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Mathematics 7 Online
OpenStudy (anonymous):

the plane through the points p1(-2,1,4) , p2(1,0,3) that is perpendicular to the plane 4x-y+3z=2

ganeshie8 (ganeshie8):

two planes are perpendicular <=> normals to planes are perpendicular

ganeshie8 (ganeshie8):

also, normal to a plane is perpendicular to every line that lies within the plane

ganeshie8 (ganeshie8):

so you could cross p1p2 with the normal of given plane, that gives normal to the plane u want

ganeshie8 (ganeshie8):

normal to the required plane = \(\overrightarrow{p_1p_2} \times \left< 4, -1, 3\right >\)

OpenStudy (anonymous):

thanks .. :)

ganeshie8 (ganeshie8):

once u have normal, use one of the given points to plug in a(x-x1) + b(y-y1) + c(z-z1) = 0

ganeshie8 (ganeshie8):

^thats one way

ganeshie8 (ganeshie8):

another way is to simply setup two equations in a, b, c and solve them

ganeshie8 (ganeshie8):

pick whichever stone u like

OpenStudy (anonymous):

okay

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