Planes II

Interaction of a Line with a Plane

Line and Plane Parallelism

Consider a line given in vector form as

\[ \vec{r} = \vec{a} + \lambda \vec{b}, \quad \lambda \in \mathbb{R}, \]

and a plane given by

\[ \vec{r} \cdot \vec{n} = d. \]

We wish to determine the condition under which the line and the plane are parallel.

A line is said to be parallel to a plane if it does not intersect the plane. This happens precisely when the direction vector \( \vec{b} \) of the line is perpendicular to the normal vector \( \vec{n} \) of the plane.

That is, the line and plane are parallel if and only if:

\[ \vec{b} \cdot \vec{n} = 0. \]

Now, since the line is parallel to the plane, the distance between the line and the plane is the perpendicular distance from any point on the line to the plane.

We may choose the point with position vector \( \vec{a} \), which lies on the line

\[ \vec{r} = \vec{a} + \lambda \vec{b}. \]

Then the perpendicular distance from \( \vec{a} \) to the plane \( \vec{r} \cdot \vec{n} = d \) is given by the standard formula:

\[ \text{Distance} = \frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}. \]

Thus, the distance between the line and the plane is

\[ \boxed{\frac{|\vec{a} \cdot \vec{n} - d|}{|\vec{n}|}}. \]

Line and Plane Perpendicularity

A line given by

\[ \vec{r} = \vec{a} + \lambda \vec{b} \]

is said to be perpendicular to the plane

\[ \vec{r} \cdot \vec{n} = d \]

if its direction vector \( \vec{b} \) is collinear with the normal vector \( \vec{n} \) of the plane.

That is,

\[ \vec{b} \parallel \vec{n} \quad \text{or} \quad \vec{b} = k\vec{n} \quad \text{for some } k \in \mathbb{R}. \]

This means the line passes through the plane in a direction perpendicular to it, or is entirely contained within the normal line of the plane.

Angle Between a Line and a Plane

Suppose a line is given by

\[ \vec{r} = \vec{a} + \lambda \vec{b}, \]

and a plane is given by

\[ \vec{r} \cdot \vec{n} = d. \]

Assume the line intersects the plane at a point \( P \). From any point \( Q \) on the line, drop a perpendicular onto the plane, meeting the plane at point \( N \). The angle \( \theta \) between the line and the plane is defined as the angle \( \angle QPN \).

Now observe that the vector \( \vec{b} \), which is the direction of the line, and the normal vector \( \vec{n} \) to the plane form a right triangle at point \( P \), with angle between them equal to \( 90^\circ - \theta \).

Hence,

\[ \angle (\vec{b},\ \vec{n}) = 90^\circ - \theta. \]

Taking cosine on both sides, we get:

\[ \cos(90^\circ - \theta) = \sin\theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}|\,|\vec{n}|}. \]

Therefore, the angle \( \theta \) between a line and a plane is given by

\[ \boxed{ \sin\theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}|\,|\vec{n}|} }. \]

This formula captures the acute angle between the line and the plane at the point of intersection.

Intersection Point of a Line and a Plane

This is straightforward and has already appeared in many earlier contexts.

Consider a line given by

\[ \vec{r} = \vec{a} + \lambda \vec{b}, \]

and a plane given by

\[ \vec{r} \cdot \vec{n} = d. \]

To find the point of intersection of the line and the plane, substitute the expression for \( \vec{r} \) from the line into the plane equation:

\[ (\vec{a} + \lambda \vec{b}) \cdot \vec{n} = d. \]

Expanding the dot product:

\[ \vec{a} \cdot \vec{n} + \lambda\, \vec{b} \cdot \vec{n} = d \implies \lambda = \frac{d - \vec{a} \cdot \vec{n}}{\vec{b} \cdot \vec{n}}. \]

Thus, the value of \( \lambda \) at the point of intersection is known. Substituting this back into the line equation gives the intersection point:

\[ \vec{r} = \vec{a} + \left( \frac{d - \vec{a} \cdot \vec{n}}{\vec{b} \cdot \vec{n}} \right) \vec{b}. \]

This expression gives the position vector of the point where the line intersects the plane, provided \( \vec{b} \cdot \vec{n} \ne 0 \). If \( \vec{b} \cdot \vec{n} = 0 \), then the line is parallel to the plane.

If the line and the plane are both given in Cartesian form, the process of finding their intersection is just as systematic.

Let the line be given as

\[ \frac{x - x_0}{l} = \frac{y - y_0}{m} = \frac{z - z_0}{n} = \lambda, \]

and the plane be given by the equation

\[ ax + by + cz + d = 0. \]

Let the point of intersection be the point on the line corresponding to parameter \( \lambda \). The coordinates of this point are:

\[ x = l\lambda + x_0, \quad y = m\lambda + y_0, \quad z = n\lambda + z_0. \]

Substitute these into the plane equation:

\[ a(l\lambda + x_0) + b(m\lambda + y_0) + c(n\lambda + z_0) + d = 0. \]

Expanding:

\[ \lambda(al + bm + cn) + (a x_0 + b y_0 + c z_0 + d) = 0. \]

Solving for \( \lambda \):

\[ \lambda = -\frac{a x_0 + b y_0 + c z_0 + d}{a l + b m + c n}. \]

This gives the parameter value corresponding to the point of intersection. Substituting back, the coordinates of the intersection point are:

\[ x = l\lambda + x_0, \quad y = m\lambda + y_0, \quad z = n\lambda + z_0, \]

with \( \lambda \) as above.

