Problems on Lines in 3D with Detailed Solutions

Given: \[ \langle x, y, z \rangle = \langle -2, 3, 0 \rangle + t \langle 3, 2, 5 \rangle \] Equality of vector components of the above vector equation gives: \[ x = -2 + 3t, \quad y = 3 + 2t, \quad z = 0 + 5t \] Solve for \( t \) from each of the above: \[ t = \dfrac{x + 2}{3}, \quad t = \dfrac{y - 3}{2}, \quad t = \dfrac{z}{5} \] All equal to \( t \), hence the symmetric form of the equation: \[ \dfrac{x + 2}{3} = \dfrac{y - 3}{2} = \dfrac{z}{5} \]

\[ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + t \langle 2, 0, -3 \rangle \] \[ \text{Equality of vector components:} \quad x = 1 + 2t, \quad y = -2, \quad z = 3 - 3t \] \[ \text{Solve for } t: \quad t = \dfrac{x - 1}{2}, \quad t = \dfrac{z - 3}{-3} \] \[ \text{Symmetric form:} \quad \dfrac{x - 1}{2} = \dfrac{z - 3}{-3}, \quad y = -2 \]

Direction vector \( \vec{PQ} = \langle 0 - 1 , -2 - 2 , 1 - 3 \rangle = \langle -1 , -4 , -2 \rangle \) \[ \langle x , y , z \rangle = \langle 1 , 2 , 3 \rangle + t \langle -1 , -4 , -2 \rangle \] Equality of vector components of the above vector equation gives: \[ x = 1 - t, \quad y = 2 - 4t, \quad z = 3 - 2t \]

Write given line \( x = 2t + 5 \), \( y = -4t \) and \( z = -t + 3 \). In symmetric form: \[ \dfrac{x - 5}{2} = \dfrac{y}{-4} = \dfrac{z - 3}{-1} \] The direction vector is: \( \langle 2, -4, -1 \rangle \) The line through point \( P(-3, 5, 2) \) is parallel to the given line and hence they have the same direction vector. Hence the vector equation of the line through \( P \): \[ x = -3 + 2t, \quad y = 5 - 4t, \quad z = 2 - t \]

Line \( L \) (to find) is perpendicular to lines \( L_1 \) and \( L_2 \), and is therefore perpendicular to their direction vectors \( \vec{d}_1 \) and \( \vec{d}_2 \), respectively. Hence, the cross product of \( \vec{d}_1 \) and \( \vec{d}_2 \) gives the direction vector of the line \( L \). From the symmetric equation of \( L_1 \), the direction vector is: \[ \vec{d}_1 = \langle 3, -4, 4 \rangle \] Write the equation of \( L_2 \) in symmetric form: \[ \dfrac{x + 4}{3} = \dfrac{y - 6}{-1} = \dfrac{z}{5} \] The direction vector of \( L_2 \) is: \[ \vec{d}_2 = \langle 3, -1, 5 \rangle \] The direction vector of line \( L \) may be taken as the cross product: \[ \vec{d} = \vec{d}_1 \times \vec{d}_2 = \langle 3, -4, 4 \rangle \times \langle 3, -1, 5 \rangle = \langle -16, -3, 9 \rangle \] The equation for line \( L \) through point \( P(1, -2, 3) \) and parallel to vector \( \vec{d} \) is given by: \[ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + t \langle -16, -3, 9 \rangle \]

Line \( L_1 \) is given by the parametric equations:
\[ x = 2t - 1, \quad y = -3t + 2, \quad z = 4t - 3 \]

Write the equations of line \( L_2 \) in parametric form using the parameter \( s \) as follows:
\[ x = 4s + 7, \quad y = 2s - 2, \quad z = -3s + 2 \]

Let \( A(x, y, z) \) be the point of intersection of the two lines. To be a point of intersection, the coordinates of \( A \) must satisfy the equations of both lines simultaneously. Hence:

Equating the \( x \)-coordinates gives the equation:
\[ 2t - 1 = 4s + 7 \quad (1) \]

Equating the \( y \)-coordinates gives the equation:
\[ -3t + 2 = 2s - 2 \quad (2) \]

Equating the \( z \)-coordinates gives the equation:
\[ 4t - 3 = -3s + 2 \quad (3) \]

Rewrite equations (1) and (2) in terms of \( t \) and \( s \) as follows:
\[ 2t - 4s = 8 \quad \text{and} \quad -3t - 2s = -4 \]

Solve for \( t \) and \( s \) to get:
\[ t = 2 \quad \text{and} \quad s = -1 \]

To have an intersection, \( t = 2 \) and \( s = -1 \) must also satisfy equation (3):
\[ 4t - 3 = -3s + 2 \]

Check:
\[ \text{Left side of equation (3):} \quad 4(2) - 3 = 5 \] \[ \text{Right side of equation (3):} \quad -3(-1) + 2 = 5 \]
Thus, \( t = 2 \) and \( s = -1 \) are solutions to all three equations.

The point of intersection is obtained by substituting into one of the parametric equations of \( L_1 \) or \( L_2 \).

