Problems on Lines in 3D with Detailed Solutions

Write the equation of the line given in vector form by $$ \langle x, y, z \rangle = \langle -2, 3, 0 \rangle + t \langle 3, 2, 5 \rangle $$ into parametric and symmetric forms.

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Solution:

Equality of vector components of the above vector equation gives the parametric form:

$$ x = -2 + 3t, \quad y = 3 + 2t, \quad z = 0 + 5t $$

Solve for $t$ from each of the above:

$$ t = \frac{x + 2}{3}, \quad t = \frac{y - 3}{2}, \quad t = \frac{z}{5} $$

All equal to $t$, hence the symmetric form of the equation:

$$ \frac{x + 2}{3} = \frac{y - 3}{2} = \frac{z}{5} $$

Find the symmetric form of the equation of the line through the point $P(1, -2, 3)$ and parallel to the vector $\vec{n} = \langle 2, 0, -3 \rangle$.

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Solution:

Vector equation: $$ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + t \langle 2, 0, -3 \rangle $$

Equality of vector components: $$ x = 1 + 2t, \quad y = -2, \quad z = 3 - 3t $$

Solve for $t$: $$ t = \frac{x - 1}{2}, \quad t = \frac{z - 3}{-3} $$

Because the $y$-direction is $0$, $y$ is constantly $-2$. The symmetric form is:

$$ \frac{x - 1}{2} = \frac{z - 3}{-3}, \quad y = -2 $$

Find the parametric equations of the line through the two points $P(1, 2, 3)$ and $Q(0, -2, 1)$.

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Solution:

Direction vector $\vec{PQ} = \langle 0 - 1 , -2 - 2 , 1 - 3 \rangle = \langle -1 , -4 , -2 \rangle$

Vector equation: $$ \langle x , y , z \rangle = \langle 1 , 2 , 3 \rangle + t \langle -1 , -4 , -2 \rangle $$

Equality of vector components of the above vector equation gives:

$$ x = 1 - t, \quad y = 2 - 4t, \quad z = 3 - 2t $$

Find the parametric equations of the line that passes through the point $P(-3, 5, 2)$ and is parallel to the line with equations: $$ x = 2t + 5, \quad y = -4t, \quad z = -t + 3 $$

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Solution:

Write given line in symmetric form:

$$ \frac{x - 5}{2} = \frac{y}{-4} = \frac{z - 3}{-1} $$

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. The parametric equations of the line through $P$ are:

$$ x = -3 + 2t, \quad y = 5 - 4t, \quad z = 2 - t $$

Find the equation of a line through $P(1 , -2 , 3)$ and perpendicular to the two lines $L_1$ and $L_2$ given by:

Line $L_1$: $$ \frac{x - 2}{3} = \frac{y + 1}{-4} = \frac{z + 9}{4} $$

Line $L_2$: $$ x = 3t - 4, \quad y = -t + 6, \quad z = 5t $$

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Solution:

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 $\vec{d}_2$. 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: $$ \vec{d}_1 = \langle 3, -4, 4 \rangle $$

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:

$$ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + t \langle -16, -3, 9 \rangle $$

Find the point of intersection of the lines $L_1$ and $L_2$ in 3D defined by:

Line $L_1$: $$ x = 2t - 1, \quad y = -3t + 2, \quad z = 4t - 3 $$

Line $L_2$: $$ \frac{x - 7}{4} = \frac{y + 2}{2} = \frac{z - 2}{-3} $$

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Solution:

Write the equations of line $L_2$ in parametric form using a different parameter $s$:

$$ x = 4s + 7, \quad y = 2s - 2, \quad z = -3s + 2 $$

Let $A(x, y, z)$ be the point of intersection. Equating the coordinates gives the system:

$$ 2t - 1 = 4s + 7 \quad (1) $$

$$ -3t + 2 = 2s - 2 \quad (2) $$

$$ 4t - 3 = -3s + 2 \quad (3) $$

Rewrite equations (1) and (2):

$$ 2t - 4s = 8 \quad \text{and} \quad -3t - 2s = -4 $$

Solve for $t$ and $s$: $$ t = 2 \quad \text{and} \quad s = -1 $$

Check these in equation (3):

Left side: $4(2) - 3 = 5$. Right side: $-3(-1) + 2 = 5$. They match, so they intersect.