This method works as long as the denominator \( al + bm + cn \ne 0 \); if it is zero, the line is parallel to the plane, and there may be either no intersection.

A Plane Contains a Line

A plane contains a line if the entire line lies on the plane—that is, every point of the line satisfies the plane equation. In this case, the line intersects the plane at infinitely many points, not just one.

Let the line be given in vector form as

\[ \vec{r} = \vec{a} + \lambda \vec{b}, \]

and the plane be

\[ \vec{r} \cdot \vec{n} = d. \]

Then the plane contains the line if and only if both of the following conditions hold:

\( \vec{b} \cdot \vec{n} = 0 \),

so that the direction vector of the line is perpendicular to the normal vector of the plane, meaning the line is parallel to the plane.

\( \vec{a} \cdot \vec{n} = d \),

so that the point \( \vec{a} \) (through which the line passes) lies on the plane.

These two together ensure that not only is the line parallel to the plane, but it also passes through a point on the plane, and hence lies entirely on the plane.

Projection of a Line on a Plane

Let us understand the idea of projecting a line onto a plane.

By the projection of a line on a plane, we mean the following: if we drop perpendiculars from every point on the given line onto the plane, then the feet of these perpendiculars trace out a new line lying entirely on the plane. This new line is called the projection of the given line onto the plane.

Case I : Line is Parallel to the Plane

Consider a line in symmetric form:

\[ \frac{x - x_1}{p} = \frac{y - y_1}{q} = \frac{z - z_1}{r}, \]

and a plane:

\[ ax + by + cz + d = 0. \]

Assume the line is parallel to the plane, which means its direction vector \( (p, q, r) \) is perpendicular to the normal vector of the plane \( (a, b, c) \), i.e.,

\[ ap + bq + cr = 0. \]

To write the equation of the projected line, we need:

A point on the projected line,
Its direction.

The direction is immediate: it is the same as that of the original line since the projection preserves direction when the line is parallel to the plane. So, the direction ratios of the projected line are \( (p, q, r) \).

To find a point on the projected line, we project any one point from the original line onto the plane. The simplest choice is the point \( (x_1, y_1, z_1) \), which lies on the line.

We now find the foot of the perpendicular from \( (x_1, y_1, z_1) \) to the plane using the formula:

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{a x_1 + b y_1 + c z_1 + d}{a^2 + b^2 + c^2}. \]

Let the right-hand side be \( \lambda \), then the coordinates of the foot (say, point \( P \)) are:

\[ x = x_1 + a\lambda, \quad y = y_1 + b\lambda, \quad z = z_1 + c\lambda. \]

Substituting:

\[ \lambda = -\frac{a x_1 + b y_1 + c z_1 + d}{a^2 + b^2 + c^2}. \]

This gives a point on the projected line.

Thus, the projection of the given line onto the plane is the line that:

– passes through the point

\[
\left(x_1 + a\lambda,\ y_1 + b\lambda,\ z_1 + c\lambda\right),
\]

with \( \lambda = -\dfrac{a x_1 + b y_1 + c z_1 + d}{a^2 + b^2 + c^2} \),

– and has direction ratios \( (p, q, r) \).

That is the required projected line lying entirely on the plane.

Case 2: Line Is Not Parallel to the Plane

Now consider the case when the given line is not parallel to the plane. That is, the direction vector of the line is not orthogonal to the normal vector of the plane. So the line intersects the plane at exactly one point.

Let the line be

\[ \frac{x - x_1}{p} = \frac{y - y_1}{q} = \frac{z - z_1}{r} \quad \text{or in vector form} \quad \vec{r} = \vec{a} + \lambda \vec{b}, \]

and let the plane be

\[ ax + by + cz + d = 0 \quad \text{or in vector form} \quad \vec{r} \cdot \vec{n} = d. \]

In this case, the projection of the line onto the plane is again the locus of the feet of perpendiculars dropped from every point on the line to the plane.

To describe this projection, we need:

– A point on the projected line,
– A direction vector of the projected line.

We can easily get two points lying on the projected line:

The intersection point of the line and the plane. This is obtained by substituting \( \vec{r} = \vec{a} + \lambda \vec{b} \) into the plane equation and solving for \( \lambda \), as done earlier.
The foot of the perpendicular from the point \( (x_1, y_1, z_1) \) to the plane, using the formula

\[ \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = -\frac{a x_1 + b y_1 + c z_1 + d}{a^2 + b^2 + c^2}. \]

Having these two points, we can write the equation of the projected line in symmetric form.

But there is a far more elegant way to directly obtain the direction of the projected line—using the vector triple product.

Observe that the direction vector of the projected line:

– lies in the same plane as the direction vector \( \vec{b} \) of the given line and the normal vector \( \vec{n} \) of the plane (since projection of a line is obtained by dropping perpendiculars from each point of the line),
– and is also perpendicular to \( \vec{n} \), because the projected line lies on the plane.

