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 $Q$.




geometry vectors




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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





















      up vote
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      up vote
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      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











      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





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      share|cite|improve this question




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      asked Jul 26 at 12:22









      ilovewt

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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
          $x+2y-3z=pm fracsqrt142$.






          share|cite|improve this answer





















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            1 Answer
            1






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            active

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            up vote
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            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

























              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























                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













                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













                share|cite|improve this answer



                share|cite|improve this answer











                answered Jul 26 at 12:39









                mechanodroid

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