Section 12.5: Lines & Planes

A line in the xy-plane is determined when a point on the line and the direction of the line (its slope or angle of inclination) are given. The equation of the line can then be written using the point-slope form.

Likewise, a line L in three-dimensional space is determined when we know a point \(P_0(x_0, y_0, z_0)\) on L and the direction of L. In three dimensions the direction of a line is conveniently described by a vector, so we let v be a vector parallel to L. Let \(P(x, y, z)\) be an arbitrary point on L and let \(\mathbf{r}_0\) and r be the position vectors of \(P_0\) and \(P\) (that is, they have representations \(\vec{OP_0}\) and \(\vec{OP}\)). If a is the vector with representation \(\vec{P_0P}\), as in Figure 1, then the Triangle Law for vector addition gives \(\mathbf{r} = \mathbf{r}_0 + \mathbf{a}\). But, since a and v are parallel vectors, there is a scalar \(t\) such that \(\mathbf{a} = t\mathbf{v}\). Thus \[ \mathbf{r} = \mathbf{r}_0 + t\mathbf{v} \tag{1} \] which is a vector equation of L. Each value of the parameter \(t\) gives the position vector r of a point on L. In other words, as \(t\) varies, the line is traced out by the tip of the vector r. As Figure 2 indicates, positive values of \(t\) correspond to points on L that lie on one side of \(P_0\), whereas negative values of \(t\) correspond to points that lie on the other side of \(P_0\).

If the vector v that gives the direction of the line L is written in component form as \(\mathbf{v} = \langle a, b, c \rangle\), then we have \(t\mathbf{v} = \langle ta, tb, tc \rangle\). We can also write \(\mathbf{r} = \langle x, y, z \rangle\) and \(\mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle\), so the vector equation (1) becomes \[ \langle x, y, z \rangle = \langle x_0 + ta, y_0 + tb, z_0 + tc \rangle \] Two vectors are equal if and only if corresponding components are equal. Therefore we have the three scalar equations: \[ x = x_0 + at \quad y = y_0 + bt \quad z = z_0 + ct \] where \(t \in \mathbb{R}\). These equations are called parametric equations of the line L through the point \(P_0(x_0, y_0, z_0)\) and parallel to the vector \(\mathbf{v} = \langle a, b, c \rangle\). Each value of the parameter \(t\) gives a point \((x, y, z)\) on L.

Parametric equations for a line through the point \((x_0, y_0, z_0)\) and parallel to the direction vector \(\langle a, b, c \rangle\) are \[ x = x_0 + at \quad y = y_0 + bt \quad z = z_0 + ct \tag{2} \]

EXAMPLE 1 (a) Find a vector equation and parametric equations for the line that passes through the point \((5, 1, 3)\) and is parallel to the vector \(\mathbf{i} + 4\mathbf{j} - 2\mathbf{k}\). (b) Find two other points on the line.

The vector equation and parametric equations of a line are not unique. If we change the point or the parameter or choose a different parallel vector, then the equations change. For instance, if, instead of \((5, 1, 3)\), we choose the point \((6, 5, 1)\) in Example 1, then the parametric equations of the line become \[ x = 6 + t \quad y = 5 + 4t \quad z = 1 - 2t \] Or, if we stay with the point \((5, 1, 3)\) but choose the parallel vector \(2\mathbf{i} + 8\mathbf{j} - 4\mathbf{k}\), we arrive at the equations \[ x = 5 + 2t \quad y = 1 + 8t \quad z = 3 - 4t \] In general, if a vector \(\mathbf{v} = \langle a, b, c \rangle\) is used to describe the direction of a line L, then the numbers \(a, b,\) and \(c\) are called direction numbers of L. Since any vector parallel to v could also be used, we see that any three numbers proportional to \(a, b,\) and \(c\) could also be used as a set of direction numbers for L.