Substitute $t=2$ back into $L_1$ equations to find the intersection point:

$$ x = 3, \quad y = -4, \quad z = 5 $$

Find the angle between the lines $L_1$ and $L_2$ with symmetric equations:

$$ L_1: \frac{x - 1}{2} = \frac{y + 2}{-2} = \frac{z}{-4} $$

$$ L_2: \frac{x + 3}{6} = \frac{y + 2}{2} = \frac{z - 1}{2} $$

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Solution:

Let $\mathbf{d}_1$ and $\mathbf{d}_2$ be the direction vectors of lines $L_1$ and $L_2$.

$$ \mathbf{d}_1 = \langle 2, -2, -4 \rangle \quad \text{and} \quad \mathbf{d}_2 = \langle 6, 2, 2 \rangle $$

The angle $\theta$ between the lines $L_1$ and $L_2$ is equal to the angle between their direction vectors, given by:

$$ \theta = \arccos\left( \frac{ \mathbf{d}_1 \cdot \mathbf{d}_2 }{ |\mathbf{d}_1| \, |\mathbf{d}_2| } \right) $$

Calculate the dot product:

$$ \mathbf{d}_1 \cdot \mathbf{d}_2 = (2)(6) + (-2)(2) + (-4)(2) = 12 - 4 - 8 = 0 $$

Since the dot product is $0$ and magnitudes are nonzero:

$$ \theta = \arccos(0) = 90^\circ $$

Show that the symmetric equations given below represent the same line.

Line $L_1$: $$ \frac{x - 2}{-1} = \frac{y}{2} = \frac{z + 1}{4} $$

Line $L_2$: $$ \frac{x - 1}{-2} = \frac{y - 2}{4} = \frac{z - 3}{8} $$

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Solution:

We can show that the two lines have two points in common. Rewrite the two equations in parametric form as:

$$ L_1: \quad x = 2 - t, \quad y = 2t, \quad z = -1 + 4t $$

$$ L_2: \quad x = 1 - 2s, \quad y = 2 + 4s, \quad z = 3 + 8s $$

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 $L_2$:

$$ \frac{2 - 1}{-2} = \frac{0 - 2}{4} = \frac{-1 - 3}{8} = -\frac{1}{2} $$

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 $L_1$:

$$ \frac{1 - 2}{-1} = \frac{2}{2} = \frac{3 + 1}{4} = 1 $$

All terms are equal, so $P_2$ is on $L_1$.

Since they share at least two distinct points, the two equations represent the same line.

Find the distance between the point $P_0(1, -2, 3)$ and the line given by the vector equation: $$ \langle x, y, z \rangle = \langle 2, 3, 0 \rangle + t \langle -2, 3, 1 \rangle $$

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Solution:

According to the equation, 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 $P_0$ is given by:

$$ D = \frac{|\vec{P_0P} \times \vec{d}|}{|\vec{d}|} $$

Let $\vec{P_0P} = \langle 2-1, 3-(-2), 0-3 \rangle = \langle 1, 5, -3 \rangle$.

Then, $\vec{P_0P} \times \vec{d} = \langle 1, 5, -3 \rangle \times \langle -2, 3, 1 \rangle = \langle 14, 5, 13 \rangle$.

$$ |\vec{P_0P} \times \vec{d}| = \sqrt{14^2 + 5^2 + 13^2} = \sqrt{196 + 25 + 169} = \sqrt{390} $$

$$ |\vec{d}| = \sqrt{(-2)^2 + 3^2 + 1^2} = \sqrt{14} $$

$$ D = \frac{\sqrt{390}}{\sqrt{14}} = \frac{\sqrt{195}}{\sqrt{7}} $$

Find the shortest distance between the two lines $L_1$ and $L_2$ defined by their equations:

$$ L_1: \langle x, y, z \rangle = \langle 2, 0, -1 \rangle + t \langle -1, 4, -4 \rangle $$

$$ L_2: \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + m \langle -5, 2, -2 \rangle $$

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Solution:

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:

$$ \mathbf{d}_1 = \langle -1, 4, -4 \rangle \quad \text{and} \quad \mathbf{d}_2 = \langle -5, 2, -2 \rangle $$

The shortest distance $D$ between the two lines is given by the formula:

$$ D = \frac{|\mathbf{n} \cdot \overrightarrow{P_1P_2}|}{|\mathbf{n}|} $$

where $\mathbf{n}$ is the cross product of $\mathbf{d}_1$ and $\mathbf{d}_2$.

