Finding parameters for which the line lies in the plane
I tried to solve the following task:
A line L has equation: $frac{x-2}{p} = frac{y-q}{2}= z-1$, where $p,q in mathbb{R} $. A plane P has equation: $ x +y +3z = 9$. Given that line L lies in the plane P, find the value of $p$ and the value of $q$.
In order to find $p$ and $q$, I did following: I found the parametric form of the line: $x = t p +2$, $ y = 2t +q$, $ z= t+1$, where $t in mathbb{R}$ and then, I inserted these values in the equation of P. I got this:
$(tp +2) + (2t +q) + 3(t+1) = 9$ or
$q +tp + 5t = 4$
Can I just insert now 2 arbitrary real values of $t$ and then solve the system of two linear equations to find $p$ and $q$? Is that correct approach? Thanks in advance.
analytic-geometry plane-geometry
asked Dec 8 at 8:12
user121
273
add a comment |
I tried to solve the following task:
A line L has equation: $frac{x-2}{p} = frac{y-q}{2}= z-1$, where $p,q in mathbb{R} $. A plane P has equation: $ x +y +3z = 9$. Given that line L lies in the plane P, find the value of $p$ and the value of $q$.
In order to find $p$ and $q$, I did following: I found the parametric form of the line: $x = t p +2$, $ y = 2t +q$, $ z= t+1$, where $t in mathbb{R}$ and then, I inserted these values in the equation of P. I got this:
$(tp +2) + (2t +q) + 3(t+1) = 9$ or
$q +tp + 5t = 4$
Can I just insert now 2 arbitrary real values of $t$ and then solve the system of two linear equations to find $p$ and $q$? Is that correct approach? Thanks in advance.
analytic-geometry plane-geometry
asked Dec 8 at 8:12
user121
273
Isn't there more than one line having this equation that lie in the plane? Infinetly many maybe?
– Fareed AF
Dec 8 at 8:25
add a comment |
I tried to solve the following task:
A line L has equation: $frac{x-2}{p} = frac{y-q}{2}= z-1$, where $p,q in mathbb{R} $. A plane P has equation: $ x +y +3z = 9$. Given that line L lies in the plane P, find the value of $p$ and the value of $q$.
In order to find $p$ and $q$, I did following: I found the parametric form of the line: $x = t p +2$, $ y = 2t +q$, $ z= t+1$, where $t in mathbb{R}$ and then, I inserted these values in the equation of P. I got this:
$(tp +2) + (2t +q) + 3(t+1) = 9$ or
$q +tp + 5t = 4$
Can I just insert now 2 arbitrary real values of $t$ and then solve the system of two linear equations to find $p$ and $q$? Is that correct approach? Thanks in advance.
analytic-geometry plane-geometry
asked Dec 8 at 8:12
user121
273
I tried to solve the following task:
A line L has equation: $frac{x-2}{p} = frac{y-q}{2}= z-1$, where $p,q in mathbb{R} $. A plane P has equation: $ x +y +3z = 9$. Given that line L lies in the plane P, find the value of $p$ and the value of $q$.
In order to find $p$ and $q$, I did following: I found the parametric form of the line: $x = t p +2$, $ y = 2t +q$, $ z= t+1$, where $t in mathbb{R}$ and then, I inserted these values in the equation of P. I got this:
$(tp +2) + (2t +q) + 3(t+1) = 9$ or
$q +tp + 5t = 4$
Can I just insert now 2 arbitrary real values of $t$ and then solve the system of two linear equations to find $p$ and $q$? Is that correct approach? Thanks in advance.
analytic-geometry plane-geometry
analytic-geometry plane-geometry
asked Dec 8 at 8:12
user121
273
asked Dec 8 at 8:12
user121
273
asked Dec 8 at 8:12
user121
273
asked Dec 8 at 8:12
user121
273
asked Dec 8 at 8:12
user121
273
273
Isn't there more than one line having this equation that lie in the plane? Infinetly many maybe?
– Fareed AF
Dec 8 at 8:25
add a comment |
Isn't there more than one line having this equation that lie in the plane? Infinetly many maybe?
– Fareed AF
Dec 8 at 8:25
Isn't there more than one line having this equation that lie in the plane? Infinetly many maybe?