Using \( L_1 \): Set \( t = 2 \) in the equations: \[ x = 2t - 1, \quad y = -3t + 2, \quad z = 4t - 3 \] to obtain the coordinates of the point of intersection: \[ x = 3, \quad y = -4, \quad z = 5 \]

Using \( L_2 \) (to verify): Substitute \( s = -1 \) in
\[ x = 4s + 7, \quad y = 2s - 2, \quad z = -3s + 2 \] giving the coordinates of the point of intersection: \[ x = 3, \quad y = -4, \quad z = 5 \]

Let \( \mathbf{d}_1 \) and \( \mathbf{d}_2 \) be the direction vectors of lines \( L_1 \) and \( L_2 \). For \( L_1 \): \[ \mathbf{d}_1 = \langle 2, -2, -4 \rangle \] For \( L_2 \): \[ \mathbf{d}_2 = \langle 6, 2, 2 \rangle \] The angle \( \theta \) between the lines \( L_1 \) and \( L_2 \) is equal to the angle between their direction vectors \( \mathbf{d}_1 \) and \( \mathbf{d}_2 \), given by \[ \theta = \arccos\left( \dfrac{ \mathbf{d}_1 \cdot \mathbf{d}_2 }{ |\mathbf{d}_1| \, |\mathbf{d}_2| } \right) \] where \( \mathbf{d}_1 \cdot \mathbf{d}_2 \) is the scalar (dot) product of vectors \( \mathbf{d}_1 \) and \( \mathbf{d}_2 \).

\( |\mathbf{d}_1| \) is the magnitude of vector \( \mathbf{d}_1 \), and \( |\mathbf{d}_2| \) is the magnitude of vector \( \mathbf{d}_2 \). \[ \mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(6) + (-2)(2) + (-4)(2) = 0 \] Since both \( |\mathbf{d}_1| \) and \( |\mathbf{d}_2| \) are nonzero, \[ \theta = \arccos(0) = 90^\circ \] If the lines intersect, they form a right angle of \( 90^\circ \). If they do not intersect, the skew or parallel lines aligned with these directions also form a \( 90^\circ \) angle.

One way to approach this problem is to show that the two lines \( L_1 \) and \( L_2 \) have two points in common. Rewrite the two equations in parametric form as: \[ L_1: \quad x = 2 - t, \quad y = 2t, \quad z = -1 + 4t \] \[ L_2: \quad x = 1 - 2t, \quad y = 2 + 4t, \quad z = 3 + 8t \] Point \( P_1 \) on \( L_1 \): \( (2, 0, -1) \)

Point \( P_2 \) on \( L_2 \): \( (1, 2, 3) \)

Check that \( P_1(2, 0, -1) \) is on \( L_2 \) using the symmetric equations of \( L_2 \): \[ \dfrac{2 - 1}{-2} = \dfrac{0 - 2}{4} = \dfrac{-1 - 3}{8} = -\dfrac{1}{2} \] All terms are equal, so \( P_1 \) is on \( L_2 \). Check that \( P_2(1, 2, 3) \) is on \( L_1 \) using the symmetric equations of \( L_1 \): \[ \dfrac{1 - 2}{-1} = \dfrac{2}{2} = \dfrac{3 + 1}{4} = 1 \] All terms are equal, so \( P_2 \) is on \( L_1 \). \( P_1 \) is on \( L_1 \) and \( L_2 \), and \( P_2 \) is also on \( L_1 \) and \( L_2 \). The two equations represent the same line.

According to the equation of the line, point \( P(2, 3, 0) \) lies on the line. The direction vector is \( \vec{d} = \langle -2, 3, 1 \rangle \).

The distance \( D \) from the line to the point \( P_0(1, -2, 3) \) is given by Distance from a point to a line : \[ D = \dfrac{|\vec{P_0P} \times \vec{d}|}{|\vec{d}|} \] Let \( \vec{P_0P} = \langle 1, 5, -3 \rangle \), and \( \vec{d} = \langle -2, 3, 1 \rangle \).

Then, \( \vec{P_0P} \times \vec{d} = \langle 14, 5, 13 \rangle \). \[ |\vec{P_0P} \times \vec{d}| = \sqrt{14^2 + 5^2 + 13^2} = \sqrt{390} \] \[ |\vec{d}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{14} \] \[ D = \dfrac{\sqrt{390}}{\sqrt{14}} = \dfrac{\sqrt{195}}{\sqrt{7}} \]

Points \( P_1(2, 0, -1) \) and \( P_2(1, -2, 3) \) are points on lines \( L_1 \) and \( L_2 \) respectively. The direction vectors are \[ \mathbf{d}_1 = \langle -1, 4, -4 \rangle \quad \text{and} \quad \mathbf{d}_2 = \langle -5, 2, -2 \rangle \] for lines \( L_1 \) and \( L_2 \).

