Mixing
Show that the line with parametric equations don't intersect
$begingroup$
Show that the line with parametric equations $x = 6 + 8t$, $y = −5 + t$, $z = 2 + 3t$ does not intersect the plane with equation $2x − y − 5z − 2 = 0$.
To answer this do i just plug in the $x$, $y$, and $z$ equation into $2x − y − 5z − 2 = 0$? So $2(6 + 8t) − y − 5(−5 + t) − 2(2 + 3t) = 0 $
calculus multivariable-calculus parametric
edited Mar 26 '16 at 23:38
Christopher Carl Heckman
3,9271020
asked Mar 26 '16 at 23:36
Kimmy.JKimmy.J
487
$endgroup$
add a comment |
$begingroup$
Show that the line with parametric equations $x = 6 + 8t$, $y = −5 + t$, $z = 2 + 3t$ does not intersect the plane with equation $2x − y − 5z − 2 = 0$.
To answer this do i just plug in the $x$, $y$, and $z$ equation into $2x − y − 5z − 2 = 0$? So $2(6 + 8t) − y − 5(−5 + t) − 2(2 + 3t) = 0 $
calculus multivariable-calculus parametric
edited Mar 26 '16 at 23:38
Christopher Carl Heckman
3,9271020
asked Mar 26 '16 at 23:36
Kimmy.JKimmy.J
487
$endgroup$
$begingroup$
Yes, except your substitution is incorrect; you should get $$2(6+8t)-(-5+t)-5(2+3t)-2=0.$$
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:38
$begingroup$
I get the solution 0t=5. Does that mean there is no intersection?
$endgroup$
– Kimmy.J
Mar 26 '16 at 23:42
$begingroup$
Yes; there's no value of $t$ that makes it true.
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:42
add a comment |
$begingroup$
Show that the line with parametric equations $x = 6 + 8t$, $y = −5 + t$, $z = 2 + 3t$ does not intersect the plane with equation $2x − y − 5z − 2 = 0$.
To answer this do i just plug in the $x$, $y$, and $z$ equation into $2x − y − 5z − 2 = 0$? So $2(6 + 8t) − y − 5(−5 + t) − 2(2 + 3t) = 0 $
calculus multivariable-calculus parametric
edited Mar 26 '16 at 23:38
Christopher Carl Heckman
3,9271020
asked Mar 26 '16 at 23:36
Kimmy.JKimmy.J
487
$endgroup$
Show that the line with parametric equations $x = 6 + 8t$, $y = −5 + t$, $z = 2 + 3t$ does not intersect the plane with equation $2x − y − 5z − 2 = 0$.
To answer this do i just plug in the $x$, $y$, and $z$ equation into $2x − y − 5z − 2 = 0$? So $2(6 + 8t) − y − 5(−5 + t) − 2(2 + 3t) = 0 $
calculus multivariable-calculus parametric
calculus multivariable-calculus parametric
edited Mar 26 '16 at 23:38
Christopher Carl Heckman
3,9271020
asked Mar 26 '16 at 23:36
Kimmy.JKimmy.J
487
edited Mar 26 '16 at 23:38
Christopher Carl Heckman
3,9271020
asked Mar 26 '16 at 23:36
Kimmy.JKimmy.J
487
edited Mar 26 '16 at 23:38
Christopher Carl Heckman
3,9271020
edited Mar 26 '16 at 23:38
Christopher Carl Heckman
3,9271020
edited Mar 26 '16 at 23:38
Christopher Carl Heckman
3,9271020
3,9271020
asked Mar 26 '16 at 23:36
Kimmy.JKimmy.J
487
asked Mar 26 '16 at 23:36
Kimmy.JKimmy.J
487
asked Mar 26 '16 at 23:36
Kimmy.JKimmy.J
487
487
$begingroup$
Yes, except your substitution is incorrect; you should get $$2(6+8t)-(-5+t)-5(2+3t)-2=0.$$
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:38
$begingroup$
I get the solution 0t=5. Does that mean there is no intersection?
