1.5 Equations of Lines in 3d

What's fishy about those normal vectors?

There are infinitely many planes satisfying the condition described, but we're asked for “the" line.

We need two normal directions to figure out the direction of a line in $\mathbb R^3$. Since the given normal vectors are parallel to each other, they really only specify one normal direction.

Note $12\llt \frac13,-\frac12,\frac16 \rgt=\llt 4,-6,2 \rgt$. So, we actually only know one normal direction to the line we're supposed to be describing. That means there are actually infinitely many lines satisfying the given conditions.

The two lines $\llt x-x_0\,,\,y-y_0\,,\,z-z_0\rgt =t\vd$ and $\llt x-x'_0\,,\,y-y'_0\,,\,z-z'_0\rgt =t\vd'$ are not parallel if $\vd\times\vd'\ne\vZero$.

There are infinitely many correct answers. One is

There are infinitely many correct answers. One is

No two of the lines are parallel because

No two of the lines intersect because

Review Example 1.5.2 in the CLP-3 text.

(a) $\llt x,y,z\rgt = \llt 3,5,0\rgt+\llt 2,-\half,1\rgt t$

(b) $\llt x,y,z\rgt = \llt -2,-1,0\rgt +\llt \frac{3}{2},1,1\rgt t$

(a) The point $(x,y,z)$ obeys both $x-2z=3$ and $y+\half z=5$ if and only if $\llt x,y,z\rgt = \llt 3+2z, 5-\half z, z\rgt = \llt 3,5,0\rgt+\llt 2,-\half,1\rgt z$. So, introducing a new variable $t$ obeying $t=z$, we get the vector parametric equation $\llt x,y,z\rgt = \llt 3,5,0\rgt+\llt 2,-\half,1\rgt t$.

(b) The point $(x,y,z)$ obeys

Hence the point $(x,y,z)$ is on the line if and only if $\llt x,y,z\rgt = \llt \frac{1}{4}(2y+4z-6), z-1, z\rgt= \llt \frac{3}{2}z-2, z-1, z\rgt = \llt -2,-1,0\rgt+\llt \frac{3}{2},1,1\rgt z$. So, introducing a new variable $t$ obeying $t=z$, we get the vector parametric equation $\llt x,y,z\rgt = \llt -2,-1,0\rgt +\llt \frac{3}{2},1,1\rgt t$.

Review Example 1.5.2 in the CLP-3 text.

(a) $\llt x,y,z\rgt=\llt -1,4,0\rgt+t\llt 1,-2,1\rgt$

(b) The two planes are parallel and do not intersect.

(a) The normals to the two planes are $\llt 1,1,1\rgt$ and $\llt 1,2,3\rgt$ respectively. The line of intersection must have direction perpendicular to both of these normals. Its direction vector is

Substituting $z=0$ into the equations of the two planes and solving

we see that $z=0,\ y=4, x=-1$ lies on both planes. The line of intersection is $\llt x,y,z\rgt=\llt -1,4,0\rgt+t\llt 1,-2,1\rgt$. This can be checked by verifying that, for all values of $t$, $\llt x,y,z\rgt=\llt -1,4,0\rgt+t\llt 1,-2,1\rgt$ satisfies both $x+y+z=3$ and $x+2y+3z=7$.

(b) The equation $x+y+z=3$ is equivalent to $2x+2y+2z=6$. So the two equations $x+y+z=3$ and $2x+2y+2z=7$ are mutually contradictory. They have no solution. The two planes are parallel and do not intersect.

(a) $(1,0,3)$ lies on both lines. $x+y+2z=7$ is the only plane containing both lines.

(b) The two lines do not intersect. No plane contains the two lines.

(c) The two lines do not intersect. $x+z=1$ is the only plane containing both lines.

(d) The two lines are identical. For arbitrary $a$ and $b$ (not both zero) the plane $ax+by+(a+b)z=a$ contains both lines.

(a) Note that the value of the parameter $t$ in the equation $\llt x,y,z\rgt = \llt -3,2,4\rgt+t\llt -4,2,1\rgt$ need not have the same value as the parameter $t$ in the equation $\llt x,y,z\rgt = \llt 2,1,2\rgt+t\llt 1,1,-1\rgt$. So it is much safer to change the name of the parameter in the first equation from $t$ to $s$. In order for a point $(x,y,z)$ to lie on both lines we need

or equivalently, writing out the three component equations and moving all $s$'s and $t$'s to the left and constants to the right,

