Line parallel to the plane
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We are given plane with equation: $4mx+(m−1)y−(m−4)z=−2m−2$. Find m so that line with vector $(4,4,1)$ will be parallel.
Now I tried $a*n=0$ and that gave me $-21m=-2$ so in that case solution would be $m=frac{2}{21}$ but right solution is $m=0$. So what is correct procedure to solve this?
linear-algebra
asked Jan 8 at 16:19
MikoMiko
243
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add a comment |
$begingroup$
We are given plane with equation: $4mx+(m−1)y−(m−4)z=−2m−2$. Find m so that line with vector $(4,4,1)$ will be parallel.
Now I tried $a*n=0$ and that gave me $-21m=-2$ so in that case solution would be $m=frac{2}{21}$ but right solution is $m=0$. So what is correct procedure to solve this?
linear-algebra
asked Jan 8 at 16:19
MikoMiko
243
$endgroup$
add a comment |
$begingroup$
We are given plane with equation: $4mx+(m−1)y−(m−4)z=−2m−2$. Find m so that line with vector $(4,4,1)$ will be parallel.
Now I tried $a*n=0$ and that gave me $-21m=-2$ so in that case solution would be $m=frac{2}{21}$ but right solution is $m=0$. So what is correct procedure to solve this?
linear-algebra
asked Jan 8 at 16:19
MikoMiko
243
$endgroup$
We are given plane with equation: $4mx+(m−1)y−(m−4)z=−2m−2$. Find m so that line with vector $(4,4,1)$ will be parallel.
Now I tried $a*n=0$ and that gave me $-21m=-2$ so in that case solution would be $m=frac{2}{21}$ but right solution is $m=0$. So what is correct procedure to solve this?
linear-algebra
linear-algebra
asked Jan 8 at 16:19
MikoMiko
243
asked Jan 8 at 16:19
MikoMiko
243
asked Jan 8 at 16:19
MikoMiko
243
asked Jan 8 at 16:19
MikoMiko
243
asked Jan 8 at 16:19
MikoMiko
243
243
add a comment |
add a comment |
1 Answer
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I can't reconstruct how you got $m=tfrac{2}{21}$, but did you take the (red) minus sign into account?
$4mx+(m−1)ycolor{red}{−}(m−4)z=−2m−2$
Because with $color{red}{−}(m−4) = color{blue}{4-m}$, that gives:
$$(4m,m-1,color{blue}{4-m})cdot(4,4,1) = 0 iff 19m = 0 iff m = 0$$
answered Jan 8 at 16:25
StackTDStackTD
22.9k2152
$endgroup$
$begingroup$
So where do I move $−2m−2$? I got 21 by moving -2m to other side and adding 19m so that resulted into 21. Or can I just ignore right side and turn it into 0 ?
$endgroup$
– Miko
Jan 8 at 16:44
$begingroup$
You don't need the right-hand side because it doesn't influence the orientation of the plane. You only need the plane's normal vector (and get that to be perpendicular to the line's direction vector).
$endgroup$
– StackTD
Jan 8 at 16:46
add a comment |
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1 Answer
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1 Answer
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$begingroup$
I can't reconstruct how you got $m=tfrac{2}{21}$, but did you take the (red) minus sign into account?
$4mx+(m−1)ycolor{red}{−}(m−4)z=−2m−2$
Because with $color{red}{−}(m−4) = color{blue}{4-m}$, that gives:
$$(4m,m-1,color{blue}{4-m})cdot(4,4,1) = 0 iff 19m = 0 iff m = 0$$
answered Jan 8 at 16:25
StackTDStackTD
22.9k2152
$endgroup$
$begingroup$
So where do I move $−2m−2$? I got 21 by moving -2m to other side and adding 19m so that resulted into 21. Or can I just ignore right side and turn it into 0 ?
$endgroup$
– Miko
Jan 8 at 16:44
$begingroup$
You don't need the right-hand side because it doesn't influence the orientation of the plane. You only need the plane's normal vector (and get that to be perpendicular to the line's direction vector).
