$begingroup$
I understand that their dot product is zero but I’m not quite sure what I should do with that fact
$endgroup$
– Alex Maslach
Jan 27 at 4:31
$begingroup$
I understand that their dot product is zero but I’m not quite sure what I should do with that fact
$endgroup$
– Alex Maslach
Jan 27 at 4:31
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First find a vector perpendicular to $(-3,5)$. You want a vector that has a dot product of zero with this vector. Inspection should quickly show that $(5,3)$ works fine.
Now set $lambda(-3,5) + mu(5,3)= (-5,5)$, where $lambda$ and $mu$ are scalars.
By splitting the components, you can now get two simple linear simultaneous equations ($-3lambda + 5mu = -5$ and $5lambda + 3mu = 5$) to solve for the scalars.
answered Jan 27 at 4:35
DeepakDeepak
17.5k11539
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First find a vector perpendicular to $(-3,5)$. You want a vector that has a dot product of zero with this vector. Inspection should quickly show that $(5,3)$ works fine.
Now set $lambda(-3,5) + mu(5,3)= (-5,5)$, where $lambda$ and $mu$ are scalars.
By splitting the components, you can now get two simple linear simultaneous equations ($-3lambda + 5mu = -5$ and $5lambda + 3mu = 5$) to solve for the scalars.
answered Jan 27 at 4:35
DeepakDeepak
17.5k11539
$endgroup$
add a comment |
$begingroup$
First find a vector perpendicular to $(-3,5)$. You want a vector that has a dot product of zero with this vector. Inspection should quickly show that $(5,3)$ works fine.
Now set $lambda(-3,5) + mu(5,3)= (-5,5)$, where $lambda$ and $mu$ are scalars.
By splitting the components, you can now get two simple linear simultaneous equations ($-3lambda + 5mu = -5$ and $5lambda + 3mu = 5$) to solve for the scalars.
answered Jan 27 at 4:35
DeepakDeepak
17.5k11539
$endgroup$
add a comment |
$begingroup$
First find a vector perpendicular to $(-3,5)$. You want a vector that has a dot product of zero with this vector. Inspection should quickly show that $(5,3)$ works fine.
Now set $lambda(-3,5) + mu(5,3)= (-5,5)$, where $lambda$ and $mu$ are scalars.
By splitting the components, you can now get two simple linear simultaneous equations ($-3lambda + 5mu = -5$ and $5lambda + 3mu = 5$) to solve for the scalars.
answered Jan 27 at 4:35
DeepakDeepak
17.5k11539
$endgroup$
First find a vector perpendicular to $(-3,5)$. You want a vector that has a dot product of zero with this vector. Inspection should quickly show that $(5,3)$ works fine.
Now set $lambda(-3,5) + mu(5,3)= (-5,5)$, where $lambda$ and $mu$ are scalars.
By splitting the components, you can now get two simple linear simultaneous equations ($-3lambda + 5mu = -5$ and $5lambda + 3mu = 5$) to solve for the scalars.
answered Jan 27 at 4:35
DeepakDeepak
17.5k11539
answered Jan 27 at 4:35
DeepakDeepak
17.5k11539
answered Jan 27 at 4:35
DeepakDeepak
17.5k11539
answered Jan 27 at 4:35
DeepakDeepak
17.5k11539
17.5k11539
add a comment |
add a comment |
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$begingroup$
What do we know about two vectors if they are perpendicular?
$endgroup$
– Aniruddh Venkatesan
Jan 27 at 4:27
$begingroup$
I understand that their dot product is zero but I’m not quite sure what I should do with that fact
$endgroup$
– Alex Maslach
Jan 27 at 4:31