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Problem in several editions of Larsen calculus texts:

Plane passes through $(1,-2,-1) $ and $(2,5,6)$ and is parallel to $x$- axis.

Solution given takes cross-product of vector between the points and vector $u = i$, claiming both lie in plane. How could unit vector $ (1,0,0) $ lie in plane parallel to $x$-axis?

Thank you.

MickG
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Mark
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    It has to do with translation invariance of vectors. The vectors have the same properties at any position in space, so a vector travelling "along" the axis is always parallel to it, even if it is placed somewhere else in space. – Terra Hyde Sep 03 '15 at 20:35

4 Answers4

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By definition,the vector $i$ is parallel to all planes parallel to $ XY$ and $XZ$ planes, or any other plane rotated about x-axis and translated arbitrarily thereafter.

Narasimham
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Well just think, point $(1,0,0)$ lies on the $x$-axis, not on $y$ or $z$-axis. Hence your vector is parallel to $x$-axis and therefore lies in an arbitrary plane parallel to $x$-axis. 3D is just the working of your brain :)

zar
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Kismat
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Probably you are confusing an Orientated Segment with a Vector.

An orientated segment (sometimes called applied vector in physics) is defined by the ordered couple of two points (A,B), which are the ends of the segment, and in which the direction from $A$ to $B$ is chosen.

A vector $\mathbf v$, i.e. an ordered D-uple, can be thought as individuating the "class" of all the orientated segments such that $$ {\bf v} = \mathop {OB}\limits^ \to - \mathop {OA}\limits^ \to $$ which can be otherwise written as $$ \mathop {OB}\limits^ \to = \mathop {OA}\limits^ \to + {\bf v} $$ (vector $\bf v$ "applied" in $A$).

Thus:
- the or. segment $((0,0,0),(1,0,0))$ lies on the $x$ axis but does not lie on a plane parallel (and distant) from it;
- the vector $(1,0,0)$ is parallel to the $x$ axis and to any a plane parallel from it.

G Cab
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Let $A=(1,-2,-1)$ and $B=(2,5,6)$.

The vectors $\overrightarrow{UV}$ where $U=A$ or $B$ and $V=U+(1,0,0)$ will lie in a plane that is parallel to the $x-$axis.

The line through $A$ and $B$ can be expresses as

$$L(t)= A + t(B-A)=(1,-2,-1) + t(1,7,7)$$

Note $L_{A,B}(0)=A$ and $L_{A,B}(1)=B$.

The $x-$axis can be thought of as the line through the points $(0,0,0)$ and $(1,0,0)$.

$$L_x(s)=s(1,0,0)$$

The "direction" of $L_{A,B}$ is $B-A = (1,7,7)$ and the "direction" of the $x-$axis is $(1,0,0)$.

So the cross-product $(1,7,7) \times (1,0,0)=(0,7,-7)=7(0,1,-1)$ will be perpendicular to both lines.

The plane we seek must therefore be perpendicular to $(0,1,-1)$ and must pass through the point $A=(1,-2,-1)$

\begin{align} (x,y,z)\circ(0,1,-1) &= A\circ (0,1,-1) \\ y-z &= -1 \\ \end{align}

Note that both points $A$ and $B$ and none of the points on $L_x(s)$ satisfy the equation $y-z=-1$.

Steven Alexis Gregory
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