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I am asked to find the plane that contains the point $(-4, 7, -3)$ and is perpendicular to the line defined by the parametric equations $x = -4 + t$, $y = 7 + 3t$ and $z = -2t$. Now, I know that if $n = (a, b, c)$ is a normal vector to a plane $P$ and if $r_{0} = (x_{0}, y_{0}, z_{0})$ is a point in the plane, then $$P = \{ r: n \cdot (r - r_{0}) = 0 \} = \{ (x, y, z): a (x - x_{0}) + b (y - y_{0}) + c (z - z_{0}) = 0 \},$$ so I thought I just had to take a particular point in the line and call it $n$, then call $r_{0} = (-4, 7, -3)$ and apply the aforementioned equation, but I think there is something wrong with this approach since the four possible answers to the exercise are:

\begin{equation} x - 3y - 2z = 21 \end{equation}

\begin{equation} 3x +y - 2z = 23 \end{equation}

\begin{equation} x + 3y - 2z = 23 \end{equation}

\begin{equation} x + 3y +z = 21, \end{equation}

leaving the possible choices of $n$ to $(1, -3, -2)$, $(3, 1, -2)$, $(1, 3, -2)$ or $(1, 3, 2)$. The problem is that none of those points are in the given line.

I would like to know what is wrong with my strategy to find the equation of the plane.

Chris Ruzz
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  • Hi. I would like to know how to find the answer and to identify what's wrong with my approach. Thanks. – Chris Ruzz Feb 09 '14 at 04:09
  • The problem says the plane contains the point $r_{0} = (-4, 7, -3)$, not $r_{0} = (-4, 7, 0)$ or $r_{0} = (-4, 7, 3)$, that's why I'm using it. – Chris Ruzz Feb 09 '14 at 04:13
  • Vector along the line will be perpendicular to plane in this case,ie.that vector can be consider normal vector.vector along line can be get from equation of line – TheDragonReborn May 27 '14 at 12:31

2 Answers2

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If the plane is perpendicular to the line, then the direction vector of the line is normal to the plane. The direction vector of your line is $\vec{n} = (1,3,-2)$. So, using $\vec{r}_0 = (-4,7,-3)$, the plane equation $\vec{n} \cdot (\vec{r} - \vec{r}_0) = 0$ becomes: $$ (1,3,-2) \cdot (x+4, y-7, z+3) = 0 $$ This simplifies to $$ x + 3y -2z = 23 $$ Your approach didn't work because you used a point on the line to define $\vec{n}$; you need to use the direction vector of the line, instead.

bubba
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  • I am not any questions. But it seems my comments have been extremely confusing, so I will remove them. Also, I am not the OP – IAmNoOne Feb 09 '14 at 04:33
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A more simplified method would be to sort what you are given first, so you are given a point, let us call this point P(-4,7,-3), and parametric equations for a line perpendicular to the line you are solving for

x= -4 + t

y= 7 + 3t

z= -2t

A simple trick for solving for the vector (n) is to simply look at the coefficients of t in each equation, these are the (a, b, c) of the vector.

Therefore let us say vector n = (1,3,-2). That is step 1.

Next, the line needed is perpendicular to the given line, so the assumption, to be perpendicular the vector given from the product of point P and another point, let us call it Q, needs to equal 0. The equation for vector PQ is also a(x-x0) + b(y-y0) + c(z-z0) which should equal 0.

Then by substituting in the solved vector n = (1,3,-2) and the given point P for (x0,y0,z0) = (-4,7-3) you get:

[1(x+4) + 3(y-7) - 2(z+3)] = 0

which simplified is:

[x + 4 + 3y - 21 - 2z - 6] = 0

[x + 3y - 2z - 23] = 0

Final Answer = x + 3y - 2z = 23

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AKaushal
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