Another way of describing a line L is to eliminate the parameter \(t\) from Equations 2. If none of \(a, b,\) or \(c\) is 0, we can solve each of these equations for \(t\): \[ t = \frac{x - x_0}{a} \quad t = \frac{y - y_0}{b} \quad t = \frac{z - z_0}{c} \] Equating the results, we obtain \[ \frac{x - x_0}{a} = \frac{y - y_0}{b} = \frac{z - z_0}{c} \tag{3} \] These equations are called symmetric equations of L. Notice that the numbers \(a, b,\) and \(c\) that appear in the denominators of Equations 3 are direction numbers of L, that is, components of a vector parallel to L. If one of \(a, b,\) or \(c\) is 0, we can still eliminate \(t\). For instance, if \(a = 0\), we could write the equations of L as \[ x = x_0 \quad \frac{y - y_0}{b} = \frac{z - z_0}{c} \] This means that L lies in the vertical plane \(x = x_0\).

EXAMPLE 2 (a) Find parametric equations and symmetric equations of the line that passes through the points \(A(2, 4, -3)\) and \(B(3, -1, 1)\). (b) At what point does this line intersect the xy-plane?

Often, we need a description, not of an entire line, but of just a line segment. How, for instance, could we describe the line segment AB in Example 2? If we put \(t = 0\) in the parametric equations in Example 2(a), we get the point \((2, 4, -3)\) and if we put \(t = 1\) we get \((3, -1, 1)\). So the line segment AB is described by the parametric equations \[ x = 2 + t \quad y = 4 - 5t \quad z = -3 + 4t \quad 0 \le t \le 1 \] or by the corresponding vector equation \[ \mathbf{r}(t) = \langle 2 + t, 4 - 5t, -3 + 4t \rangle \quad 0 \le t \le 1 \] In general, we know from Equation 1 that the vector equation of a line through the (tip of the) vector \(\mathbf{r}_0\) in the direction of a vector v is \(\mathbf{r} = \mathbf{r}_0 + t\mathbf{v}\). If the line also passes through (the tip of) \(\mathbf{r}_1\), then we can take \(\mathbf{v} = \mathbf{r}_1 - \mathbf{r}_0\) and so its vector equation is \[ \mathbf{r} = \mathbf{r}_0 + t(\mathbf{r}_1 - \mathbf{r}_0) = (1 - t)\mathbf{r}_0 + t\mathbf{r}_1 \] The line segment from \(\mathbf{r}_0\) to \(\mathbf{r}_1\) is given by the parameter interval \(0 \le t \le 1\).

The line segment from \(\mathbf{r}_0\) to \(\mathbf{r}_1\) is given by the vector equation \[ \mathbf{r}(t) = (1 - t)\mathbf{r}_0 + t\mathbf{r}_1 \quad 0 \le t \le 1 \tag{4} \]

EXAMPLE 3 Show that the lines \(L_1\) and \(L_2\) with parametric equations \[ L_1: x = 1 + t \quad y = -2 + 3t \quad z = 4 - t \] \[ L_2: x = 2s \quad y = 3 + s \quad z = -3 + 4s \] are skew lines; that is, they do not intersect and are not parallel (and therefore do not lie in the same plane).

Although a line in space is determined by a point and a direction, a plane in space is more difficult to describe. A single vector parallel to a plane is not enough to convey the “direction” of the plane, but a vector perpendicular to the plane does completely specify its direction. Thus a plane in space is determined by a point \(P_0(x_0, y_0, z_0)\) in the plane and a vector n that is orthogonal to the plane. This orthogonal vector n is called a normal vector.

Let \(P(x, y, z)\) be an arbitrary point in the plane, and let \(\mathbf{r}_0\) and r be the position vectors of \(P_0\) and \(P\). Then the vector \(\mathbf{r} - \mathbf{r}_0\) is represented by \(\vec{P_0P}\). The normal vector n is orthogonal to every vector in the given plane. In particular, n is orthogonal to \(\mathbf{r} - \mathbf{r}_0\) and so we have \[ \mathbf{n} \cdot (\mathbf{r} - \mathbf{r}_0) = 0 \tag{5} \] which can be rewritten as \[ \mathbf{n} \cdot \mathbf{r} = \mathbf{n} \cdot \mathbf{r}_0 \tag{6} \] Either Equation 5 or Equation 6 is called a vector equation of the plane.

