Coplanarity and Geometric Relationships

Coplanarity of Two Lines

Two lines in three-dimensional space are said to be coplanar if they lie in the same plane. Geometrically, this happens if the lines either intersect or are parallel (including coincident lines).

Vector Condition

Let the two lines be given by: $L_1: \vec{r} = \vec{a}_1 + \lambda\vec{b}_1$ $L_2: \vec{r} = \vec{a}_2 + \mu\vec{b}_2$ For these lines to be coplanar, the vector connecting their known points $(\vec{a}_2 - \vec{a}_1)$ must lie in the same plane as their direction vectors $\vec{b}_1$ and $\vec{b}_2$. Mathematically, their scalar triple product must be zero: $$(\vec{a}_2 - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0$$ This means the three vectors do not form a three-dimensional volume.

Cartesian Condition If the lines are given in Cartesian form: $L_1: \frac{x - x_1}{a_1} = \frac{y - y_1}{b_1} = \frac{z - z_1}{c_1}$ $L_2: \frac{x - x_2}{a_2} = \frac{y - y_2}{b_2} = \frac{z - z_2}{c_2}$ then the lines are coplanar if $$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$$

Equation of the Plane Containing Two Coplanar Lines

If two lines are coplanar and intersecting, they uniquely define a plane.

Vector Form

If the direction vectors are $\vec{b}_1$ and $\vec{b}_2$, then a normal to the plane is $$\vec{n} = \vec{b}_1 \times \vec{b}_2$$ So the plane through the point with position vector $\vec{a}_1$ is $$(\vec{r} - \vec{a}_1) \cdot (\vec{b}_1 \times \vec{b}_2) = 0$$

Cartesian Form

Using a general point $(x, y, z)$, the plane containing the two intersecting lines is $$\begin{vmatrix} x - x_1 & y - y_1 & z - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0$$

Important Note for Parallel Lines:

If the lines are parallel, then $\vec{b}_1 \times \vec{b}_2 = \vec{0}$, so the above normal cannot be used. In that case, take one direction vector $\vec{b}$ of the lines and the vector joining a point on one line to a point on the other, $(\vec{a}_2 - \vec{a}_1)$. Then a normal to the required plane is $$(\vec{a}_2 - \vec{a}_1) \times \vec{b}.$$

Angle Between Two Planes

The angle between two planes is defined as the acute angle between their normal vectors.

Vector Form

Let the planes be $\vec{r} \cdot \vec{n}_1 = d_1$ and $\vec{r} \cdot \vec{n}_2 = d_2$. Then $$\cos\theta = \left| \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1| |\vec{n}_2|} \right|$$

Cartesian Form

If the planes are $A_1 x + B_1 y + C_1 z + D_1 = 0$ and $A_2 x + B_2 y + C_2 z + D_2 = 0$, then $$\cos\theta = \left| \frac{A_1 A_2 + B_1 B_2 + C_1 C_2}{\sqrt{A_1^2 + B_1^2 + C_1^2} \sqrt{A_2^2 + B_2^2 + C_2^2}} \right|$$

Conditions:

• Perpendicular Planes: $\vec{n}_1 \cdot \vec{n}_2 = 0$, or equivalently $A_1 A_2 + B_1 B_2 + C_1 C_2 = 0$.
• Parallel Planes: $\vec{n}_1 = k\vec{n}_2$, or equivalently $\frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.

Angle Between a Line and a Plane

The angle $\phi$ between a line and a plane is the complement of the angle between the line and the normal to the plane.

If $\theta$ is the angle between the line and the normal, then $$\phi = 90^\circ - \theta$$ and therefore $$\sin\phi = \cos\theta.$$

Vector Form

If the line is $\vec{r} = \vec{a} + \lambda\vec{b}$ and the plane is $\vec{r} \cdot \vec{n} = d$, then $$\sin\phi = \left| \frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|} \right|$$

Cartesian Form

If the line is $$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$ and the plane is $$Ax + By + Cz + D = 0,$$ then $$\sin\phi = \left| \frac{aA + bB + cC}{\sqrt{a^2 + b^2 + c^2} \sqrt{A^2 + B^2 + C^2}} \right|$$

Conditions:

• Line Parallel to Plane: $aA + bB + cC = 0$.
• Line Perpendicular to Plane: $\frac{a}{A} = \frac{b}{B} = \frac{c}{C}$.

