Question

Find the equation of \(a l l\) planes: a. Perpendicular to the line \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=\left[\begin{array}{r}2 \\ -1 \\\ 3\end{array}\right]+t\left[\begin{array}{l}2 \\ 1 \\ 3\end{array}\right]\) b. Perpendicular to the line \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=\left[\begin{array}{r}1 \\ 0 \\\ -1\end{array}\right]+t\left[\begin{array}{l}3 \\ 0 \\ 2\end{array}\right]\) c. Containing the origin. d. Containing \(P(3,2,-4)\). e. Containing \(P(1,1,-1)\) and \(Q(0,1,1)\). f. Containing \(P(2,-1,1)\) and \(Q(1,0,0)\). g. Containing the line \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=\left[\begin{array}{l}2 \\ 1 \\\ 0\end{array}\right]+t\left[\begin{array}{r}1 \\ -1 \\ 0\end{array}\right]\) h. Containing the line \(\left[\begin{array}{l}x \\ y \\\ z\end{array}\right]=\left[\begin{array}{l}3 \\ 0 \\\ 2\end{array}\right]+t\left[\begin{array}{r}1 \\ -2 \\ -1\end{array}\right]\)

Short answer

Find planes using perpendicular vectors and points for specifics, combining line directions and points' conditions for solutions.

Step-by-step solution

Understand the general plane equation

The general equation of a plane in 3D space is given by \( ax + by + cz = d \), where \( a, b, c \) are the normal vector components to the plane, and \( d \) is a constant.

Determine planes perpendicular to given lines

For part a, the line has a direction vector \( \begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \). Planes perpendicular to this line will have a normal vector in the direction of the line, hence the plane equation is \( 2x + y + 3z = d \). Similarly, for part b, the line has a direction vector \( \begin{bmatrix} 3 & 0 & 2 \end{bmatrix} \), so the planes will have equations in the form \( 3x + 2z = d \).

Find planes containing specific points

For part c, the plane must contain the origin, so substitute \( (0,0,0) \) into the equation \( 2x + y + 3z = d \) to get \( d=0 \), hence \( 2x + y + 3z = 0 \). For part d, substitute point \( P(3,2,-4) \) to find \( 2(3) + 2 + 3(-4) = d \), giving \( d=-4 \), thus \( 2x + y + 3z = -4 \).

Plane through two points

For parts e and f, use the vector formed by the points \( \vec{PQ} \) as a direction vector on the plane and find the normal. Part e: points (1,1,-1) and (0,1,1) give direction vector \( \begin{bmatrix} -1 & 0 & 2 \end{bmatrix} \). Cross it with a normal vector if necessary and use the point condition to find \( d \). Part f similarly derives normal and plane equation from given points.

Plane containing a line

For parts g and h, a plane containing a line must be perpendicular to the line's direction vector. For part g, direction is \( \begin{bmatrix} 1 & -1 & 0 \end{bmatrix} \), so equation will accommodate both line direction and origin. For h, handle line's direction to find resulting plane. Combined directions should determine if it's consistent and if requiring specific points leads to suitable \( d \).

Key concepts

Plane equation

In 3D geometry, a plane can be represented by the equation \( ax + by + cz = d \). Here, \( a, b, \) and \( c \) are the components of the normal vector of the plane, and \( d \) is a constant. The normal vector, perpendicular to the plane, plays a crucial role in defining the plane's orientation in space. This basic equation helps determine how a plane interacts with other geometric entities like points and lines. For example, if you need the plane to go through a specific point, like in some of the exercise's sub-parts, you'd substitute the coordinates of the point into the equation to find the value of \( d \). If a line is perpendicular to the plane, the direction vector of the line will align with the plane's normal vector.

Normal vector

The normal vector is a fundamental concept in 3D geometry, especially in defining planes. It is represented by the vector \( \begin{bmatrix} a & b & c \end{bmatrix} \), where \( a, b, \) and \( c \) are the coefficients in the plane equation \( ax + by + cz = d \). The role of the normal vector is to signify the direction perpendicular to the plane. This makes it the key element for understanding how planes interact with various geometric entities. In the context of the given exercise, finding planes perpendicular to a line involves using the line's direction vector as the normal vector of the plane. If the line's direction vector is \( \begin{bmatrix} 2 & 1 & 3 \end{bmatrix} \), this becomes the normal vector, leading to the plane equation \( 2x + y + 3z = d \). This approach is consistently used to find planes perpendicular to given lines.

Point-plane relationship

To determine the point-plane relationship, one primarily focuses on how specific points satisfy the plane equation. For a plane given by \( ax + by + cz = d \), any point \((x_0, y_0, z_0)\) lies on the plane if it satisfies the equation with such values plugged in. If inserting these coordinates into the equation results in a true statement, the point is on the plane. This concept is essential for solving parts of the exercise, particularly where planes must contain specific points. For instance, if a plane must pass through the origin \((0,0,0)\), substitution into the equation simplifies to \( 0 = d \). Therefore, the constant \( d \) becomes zero, resulting in \( ax + by + cz = 0 \), ensuring the plane passes through the origin.

Line-plane interaction

Exploring the interaction between a line and a plane involves understanding both intersection and alignment concepts. If a plane contains a line, the line is parallel to the plane, effectively lying within it. To ensure this condition, both the line's direction vector and a point on the line need to satisfy the plane equation. This means the direction vector of the line should be parallel (or a scalar multiple) to the normal vector of another plane containing the line. In the exercise, part g illustrates this concept. Here, the line has a direction vector \( \begin{bmatrix} 1 & -1 & 0 \end{bmatrix} \). For the plane to accommodate this line, its equation, derived using the direction vector, should hold for the entire range of points described by the line's vector equation. Managing these relationships helps in solving problems where geometric entities must coexist with certain orientations and positions.