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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.

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1 Answer 1

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$$\frac{x-1}{2}=\frac{y-2}{1}=\frac{z+1}{4}$$ the normal vector is $$n=2i+1j+4k$$

the equation of plane is $$v_x(x-x_0)+v_y(y-y_0)+v_z(z-z_0)=0$$ $$2(x+1)+1(y-3)+4(z+1)=0$$ $$2x+y+4z+3=0$$

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  • $\begingroup$ ah I see my mistake, thank you. $\endgroup$
    – bru1987
    Commented Sep 27, 2015 at 22:01
  • $\begingroup$ @bru1987 you r welcome $\endgroup$
    – E.H.E
    Commented Sep 27, 2015 at 22:03

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