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

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

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

$7y+4z-5=0$

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