The unique vector satisfying these two conditions is

\[ \vec{n} \times (\vec{n} \times \vec{b}). \]

Indeed, the cross product \( \vec{n} \times \vec{b} \) gives a vector perpendicular to both \( \vec{n} \) and \( \vec{b} \). Taking the cross product of \( \vec{n} \) again with this vector gives a vector lying in the plane spanned by \( \vec{n} \) and \( \vec{b} \), and also perpendicular to \( \vec{n} \). That is precisely the direction of the projection of \( \vec{b} \) onto the plane.

So the direction vector of the projection is

\[ \boxed{ \vec{n} \times (\vec{n} \times \vec{b}) }. \]

Thus, the projection of the line onto the plane is a line:

– passing through any point of projection (intersection point or foot),
– with direction vector \( \vec{n} \times (\vec{n} \times \vec{b}) \).

System of Planes

Angle of Intersection of Two Planes

Let two planes be given in vector form as

\[ \vec{r} \cdot \vec{n}_1 = d_1 \quad \text{and} \quad \vec{r} \cdot \vec{n}_2 = d_2. \]

The angle of intersection between the two planes is defined as the acute angle between them where they meet. Geometrically, this angle is equal to the angle between their normal vectors \( \vec{n}_1 \) and \( \vec{n}_2 \).

Let the angle between the planes be \( \theta \). Then

\[ \cos \theta = \frac{|\vec{n}_1 \cdot \vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}. \]

The absolute value ensures that \( \theta \) is the acute angle between the planes.

So, the angle between two planes is the same as the angle between their normal vectors.

Special Cases

1. Parallel Planes

Two planes are parallel if and only if their normal vectors are parallel, i.e.,

\[ \vec{n}_1 = k \vec{n}_2 \quad \text{for some } k \in \mathbb{R}. \]

Equivalently,

\[ \vec{n}_1 \times \vec{n}_2 = \vec{0}. \]

This implies that the angle between the normal vectors is either \( 0^\circ \) or \( 180^\circ \), and hence the planes are either identical or distinct but parallel.

In Cartesian form, planes

\[ a_1x + b_1y + c_1z + d_1 = 0 \quad \text{and} \quad a_2x + b_2y + c_2z + d_2 = 0 \]

are parallel if

\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}. \]

2. Perpendicular Planes

Two planes are perpendicular if and only if their normal vectors are perpendicular, i.e.,

\[ \vec{n}_1 \cdot \vec{n}_2 = 0. \]

In Cartesian form, if

\[ \vec{n}_1 = a_1 \mathbf{i} + b_1 \mathbf{j} + c_1 \mathbf{k}, \quad \vec{n}_2 = a_2 \mathbf{i} + b_2 \mathbf{j} + c_2 \mathbf{k}, \]

then the condition becomes:

\[ a_1a_2 + b_1b_2 + c_1c_2 = 0. \]

This ensures the planes intersect at a right angle.

Writing the Equation of a Plane Parallel to Another

Suppose a plane is given by

\[ ax + by + cz + d = 0. \]

Any plane parallel to this plane must have the same normal vector \( (a, b, c) \), since parallel planes share the same orientation in space.

Hence, the equation of a plane parallel to this one is

\[ ax + by + cz + k = 0, \]

where \( k \in \mathbb{R} \) is some constant.

Observe that parallel planes differ only by their constant term. The coefficients of \( x, y, z \) remain unchanged, preserving the direction of the normal vector.

To determine the specific value of \( k \), we must be given some additional information, such as a point \( (x_0, y_0, z_0) \) that lies on the required plane. Substituting this point into the general equation

\[ ax + by + cz + k = 0 \]

gives:

\[ a x_0 + b y_0 + c z_0 + k = 0 \implies k = - (a x_0 + b y_0 + c z_0). \]

Thus, the specific plane parallel to \( ax + by + cz + d = 0 \) and passing through \( (x_0, y_0, z_0) \) is

\[ ax + by + cz - (a x_0 + b y_0 + c z_0) = 0. \]

Distance Between Two Parallel Planes

Let two planes be given as

\[ \Pi_1 : ax + by + cz + d_1 = 0, \quad \Pi_2 : ax + by + cz + d_2 = 0. \]

Since the coefficients of \( x, y, z \) are the same, the two planes are parallel (or identical if \( d_1 = d_2 \)). Their normal vector is \( \vec{n} = a\mathbf{i} + b\mathbf{j} + c\mathbf{k} \).

Let \( P(\alpha, \beta, \gamma) \) be any point on the first plane \( \Pi_1 \). Then it satisfies:

\[ a\alpha + b\beta + c\gamma + d_1 = 0. \tag{1} \]

We now drop a perpendicular from \( P \) to the second plane \( \Pi_2 \). Let \( N \) be the foot of this perpendicular. Then the segment \( PN \) is perpendicular to both planes (since they are parallel), and its length gives the distance between the two planes.

Using the standard formula for the perpendicular distance from a point to a plane, the distance from \( P \) to \( \Pi_2 \) is:

\[ \frac{|a\alpha + b\beta + c\gamma + d_2|}{\sqrt{a^2 + b^2 + c^2}}. \]

But from equation (1), we know

\[ a\alpha + b\beta + c\gamma = -d_1, \]

so the numerator becomes

\[ |-d_1 + d_2| = |d_1 - d_2|. \]

Therefore, the distance between the two parallel planes is

\[ \boxed{ \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} }. \]

This is valid for any pair of parallel planes having the same normal vector \( (a, b, c) \).

