Find 3D line equation in simplest form.
The line parallel to planes $-5x + y = 0$ and $x + 6y = 0,$ passes through the point $(3,0,2).$
My solution:.
Direction vector of the line: $v (x,y,z).$
Orthogonal vectors of planes: $;p_1(-5,1,0),; p_2(1,6,0).$
$v cdot p_1 = |v| cdot |p_1| cdot cos 90 = -5x + y = 0. tag 1$
$v cdot p_2 = |v| cdot |p_2| cdot cos 90 = x + 6y = 0. tag 2$
Solve $(1)(2) => x = 0, ;y = 0.$
But answer in my book is $v(6, 30, -1).$
What have I done wrong? Thanks!
geometry
edited Nov 29 '18 at 10:08
user376343
2,9132823
asked Nov 29 '18 at 9:34
WillyWilly
414
add a comment |
The line parallel to planes $-5x + y = 0$ and $x + 6y = 0,$ passes through the point $(3,0,2).$
My solution:.
Direction vector of the line: $v (x,y,z).$
Orthogonal vectors of planes: $;p_1(-5,1,0),; p_2(1,6,0).$
$v cdot p_1 = |v| cdot |p_1| cdot cos 90 = -5x + y = 0. tag 1$
$v cdot p_2 = |v| cdot |p_2| cdot cos 90 = x + 6y = 0. tag 2$
Solve $(1)(2) => x = 0, ;y = 0.$
But answer in my book is $v(6, 30, -1).$
What have I done wrong? Thanks!
geometry
edited Nov 29 '18 at 10:08
user376343
2,9132823
asked Nov 29 '18 at 9:34
WillyWilly
414
2
It looks like your book is using the planes $-5x+y=0$ and $x+6z=0$. Notice that the second equation has replaced $y$ by $z$.
– Michael Burr
Nov 29 '18 at 9:39
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
– Willy
Dec 7 '18 at 9:06
add a comment |
The line parallel to planes $-5x + y = 0$ and $x + 6y = 0,$ passes through the point $(3,0,2).$
My solution:.
Direction vector of the line: $v (x,y,z).$
Orthogonal vectors of planes: $;p_1(-5,1,0),; p_2(1,6,0).$
$v cdot p_1 = |v| cdot |p_1| cdot cos 90 = -5x + y = 0. tag 1$
$v cdot p_2 = |v| cdot |p_2| cdot cos 90 = x + 6y = 0. tag 2$
Solve $(1)(2) => x = 0, ;y = 0.$
But answer in my book is $v(6, 30, -1).$
What have I done wrong? Thanks!
geometry
edited Nov 29 '18 at 10:08
user376343
2,9132823
asked Nov 29 '18 at 9:34
WillyWilly
414
The line parallel to planes $-5x + y = 0$ and $x + 6y = 0,$ passes through the point $(3,0,2).$
My solution:.
Direction vector of the line: $v (x,y,z).$
Orthogonal vectors of planes: $;p_1(-5,1,0),; p_2(1,6,0).$
$v cdot p_1 = |v| cdot |p_1| cdot cos 90 = -5x + y = 0. tag 1$
$v cdot p_2 = |v| cdot |p_2| cdot cos 90 = x + 6y = 0. tag 2$
Solve $(1)(2) => x = 0, ;y = 0.$
But answer in my book is $v(6, 30, -1).$
What have I done wrong? Thanks!
geometry
geometry
edited Nov 29 '18 at 10:08
user376343
2,9132823
asked Nov 29 '18 at 9:34
WillyWilly
414
edited Nov 29 '18 at 10:08
user376343
2,9132823
asked Nov 29 '18 at 9:34
WillyWilly
414
edited Nov 29 '18 at 10:08
user376343
2,9132823
edited Nov 29 '18 at 10:08
user376343
2,9132823
edited Nov 29 '18 at 10:08
user376343
2,9132823
2,9132823
asked Nov 29 '18 at 9:34
WillyWilly
414
asked Nov 29 '18 at 9:34
WillyWilly
414
asked Nov 29 '18 at 9:34
WillyWilly
414
414
2
It looks like your book is using the planes $-5x+y=0$ and $x+6z=0$. Notice that the second equation has replaced $y$ by $z$.
– Michael Burr
Nov 29 '18 at 9:39
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
– Willy
Dec 7 '18 at 9:06
add a comment |
2
It looks like your book is using the planes $-5x+y=0$ and $x+6z=0$. Notice that the second equation has replaced $y$ by $z$.
