Find 3D line equation in simplest form.
$begingroup$
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
3,3382825
asked Nov 29 '18 at 9:34
WillyWilly
414
$endgroup$
add a comment |
$begingroup$
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
3,3382825
asked Nov 29 '18 at 9:34
WillyWilly
414
$endgroup$
2
$begingroup$
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 $z$.
$endgroup$
– Michael Burr
Nov 29 '18 at 9:39
$begingroup$
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
$endgroup$
– Willy
Dec 7 '18 at 9:06
add a comment |
$begingroup$
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
3,3382825
asked Nov 29 '18 at 9:34
WillyWilly
414
$endgroup$
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
3,3382825
asked Nov 29 '18 at 9:34
WillyWilly
414
edited Nov 29 '18 at 10:08
user376343
3,3382825
asked Nov 29 '18 at 9:34
WillyWilly
414
edited Nov 29 '18 at 10:08
user376343
3,3382825
edited Nov 29 '18 at 10:08
user376343
3,3382825
edited Nov 29 '18 at 10:08
user376343
3,3382825
3,3382825
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
$begingroup$
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 $z$.
$endgroup$
– Michael Burr
Nov 29 '18 at 9:39
$begingroup$
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
$endgroup$
– Willy
Dec 7 '18 at 9:06
add a comment |
2
$begingroup$
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 $z$.
$endgroup$
– Michael Burr
Nov 29 '18 at 9:39
$begingroup$
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
$endgroup$
– Willy
Dec 7 '18 at 9:06
2
2
$begingroup$
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 $z$.
$endgroup$
– Michael Burr
Nov 29 '18 at 9:39
$begingroup$
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 $z$.
$endgroup$
– Michael Burr
Nov 29 '18 at 9:39
$begingroup$
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
$endgroup$
– Willy
Dec 7 '18 at 9:06
$begingroup$
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
$endgroup$
– Willy
Dec 7 '18 at 9:06
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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
15.3k3939
$endgroup$
$begingroup$
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.
$endgroup$
– Willy
Dec 4 '18 at 11:11
$begingroup$
Nice! Good luck!!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
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$begingroup$
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
15.3k3939
$endgroup$
$begingroup$
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.
$endgroup$
– Willy
Dec 4 '18 at 11:11
$begingroup$
Nice! Good luck!!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
$begingroup$
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
15.3k3939
$endgroup$
$begingroup$
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.
$endgroup$
– Willy
Dec 4 '18 at 11:11
$begingroup$
Nice! Good luck!!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
$begingroup$
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
15.3k3939
$endgroup$
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
15.3k3939
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
15.3k3939
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
15.3k3939
answered Nov 29 '18 at 9:56
Mostafa AyazMostafa Ayaz
15.3k3939
15.3k3939
$begingroup$
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.
$endgroup$
– Willy
Dec 4 '18 at 11:11
$begingroup$
Nice! Good luck!!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
$begingroup$
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.
$endgroup$
– Willy
Dec 4 '18 at 11:11
$begingroup$
Nice! Good luck!!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 11:15
$begingroup$
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.
$endgroup$
– Willy
Dec 4 '18 at 11:11
$begingroup$
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.
$endgroup$
– Willy
Dec 4 '18 at 11:11
$begingroup$
Nice! Good luck!!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 11:15
$begingroup$
Nice! Good luck!!
$endgroup$
– Mostafa Ayaz
Dec 4 '18 at 11:15
add a comment |
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2
$begingroup$
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 $z$.
$endgroup$
– Michael Burr
Nov 29 '18 at 9:39
$begingroup$
Your right. My eyes were fooled by my mind. It is caused by an interesting neurological phenomenon: BRAIN FILTERING.
$endgroup$
– Willy
Dec 7 '18 at 9:06