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the plane through the points p1(-2,1,4) , p2(1,0,3) that is perpendicular to the plane 4x-y+3z=2
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two planes are perpendicular <=> normals to planes are perpendicular
also, normal to a plane is perpendicular to every line that lies within the plane
so you could cross p1p2 with the normal of given plane, that gives normal to the plane u want
normal to the required plane = \(\overrightarrow{p_1p_2} \times \left< 4, -1, 3\right >\)
thanks .. :)
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once u have normal, use one of the given points to plug in a(x-x1) + b(y-y1) + c(z-z1) = 0
^thats one way
another way is to simply setup two equations in a, b, c and solve them
pick whichever stone u like
okay
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