Question

In each case, find a vector equation of the line. a. Passing through \(P(3,-1,4)\) and perpendicular to the plane \(3 x-2 y-z=0\) b. Passing through \(P(2,-1,3)\) and perpendicular to the plane \(2 x+y=1\) c. Passing through \(P(0,0,0)\) and perpendicular $$ \begin{array}{l} \text { to the lines }\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 1 \\ 1 \\ 0 \end{array}\right]+t\left[\begin{array}{r} 2 \\ 0 \\ -1 \end{array}\right] \text { and } \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 2 \\ 1 \\ -3 \end{array}\right]+t\left[\begin{array}{r} 1 \\ -1 \\ 5 \end{array}\right]} \end{array} $$ d. Passing through \(P(1,1,-1)\), and perpendicular to the lines $$ \begin{array}{l} {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{l} 2 \\ 0 \\ 1 \end{array}\right]+t\left[\begin{array}{r} 1 \\ 1 \\ -2 \end{array}\right] \text { an }} \\ {\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{r} 5 \\ 5 \\ -2 \end{array}\right]+t\left[\begin{array}{r} 1 \\ 2 \\ -3 \end{array}\right]} \end{array} $$ e. Passing through \(P(2,1,-1)\), intersecting the line \(\left[\begin{array}{l}x \\\ y \\ z\end{array}\right]=\left[\begin{array}{r}1 \\ 2 \\\ -1\end{array}\right]+t\left[\begin{array}{l}3 \\ 0 \\ 1\end{array}\right],\) and perpendicular to that line. f. Passing through \(P(1,1,2)\), intersecting the line \(\left[\begin{array}{l}x \\\ y \\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\\ 0\end{array}\right]+t\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right],\) and perpendicular to line.

Short answer

Find normal or use cross product for direction vector; write vector equation with point and direction.

Step-by-step solution

Identifying Normal Vector for Part a

Given the plane equation \(3x - 2y - z = 0\), the normal vector is \(\mathbf{n} = \langle 3, -2, -1 \rangle\). The direction vector of the line must be this normal vector since the line is perpendicular to the plane.

Writing Vector Equation for Part a

Since the line passes through \(P(3, -1, 4)\) and has direction vector \(\mathbf{n} = \langle 3, -2, -1 \rangle\), the vector equation of the line is: \[ \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 3 \ -1 \ 4 \end{pmatrix} + t\begin{pmatrix} 3 \ -2 \ -1 \end{pmatrix} \]

Identifying Normal Vector for Part b

For the plane \(2x + y = 1\), the normal vector is \(\mathbf{n} = \langle 2, 1, 0 \rangle\). The line's direction vector should be this normal vector.

Writing Vector Equation for Part b

Using point \(P(2, -1, 3)\) and direction vector \(\mathbf{n} = \langle 2, 1, 0 \rangle\), the equation is: \[ \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 2 \ -1 \ 3 \end{pmatrix} + t\begin{pmatrix} 2 \ 1 \ 0 \end{pmatrix} \]

Find Cross Product for Direction in Part c

The direction vectors of the given lines are \(\langle 2, 0, -1 \rangle\) and \(\langle 1, -1, 5 \rangle\). The direction of a line perpendicular to both will be the cross product of these: \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 2 & 0 & -1 \ 1 & -1 & 5 \end{vmatrix} = \langle -5, -7, -2 \rangle \]

Writing Vector Equation for Part c

The line passes through \(P(0, 0, 0)\) and has direction vector \(\langle -5, -7, -2 \rangle\). The vector equation is: \[ \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 0 \ 0 \ 0 \end{pmatrix} + t\begin{pmatrix} -5 \ -7 \ -2 \end{pmatrix} \]

Find Cross Product for Direction in Part d

The direction vectors for the lines are \(\langle 1, 1, -2 \rangle\) and \(\langle 1, 2, -3 \rangle\). Calculate cross product: \[ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & -2 \ 1 & 2 & -3 \end{vmatrix} = \langle 1, 1, 1 \rangle \]

Writing Vector Equation for Part d

The line passes through point \(P(1, 1, -1)\) with the direction vector \(\langle 1, 1, 1 \rangle\). The equation is: \[ \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ 1 \ -1 \end{pmatrix} + t\begin{pmatrix} 1 \ 1 \ 1 \end{pmatrix} \]

Finding Perpendicular Vector for Part e

The direction vector of the line is \(\langle 3, 0, 1 \rangle\). A line perpendicular to this should have a direction vector \(\langle a, b, c \rangle\) such that \(3a + 0b + 1c = 0\). Choose \(a = 0\) and \(b = 1\) to get \(c = -3\), thus the direction vector is \(\langle 0, 1, -3 \rangle\).