$$ \mathbf{n} = \mathbf{d}_1 \times \mathbf{d}_2 = \langle -1, 4, -4 \rangle \times \langle -5, 2, -2 \rangle = \langle 0, 18, 18 \rangle $$

$$ \overrightarrow{P_1P_2} = \langle 1-2, -2-0, 3-(-1) \rangle = \langle -1, -2, 4 \rangle $$

$$ D = \frac{| \langle 0, 18, 18 \rangle \cdot \langle -1, -2, 4 \rangle |}{\sqrt{0^2 + 18^2 + 18^2}} = \frac{| -36 + 72 |}{18\sqrt{2}} = \frac{36}{18\sqrt{2}} = \sqrt{2} $$

Find the value of $b$ so that the lines $L_1$ and $L_2$ given by their equations below are parallel.

$$ L_1: \langle x, y, z \rangle = \langle 2, 0, -1 \rangle + t \langle 10, b, 4 \rangle $$

$$ L_2: \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + m \langle -5, 2, -2 \rangle $$

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Solution:

Direction vectors must be proportional: $$ \langle 10, b, 4 \rangle = k \langle -5, 2, -2 \rangle $$

From the first component: $$ 10 = -5k \implies k = -2 $$

From the third component (to verify): $$ 4 = k(-2) \implies k = -2 $$

From the second component: $$ b = k(2) = -2 \times 2 = -4 $$

Find the equation of a line passing through the point $P(1, -2, 3)$ that intersects and is perpendicular to the line given by the parametric equations $$ x = -3 + t, \quad y = 3 + t, \quad z = -1 + t $$ Also, find the point of intersection of the two lines.

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Solution:

Equation of line to be found: $$ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + s \langle a, b, c \rangle $$

Perpendicular to: $$ \langle x, y, z \rangle = \langle -3, 3, -1 \rangle + t \langle 1, 1, 1 \rangle $$

Intersection requires finding a specific point $M(x_0, y_0, z_0)$ on the given line such that $\vec{PM}$ is perpendicular to the given line's direction $\langle 1,1,1 \rangle$.

$\vec{PM} = \langle -3+t - 1, 3+t - (-2), -1+t - 3 \rangle = \langle t-4, t+5, t-4 \rangle$.

Since $\vec{PM}$ is perpendicular to $\langle 1,1,1 \rangle$, their dot product is zero:

$$ (t-4)(1) + (t+5)(1) + (t-4)(1) = 0 $$

$$ 3t - 3 = 0 \implies t = 1 $$

The point of intersection $M$ (when $t=1$) is: $$ M(-3+1, 3+1, -1+1) = (-2, 4, 0) $$

The direction vector $\vec{PM}$ is $\langle 1-4, 1+5, 1-4 \rangle = \langle -3, 6, -3 \rangle$. This can be scaled to $\langle 1, -2, 1 \rangle$.

Equation of the new line: $$ \langle x, y, z \rangle = \langle 1, -2, 3 \rangle + s \langle 1, -2, 1 \rangle $$

Which of the points $A(3, 4, 4)$, $B(0, 5, 3)$, and $C(6, 3, 7)$ lies on the line defined by the parametric equations $$ x = 3t + 3, \quad y = -t + 4, \quad z = 2t + 5 $$

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Solution:

Write the symmetric equations form of the line: $$ \frac{x - 3}{3} = \frac{y - 4}{-1} = \frac{z - 5}{2} $$

Point A $(3, 4, 4)$: $$ \frac{3 - 3}{3} = 0, \quad \frac{4 - 4}{-1} = 0, \quad \frac{4 - 5}{2} = -\frac{1}{2} $$
The last expression is not equal to the first two, hence $A$ is not on the line.

Point B $(0, 5, 3)$: $$ \frac{0 - 3}{3} = -1, \quad \frac{5 - 4}{-1} = -1, \quad \frac{3 - 5}{2} = -1 $$
All expressions are equal, hence $B$ is on the line.

Point C $(6, 3, 7)$: $$ \frac{6 - 3}{3} = 1, \quad \frac{3 - 4}{-1} = 1, \quad \frac{7 - 5}{2} = 1 $$
All expressions are equal, hence $C$ is on the line.