– Michael Burr
Nov 29 '18 at 9:39
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
– Willy
Dec 7 '18 at 9:06
2
2
It looks like your book is using the planes $-5x+y=0$ and $x+6z=0$. Notice that the second equation has replaced $y$ by $z$.
– Michael Burr
Nov 29 '18 at 9:39
It looks like your book is using the planes $-5x+y=0$ and $x+6z=0$. Notice that the second equation has replaced $y$ by $z$.
– Michael Burr
Nov 29 '18 at 9:39
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
– Willy
Dec 7 '18 at 9:06
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
– Willy
Dec 7 '18 at 9:06
add a comment |
1 Answer
1
active
oldest
votes
Your solution is wrong. In fact$$v=p_1times p_2$$which shows the orthogonality therefore$$v=(-5hat i+hat j)times (hat i+6hat j)=-31 hat k$$hence the line is parallel to z-axis (which is expected) and the equation of the line would be$$begin{cases}x=3\y=0\z=tend{cases}$$
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
14.1k3937
Thanks! Your answer & mine are both right. Alter solve(1), (2), find out v = (0,0,t), I can conclude the line equation like yours.
– Willy
Dec 4 '18 at 11:11
Nice! Good luck!!
– Mostafa Ayaz
Dec 4 '18 at 11:15
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
Your solution is wrong. In fact$$v=p_1times p_2$$which shows the orthogonality therefore$$v=(-5hat i+hat j)times (hat i+6hat j)=-31 hat k$$hence the line is parallel to z-axis (which is expected) and the equation of the line would be$$begin{cases}x=3\y=0\z=tend{cases}$$
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
14.1k3937
Thanks! Your answer & mine are both right. Alter solve(1), (2), find out v = (0,0,t), I can conclude the line equation like yours.
– Willy
Dec 4 '18 at 11:11
Nice! Good luck!!
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
Your solution is wrong. In fact$$v=p_1times p_2$$which shows the orthogonality therefore$$v=(-5hat i+hat j)times (hat i+6hat j)=-31 hat k$$hence the line is parallel to z-axis (which is expected) and the equation of the line would be$$begin{cases}x=3\y=0\z=tend{cases}$$
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
14.1k3937
Thanks! Your answer & mine are both right. Alter solve(1), (2), find out v = (0,0,t), I can conclude the line equation like yours.
– Willy
Dec 4 '18 at 11:11
Nice! Good luck!!
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
Your solution is wrong. In fact$$v=p_1times p_2$$which shows the orthogonality therefore$$v=(-5hat i+hat j)times (hat i+6hat j)=-31 hat k$$hence the line is parallel to z-axis (which is expected) and the equation of the line would be$$begin{cases}x=3\y=0\z=tend{cases}$$
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
14.1k3937
Your solution is wrong. In fact$$v=p_1times p_2$$which shows the orthogonality therefore$$v=(-5hat i+hat j)times (hat i+6hat j)=-31 hat k$$hence the line is parallel to z-axis (which is expected) and the equation of the line would be$$begin{cases}x=3\y=0\z=tend{cases}$$
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
14.1k3937
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
14.1k3937
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
14.1k3937
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
14.1k3937
14.1k3937
Thanks! Your answer & mine are both right. Alter solve(1), (2), find out v = (0,0,t), I can conclude the line equation like yours.
– Willy
Dec 4 '18 at 11:11
Nice! Good luck!!
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
Thanks! Your answer & mine are both right. Alter solve(1), (2), find out v = (0,0,t), I can conclude the line equation like yours.
– Willy
Dec 4 '18 at 11:11
Nice! Good luck!!
– Mostafa Ayaz
Dec 4 '18 at 11:15
Thanks! Your answer & mine are both right. Alter solve(1), (2), find out v = (0,0,t), I can conclude the line equation like yours.
– Willy
Dec 4 '18 at 11:11
Thanks! Your answer & mine are both right. Alter solve(1), (2), find out v = (0,0,t), I can conclude the line equation like yours.
– Willy
Dec 4 '18 at 11:11
Nice! Good luck!!
– Mostafa Ayaz
Dec 4 '18 at 11:15
Nice! Good luck!!
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
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2
It looks like your book is using the planes $-5x+y=0$ and $x+6z=0$. Notice that the second equation has replaced $y$ by $z$.
– Michael Burr
Nov 29 '18 at 9:39
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
– Willy
Dec 7 '18 at 9:06