To obtain a scalar equation for the plane, we write \(\mathbf{n} = \langle a, b, c \rangle, \mathbf{r} = \langle x, y, z \rangle,\) and \(\mathbf{r}_0 = \langle x_0, y_0, z_0 \rangle\). Then the vector equation (5) becomes \[ \langle a, b, c \rangle \cdot \langle x - x_0, y - y_0, z - z_0 \rangle = 0 \] or \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \]

A scalar equation of the plane through point \(P_0(x_0, y_0, z_0)\) with normal vector \(\mathbf{n} = \langle a, b, c \rangle\) is \[ a(x - x_0) + b(y - y_0) + c(z - z_0) = 0 \tag{7} \]

EXAMPLE 4 Find an equation of the plane through the point \((2, 4, -1)\) with normal vector \(\mathbf{n} = \langle 2, 3, 4 \rangle\). Find the intercepts and sketch the plane.

By collecting terms in Equation 7 as we did in Example 4, we can rewrite the equation of a plane as \[ ax + by + cz + d = 0 \tag{8} \] where \(d = -(ax_0 + by_0 + cz_0)\). Equation 8 is called a linear equation in \(x, y,\) and \(z\). Conversely, it can be shown that if \(a, b,\) and \(c\) are not all 0, then the linear equation (8) represents a plane with normal vector \(\langle a, b, c \rangle\).

EXAMPLE 5 Find an equation of the plane that passes through the points \(P(1, 3, 2), Q(3, -1, 6),\) and \(R(5, 2, 0)\).

EXAMPLE 6 Find the point at which the line with parametric equations \(x = 2 + 3t, y = -4t, z = 5 + t\) intersects the plane \(4x + 5y - 2z = 18\).

Two planes are parallel if their normal vectors are parallel. For instance, the planes \(x + 2y - 3z = 4\) and \(2x + 4y - 6z = 3\) are parallel because their normal vectors are \(\mathbf{n}_1 = \langle 1, 2, -3 \rangle\) and \(\mathbf{n}_2 = \langle 2, 4, -6 \rangle\) and \(\mathbf{n}_2 = 2\mathbf{n}_1\). If two planes are not parallel, then they intersect in a straight line and the angle between the two planes is defined as the acute angle between their normal vectors (see angle \(\theta\) in Figure 9).

EXAMPLE 7 (a) Find the angle between the planes \(x + y + z = 1\) and \(x - 2y + 3z = 1\). (b) Find symmetric equations for the line of intersection L of these two planes.

EXAMPLE 8 Find a formula for the distance \(D\) from a point \(P_1(x_1, y_1, z_1)\) to the plane \(ax + by + cz + d = 0\).

EXAMPLE 9 Find the distance between the parallel planes \(10x + 2y - 2z = 5\) and \(5x + y - z = 1\).

EXAMPLE 10 In Example 3 we showed that the lines \[ L_1: x = 1 + t \quad y = -2 + 3t \quad z = 4 - t \] \[ L_2: x = 2s \quad y = 3 + s \quad z = -3 + 4s \] are skew. Find the distance between them.

Determine whether each statement is true or false in \(\mathbb{R}^3\). (a) Two lines parallel to a third line are parallel. (b) Two lines perpendicular to a third line are parallel. (c) Two planes parallel to a third plane are parallel. (d) Two planes perpendicular to a third plane are parallel. (e) Two lines parallel to a plane are parallel. (f) Two lines perpendicular to a plane are parallel. (g) Two planes parallel to a line are parallel. Two planes perpendicular to a line are parallel. (i) Two planes either intersect or are parallel. (j) Two lines either intersect or are parallel. (k) A plane and a line either intersect or are parallel.

Find a vector equation and parametric equations for the line through the point \((6, -5, 2)\) and parallel to the vector \(\langle 1, 3, -\frac{2}{3} \rangle\).

Find a vector equation and parametric equations for the line through the point \((2, 2.4, 3.5)\) and parallel to the vector \(3\mathbf{i} + 2\mathbf{j} - \mathbf{k}\).

Find a vector equation and parametric equations for the line through the point \((0, 14, -10)\) and parallel to the line \(x = -1 + 2t, y = 6 - 3t, z = 3 + 9t\).

Find a vector equation and parametric equations for the line through the point \((1, 0, 6)\) and perpendicular to the plane \(x + 3y + z = 5\).

Find parametric equations and symmetric equations for the line through the origin and the point \((4, 3, -1)\).

Find parametric equations and symmetric equations for the line through the points \((0, \frac{1}{2}, 1)\) and \((2, 1, -3)\).

