1.4 Equations of Planes in 3d

You are looking for a vector that is perpendicular to $z=0$ and hence is parallel to $\hk$.

Any vector of the form $c\,\hk$ with $c\ne 0$ and $c\ne 1$ works. Three possible choices are $-\hk$, $2\,\hk$, $7.12345\,\hk$.

We are looking for a vector that is perpendicular to $z=0$ and hence is parallel to $\hk$. To be parallel of $\hk$, the vector has to be of the form $c\,\hk$ for some real number $c$. For the vector to be nonzero, we need $c\ne 0$ and for the vector to be different from $\hk$, we need $c\ne 1$. So three possible choices are $-\hk$, $2\,\hk$, $7.12345\,\hk$.

A point $(x,y,z)$ is on the $y$-axis if and only if $x=z=0$. Similarly, a point $(x,y,z)$ is on the $z$-axis if and only if $x=y=0$.

(a) $(0,8,0)$ (b) $(0,0,4)$

(c)

(a) $x+2y+3z=0$ (b) $x+y+3z=3$

(c) There is no plane that passes through both $(1,2,3)$ and $(1,0,0)$ and has normal vector $\llt 4,5,6\rgt$.

(d) $2x+y+z=7$

(a) If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(0,0,0)$, namely $\llt x,y,z\rgt$, lie in the plane. So the vector $\llt x,y,z\rgt$ must be perpendicular to $\llt 1,2,3\rgt$ and

(b) If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(0,0,1)$, namely $\llt x,y,z-1\rgt$, lie in the plane. So the vector $\llt x,y,z-1\rgt$ must be perpendicular to $\llt 1,1,3\rgt$ and

(c) If both $(1,2,3)$ and $(1,0,0)$ are on the plane, then both the head and the tail of the vector from $(1,2,3)$ to $(1,0,0)$, namely $\llt 0,2,3\rgt$, lie in the plane. So the vector $\llt 0,2,3\rgt$ must be perpendicular to $\llt 4,5,6\rgt$. As

the vector $\llt 0,2,3\rgt$ is not perpendicular to $\llt 4,5,6\rgt$. So there is no plane that passes through both $(1,2,3)$ and $(1,0,0)$ and has normal vector $\llt 4,5,6\rgt$.

(d) If both $(1,2,3)$ and $(0,3,4)$ are on the plane, then both the head and the tail of the vector from $(1,2,3)$ to $(0,3,4)$, namely $\llt 1,-1,-1\rgt$, lie in the plane. So the vector $\llt 1,-1,-1\rgt$ must be perpendicular to $\llt 2,1,1\rgt$. As

the vector $\llt 1,-1,-1\rgt$ is indeed perpendicular to $\llt 2,1,1\rgt$. So there is a plane that passes through both $(1,2,3)$ and $(0,3,4)$ and has normal vector $\llt 2,1,1\rgt$. We now just have to build its equation.

If $(x,y,z)$ is any point on the plane, then both the head and the tail of the vector from $(x,y,z)$ to $(1,2,3)$, namely $\llt x-1,y-2,z-3\rgt$, lie in the plane. So the vector $\llt x-1,y-2,z-3\rgt$ must be perpendicular to $\llt 2,1,1\rgt$ and

As a check, note that both $(x,y,z)=(1,2,3)$ and $(x,y,z)=(0,3,4)$ obey the equation $2x+y+z=7$.

Guess.

$x+y+z=1$

Solution 1: That's too easy. We just guess. The plane $x+y+z=1$ contains all three given points.

Solution 2: The plane does not pass through the origin. (You can see this by just making a quick sketch.) So the plane has an equation of the form $ax+by+cz=1$.

So the plane is $x+y+z=1$.

Solution 3:
Both the head and the tail of the vector from $(1,0,0)$ to $(0,1,0)$, namely $\llt -1,1,0\rgt$, lie in the plane. Similarly, both the head and the tail of the vector from $(1,0,0)$ to $(0,0,1)$, namely $\llt -1,0,1\rgt$, lie in the plane. So the vector

is a normal vector for the plane. As $(1,0,0)$ is a point in the plane,

is an equation for the plane.

(a) See Question 4 — or just have a guess!

(a) $x+y+z=2$ (b) No. (c) No. (d) Yes.

(a) Solution 1: That's too easy. We just guess. The plane $x+y+z=2$ contains all three given points.

