Find the equation of the plane through #(2, 3, -4)# and #(1, -1, 3)# parallel to the x-axis.?

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Answer 1 · 2017-07-09T09:42:43

The equation of the plane is #7y+4z-5=0#

A vector in the plane is

#vecu=((1),(-1),(3))-((2),(3),(-4))=((-1),(-4),(7))#

A vector direction for the x-axis is

#vecv=((1),(0),(0))#

So,

A normal vector to the plane is

#vecn=vecu xx vecv#

#=|(hati,hatj,hatk),(-1,-4,7),(1,0,0)|#

#=hati(0-0)-hatj(0-7)+hatk(0+4)#

#= <0,7,4>#

The equation of the plane is

#0(x-2)+7(y-3)+4(z+4)=0#

#7y-21+4z+16=0#

#7y+4z-5=0#