Find parametric equations and symmetric equations for the line through the points \((1, 2.4, 4.6)\) and \((2.6, 1.2, 0.3)\).

Find parametric equations and symmetric equations for the line through the points \((-8, 1, 4)\) and \((3, -2, 4)\).

Find parametric equations and symmetric equations for the line through \((2, 1, 0)\) and perpendicular to both \(\mathbf{i} + \mathbf{j}\) and \(\mathbf{j} + \mathbf{k}\).

Find parametric equations and symmetric equations for the line through \((-6, 2, 3)\) and parallel to the line \(\frac{x}{2} = \frac{y}{3} = z + 1\).

Find parametric equations and symmetric equations for the line of intersection of the planes \(x + 2y + 3z = 1\) and \(x - y + z = 1\).

Is the line through \((-4, -6, 1)\) and \((-2, 0, -3)\) parallel to the line through \((10, 18, 4)\) and \((5, 3, 14)\)?

Is the line through \((-2, 4, 0)\) and \((1, 1, 1)\) perpendicular to the line through \((2, 3, 4)\) and \((3, -1, -8)\)?

1. Find symmetric equations for the line that passes through the point \((1, -5, 6)\) and is parallel to the vector \(\langle -1, 2, -3 \rangle\).
2. Find the points in which the required line in part (a) intersects the coordinate planes.

1. Find parametric equations for the line through \((2, 4, 6)\) that is perpendicular to the plane \(x - y + 3z = 7\).
2. In what points does this line intersect the coordinate planes?

Find a vector equation for the line segment from \((6, -1, 9)\) to \((7, 6, 0)\).

Find parametric equations for the line segment from \((-2, 18, 31)\) to \((11, -4, 48)\).

Determine whether the lines \(L_1\) and \(L_2\) are parallel, skew, or intersecting. If they intersect, find the point of intersection. \(L_1: x = -3 + 2t, y = 4 - t, z = 1 + 3t\) \(L_2: x = 1 + 4s, y = 3 - 2s, z = 4 + 5s\)

Determine whether the lines \(L_1\) and \(L_2\) are parallel, skew, or intersecting. If they intersect, find the point of intersection. \(L_1: x = 5 - 12t, y = 3 + 9t, z = 1 - 3t\) \(L_2: x = 3 + 8s, y = -6s, z = 7 + 2s\)

Determine whether the lines \(L_1\) and \(L_2\) are parallel, skew, or intersecting. If they intersect, find the point of intersection. \(L_1: \frac{x-2}{1} = \frac{y-3}{-2} = \frac{z-1}{-3}\) \(L_2: \frac{x-3}{1} = \frac{y+4}{3} = \frac{z-2}{-7}\)

Determine whether the lines \(L_1\) and \(L_2\) are parallel, skew, or intersecting. If they intersect, find the point of intersection. \(L_1: \frac{x}{1} = \frac{y-1}{-1} = \frac{z-2}{3}\) \(L_2: \frac{x-2}{2} = \frac{y-3}{-2} = \frac{z}{7}\)

Find an equation of the plane through the origin and perpendicular to the vector \(\langle 1, -2, 5 \rangle\).

Find an equation of the plane through the point \((5, 3, 5)\) and with normal vector \(2\mathbf{i} + \mathbf{j} - \mathbf{k}\).

Find an equation of the plane through the point \((-1, \frac{1}{2}, 3)\) and with normal vector \(\mathbf{i} + 4\mathbf{j} + \mathbf{k}\).

Find an equation of the plane through the point \((2, 0, 1)\) and perpendicular to the line \(x = 3t, y = 2 - t, z = 3 + 4t\).

Find an equation of the plane through the point \((1, -1, -1)\) and parallel to the plane \(5x - y - z = 6\).

Find an equation of the plane through the point \((3, -2, 8)\) and parallel to the plane \(z = x + y\).

Find an equation of the plane through the point \((1, \frac{1}{2}, \frac{1}{3})\) and parallel to the plane \(x + y + z = 0\).

Find an equation of the plane that contains the line \(x = 1 + t, y = 2 - t, z = 4 - 3t\) and is parallel to the plane \(5x + 2y + z = 1\).

Find an equation of the plane through the points \((0, 1, 1), (1, 0, 1),\) and \((1, 1, 0)\).