Equation of a Plane at a Certain Distance from a Given Plane

Let a plane be given in Cartesian form as

\[ ax + by + cz + d = 0. \]

We are asked to find the equation(s) of plane(s) parallel to it at a distance \( D \).

Since the required plane is parallel, it must have the same normal vector \( (a, b, c) \). Therefore, its equation must be of the form

\[ ax + by + cz + k = 0, \]

where \( k \in \mathbb{R} \) is to be determined.

Now, the distance between two parallel planes

\[ ax + by + cz + d = 0 \quad \text{and} \quad ax + by + cz + k = 0 \]

is given by:

\[ \frac{|k - d|}{\sqrt{a^2 + b^2 + c^2}}. \]

We are told that this distance equals \( D \). Therefore:

\[ \frac{|k - d|}{\sqrt{a^2 + b^2 + c^2}} = D \implies |k - d| = D \sqrt{a^2 + b^2 + c^2}. \]

Hence,

\[ k - d = \pm D \sqrt{a^2 + b^2 + c^2} \implies k = d \pm D \sqrt{a^2 + b^2 + c^2}. \]

This gives us two values of \( k \), and hence two parallel planes, one on each side of the original plane at distance \( D \).

Thus, the equations of the required planes are:

\[ \boxed{ ax + by + cz + d \pm D\sqrt{a^2 + b^2 + c^2} = 0 }. \]

These represent the two planes that are parallel to \( ax + by + cz + d = 0 \) and lie at a distance \( D \) from it.

Line of Intersection of Two Planes

Let the two planes be given by the equations

\[ \pi_1:\quad a_1x + b_1y + c_1z = d_1, \qquad \pi_2:\quad a_2x + b_2y + c_2z = d_2 \]

and suppose they are not parallel. Then the intersection is a straight line which lies on both planes.

A point on this line must satisfy both plane equations simultaneously. Thus, the line is the set of points \((x, y, z)\in \mathbb{R}^3\) satisfying both equations. Geometrically, the line lies entirely in each plane.

To describe this line, we need:

A point on it, say \((x_0, y_0, z_0)\),
A direction vector \((A, B, C)\) along the line.

Let

\[ \vec{n}_1 = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}, \qquad \vec{n}_2 = a_2\hat{i} + b_2\hat{j} + c_2\hat{k} \]

be the normal vectors to \(\pi_1\) and \(\pi_2\) respectively.

Since the line lies in both planes, it must be perpendicular to both normal vectors. Hence, its direction vector must be orthogonal to both \(\vec{n}_1\) and \(\vec{n}_2\). The vector that satisfies this condition is their cross product:

\[ \vec{d} = \vec{n}_1 \times \vec{n}_2 \]

To find a point on the line, assign a convenient value to one of the variables—say \(z = 0\)—as long as it appears with nonzero coefficient in at least one of the equations. Substitute into both plane equations to reduce the system to two equations in two variables. Solving this yields a point \((x_0, y_0, z_0)\) on the line.

Once we have a point and a direction vector, the symmetric form of the line is:

\[ \frac{x - x_0}{A} = \frac{y - y_0}{B} = \frac{z - z_0}{C} \]

Example

Consider the two planes

\[ \pi_1: \quad x + y - 2z = 3, \qquad \pi_2: \quad 2x - y + 4z = 4 \]

We seek to find the line of their intersection.

Let us denote by \(\vec{n}_1 = \hat{i} + \hat{j} - 2\hat{k}\) and \(\vec{n}_2 = 2\hat{i} - \hat{j} + 4\hat{k}\) the normal vectors to the two planes. Since the line of intersection lies in both planes, its direction vector must be orthogonal to both \(\vec{n}_1\) and \(\vec{n}_2\). Hence, the direction vector is given by their cross product:

\[ \vec{d} = \vec{n}_1 \times \vec{n}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & -2 \\ 2 & -1 & 4 \end{vmatrix} = (4 - 2)\hat{i} - (4 + 4)\hat{j} + (-1 - 2)\hat{k} = 2\hat{i} - 8\hat{j} - 3\hat{k} \]

Thus, the direction vector of the line is \((2, -8, -3)\).

To find a point on the line, we assign a convenient value to one of the variables. Since \(z\) appears in both equations, we set \(z = 0\). Substituting into the plane equations gives

\[ x + y = 3 \tag{1} \]

\[ 2x - y = 4 \tag{2} \]

Adding (1) and (2) yields \(3x = 7 \implies x = \frac{7}{3}\), and substituting into (1) gives \(y = 3 - \frac{7}{3} = \frac{2}{3}\). Thus, a point on the line is \(\left(\frac{7}{3}, \frac{2}{3}, 0\right)\).

We now have both a point and a direction vector. Therefore, the symmetric form of the line is

\[ \frac{x - \frac{7}{3}}{2} = \frac{y - \frac{2}{3}}{-8} = \frac{z}{-3} \]

which is the required equation of the line of intersection.