Adding the last two equations together gives $3s=-3$ or $s=-1$. Substituting this into the last equation gives $t=-1$. Note that $s=t=-1$ does indeed satisfy all three equations so that $\llt x,y,z\rgt=\llt -3,2,4\rgt-\llt -4,2,1\rgt=\llt 1,0,3\rgt$ lies on both lines. Any plane that contains the two lines must be parallel to both direction vectors $\llt -4,2,1\rgt$ and $\llt 1,1,-1\rgt$. So its normal vector must be perpendicular to them, i.e. must be parallel to $\llt -4,2,1\rgt\times\llt 1,1,-1\rgt=\llt -3,-3,-6\rgt=-3\llt 1,1,2\rgt$. The plane must contain $(1,0,3)$ and be perpendicular to $\llt 1,1,2\rgt$. Its equation is $\llt 1,1,2\rgt\cdot\llt x-1,y,z-3\rgt=0$ or $x+y+2z=7$. This can be checked by verifying that $\llt -3,2,4\rgt+s\llt -4,2,1\rgt$ and $\llt 2,1,2\rgt+t\llt 1,1,-1\rgt$ obey $x+y+2z=7$ for all $s$ and $t$ respectively.

(b) In order for a point $(x,y,z)$ to lie on both lines we need

or equivalently, writing out the three component equations and moving all $s$'s and $t$'s to the left and constants to the right,

Adding the last two equations together gives $3s=-6$ or $s=-2$. Substituting this into the last equation gives $t=-3$. However, substituting $s=-2,\ t=-3$ into the first equation gives $11=5$, which is impossible. The two lines do not intersect. In order for two lines to lie in a common plane and not intersect, they must be parallel. So, in this case no plane contains the two lines.

(c) In order for a point $(x,y,z)$ to lie on both lines we need

or equivalently, writing out the three component equations and moving all $s$'s and $t$'s to the left and constants to the right,

The first two equations are obviously contradictory. The two lines do not intersect. Any plane containing the two lines must be parallel to $\llt 1,1,-1\rgt$ (and hence automatically parallel to $\llt -2,-2,2\rgt=-2\llt 1,1,-1\rgt$) and must also be parallel to the vector from the point $(-3,2,4)$, which lies on the first line, to the point $(2,1,-1)$, which lies on the second. The vector is $\llt 5,-1,-5\rgt$. Hence the normal to the plane is $\llt 5,-1,-5\rgt\times\llt 1,1,-1\rgt=\llt 6,0,6\rgt=6\llt 1,0,1\rgt$. The plane perpendicular to $\llt 1,0,1\rgt$ containing $(2,1,-1)$ is $\llt 1,0,1\rgt\cdot\llt x-2,y-1,z+1\rgt=0$ or $x+z=1$.

(d) Again the two lines are parallel, since $\llt -2,-2,2\rgt=-2\llt 1,1,-1\rgt$. Furthermore the point $\llt 3,2,-2\rgt=\llt 3,2,-2\rgt+0\llt -2,-2,2\rgt =\llt 2,1,-1\rgt+1\llt 1,1,-1\rgt$ lies on both lines. So the two lines not only intersect but are identical. Any plane that contains the point $(3,2,-2)$ and is parallel to $\llt 1,1,-1\rgt$ contains both lines. In general, the plane $ax+by+cz=d$ contains $(3,2,-2)$ if and only if $d=3a+2b-2c$ and is parallel to $\llt 1,1,-1\rgt$ if and only if $\llt a,b,c\rgt\cdot\llt 1,1,-1\rgt=a+b-c=0$. So, for arbitrary $a$ and $b$ (not both zero) $ax+by+(a+b)z=a$ works.

vector parametric equation: $\llt x-2,y+1,z+1\rgt= t\llt 1,-1,-1\rgt$

scalar parametric equation: $x=2+t$, $y=-1-t$, $z=-1-t$

symmetric equation: $x-2=-y-1=-z-1$

First observe that

Consequently

is perpendicular to both $\llt 1,1,0\rgt$ and $\llt 1,-1,2\rgt$. So $\frac{1}{2}\llt 2 \,,\, -2 \,,\, -2 \rgt=\llt 1,-1,-1\rgt\>$ is also perpendicular to both $\llt 1,1,0\rgt$ and $\llt 1,-1,2\rgt$ and hence is parallel to the line. As the point $(2,-1,-1)$ is on the line, the vector equation of the line is

The scalar parametric equations for the line are

The symmetric equations for the line are

Review Example 1.5.2 in the CLP-3 text.

$\llt x,y,z\rgt = \llt 1,0,2\rgt +t\llt-1,1,-1 \rgt$

Let's parametrize $L$ using $y$, renamed to $t$, as the parameter. Then $y=t$, so that

and

and

is a vector parametric equation for $L$.

Review Example 1.5.4 in the CLP-3 text.

(a) $\llt x,y,z\rgt=\llt 5,3,0\rgt+t\llt 4,1,-2\rgt =\llt 5+4t,3+t,-2t\rgt$. (b) $\sqrt{5}$

(a) The normal vectors to the two given planes are $\llt 1,2,3\rgt$ and $\llt 1,-2,1\rgt$ respectively. Since the line is to be contained in both planes, its direction vector must be perpendicular to both $\llt 1,2,3\rgt$ and $\llt 1,-2,1\rgt$, and hence must be parallel to

or to $\llt 4,1,-2\rgt$. Setting $z=0$ in $x+2y+3z=11,\ x-2y+z=-1$ and solving

we see that $(5,3,0)$ is on the line. So the vector parametric equation of the line is $\llt x,y,z\rgt=\llt 5,3,0\rgt+t\llt 4,1,-2\rgt =\llt 5+4t,3+t,-2t\rgt$.

