State the possible Cartesian equations of $p_2$.

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Given a plane $p_1$ to have the equation $x+2y-3z=0$ with points $B = (0,3,2), M = (1.5,3,2.5)$. Another plane $p_2$ has a perpendicular distance of $0.5$ with $p_1$. State the possible Cartesian equations of $p_2$.

I think I used a wrong way but got the correct answer, I let $Q$ be a point $(x,y,z)$ on $p_2$, and I know $BQ = (x,y-3,z-2)$ then i use length of projection formula to get $$|x+2y-3z| = frac2sqrt14$$

Then I let $x = frac2sqrt14, y = 0, z=0$ to get a point $Q$.

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asked Jul 26 at 12:22

ilovewt

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Given a plane $p_1$ to have the equation $x+2y-3z=0$ with points $B = (0,3,2), M = (1.5,3,2.5)$. Another plane $p_2$ has a perpendicular distance of $0.5$ with $p_1$. State the possible Cartesian equations of $p_2$.

I think I used a wrong way but got the correct answer, I let $Q$ be a point $(x,y,z)$ on $p_2$, and I know $BQ = (x,y-3,z-2)$ then i use length of projection formula to get $$|x+2y-3z| = frac2sqrt14$$

Then I let $x = frac2sqrt14, y = 0, z=0$ to get a point $Q$.

geometry vectors

share|cite|improve this question

asked Jul 26 at 12:22

ilovewt

821314

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

Given a plane $p_1$ to have the equation $x+2y-3z=0$ with points $B = (0,3,2), M = (1.5,3,2.5)$. Another plane $p_2$ has a perpendicular distance of $0.5$ with $p_1$. State the possible Cartesian equations of $p_2$.

I think I used a wrong way but got the correct answer, I let $Q$ be a point $(x,y,z)$ on $p_2$, and I know $BQ = (x,y-3,z-2)$ then i use length of projection formula to get $$|x+2y-3z| = frac2sqrt14$$

Then I let $x = frac2sqrt14, y = 0, z=0$ to get a point $Q$.

geometry vectors

share|cite|improve this question

asked Jul 26 at 12:22

ilovewt

821314

Given a plane $p_1$ to have the equation $x+2y-3z=0$ with points $B = (0,3,2), M = (1.5,3,2.5)$. Another plane $p_2$ has a perpendicular distance of $0.5$ with $p_1$. State the possible Cartesian equations of $p_2$.

I think I used a wrong way but got the correct answer, I let $Q$ be a point $(x,y,z)$ on $p_2$, and I know $BQ = (x,y-3,z-2)$ then i use length of projection formula to get $$|x+2y-3z| = frac2sqrt14$$

Then I let $x = frac2sqrt14, y = 0, z=0$ to get a point $Q$.

geometry vectors

share|cite|improve this question

asked Jul 26 at 12:22

ilovewt

821314

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 12:22

ilovewt

821314

asked Jul 26 at 12:22

ilovewt

821314

asked Jul 26 at 12:22

ilovewt

821314

821314

add a comment | 

add a comment | 

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The plane $p_2$ is parallel to $p_1$ so it is of the form $x+2y-3z=c$.

The plane $p_1$ passes through the origin so the distance of $p_1$ and $p_2$ is equal to the distance of the origin and $p_2$.

Distance of a plane $Ax+By+Cz = D$ from the origin is given by $fracDsqrtA^2+B^2+C^2$ so in this case we have

$$0.5 = d(0, p_2) = fracsqrt1^2+2^2+3^2 = fracsqrt14$$

so $c = pm fracsqrt142$. Therefore $p_2$ is given by
$x+2y-3z=pm fracsqrt142$.

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answered Jul 26 at 12:39

mechanodroid

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

The plane $p_2$ is parallel to $p_1$ so it is of the form $x+2y-3z=c$.

The plane $p_1$ passes through the origin so the distance of $p_1$ and $p_2$ is equal to the distance of the origin and $p_2$.

Distance of a plane $Ax+By+Cz = D$ from the origin is given by $fracDsqrtA^2+B^2+C^2$ so in this case we have

$$0.5 = d(0, p_2) = fracsqrt1^2+2^2+3^2 = fracsqrt14$$

so $c = pm fracsqrt142$. Therefore $p_2$ is given by
$x+2y-3z=pm fracsqrt142$.

share|cite|improve this answer

answered Jul 26 at 12:39

mechanodroid

22.2k52041

add a comment | 

up vote
1
down vote



accepted

The plane $p_2$ is parallel to $p_1$ so it is of the form $x+2y-3z=c$.

The plane $p_1$ passes through the origin so the distance of $p_1$ and $p_2$ is equal to the distance of the origin and $p_2$.

Distance of a plane $Ax+By+Cz = D$ from the origin is given by $fracDsqrtA^2+B^2+C^2$ so in this case we have

$$0.5 = d(0, p_2) = fracsqrt1^2+2^2+3^2 = fracsqrt14$$

so $c = pm fracsqrt142$. Therefore $p_2$ is given by
$x+2y-3z=pm fracsqrt142$.

share|cite|improve this answer

answered Jul 26 at 12:39

mechanodroid

22.2k52041

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

The plane $p_2$ is parallel to $p_1$ so it is of the form $x+2y-3z=c$.

The plane $p_1$ passes through the origin so the distance of $p_1$ and $p_2$ is equal to the distance of the origin and $p_2$.

Distance of a plane $Ax+By+Cz = D$ from the origin is given by $fracDsqrtA^2+B^2+C^2$ so in this case we have

$$0.5 = d(0, p_2) = fracsqrt1^2+2^2+3^2 = fracsqrt14$$

so $c = pm fracsqrt142$. Therefore $p_2$ is given by
$x+2y-3z=pm fracsqrt142$.

share|cite|improve this answer

answered Jul 26 at 12:39

mechanodroid

22.2k52041

The plane $p_2$ is parallel to $p_1$ so it is of the form $x+2y-3z=c$.

The plane $p_1$ passes through the origin so the distance of $p_1$ and $p_2$ is equal to the distance of the origin and $p_2$.

Distance of a plane $Ax+By+Cz = D$ from the origin is given by $fracDsqrtA^2+B^2+C^2$ so in this case we have

$$0.5 = d(0, p_2) = fracsqrt1^2+2^2+3^2 = fracsqrt14$$

so $c = pm fracsqrt142$. Therefore $p_2$ is given by
$x+2y-3z=pm fracsqrt142$.

share|cite|improve this answer

answered Jul 26 at 12:39

mechanodroid

22.2k52041

share|cite|improve this answer

share|cite|improve this answer

answered Jul 26 at 12:39

mechanodroid

22.2k52041

answered Jul 26 at 12:39

mechanodroid

22.2k52041

answered Jul 26 at 12:39

mechanodroid

22.2k52041

22.2k52041

add a comment | 

add a comment | 

 

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