An alternative method to find the line of intersection of two planes involves treating one variable as a parameter and solving the resulting system in terms of that parameter. This approach avoids the need to choose a specific numerical value and instead yields a fully parametric representation directly.

Consider the two planes:

\[ \pi_1: \quad x + y - 2z = 3, \qquad \pi_2: \quad 2x - y + 4z = 4 \]

Let us set

\[ z = \lambda \]

which is valid since \(z\) appears with nonzero coefficient in both equations. Substituting into the plane equations yields two equations in \(x\) and \(y\) in terms of \(\lambda\):

\[ x + y = 3 + 2\lambda \tag{1} \]

\[ 2x - y = 4 - 4\lambda \tag{2} \]

Adding (1) and (2) gives:

\[ 3x = 7 - 2\lambda \implies x = \frac{7 - 2\lambda}{3} \]

Substituting back into (1):

\[ \frac{7 - 2\lambda}{3} + y = 3 + 2\lambda \implies y = 3 + 2\lambda - \frac{7 - 2\lambda}{3} \]

Simplify:

\[ y = \frac{9 + 6\lambda - 7 + 2\lambda}{3} = \frac{2 + 8\lambda}{3} \]

Thus, the coordinates of a general point on the line are:

\[ x = \frac{7 - 2\lambda}{3}, \quad y = \frac{2 + 8\lambda}{3}, \quad z = \lambda \]

To express this in symmetric form, islolate \(\lambda\) from each equation:

From \(x = \frac{7 - 2\lambda}{3} \implies 3x = 7 - 2\lambda \implies \lambda = \frac{7 - 3x}{2}\)

From \(y = \frac{2 + 8\lambda}{3} \implies 3y = 2 + 8\lambda \implies \lambda = \frac{3y - 2}{8}\)

From \(z = \lambda\)

Thus, the symmetric form is:

\[ \frac{7 - 3x}{2} = \frac{3y - 2}{8} = z \]

Equivalently, to match the usual structure:

\[ \frac{3x - 7}{-2} = \frac{3y - 2}{8} = z \]

This is the required equation of the line of intersection.

If we choose a different variable—say \(x = \lambda\)—the parametric expressions for \(y\) and \(z\) will look different, but the resulting line remains the same, as it represents the same set of points in space.

Let us again consider the planes:

\[ \pi_1: \quad x + y - 2z = 3, \qquad \pi_2: \quad 2x - y + 4z = 4 \]

Let

\[ x = \lambda \]

Substituting into both equations:

From \(\pi_1:\)

\[ \lambda + y - 2z = 3 \implies y - 2z = 3 - \lambda \tag{1} \]

From \(\pi_2:\)

\[ 2\lambda - y + 4z = 4 \implies -y + 4z = 4 - 2\lambda \tag{2} \]

Add (1) and (2):

\[ (-y + 4z) + (y - 2z) = (4 - 2\lambda) + (3 - \lambda) \implies 2z = 7 - 3\lambda \implies z = \frac{7 - 3\lambda}{2} \]

Substitute into (1):

\[ y - 2\left(\frac{7 - 3\lambda}{2}\right) = 3 - \lambda \implies y - (7 - 3\lambda) = 3 - \lambda \implies y = 10 - 4\lambda \]

Thus, we have:

\[ x = \lambda, \quad y = 10 - 4\lambda, \quad z = \frac{7 - 3\lambda}{2} \]

Now eliminate \(\lambda\) to get the symmetric form:

From \(x = \lambda\), we have \(\lambda = x\)

From \(y = 10 - 4\lambda\), we get \(\lambda = \frac{10 - y}{4}\)

From \(z = \frac{7 - 3\lambda}{2} \implies 2z = 7 - 3\lambda \implies \lambda = \frac{7 - 2z}{3}\)

Thus, the symmetric form is:

\[ x = \frac{10 - y}{4} = \frac{7 - 2z}{3} \]

Or rearranged into the usual form:

\[ x = \frac{y - 10}{-4} = \frac{2z - 7}{-3} \]

This equation represents the same line as before. Though the expressions differ, they describe the same geometric object—the line of intersection of the two given planes.

Family of Planes Passing Through the Line of Intersection of Two Given Planes

Let \(\pi_1: a_1x + b_1y + c_1z + d_1 = 0\) and \(\pi_2: a_2x + b_2y + c_2z + d_2 = 0\) be two non-parallel planes. The intersection of \(\pi_1\) and \(\pi_2\) is a straight line. We are interested in finding the family of all planes that pass through this common line.

Any such plane can be written in the form:

\[ \pi: \quad (a_1x + b_1y + c_1z + d_1) + \lambda(a_2x + b_2y + c_2z + d_2) = 0, \quad \lambda \in \mathbb{R} \]

This is usually written compactly as:

\[ \pi = \pi_1 + \lambda \pi_2 = 0 \]

Let us understand why this form always defines a plane and why it necessarily contains the line of intersection of \(\pi_1\) and \(\pi_2\).

First, observe that \(\pi_1 + \lambda \pi_2 = 0\) is of the form:

\[ (a_1 + \lambda a_2)x + (b_1 + \lambda b_2)y + (c_1 + \lambda c_2)z + (d_1 + \lambda d_2) = 0 \]

which is a linear equation in \(x, y, z\), hence represents a plane in \(\mathbb{R}^3\) for each fixed \(\lambda\).