Writing Vector Equation for Part e

The line passing through \(P(2, 1, -1)\) with direction \(\langle 0, 1, -3 \rangle\) has the equation: \[ \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 2 \ 1 \ -1 \end{pmatrix} + t\begin{pmatrix} 0 \ 1 \ -3 \end{pmatrix} \]

Finding Perpendicular Vector for Part f

Given line has direction \(\langle 1, 1, 1 \rangle\). Choose a direction \(\langle a, b, c \rangle\) such that \(a+b+c=0\); try \(1, -1, 0\) since 1 - 1 + 0 = 0.

Writing Vector Equation for Part f

The line through \(P(1, 1, 2)\) with direction \(\langle 1, -1, 0 \rangle\) is: \[ \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} 1 \ 1 \ 2 \end{pmatrix} + t\begin{pmatrix} 1 \ -1 \ 0 \end{pmatrix} \]

Key concepts

lines and planes in 3D

In three-dimensional geometry, lines and planes are fundamental elements. A **plane** in 3D can be represented by a linear equation of the form \(ax + by + cz = d\), where \(a\), \(b\), and \(c\) are constants that are proportional to the coordinates of a vector, known as the *normal vector*. The normal vector is perpendicular to the plane.

A **line** in 3D can be uniquely defined by a point on the line and a direction vector. The vector equation of a line is typically given as \( \begin{pmatrix} x \ y \ z \end{pmatrix} = \begin{pmatrix} x_0 \ y_0 \ z_0 \end{pmatrix} + t \begin{pmatrix} a \ b \ c \end{pmatrix} \), where \( \begin{pmatrix} x_0 \ y_0 \ z_0 \end{pmatrix} \) is a point on the line, \( \begin{pmatrix} a \ b \ c \end{pmatrix} \) is the direction vector, and \(t\) is a scalar parameter.

The line's direction vector can also be derived, in some scenarios, from the parameters given by two or more related planes or from calculations such as cross products as needed in exercises to find perpendicularity.

normal vectors

Normal vectors play a crucial role in defining geometric properties in 3D space. Specifically, a **normal vector** to a plane is a vector that is perpendicular to every line lying in that plane. For a plane of equation \(ax + by + cz = d\), the normal vector is expressed as \(\langle a, b, c \rangle\).

They are used to help determine various orientations and intersections with other geometric entities such as lines. For example, a line that is perpendicular to a plane must have a direction vector that aligns with the normal vector of that plane. Normal vectors are fundamental because they give insight into the shape and position of planes in space.

Thus, when given a plane, identifying the normal vector is the first step towards solving problems related to perpendicularity or intersections with lines.

cross products

**Cross products** are mathematical operations used to find a vector perpendicular to two other vectors in 3D space. Suppose you have two vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \). The cross product is computed by the determinant of a 3x3 matrix: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = \langle a_2b_3 - a_3b_2, a_3b_1 - a_1b_3, a_1b_2 - a_2b_1 \rangle \]
This operation is pivotal when we need to find a direction vector for a line that should be perpendicular to two other given lines in geometry problems.

In such cases, the cross product provides a new vector that is orthogonal to both input vectors. Cross products are essential in vector calculus and physics, especially when calculating torques and rotational forces, making them very handy in various real-world applications beyond just geometric calculations.

perpendicular vectors

In vector geometry, two vectors are said to be **perpendicular** (or orthogonal) if their dot product equals zero. For vectors \( \mathbf{a} = \langle a_1, a_2, a_3 \rangle \) and \( \mathbf{b} = \langle b_1, b_2, b_3 \rangle \), they are perpendicular if:

• \(a_1b_1 + a_2b_2 + a_3b_3 = 0 \)

Perpendicularity is a crucial property in many geometric contexts, particularly when dealing with lines and planes. When finding lines perpendicular to planes or other lines, the condition of perpendicularity guides us in choosing or confirming the direction vectors.

Knowing when two vectors are perpendicular is beneficial not just in geometry, but in physics as it helps to understand forces and motion paths. By checking the dot product, mathematicians and scientists can easily identify relationships and properties of geometric forms and structures.