– Fareed AF
Dec 8 at 8:25
Isn't there more than one line having this equation that lie in the plane? Infinetly many maybe?
– Fareed AF
Dec 8 at 8:25
add a comment |
1 Answer
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Yes, you may do that. You can also note more directly, by comparing the coefficients of $t$, that $p,q$ are real constants, so the only way $t(p+5)=4-q, forall tinBbb R$ is when $p=-5, q=4$.
Another way to find $p,q$ would be to note that $(2,q,1)$ is a point on the line and hence on the plane $implies 2+q+3=9implies q=4$. Also note that the line is parallel to the free vector $(p,2,1)$, that must be perpendicular to the normal of the plane in order for the line to lie in the plane $implies (p,2,1)cdot(1,1,3)=0implies p=-5$.
answered Dec 8 at 9:22
Shubham Johri
3,861716
Great, thanks a lot!
– user121
Dec 8 at 9:34
add a comment |
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1 Answer
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Yes, you may do that. You can also note more directly, by comparing the coefficients of $t$, that $p,q$ are real constants, so the only way $t(p+5)=4-q, forall tinBbb R$ is when $p=-5, q=4$.
Another way to find $p,q$ would be to note that $(2,q,1)$ is a point on the line and hence on the plane $implies 2+q+3=9implies q=4$. Also note that the line is parallel to the free vector $(p,2,1)$, that must be perpendicular to the normal of the plane in order for the line to lie in the plane $implies (p,2,1)cdot(1,1,3)=0implies p=-5$.
answered Dec 8 at 9:22
Shubham Johri
3,861716
Great, thanks a lot!
– user121
Dec 8 at 9:34
add a comment |
Yes, you may do that. You can also note more directly, by comparing the coefficients of $t$, that $p,q$ are real constants, so the only way $t(p+5)=4-q, forall tinBbb R$ is when $p=-5, q=4$.
Another way to find $p,q$ would be to note that $(2,q,1)$ is a point on the line and hence on the plane $implies 2+q+3=9implies q=4$. Also note that the line is parallel to the free vector $(p,2,1)$, that must be perpendicular to the normal of the plane in order for the line to lie in the plane $implies (p,2,1)cdot(1,1,3)=0implies p=-5$.
answered Dec 8 at 9:22
Shubham Johri
3,861716
Great, thanks a lot!
– user121
Dec 8 at 9:34
add a comment |
Yes, you may do that. You can also note more directly, by comparing the coefficients of $t$, that $p,q$ are real constants, so the only way $t(p+5)=4-q, forall tinBbb R$ is when $p=-5, q=4$.
Another way to find $p,q$ would be to note that $(2,q,1)$ is a point on the line and hence on the plane $implies 2+q+3=9implies q=4$. Also note that the line is parallel to the free vector $(p,2,1)$, that must be perpendicular to the normal of the plane in order for the line to lie in the plane $implies (p,2,1)cdot(1,1,3)=0implies p=-5$.
answered Dec 8 at 9:22
Shubham Johri
3,861716
Yes, you may do that. You can also note more directly, by comparing the coefficients of $t$, that $p,q$ are real constants, so the only way $t(p+5)=4-q, forall tinBbb R$ is when $p=-5, q=4$.
Another way to find $p,q$ would be to note that $(2,q,1)$ is a point on the line and hence on the plane $implies 2+q+3=9implies q=4$. Also note that the line is parallel to the free vector $(p,2,1)$, that must be perpendicular to the normal of the plane in order for the line to lie in the plane $implies (p,2,1)cdot(1,1,3)=0implies p=-5$.
answered Dec 8 at 9:22
Shubham Johri
3,861716
edited Dec 8 at 9:30
answered Dec 8 at 9:22
Shubham Johri
3,861716
answered Dec 8 at 9:22
Shubham Johri
3,861716
answered Dec 8 at 9:22
Shubham Johri
3,861716
3,861716
Great, thanks a lot!
– user121
Dec 8 at 9:34
add a comment |
Great, thanks a lot!
– user121
Dec 8 at 9:34
Great, thanks a lot!
– user121
Dec 8 at 9:34
Great, thanks a lot!
– user121
Dec 8 at 9:34
add a comment |
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Isn't there more than one line having this equation that lie in the plane? Infinetly many maybe?
– Fareed AF
Dec 8 at 8:25