# Normal Form Vector Equation for Plane Containing $A(-2,1,5)$ and $\text{Line: } (2-\mu)i+(3+\mu)j+(4+\mu)k$

I am doing part (ii) of

Find normal form equation of plane containing $A(-2,1,5)$ and $\text{Line: } (2-\mu)i+(3+\mu)j+(4+\mu)k$

I did:

$AB=4i+2j-k$

$\text{normal} = (4i+2j-k)\times(=2i+j+5k)=11i-18j+11k$

this aleary doesnt seem right. Answer is $r\cdot (i-j+2k)=7$

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Take any two points on the line; with the other given point, you now have three points. You can now use the method of the previous question. –  Ｊ. Ｍ. Nov 24 '11 at 10:10

In general, the equation of the plane normal to $\bf n$ containing $(x_0,y_0,z_0)$ is $${\bf n}\cdot[ (x-x_0) i +(y-y_0) j + (z-z_0)k]=0.\qquad\qquad(1)$$

In your problem, we need to find a normal vector for the plane.

To do this, we need two independent vectors in the plane. The direction vector of the line, $-i+j+k$, will serve as one. For the other, take the vector formed by the difference of $A$ and any point on the line. Since the line is in the plane $B=2i+3j+4k$ will serve. Now $AB= 4i +2j -k$ is your second vector in the plane.

To obtain the vector normal to the plane, take the cross product of $AB$ and the direction vector of the line.

$${\bf n}=AB\times( -i+j+k )=\Biggl|\matrix{ i&j&k\cr 4&2&-1\cr -1&1&1\cr}\Biggr|= 3 i - 3j +6k.$$

The equation of the plane is, using (1) with $B$ as the point: $${\bf n}\cdot[ (x-2) i +(y-3) j + (z-4)k]=0.$$

To see that this gives the same result of the author: \eqalign{ &{\bf n}\cdot[ (x-2) i +(y-3) j + (z-4)k]=0 \cr \iff&(3i-3j+6k)\cdot[ (x-2) i +(y-3) j + (z-4)k]=0 \cr \iff&(3i-3j+6k)\cdot[ (x i +yj + zk) +( -2i-3j-4k)]=0 \cr \iff&(3i-3j+6k)\cdot (x i +yj + zk) =(3i-3j+6k)\cdot( 2i+3j+4k) \cr \iff&(3i-3j+6k)\cdot (x i +yj + zk) =21 \cr \iff&(i-j+2k)\cdot (x i +yj + zk) =7 \cr}
Final edit:

More simply, the author is using the vector plane equation $$\tag{2}{\bf n}\cdot {\bf r} ={\bf a}\cdot{\bf n}$$ where $\bf n$ is the normal vector, $\bf r$ is the position vector to the plane and $\bf a$ is the position vector of a point on the plane. So, once you find $\bf n$, take ${\bf a}=2i+3j+4k$, and substitute into (2).

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