Solution 2:
Both the head and the tail of the vector from $(1,0,1)$ to $(0,1,1)$, namely $\llt 1,-1,0\rgt$, lie in the plane. Similarly, both the head and the tail of the vector from $(1,1,0)$ to $(0,1,1)$, namely $\llt 1,0,-1\rgt$, lie in the plane. So the vector

is a normal vector for the plane. As $(0,1,1)$ is a point in the plane,

is an equation for the plane.

(b) Since

the point $(1,1,1)$ is not on $x+y+z=2$.

(c) Since

the origin is not on $x+y+z=2$.

(d) Since

the point $(4,-1,-1)$ is on $x+y+z=2$.

Three points don't always determine a plane — why?

All three points $(1,2,3),\ (2,3,4)$ and $(3,4,5)$ are on the line $\vx(t)=(1,2,3)+t(1,1,1)$. There are many planes through that line.

The vector from $(1,2,3)$ to $(2,3,4)$, namely $\llt 1,1,1\rgt$ is parallel to the vector from $(1,2,3)$ to $(3,4,5)$, namely $\llt 2,2,2\rgt$. So the three given points are collinear. Precisely, all three points $(1,2,3),\ (2,3,4)$ and $(3,4,5)$ are on the line $\llt x-1,y-2,z-3\rgt=t\llt 1,1,1\rgt$. There are many planes through that line.

(a) $9x-y-z=8$ (b) $14x-7y-8z=52$

(c) For any real numbers $a$ and $b$, the plane
$ax+by-(a+b)z=4a+b$ contains the three given points.

(a) The plane must be parallel to $\llt 2,4,6\rgt -\llt 1,0,1\rgt =\llt 1,4,5\rgt$ and to $\llt 1,2,-1\rgt -\llt 1,0,1\rgt =\llt 0,2,-2\rgt$. So its normal vector must be perpendicular to both $\llt 1,4,5\rgt$ and $\llt 0,2,-2\rgt$ and hence parallel to

The plane is $9(x-1)-y-(z-1)=0$ or $9x-y-z=8$.

We can check this by observing that $(1,0,1),\ (2,4,6)$ and $(1,2,-1)$ all satisfy $9x-y-z=8$.

(b) The plane must be parallel to $\llt 4,-4,4\rgt -\llt 1,-2,-3\rgt =\llt 3,-2,7\rgt$ and to $\llt 3,2,-3\rgt -\llt 1,-2,-3\rgt =\llt 2,4,0\rgt$. So its normal vector must be perpendicular to both $\llt 3,-2,7\rgt$ and $\llt 2,4,0\rgt$ and hence parallel to

The plane is $14(x-1)-7(y+2)-8(z+3)=0$ or $14x-7y-8z=52$.

We can check this by observing that $(1,-2,-3),\ (4,-4,4)$ and $(3,2,-3)$ all satisfy $14x-7y-8z=52$.

(c) The plane must be parallel to $\llt 5,2,1\rgt -\llt 1,-2,-3\rgt =\llt 4,4,4\rgt$ and to $\llt -1,-4,-5\rgt -\llt 1,-2,-3\rgt =\llt -2,-2,-2\rgt$. My, my. These two vectors are parallel. So the three points are all on the same straight line. Any plane containing the line contains all three points. If $\llt a,b,c\rgt$ is any vector perpendicular to $\llt 1,1,1\rgt$ (i.e. which obeys $a+b+c=0$) then the plane
$a(x-1)+b(y+2)+c(z+3)=0$ or $a(x-1)+b(y+2)-(a+b)(z+3)=0$ or $ax+by-(a+b)z=4a+b$ contains the three given points.

We can check this by observing that $(1,-2,-3),\ (5,2,1)$ and $(-1,-4,-5)$ all satisfy the equation $ax+by-(a+b)z=4a+b$ for all $a$ and $b$.

(a) $\sqrt{3}$ (b) $7/\sqrt{6}$

(a) One point on the plane is $(0,0,7)$. The vector from $(-1,2,3)$ to $(0,0,7)$ is $\llt 0,0,7\rgt -\llt -1,2,3\rgt =\llt 1,-2,4\rgt$. A unit vector perpendicular to the plane is $\frac{1}{\sqrt{3}}\llt 1,1,1\rgt$. The distance from
$(-1,2,3)$ to the plane is the length of the projection of $\llt 1,-2,4\rgt$ on $\frac{1}{\sqrt{3}}\llt 1,1,1\rgt$ which is

(b) One point on the plane is $(0,0,5)$. The vector from $(1,-4,3)$ to $(0,0,5)$ is $\llt 0,0,5\rgt -\llt 1,-4,3\rgt =\llt -1,4,2\rgt$. A unit vector perpendicular to the plane is $\frac{1}{\sqrt{6}}\llt 1,-2,1\rgt$. The distance from
$(1,-4,3)$ to the plane is the length of the projection of $\llt -1,4,2\rgt$ on $\frac{1}{\sqrt{6}}\llt 1,-2,1\rgt$ which is the absolute value of

or $7/\sqrt{6}$.