$endgroup$
– StackTD
Jan 8 at 16:46
add a comment |
$begingroup$
I can't reconstruct how you got $m=tfrac{2}{21}$, but did you take the (red) minus sign into account?
$4mx+(m−1)ycolor{red}{−}(m−4)z=−2m−2$
Because with $color{red}{−}(m−4) = color{blue}{4-m}$, that gives:
$$(4m,m-1,color{blue}{4-m})cdot(4,4,1) = 0 iff 19m = 0 iff m = 0$$
answered Jan 8 at 16:25
StackTDStackTD
22.9k2152
$endgroup$
$begingroup$
So where do I move $−2m−2$? I got 21 by moving -2m to other side and adding 19m so that resulted into 21. Or can I just ignore right side and turn it into 0 ?
$endgroup$
– Miko
Jan 8 at 16:44
$begingroup$
You don't need the right-hand side because it doesn't influence the orientation of the plane. You only need the plane's normal vector (and get that to be perpendicular to the line's direction vector).
$endgroup$
– StackTD
Jan 8 at 16:46
add a comment |
$begingroup$
I can't reconstruct how you got $m=tfrac{2}{21}$, but did you take the (red) minus sign into account?
$4mx+(m−1)ycolor{red}{−}(m−4)z=−2m−2$
Because with $color{red}{−}(m−4) = color{blue}{4-m}$, that gives:
$$(4m,m-1,color{blue}{4-m})cdot(4,4,1) = 0 iff 19m = 0 iff m = 0$$
answered Jan 8 at 16:25
StackTDStackTD
22.9k2152
$endgroup$
I can't reconstruct how you got $m=tfrac{2}{21}$, but did you take the (red) minus sign into account?
$4mx+(m−1)ycolor{red}{−}(m−4)z=−2m−2$
Because with $color{red}{−}(m−4) = color{blue}{4-m}$, that gives:
$$(4m,m-1,color{blue}{4-m})cdot(4,4,1) = 0 iff 19m = 0 iff m = 0$$
answered Jan 8 at 16:25
StackTDStackTD
22.9k2152
answered Jan 8 at 16:25
StackTDStackTD
22.9k2152
answered Jan 8 at 16:25
StackTDStackTD
22.9k2152
answered Jan 8 at 16:25
StackTDStackTD
22.9k2152
22.9k2152
$begingroup$
So where do I move $−2m−2$? I got 21 by moving -2m to other side and adding 19m so that resulted into 21. Or can I just ignore right side and turn it into 0 ?
$endgroup$
– Miko
Jan 8 at 16:44
$begingroup$
You don't need the right-hand side because it doesn't influence the orientation of the plane. You only need the plane's normal vector (and get that to be perpendicular to the line's direction vector).
$endgroup$
– StackTD
Jan 8 at 16:46
add a comment |
$begingroup$
So where do I move $−2m−2$? I got 21 by moving -2m to other side and adding 19m so that resulted into 21. Or can I just ignore right side and turn it into 0 ?
$endgroup$
– Miko
Jan 8 at 16:44
$begingroup$
You don't need the right-hand side because it doesn't influence the orientation of the plane. You only need the plane's normal vector (and get that to be perpendicular to the line's direction vector).
$endgroup$
– StackTD
Jan 8 at 16:46
$begingroup$
So where do I move $−2m−2$? I got 21 by moving -2m to other side and adding 19m so that resulted into 21. Or can I just ignore right side and turn it into 0 ?
$endgroup$
– Miko
Jan 8 at 16:44
$begingroup$
So where do I move $−2m−2$? I got 21 by moving -2m to other side and adding 19m so that resulted into 21. Or can I just ignore right side and turn it into 0 ?
$endgroup$
– Miko
Jan 8 at 16:44
$begingroup$
You don't need the right-hand side because it doesn't influence the orientation of the plane. You only need the plane's normal vector (and get that to be perpendicular to the line's direction vector).
$endgroup$
– StackTD
Jan 8 at 16:46
$begingroup$
You don't need the right-hand side because it doesn't influence the orientation of the plane. You only need the plane's normal vector (and get that to be perpendicular to the line's direction vector).
$endgroup$
– StackTD
Jan 8 at 16:46
add a comment |
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