Condition for a Line to Lie Entirely in a Plane

A line $$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$ lies completely in the plane $$Ax + By + Cz + D = 0$$ if and only if both conditions hold:

1. The point $(x_1, y_1, z_1)$ lies on the plane: $$Ax_1 + By_1 + Cz_1 + D = 0$$
2. The line is parallel to the plane, so its direction vector is perpendicular to the plane's normal: $$Aa + Bb + Cc = 0$$

If only the second condition holds, then the line is parallel to the plane but not contained in it.

Example 1: Checking Coplanarity

Show that the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = \frac{z}{1}$ are coplanar.

Solution: Step 1: Identify the points on the two lines. For the first line, a point is $A(1,2,3)$. For the second line, a point is $B(4,1,0)$.

Step 2: Identify direction ratios. For line 1, direction ratios are $(2,3,4)$. For line 2, direction ratios are $(5,2,1)$.

Step 3: Use the coplanarity determinant. $$\begin{vmatrix} 4 - 1 & 1 - 2 & 0 - 3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix} = \begin{vmatrix} 3 & -1 & -3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix}$$

Step 4: Expand the determinant carefully. $$= 3\begin{vmatrix} 3 & 4 \\ 2 & 1 \end{vmatrix} - (-1)\begin{vmatrix} 2 & 4 \\ 5 & 1 \end{vmatrix} + (-3)\begin{vmatrix} 2 & 3 \\ 5 & 2 \end{vmatrix}$$ $$= 3(3 - 8) + 1(2 - 20) - 3(4 - 15)$$ $$= 3(-5) - 18 + 33 = -15 - 18 + 33 = 0.$$

Step 5: Since the determinant is zero, the scalar triple product is zero. Hence the two lines are coplanar.

Answer: The lines are coplanar.

Example 2: Equation of Plane Containing Two Intersecting Lines

Find the equation of the plane containing the lines $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$.

Solution: Step 1: Both lines pass through the origin, so the plane passes through $(0,0,0)$.

Step 2: Direction vectors are $$\vec{b}_1 = (1,2,3), \qquad \vec{b}_2 = (2,3,4).$$

Step 3: Find a normal vector by cross product. $$\vec{n} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 2 & 3 & 4 \end{vmatrix}$$ $$= \hat{i}(8-9) - \hat{j}(4-6) + \hat{k}(3-4) = -\hat{i} + 2\hat{j} - \hat{k}.$$ So a normal vector is $(-1,2,-1)$.

Step 4: Use point-normal form through the origin. $$-x + 2y - z = 0.$$ Multiplying by $-1$ gives the equivalent form $$x - 2y + z = 0.$$

Step 5: Quick verification. For the first line, $x=t, y=2t, z=3t$ gives $t - 4t + 3t = 0$. For the second line, $x=2s, y=3s, z=4s$ gives $2s - 6s + 4s = 0$. So both lines lie in the plane.

Answer: $x - 2y + z = 0$

Example 3: Angle Between Two Planes

Find the angle between the planes $2x + y - 2z = 5$ and $3x - 6y - 2z = 7$.

Solution: Step 1: The normals are $$\vec{n}_1 = (2,1,-2), \qquad \vec{n}_2 = (3,-6,-2).$$

Step 2: Find their dot product. $$\vec{n}_1 \cdot \vec{n}_2 = 2\cdot 3 + 1\cdot(-6) + (-2)\cdot(-2) = 6 - 6 + 4 = 4.$$

Step 3: Find the magnitudes. $$|\vec{n}_1| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{9} = 3,$$ $$|\vec{n}_2| = \sqrt{3^2 + (-6)^2 + (-2)^2} = \sqrt{49} = 7.$$

Step 4: Apply the angle formula. $$\cos\theta = \left| \frac{4}{3\cdot 7} \right| = \frac{4}{21}.$$

Step 5: Therefore, $$\theta = \cos^{-1}\left(\frac{4}{21}\right).$$

Answer: $\cos^{-1}(4/21)$

Example 4: Angle Between a Line and a Plane

Find the angle between the line $\frac{x + 1}{2} = \frac{y}{3} = \frac{z - 3}{6}$ and the plane $10x + 2y - 11z = 3$.