The shortest distance \( D \) between the two lines is given by the formula: \[ D = \dfrac{|\mathbf{n} \cdot \overrightarrow{P_1P_2}|}{|\mathbf{n}|} \] where \(\overrightarrow{P_1P_2}\) is the vector defined by points \( P_1 \) and \( P_2 \), and \(\mathbf{n}\) is the cross product of \(\mathbf{d}_1\) and \(\mathbf{d}_2\). \[ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \langle -1, 4, -4 \rangle \times \langle -5, 2, -2 \rangle = \langle 0, 18, 18 \rangle \] \[ \overrightarrow{P_1P_2} = \langle -1, -2, 4 \rangle \] \[ D = \dfrac{| \langle 0, 18, 18 \rangle \cdot \langle -1, -2, 4 \rangle |}{\sqrt{0^2 + 18^2 + 18^2}} = \sqrt{2} \]

Direction vectors must be proportional:
\[ \langle 10, b, 4 \rangle = k \langle -5, 2, -2 \rangle \] From the first component:
\[ 10 = -5k, \quad k = -2 \]

From the third component: \[ 4 = k(-2), \quad k = -2 \] From the second component:
\[ b = k(2) = -2 \times 2 = -4 \]

a) Equation of line to be found: \[ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + t \langle a, b, c \rangle \]
Perpendicular to: \[ \langle x, y, z \rangle = \langle -3, 3, -1 \rangle + m \langle 1, 1, 1 \rangle \]
Apply conditions to find \(a\), \(b\), and \(c\).
Intersection gives 3 equations:
\[ 1 + a t = -3 + m \]
\[ -2 + b t = 3 + m \]
\[ 3 + c t = -1 + m \]
Add sides of all three equations to obtain
\[ 1 - 2 + 3 + t(a + b + c) = -1 + 3 m \]
Direction vectors \(\langle a, b, c \rangle\) and \(\langle 1, 1, 1 \rangle\) are orthogonal therefore their dot product is zero.
\[ a + b + c = 0 \]
The above equation \[ 1 - 2 + 3 + t(a + b + c) = -1 + 3 m \] becomes \[ 3 m = 3 \]
Solve for \(m\):
\[ m = 1 \]
Put \(m = 1\) in the first of the three equations: \[ 1 + a t = -3 + m \] to find \[ a t = -3 \]
Let \(a = 1\) to get \(t = -3\)
Use the equation to find \(b\): \[ -2 + b t = 3 + m \implies -2 + b(-3) = 3 + 1 \] \[ b = \dfrac{-6}{3} = -2 \]
Use the equation to find \(c\): \[ 3 + c t = -1 + m \implies 3 + c(-3) = -1 + 1 \] \[ c = 1 \]
Equation of line through the point \((1, -2, 3)\): \[ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + t \langle 1, -2, 1 \rangle \]

b) Point of intersection
Set \(t = -3\) in the equation of the line found: \[ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + t \langle 1, -2, 1 \rangle = \langle 1, -2, 3 \rangle - 3 \langle 1, -2, 1 \rangle = \langle -2, 4, 0 \rangle \]
To check, set \(m = 1\) in the given equation of the line:
\[ \langle x, y, z \rangle = \langle -3, 3, -1 \rangle + m \langle 1, 1, 1 \rangle = \langle -3, 3, -1 \rangle + 1 \langle 1, 1, 1 \rangle = \langle -2, 4, 0 \rangle \]
The is a point of intersection given by: \[ \langle -2, 4, 0 \rangle \]

Write the symmetric equations form of the line: \[ \dfrac{x - 3}{3} = \dfrac{y - 4}{-1} = \dfrac{z - 5}{2} \] Point A: Substitute \(x\), \(y\), and \(z\) in the symmetric equations by the coordinates of point \(A(3, 4, 4)\). \[ \dfrac{x - 3}{3} = \dfrac{3 - 3}{3} = 0 \] \[ \dfrac{y - 4}{-1} = \dfrac{4 - 4}{-1} = 0 \] \[ \dfrac{z - 5}{2} = \dfrac{4 - 5}{2} = -\dfrac{1}{2} \] Last expression is not equal to the first two, hence \(A\) is not on the line.

Point B: Substitute \(x\), \(y\), and \(z\) in the symmetric equations by the coordinates of point \(B(0, 5, 3)\). \[ \dfrac{x - 3}{3} = \dfrac{0 - 3}{3} = -1 \] \[ \dfrac{y - 4}{-1} = \dfrac{5 - 4}{-1} = -1 \] \[ \dfrac{z - 5}{2} = \dfrac{3 - 5}{2} = -1 \] All expressions are equal, hence \(B\) is on the line.

Point C: Substitute \(x\), \(y\), and \(z\) in the symmetric equations by the coordinates of point \(C(6, 3, 7)\). \[ \dfrac{x - 3}{3} = \dfrac{6 - 3}{3} = 1 \] \[ \dfrac{y - 4}{-1} = \dfrac{3 - 4}{-1} = 1 \] \[ \dfrac{z - 5}{2} = \dfrac{7 - 5}{2} = 1 \] All expressions are equal, hence \(C\) is on the line.

Conclusion: Points B and C lie on the line.