Next, let us check that this plane always passes through the line of intersection of \(\pi_1\) and \(\pi_2\). Let \((x_0, y_0, z_0)\) be any point lying on that line. Then:

\[ a_1x_0 + b_1y_0 + c_1z_0 + d_1 = 0 \quad \text{and} \quad a_2x_0 + b_2y_0 + c_2z_0 + d_2 = 0 \]

Now substitute \((x_0, y_0, z_0)\) into the equation of \(\pi\):

\[ (a_1x_0 + b_1y_0 + c_1z_0 + d_1) + \lambda(a_2x_0 + b_2y_0 + c_2z_0 + d_2) = 0 + \lambda \cdot 0 = 0 \]

Hence, every point on the intersection line of \(\pi_1\) and \(\pi_2\) also lies on every member of the family \(\pi_1 + \lambda \pi_2 = 0\).

Therefore, this form indeed defines the family of all planes that pass through the line of intersection of two given non-parallel planes.

Example

Find the equation of a plane that passes through the origin and also contains the line of intersection of the two planes

\[ \pi_1: x + y + z = 9, \qquad \pi_2: 2x - y - z = 3 \]

Solution:

Since the required plane passes through the line of intersection of \(\pi_1\) and \(\pi_2\), its equation must be of the form

\[ \pi: \quad \pi_1 + \lambda \pi_2 = 0 \]

which expands to:

\[ (x + y + z - 9) + \lambda(2x - y - z - 3) = 0 \]

To determine \(\lambda\), we use the additional condition: the required plane passes through the origin \((0, 0, 0)\). Substituting into the equation gives:

\[ (0 + 0 + 0 - 9) + \lambda(0 - 0 - 0 - 3) = 0 \implies -9 - 3\lambda = 0 \implies \lambda = -3 \]

Substitute this value into the general form:

\[ (x + y + z - 9) - 3(2x - y - z - 3) = 0 \]

Simplify:

\[ x + y + z - 9 - 6x + 3y + 3z + 9 = 0 \implies (-5x + 4y + 4z) = 0 \]

Hence, the required plane is:

\[ 5x - 4y - 4z = 0 \]

Usefulness of family of planes

One of the significant advantages of using the family of planes method is that it avoids the explicit computation of the line of intersection, which can often be tedious and involve solving a system of equations, parameterizing variables, and converting to symmetric form.

Instead of calculating the line of intersection, we directly assume the general form of any plane passing through it as:

\[ \pi = \pi_1 + \lambda \pi_2 = 0 \]

This is already a plane that contains the line of intersection of the given non-parallel planes \(\pi_1\) and \(\pi_2\). The parameter \(\lambda\) controls the orientation of the plane within the family, and any additional condition (such as passing through a specific point) can be used to determine the value of \(\lambda\).

In the earlier example, we were able to find the required plane passing through the origin and through the line of intersection of

\[ \pi_1: x + y + z = 9, \quad \pi_2: 2x - y - z = 3 \]

without ever needing to find the actual line of intersection. We simply enforced that the origin satisfies the combination

\[ (x + y + z - 9) + \lambda(2x - y - z - 3) = 0 \]

and determined \(\lambda = -3\), yielding the desired plane:

\[ 5x - 4y - 4z = 0 \]

Equation of Bisector Planes

Let two non-parallel planes be given by

\[ \pi_1: \quad a_1x + b_1y + c_1z + d_1 = 0, \qquad \pi_2: \quad a_2x + b_2y + c_2z + d_2 = 0 \]

These planes intersect in a straight line. The bisector planes are defined as the two planes that bisect the angle formed between \(\pi_1\) and \(\pi_2\) at each point along their line of intersection. There are always two such planes: one lies between the faces of \(\pi_1\) and \(\pi_2\) that form the acute angle, and the other lies between the faces forming the obtuse angle.

The equations of the bisector planes are given by:

\[ \frac{a_1x + b_1y + c_1z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = \pm \frac{a_2x + b_2y + c_2z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \]

Each of these planes:

contains the line of intersection of \(\pi_1\) and \(\pi_2\),
bisects the angle between them,
and is perpendicular to the other bisector plane.

Proof of the Equation of Bisector Planes

Let two non-parallel planes be given by

\[ \pi_1: \quad a_1x + b_1y + c_1z + d_1 = 0, \qquad \pi_2: \quad a_2x + b_2y + c_2z + d_2 = 0 \]

Let \((x, y, z)\) be a point lying on one of the bisector planes. By geometric definition, a bisector plane consists of all points that are equidistant from the two given planes. The perpendicular distance from a point \((x, y, z)\) to the plane \(\pi_1\) is

\[ \frac{|a_1x + b_1y + c_1z + d_1|}{\sqrt{a_1^2 + b_1^2 + c_1^2}} \]

and the perpendicular distance to \(\pi_2\) is

\[ \frac{|a_2x + b_2y + c_2z + d_2|}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \]

For the point to lie on the bisector, these distances must be equal:

\[ \frac{|a_1x + b_1y + c_1z + d_1|}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = \frac{|a_2x + b_2y + c_2z + d_2|}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \]

Removing the absolute value gives the two equations:

\[ \frac{a_1x + b_1y + c_1z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = \pm \frac{a_2x + b_2y + c_2z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \]

Each sign yields a distinct plane, and these are the two bisector planes of the angle between \(\pi_1\) and \(\pi_2\).