(a) $3x+2y+z = 8$ (b) $\big(3\,,\,-1\,,\,1\big)$

(a) The vector from $C$ to $A$, namely $\llt 1-2\,,\,1-1\,,\,3-0\rgt = \llt -1\,,\,0\,,\,3\rgt$ lies entirely inside $\Pi$. The vector from $C$ to $B$, namely $\llt 2-2\,,\,0-1\,,\,2-0\rgt = \llt 0\,,\,-1\,,\,2\rgt$ also lies entirely inside $\Pi$. Consequently, the vector

is perpendicular to $\Pi$. The equation of $\Pi$ is then

(b) Let $E$ be $(x,y,z)$. Then the vector from $D$ to $E$, namely $\llt x-6\,,\,y-1\,,\,z-2\rgt$ has to be parallel to the vector $\llt 3 \,,\, 2 \,,\, 1 \rgt$, which is perpendicular to $\Pi$. That is, there must be a number $t$ such that

As $(x,y,z)$ must be in $\Pi$,

So $(x,y,z) = \big(6+3(-1)\,,\,1+2(-1)\,,\,2+(-1)\big) = \big(3\,,\,-1\,,\,1\big)$.

(a) $x-y+z=3$ (b) $5x+y-4z=-3$

We are going to need a direction vector for $L$ in both parts (a) and (b). So we find one first.

So the vector

is a direction vector for $L$.

(a) The plane is to contain the point $(2,3,4)$ and is to have $\llt-1,1,-1 \rgt$ as a normal vector. So

does the job.

(b) The plane is to contain the points $A=(2,3,4)$ and $(1,0,2)$ (which is on $L$) so that the vector $\llt 2-1,3-0,4-2 \rgt =\llt 1,3,2 \rgt$ is to be parallel to the plane. The direction vector of $L$, namely $\llt -1 , 1 , -1 \rgt$, is also to be parallel to the plane. So the vector

is to be normal to the plane. So

does the job.

$4x + 2y - 4z=15$ and $4x + 2y - 4z=-9$

All planes that are parallel to the plane $4x + 2y - 4z = 3$ must have $\llt 4\,,\,2\,,\,-4\rgt$ as a normal vector and hence must have an equation of the form $4x + 2y - 4z = C$ for some constant $C$. We must find the $C$'s for which the distance from $4x + 2y - 4z = 3$ to $4x + 2y - 4z = C$ is $2$. One point on $4x + 2y - 4z = 3$ is $\big(0,\frac{3}{2},0\big)$. The two points $(x',y',z')$ with

are the two points that are a distance $2$ from $\big(0,\frac{3}{2},0\big)$ in the direction of the normal.The two points $(x',y',z')$ are

These two points lie on the desired planes, so the two desired planes are

and

$2$

The two vectors

both lie entirely inside the plane. So the vector

is perpendicular to the plane. The vector $\vc=\frac{1}{3}\llt 6,6,3\rgt =\llt 2,2,1\rgt$ is also perpendicular to the plane. The vector

joins the point to the plane. So, if $\theta$ is the angle between $\vd$ and $\vc$, the distance is

(a) $(x,y,z) = (-3,1,0) +t \llt 0,-1,1\rgt$ (b) $\sqrt{17}$

(a) Let's use $z$ as the parameter and rename it to $t$. That is, $z=t$. Subtracting $2$ times the $W_2$ equation from the $W_1$ equation gives

Substituting the result into the equation for $W_2$ gives

So a parametric equation is

(b) Solution 1

We can also parametrize $M$ using $z=t$:

So one point on $M$ is $(10,12,0)$ and one point on $L$ is $(-3,1,0)$ and

is one vector from a point on $M$ to a point on $L$.

The direction vectors of $L$ and $M$ are $\llt 0,-1,1\rgt$ and $\llt 2,2,1\rgt$, respectively. The vector

is then perpendicular to both $L$ and $M$.

The distance from $L$ to $M$ is then the length of the projection of $\vv$ on $\vn$, which is