Find an equation of the plane through the origin and the points \((3, -2, 1)\) and \((1, 1, 1)\).

Find an equation of the plane through the points \((2, 1, 2), (3, -8, 6),\) and \((-2, -3, 1)\).

Find an equation of the plane through the points \((3, 0, -1), (-2, -2, 3),\) and \((7, 1, -4)\).

Find an equation of the plane that passes through the point \((3, 5, -1)\) and contains the line \(x = 4 - t, y = 2t - 1, z = -3t\).

Find an equation of the plane that passes through the point \((6, -1, 3)\) and contains the line with symmetric equations \(\frac{x}{3} = y + 4 = \frac{z}{2}\).

Find an equation of the plane that passes through the point \((3, 1, 4)\) and contains the line of intersection of the planes \(x + 2y + 3z = 1\) and \(2x - y + z = -3\).

Find an equation of the plane that passes through the points \((0, -2, 5)\) and \((-1, 3, 1)\) and is perpendicular to the plane \(2z = 5x + 4y\).

Find an equation of the plane that passes through the point \((1, 5, 1)\) and is perpendicular to the planes \(2x + y - 2z = 2\) and \(x + 3z = 4\).

Find an equation of the plane that passes through the line of intersection of the planes \(x - z = 1\) and \(y + 2z = 3\) and is perpendicular to the plane \(x + y - 2z = 1\).

Use intercepts to help sketch the plane \(2x + 5y + z = 10\).

Use intercepts to help sketch the plane \(3x + y + 2z = 6\).

Use intercepts to help sketch the plane \(6x - 3y + 4z = 6\).

Use intercepts to help sketch the plane \(6x + 5y - 3z = 15\).

Find the point at which the line intersects the given plane. \(x = 2 - 2t, y = 3t, z = 1 + t; \quad x + 2y - z = 7\)

Find the point at which the line intersects the given plane. \(x = t - 1, y = 1 + 2t, z = 3 - t; \quad 3x - y + 2z = 5\)

Find the point at which the line intersects the given plane. \(5x = y/2 = z + 2; \quad 10x - 7y + 3z + 24 = 0\)

Where does the line through \((-3, 1, 0)\) and \((-1, 5, 6)\) intersect the plane \(2x + y - z = -2\)?

Find direction numbers for the line of intersection of the planes \(x + y + z = 1\) and \(x + z = 0\).

Find the cosine of the angle between the planes \(x + y + z = 0\) and \(x + 2y + 3z = 1\).

Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. (Round to one decimal place.) \(x + 4y - 3z = 1, \quad -3x + 6y + 7z = 0\)

Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. (Round to one decimal place.) \(9x - 3y + 6z = 2, \quad 2y = 6x + 4z\)

Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. (Round to one decimal place.) \(x + 2y - z = 2, \quad 2x - 2y + z = 1\)

Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. (Round to one decimal place.) \(x - y + 3z = 1, \quad 3x + y - z = 2\)

Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. (Round to one decimal place.) \(2x - 3y = z, \quad 4x = 3 + 6y + 2z\)

Determine whether the planes are parallel, perpendicular, or neither. If neither, find the angle between them. (Round to one decimal place.) \(5x + 2y + 3z = 2, \quad y = 4x - 6z\)

1. Find parametric equations for the line of intersection of the planes \(x + y + z = 1\) and \(x + 2y + 2z = 1\).
2. Find the angle between the planes.

1. Find parametric equations for the line of intersection of the planes \(3x - 2y + z = 1\) and \(2x + y - 3z = 3\).
2. Find the angle between the planes.

Find symmetric equations for the line of intersection of the planes. \(5x - 2y - 2z = 1, \quad 4x + y + z = 6\)

Find symmetric equations for the line of intersection of the planes. \(z = 2x - y - 5, \quad z = 4x + 3y - 5\)

Find an equation for the plane consisting of all points that are equidistant from the points \((1, 0, -2)\) and \((3, 4, 0)\).

Find an equation for the plane consisting of all points that are equidistant from the points \((2, 5, 5)\) and \((-6, 3, 1)\).

1. Find the point at which the given lines intersect: \(\mathbf{r} = \langle 1, 1, 0 \rangle + t\langle 1, -1, 2 \rangle\) \(\mathbf{r} = \langle 2, 0, 2 \rangle + s\langle -1, 1, 0 \rangle\)
2. Find an equation of the plane that contains these lines.