Example

Find the bisector planes of the two non-parallel planes

\[ \pi_1:\quad x + 2y + 2z = 2, \qquad \pi_2:\quad -2x + y - 2z = 1 \]

Solution:

We first bring both planes to standard form:

\[ \pi_1:\quad x + 2y + 2z - 2 = 0, \qquad \pi_2:\quad -2x + y - 2z - 1 = 0 \]

Let \(\vec{n}_1 = \hat{i} + 2\hat{j} + 2\hat{k}\) and \(\vec{n}_2 = -2\hat{i} + \hat{j} - 2\hat{k}\) be the normal vectors of \(\pi_1\) and \(\pi_2\) respectively. The angle bisector planes are defined as those planes for which any point on them is equidistant from the two given planes. Hence, their equations are given by:

\[ \frac{x + 2y + 2z - 2}{\sqrt{1^2 + 2^2 + 2^2}} = \pm \frac{-2x + y - 2z - 1}{\sqrt{(-2)^2 + 1^2 + (-2)^2}} \]

The denominators on both sides simplify to:

\[ \sqrt{1 + 4 + 4} = \sqrt{9} = 3, \qquad \sqrt{4 + 1 + 4} = \sqrt{9} = 3 \]

So the equation becomes:

\[ \frac{x + 2y + 2z - 2}{3} = \pm \frac{-2x + y - 2z - 1}{3} \implies x + 2y + 2z - 2 = \pm(-2x + y - 2z - 1) \]

We now resolve both signs.

Case 1: Positive sign

\[ x + 2y + 2z - 2 = -2x + y - 2z - 1 \]

Bring all terms to one side:

\[ x + 2y + 2z - 2 + 2x - y + 2z + 1 = 0 \implies 3x + y + 4z - 1 = 0 \]

This yields one bisector plane:

\[ 3x + y + 4z = 1 \]

Case 2: Negative sign

\[ x + 2y + 2z - 2 = 2x - y + 2z + 1 \]

Again bring all terms to one side:

\[ x + 2y + 2z - 2 - 2x + y - 2z - 1 = 0 \implies -x + 3y - 3 = 0 \]

This yields the second bisector plane:

\[ x - 3y = -3 \quad \text{or} \quad x - 3y + 3 = 0 \]

Hence, the two bisector planes are:

\[ 3x + y + 4z = 1 \quad \text{and} \quad x - 3y + 3 = 0 \]

Acute and Obtuse Angle Bisector Planes

Calculating the equations of the bisector planes between two non-parallel planes is straightforward. However, distinguishing which one bisects the acute angle and which one bisects the obtuse angle requires further analysis.

Recall that when two non-parallel planes intersect, they form two angles at their line of intersection — one acute and one obtuse. The two bisector planes divide space accordingly: one lies in the region between the acute faces of the original planes, and the other between the obtuse faces. The one that bisects the acute angle is called the acute angle bisector, and the other is the obtuse angle bisector.

This distinction cannot be made when the two planes are perpendicular, because both angles formed are right angles, and both bisector planes will be at \(45^\circ\) from the given planes.

For non-perpendicular planes, the following method allows us to determine which bisector plane corresponds to the acute angle:

Let \(\vec{b}\) be the normal vector of one of the bisector planes, and let \(\vec{n}\) be the normal vector of either of the original planes (say \(\pi_1\)). Then compute the angle \(\theta\) between them using:

\[ \cos\theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}||\vec{n}|} \]

Now interpret the result:

If \(\theta < 45^\circ\), then this bisector makes an acute angle with \(\pi_1\), and hence must be the acute angle bisector.
If \(\theta > 45^\circ\), then it is the obtuse angle bisector.

In practice, since \(\cos 45^\circ = \frac{1}{\sqrt{2}}\), we simply test:

\[ \cos\theta < \frac{1}{\sqrt{2}} \iff \theta > 45^\circ \quad \text{(obtuse bisector)} \qquad \text{or} \qquad \cos\theta > \frac{1}{\sqrt{2}} \iff \theta < 45^\circ \quad \text{(acute bisector)} \]

Thus, by checking the angle between a bisector plane and one of the original planes, we can reliably identify which bisector corresponds to the acute angle.

Alternative Method: Distinguishing Between Acute and Obtuse Angle Bisectors

Another elegant and more direct method to distinguish between the acute and obtuse angle bisector planes involves analyzing the location of the origin with respect to the two given planes. While the reasoning behind this method is deeper, it provides a clean algebraic criterion once the underlying geometry is understood.

Consider two non-parallel planes given by:

\[ \pi_1: \quad a_1x + b_1y + c_1z + d_1 = 0, \qquad \pi_2: \quad a_2x + b_2y + c_2z + d_2 = 0 \]

Assume that the origin does not lie on either of the planes. That is, \(d_1 \neq 0\), \(d_2 \neq 0\), and we further assume \(d_1 < 0\) and \(d_2 < 0\). This implies that the values of \(\pi_1(0,0,0) = d_1\) and \(\pi_2(0,0,0) = d_2\) are both negative, so the origin lies on the same side of both planes.

