I've been working on a problem on a textbook that asks for the following:
Find the plane perpendicular to $r \ \{ \frac{x-1}{2} = y - 2 = \frac{z+1}{4}$ that contains $P = (-1,3,-1)$
The options are
a) $4x+z-13 = 0$
b) $2x+y+4z-3 = 0$
c) $4y-z-13=0$
d) $2x+y+4z+13=0$
To me none of the above is correct. After using the vector from the line and placing the point to find the constant of the plane, I've found
$$ x+2z+3 = 0 $$
Is my answer correct? Or am I missing something?
Thank you.
$$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+1}{4}$$ the normal vector is $$n=2i+1j+4k$$
the equation of plane is $$v_x(x-x_0)+v_y(y-y_0)+v_z(z-z_0)=0$$ $$2(x+1)+1(y-3)+4(z+1)=0$$ $$2x+y+4z+3=0$$