(b) The vector from $(1,0,1)$ to the point $(5+4t,3+t,-2t)$ on the line is $\llt 4+4t,3+t,-1-2t\rgt$. In order for $(5+4t,3+t,-2t)$ to be the point of the line closest to $(1,0,1)$, the vector $\llt 4+4t,3+t,-1-2t\rgt$ joining those two points must be perpendicular to the direction vector $\llt 4,1,-2\rgt$ of the line. (See Example 1.5.4 in the CLP-3 text.) This is the case when

The point on the line nearest $(1,0,1)$ is thus $(5+4t,3+t,-2t)\Big|_{t=-1}=(5-4,3-1,2)=(1,2,2)$. The distance from the point to the line is the length of the vector from $(1,0,1)$ to the point on the line nearest $(1,0,1)$. That vector is $\llt 1,2,2\rgt -\llt 1,0,1\rgt =\llt 0,2,1\rgt$. So the distance is $|\llt 0,2,1\rgt|=\sqrt{5}$.

Review Example 1.4.4 in the CLP-3 text.

(a) $8x+2y-11z=59$ (b) $\frac{64}{\sqrt{189}} \approx 4.655$

(a) The plane $P$ must be parallel to both $\llt 2,3,2\rgt$ (since it contains $L_1$) and $\llt 5,2,4\rgt$ (since it is parallel to $L_2$). Hence

is normal to $P$. As the point $(1,-2,-5)$ is on $P$, the equation of $P$ is

(b) As $L_2$ is parallel to $P$, the distance from $L_2$ to $P$ is the same as the distance from any one point of $L_2$, for example $(-3,4,-1)$, to $P$. As $(1,-2,-5)$ is a point on $P$, the vector $\llt 1,-2,-5\rgt-\llt -3, 4,-1\rgt =\llt 4,-6,-4\rgt$ has its head on $P$ and tail at $(-3,4,-1)$ on $L_2$. The distance from $L_2$ to $P$ is the length of the projection of the vector $\llt 4,-6,-4\rgt$ on the normal to $P$. (See Example 1.4.5 in the CLP-3 text.) This is

(a) Any nonzero constant times $\llt 1 \,,\, -5 \,,\, -3 \rgt$. (b) $x = -4 + t$, $y = 3 - 5t$, $z = 2 - 3t$

(a) The line $L$ must be perpendicular both to $\llt 2\,,\,1\,,\,-1\rgt$, which is a normal vector for the plane $2x + y - z = 5$, and to $\llt -1\,,\,-2\,,\,3\rgt$, which is a direction vector for the line $x = 3 - t$, $y = 1 - 2t$ and $z = 3t$. Any such vector must be a nonzero constant times

(b) For the point $Q(a, b, c)$

As $a < 0$, $b > 0$, $c > 0$, the point $Q$ is $(-4, 3, 2)$ and the line $L$ is

(a) $(3,3,0)$, $(12,0,-6)$, $(0,4,2)$ (b) $x=10+t$, $y=11+t$, $z=13+t$.

(a) The line $L$ intersects the $xy$–plane when $x + y + z = 6$, $x - y + 2z = 0$, and $z=0$. When $z=0$ the equations of $L$ reduce to $x + y = 6$, $x - y = 0$. So the intersection point is $(3,3,0)$.

The line $L$ intersects the $xz$–plane when $x + y + z = 6$, $x - y + 2z = 0$, and $y=0$. When $y=0$ the equations of $L$ reduce to $x + z = 6$, $x +2z = 0$. Substituting $x=-2z$ into $x+z=6$ gives $-z=6$. So the intersection point is $(12,0,-6)$.

The line $L$ intersects the $yz$–plane when $x + y + z = 6$, $x - y + 2z = 0$, and $x=0$. When $x=0$ the equations of $L$ reduce to $y + z = 6$, $-y +2z = 0$. Substituting $y=2z$ into $y+z=6$ gives $3z=6$. So the intersection point is $(0,4,2)$.

(b) Our main job is to find a direction vector $\vd$ for the line.

So $\vd$ must be perpendicular to both $\llt 0,1,-1\rgt$ and $\llt -3,1,2 \rgt$ and so must be a nonzero constant times

We choose $\vd=\frac{1}{3}\llt 3 \,,\, 3 \,,\, 3 \rgt=\llt 1\,,\,1\,,\,1 \rgt$. So

is a vector parametric equation for the line. We can also write this as $x=10+t$, $y=11+t$, $z=13+t$.