Let us suppose that the origin lies in the acute region between the two planes. Then consider the expression

\[ a_1x + b_1y + c_1z = -d_1 \quad \text{(since \(d_1 < 0\))} \]

The right-hand side is positive, so the normal vector \(\vec{n}_1 = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}\) points from the origin toward the plane. The same logic applies to \(\pi_2\), so both normal vectors \(\vec{n}_1\) and \(\vec{n}_2\) point outward from the acute region.

Now, since the angle \(\theta\) between the planes is acute, the angle between their outward normal vectors is \(\pi - \theta\), which is obtuse. Therefore,

\[ \vec{n}_1 \cdot \vec{n}_2 = a_1a_2 + b_1b_2 + c_1c_2 < 0 \]

Hence, if the origin lies in the acute region between the two planes, then the dot product of their normal vectors is negative.

Similarly, if the origin lies in the obtuse region, then the angle between \(\vec{n}_1\) and \(\vec{n}_2\) is acute, and therefore:

\[ a_1a_2 + b_1b_2 + c_1c_2 > 0 \]

So, by computing the dot product of the normal vectors, we can determine whether the origin lies in the acute or obtuse region between the planes.

Once this is known, we use this to determine which bisector is in which region.

Consider again the two non-parallel planes:

\[ \pi_1:\quad a_1x + b_1y + c_1z + d_1 = 0, \qquad \pi_2:\quad a_2x + b_2y + c_2z + d_2 = 0 \]

Assume \(d_1 < 0\) and \(d_2 < 0\), so that the origin does not lie on either plane. The bisector planes are given by the standard equation:

\[ \frac{a_1x + b_1y + c_1z + d_1}{\sqrt{a_1^2 + b_1^2 + c_1^2}} = \pm \frac{a_2x + b_2y + c_2z + d_2}{\sqrt{a_2^2 + b_2^2 + c_2^2}} \tag{1} \]

Let us use the theory of the location of the origin to determine which of the two signs corresponds to the acute angle bisector.

Case 1: Origin lies in the acute region

Suppose the origin lies in the acute region formed by the intersection of the two planes. As we saw above, this means that

\[ \vec{n}_1 \cdot \vec{n}_2 = a_1a_2 + b_1b_2 + c_1c_2 < 0 \tag{2} \]

Let \(B_1\) denote the acute angle bisector and let \(P = (x, y, z)\) be a point on \(B_1\). Since \(B_1\) lies in the acute region, and the origin also lies in that region, the point \(P\) and the origin are on the same side of both planes. Therefore:

\[ (a_1x + b_1y + c_1z + d_1)(d_1) > 0 \implies a_1x + b_1y + c_1z + d_1 < 0 \tag{3} \]

\[ (a_2x + b_2y + c_2z + d_2)(d_2) > 0 \implies a_2x + b_2y + c_2z + d_2 < 0 \tag{4} \]

Now substitute point \(P\) into equation (1). The left-hand side and right-hand side are both negative (by (3) and (4)), and their denominators are positive. The only way equality can hold is by choosing the positive sign:

\[ \frac{\text{negative}}{\text{positive}} = \frac{\text{negative}}{\text{positive}} \]

The negative sign would produce a contradiction. Therefore, in this case the plus sign in equation (1) gives the acute angle bisector, and the minus sign gives the obtuse angle bisector.

Case 2: Origin lies in the obtuse region

Now suppose the origin lies in the obtuse region between the two planes. Then:

\[ \vec{n}_1 \cdot \vec{n}_2 = a_1a_2 + b_1b_2 + c_1c_2 > 0 \tag{5} \]

Let \(B_2\) denote the obtuse angle bisector, and let \(Q = (x, y, z)\) be a point on it. Since the origin and \(Q\) lie on the same side of both planes, the same logic as in Case 1 yields:

\[ a_1x + b_1y + c_1z + d_1 < 0, \qquad a_2x + b_2y + c_2z + d_2 < 0 \]

Again, substitution into equation (1) shows that the plus sign must be taken to preserve equality, and hence in this case the plus sign gives the obtuse angle bisector, and the minus sign gives the acute.

Summary Table:

Location of Origin	\(a_1a_2 + b_1b_2 + c_1c_2\)	‘+’ in Equation (1) Gives	‘−’ in Equation (1) Gives
Origin in acute region	\( < 0 \)	Acute angle bisector	Obtuse angle bisector
Origin in obtuse region	\( > 0 \)	Obtuse angle bisector	Acute angle bisector

This criterion allows one to determine the nature of the bisector planes purely algebraically, without computing any angles or distances.

We note that at the beginning, we assumed

\[ d_1 < 0, \qquad d_2 < 0 \]

However, the same conclusions hold if instead we had assumed

\[ d_1 > 0, \qquad d_2 > 0 \]

In that case too, the same logical deductions follow exactly as before.

Thus, before applying this method, it is necessary to ensure that \(d_1\) and \(d_2\) have the same sign, i.e., the origin must lie on the same side of both planes.