$endgroup$
– Kimmy.J
Mar 26 '16 at 23:42
$begingroup$
Yes; there's no value of $t$ that makes it true.
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:42
add a comment |
$begingroup$
Yes, except your substitution is incorrect; you should get $$2(6+8t)-(-5+t)-5(2+3t)-2=0.$$
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:38
$begingroup$
I get the solution 0t=5. Does that mean there is no intersection?
$endgroup$
– Kimmy.J
Mar 26 '16 at 23:42
$begingroup$
Yes; there's no value of $t$ that makes it true.
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:42
$begingroup$
Yes, except your substitution is incorrect; you should get $$2(6+8t)-(-5+t)-5(2+3t)-2=0.$$
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:38
$begingroup$
Yes, except your substitution is incorrect; you should get $$2(6+8t)-(-5+t)-5(2+3t)-2=0.$$
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:38
$begingroup$
I get the solution 0t=5. Does that mean there is no intersection?
$endgroup$
– Kimmy.J
Mar 26 '16 at 23:42
$begingroup$
I get the solution 0t=5. Does that mean there is no intersection?
$endgroup$
– Kimmy.J
Mar 26 '16 at 23:42
$begingroup$
Yes; there's no value of $t$ that makes it true.
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:42
$begingroup$
Yes; there's no value of $t$ that makes it true.
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:42
add a comment |
1 Answer
1
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$begingroup$
$2(6+8t)−(−5+t)−5(2+3t)−2=0$
$0t=5$
Therefore, there is no intersection.
edited Apr 17 '16 at 22:40
Yeah..
262115
answered Mar 26 '16 at 23:45
Kimmy.JKimmy.J
487
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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active
oldest
votes
$begingroup$
$2(6+8t)−(−5+t)−5(2+3t)−2=0$
$0t=5$
Therefore, there is no intersection.
edited Apr 17 '16 at 22:40
Yeah..
262115
answered Mar 26 '16 at 23:45
Kimmy.JKimmy.J
487
$endgroup$
add a comment |
$begingroup$
$2(6+8t)−(−5+t)−5(2+3t)−2=0$
$0t=5$
Therefore, there is no intersection.
edited Apr 17 '16 at 22:40
Yeah..
262115
answered Mar 26 '16 at 23:45
Kimmy.JKimmy.J
487
$endgroup$
add a comment |
$begingroup$
$2(6+8t)−(−5+t)−5(2+3t)−2=0$
$0t=5$
Therefore, there is no intersection.
edited Apr 17 '16 at 22:40
Yeah..
262115
answered Mar 26 '16 at 23:45
Kimmy.JKimmy.J
487
$endgroup$
$2(6+8t)−(−5+t)−5(2+3t)−2=0$
$0t=5$
Therefore, there is no intersection.
edited Apr 17 '16 at 22:40
Yeah..
262115
answered Mar 26 '16 at 23:45
Kimmy.JKimmy.J
487
edited Apr 17 '16 at 22:40
Yeah..
262115
edited Apr 17 '16 at 22:40
Yeah..
262115
edited Apr 17 '16 at 22:40
Yeah..
262115
262115
answered Mar 26 '16 at 23:45
Kimmy.JKimmy.J
487
answered Mar 26 '16 at 23:45
Kimmy.JKimmy.J
487
answered Mar 26 '16 at 23:45
Kimmy.JKimmy.J
487
487
add a comment |
add a comment |
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$begingroup$
Yes, except your substitution is incorrect; you should get $$2(6+8t)-(-5+t)-5(2+3t)-2=0.$$
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:38
$begingroup$
I get the solution 0t=5. Does that mean there is no intersection?
$endgroup$
– Kimmy.J
Mar 26 '16 at 23:42
$begingroup$
Yes; there's no value of $t$ that makes it true.
$endgroup$
– Christopher Carl Heckman
Mar 26 '16 at 23:42