Solution: Step 1: Direction vector of the line is $$\vec{b} = (2,3,6),$$ and the normal to the plane is $$\vec{n} = (10,2,-11).$$

Step 2: Compute the dot product. $$\vec{b} \cdot \vec{n} = 2\cdot 10 + 3\cdot 2 + 6\cdot(-11) = 20 + 6 - 66 = -40.$$

Step 3: Compute the magnitudes. $$|\vec{b}| = \sqrt{2^2+3^2+6^2} = \sqrt{49} = 7,$$ $$|\vec{n}| = \sqrt{10^2+2^2+(-11)^2} = \sqrt{225} = 15.$$

Step 4: Use the formula for the angle $\phi$ between a line and a plane: $$\sin\phi = \left| \frac{\vec{b}\cdot\vec{n}}{|\vec{b}||\vec{n}|} \right| = \frac{40}{7\cdot 15} = \frac{40}{105} = \frac{8}{21}.$$

Step 5: Hence $$\phi = \sin^{-1}\left(\frac{8}{21}\right).$$

Answer: $\sin^{-1}(8/21)$

Example 5: Finding Unknown for Coplanar Lines

Find the value of $k$ if the lines $\frac{x - 1}{2} = \frac{y + 1}{3} = \frac{z - 1}{4}$ and $\frac{x - 3}{1} = \frac{y - k}{2} = \frac{z}{1}$ are coplanar.

Solution: Step 1: Extract points and direction ratios. Line 1 passes through $A(1,-1,1)$ and has direction ratios $(2,3,4)$. Line 2 passes through $B(3,k,0)$ and has direction ratios $(1,2,1)$.

Step 2: Use the coplanarity determinant. $$\begin{vmatrix} 3-1 & k-(-1) & 0-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0$$ that is, $$\begin{vmatrix} 2 & k+1 & -1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{vmatrix} = 0.$$

Step 3: Expand the determinant. $$2\begin{vmatrix} 3 & 4 \\ 2 & 1 \end{vmatrix} - (k+1)\begin{vmatrix} 2 & 4 \\ 1 & 1 \end{vmatrix} + (-1)\begin{vmatrix} 2 & 3 \\ 1 & 2 \end{vmatrix} = 0$$ $$= 2(3-8) - (k+1)(2-4) - (4-3) = 0$$ $$= -10 + 2(k+1) - 1 = 0.$$

Step 4: Simplify. $$-10 + 2k + 2 - 1 = 0 \implies 2k - 9 = 0 \implies k = \frac{9}{2}.$$

Answer: $\frac{9}{2}$

Example 6: Point of Intersection of a Line and a Plane

Find the coordinates of the point where the line $\frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 1}{-1}$ intersects the plane $2x - y + z = 4$.

Solution: Step 1: Let the common value be $\lambda$. Then a general point on the line is $$x = 2\lambda + 1, \quad y = \lambda + 2, \quad z = -\lambda - 1.$$

Step 2: Since the point lies on the plane, substitute into $2x - y + z = 4$. $$2(2\lambda + 1) - (\lambda + 2) + (-\lambda - 1) = 4.$$

Step 3: Simplify. $$4\lambda + 2 - \lambda - 2 - \lambda - 1 = 4$$ $$2\lambda - 1 = 4$$ $$2\lambda = 5 \implies \lambda = \frac{5}{2}.$$

Step 4: Substitute back. $$x = 2\cdot\frac{5}{2} + 1 = 6,$$ $$y = \frac{5}{2} + 2 = \frac{9}{2},$$ $$z = -\frac{5}{2} - 1 = -\frac{7}{2}.$$

Answer: $\left(6, \frac{9}{2}, -\frac{7}{2}\right)$

Example 7: Line Parallel to a Plane

Show that the line $\frac{x - 1}{1} = \frac{y + 2}{2} = \frac{z - 3}{3}$ is parallel to the plane $2x - y + 5 = 0$.