Find parametric equations for the line through the point \((0, 1, 2)\) that is parallel to the plane \(x + y + z = 2\) and perpendicular to the line \(x = 1 + t, y = 1 - t, z = 2t\).

Find parametric equations for the line through the point \((0, 1, 2)\) that is perpendicular to the line \(x = 1 + t, y = 1 - t, z = 2t\) and intersects this line.

Which of the following four planes are parallel? Are any of them identical? \(P_1: 3x + 6y - 3z = 6\) \(P_2: 4x - 12y + 8z = 5\) \(P_3: 9y = 1 + 3x + 6z\) \(P_4: z = x + 2y - 2\)

Which of the following four lines are parallel? Are any of them identical? \(L_1: x = 1 + 6t, y = 1 - 3t, z = 12t + 5\) \(L_2: x = 1 + 2t, y = t, z = 1 + 4t\) \(L_3: 2x - 2 = 4 - 4y = z + 1\) \(L_4: \mathbf{r} = \langle 3, 1, 5 \rangle + t\langle 4, 2, 8 \rangle\)

Use the formula in Exercise 12.4.45 to find the distance from the point to the given line. \((4, 1, -2); \quad x = 1 + t, y = 3 - 2t, z = 4 - 3t\)

Use the formula in Exercise 12.4.45 to find the distance from the point to the given line. \((0, 1, 3); \quad x = 2t, y = 6 - 2t, z = 3 + t\)

Find the distance from the point to the given plane. \((1, -2, 4), \quad 3x + 2y + 6z = 5\)

Find the distance from the point to the given plane. \((-6, 3, 5), \quad x - 2y - 4z = 8\)

Find the distance between the given parallel planes. \(2x - 3y + z = 4, \quad 4x - 6y + 2z = 3\)

Find the distance between the given parallel planes. \(6z = 4y - 2x, \quad 9z = 1 - 3x + 6y\)

Show that the distance between the parallel planes \(ax + by + cz + d_1 = 0\) and \(ax + by + cz + d_2 = 0\) is \[ D = \frac{|d_1 - d_2|}{\sqrt{a^2 + b^2 + c^2}} \]

Find equations of the planes that are parallel to the plane \(x + 2y - 2z = 1\) and two units away from it.

Show that the lines with symmetric equations \(x = y = z\) and \(x + 1 = y/2 = z/3\) are skew, and find the distance between these lines.

Find the distance between the skew lines with parametric equations \(x = 1 + t, y = 1 + 6t, z = 2t,\) and \(x = 1 + 2s, y = 5 + 15s, z = -2 + 6s\).

Let \(L_1\) be the line through the origin and the point \((2, 0, -1)\). Let \(L_2\) be the line through the points \((1, -1, 1)\) and \((4, 1, 3)\). Find the distance between \(L_1\) and \(L_2\).

Let \(L_1\) be the line through the points \((1, 2, 6)\) and \((2, 4, 8)\). Let \(L_2\) be the line of intersection of the planes \(\mathcal{P}_1\) and \(\mathcal{P}_2\), where \(\mathcal{P}_1\) is the plane \(x - y + 2z + 1 = 0\) and \(\mathcal{P}_2\) is the plane through the points \((3, 2, -1), (0, 0, 1),\) and \((1, 2, 1)\). Calculate the distance between \(L_1\) and \(L_2\).

Two tanks are participating in a battle simulation. Tank A is at point \((325, 810, 561)\) and tank B is positioned at point \((765, 675, 599)\). (a) Find parametric equations for the line of sight between the tanks. (b) If we divide the line of sight into 5 equal segments, the elevations of the terrain at the four intermediate points from tank A to tank B are 549, 566, 586, and 589. Can the tanks see each other?

Give a geometric description of each family of planes. (a) \(x + y + z = c\) (b) \(x + y + cz = 1\) (c) \(y \cos\theta + z \sin\theta = 1\)

If a, b, and c are not all 0, show that the equation \(ax + by + cz + d = 0\) represents a plane and \(\langle a, b, c \rangle\) is a normal vector to the plane. Hint: Suppose \(a \ne 0\) and rewrite the equation in the form \(a(x + d/a) + b(y - 0) + c(z - 0) = 0\)