Solution: Step 1: The direction vector of the line is $$\vec{b} = (1,2,3).$$ The normal vector of the plane $2x - y + 5 = 0$ is $$\vec{n} = (2,-1,0).$$

Step 2: A line is parallel to a plane if its direction vector is perpendicular to the plane's normal. So compute $$\vec{b} \cdot \vec{n} = 1\cdot 2 + 2\cdot(-1) + 3\cdot 0 = 2 - 2 + 0 = 0.$$ Thus the line is either parallel to the plane or lies in it.

Step 3: Check whether a point on the line lies on the plane. The line passes through $(1,-2,3)$. Substitute into the plane: $$2(1) - (-2) + 5 = 2 + 2 + 5 = 9 \ne 0.$$ So the point is not on the plane.

Step 4: Therefore the line is not contained in the plane. Since its direction is parallel to the plane, it must be strictly parallel to the plane.

Answer: The line is parallel to the plane.

Example 8: Condition for a Line to Lie in a Plane

Find the value of $k$ for which the line $\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z - 3}{4}$ lies in the plane $x + 2y + kz = -1$.

Solution: Step 1: The line passes through the point $(1,-1,3)$ and has direction vector $(2,-1,4)$.

Step 2: For the line to lie in the plane, the point must satisfy the plane equation: $$1 + 2(-1) + 3k = -1$$ $$1 - 2 + 3k = -1$$ $$-1 + 3k = -1 \implies 3k = 0 \implies k = 0.$$

Step 3: Also, the direction vector of the line must be perpendicular to the plane's normal. When $k=0$, the plane becomes $$x + 2y = -1,$$ whose normal is $(1,2,0)$. Now $$(2,-1,4) \cdot (1,2,0) = 2 - 2 + 0 = 0.$$

Step 4: Since both conditions are satisfied, the line lies in the plane.

Answer: $k = 0$

Example 9: Equation of Plane Containing Two Parallel Lines

Find the equation of the plane containing the parallel lines $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 2}{3}$.

Solution: Step 1: Both lines have the same direction vector $$\vec{b} = (1,2,3).$$ So they are parallel.

Step 2: Take one point from each line: $$A(1,2,3), \qquad B(3,-1,2).$$ Then $$\vec{AB} = (3-1, -1-2, 2-3) = (2,-3,-1).$$

Step 3: A normal to the required plane must be perpendicular to both $\vec{b}$ and $\vec{AB}$. So $$\vec{n} = \vec{AB} \times \vec{b}$$ $$= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & -1 \\ 1 & 2 & 3 \end{vmatrix}$$ $$= \hat{i}(-9 - (-2)) - \hat{j}(6 - (-1)) + \hat{k}(4 - (-3))$$ $$= -7\hat{i} - 7\hat{j} + 7\hat{k}.$$ A simpler normal is $(1,1,-1)$.

Step 4: Use point-normal form through $A(1,2,3)$: $$(x-1) + (y-2) - (z-3) = 0.$$ Simplifying, $$x + y - z = 0.$$

Step 5: Check with point $B(3,-1,2)$: $$3 + (-1) - 2 = 0,$$ so the second line also lies in the plane.

Answer: $x + y - z = 0$

Example 10: Mutually Perpendicular Planes

Show that the planes $x + 2y - 2z = 5$ and $2x - 2y - z = 3$ are mutually perpendicular.

Solution: Step 1: The normal vectors are $$\vec{n}_1 = (1,2,-2), \qquad \vec{n}_2 = (2,-2,-1).$$

Step 2: Two planes are perpendicular if their normal vectors are perpendicular. So compute the dot product: $$\vec{n}_1 \cdot \vec{n}_2 = 1\cdot 2 + 2\cdot(-2) + (-2)\cdot(-1) = 2 - 4 + 2 = 0.$$

Step 3: Since the dot product is zero, the normals are perpendicular. Hence the planes are mutually perpendicular.

Answer: The planes are mutually perpendicular.