Chapter 3: Problem 14

Given the line \(\mathbf{r}=3 \mathbf{i}-\mathbf{i}+(2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}) t\) : (a) Find the equation of the plane containing the line and the point \((2,1,0)\). (b) Find the angle between the line and the \((y, z)\) plane. (c) Find the perpendicular distance between the line and the \(x\) axis. (d) Find the equation of the plane through the point \((2,1,0)\) and perpendicular to the line. (c) Find the equations of the line of intersection of the plane in \((\mathrm{d})\) and the plane \(y=2 z\).

Short Answer

Expert verified

Equation of plane: \(2x+ y - 2z = 5 \). Angle \( = cos^{-1}(2/3) \). Perpendicular: \( = \sqrt 2 = x\)

Step by step solution

- Identify the direction vector of the line

The given line is: \[ \mathbf{r} = 3 \mathbf{i}-\mathbf{i}+(2 \mathbf{i}+\mathbf{j}-2 \mathbf{k}) t \] To write it in a standard form: \[ \mathbf{r} = (2 + 2t) \mathbf{i} + t \mathbf{j} - 2t \mathbf{k} \] So, the direction vector is: \[ \mathbf{d} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \]

- Find the plane equation containing the line and a given point (2,1,0)

The direction vector of the line is \( \mathbf{d} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \) and a point on the line can be taken as \((2,0,0)\) when \(t=0\).The normal vector to the plane can be found using the cross product of vectors from the given point (2,1,0) to the points on the line: \( \mathbf{v_1} = 2 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k} - (2 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}) = \mathbf{j} \)and \( \mathbf{v_2} = 2 \mathbf{i} + 1 \mathbf{j} - 2 \mathbf{k} - (2 \mathbf{i} + 0 \mathbf{j} + 0 \mathbf{k}) = \mathbf{j} - 2 \mathbf{k} \)\( \mathbf{n} = \mathbf{v_1} \times \mathbf{v_2} = \mathbf{i}(0*0 - (-2)) - \mathbf{j}(2*0 - (0)) + \mathbf{k}(2*1 - 0*2) = -2 \mathbf{i} + 0 \mathbf{j} + 1 \mathbf{k} \).Thus, the equation of the plane is:\[ -2(x - 2) + k(z - 0) = 0 \] which simplifies to \[ 2x - z = 4.\]

- Angle between the line and the (y,z) plane

The normal vector of the (y,z) plane is \( \mathbf{i} \). The direction vector of the line is \( \mathbf{d} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \). The cosine of the angle between the line and the plane is given by: \[ \cos(\theta) = \frac{| \mathbf{d} \cdot \mathbf{n} |}{\| \mathbf{d} \| \| \mathbf{n} \|} \] where \( \mathbf{n} = \mathbf{i} \). \[ \cos(\theta) = \frac{|2*1 + 0*0 + (-2)*0|}{\sqrt{2^2 + 1^2 + (-2)^2} \cdot \sqrt{1}} = \frac{2}{\sqrt{9}} = \frac{2}{3} \] So the angle \( \theta = \cos^{-1}(\frac{2}{3}) \).

- Perpendicular distance to x-axis

The formula for the perpendicular distance from a point to a line in space is: \[ d = \frac{| \mathbf{a} \times \mathbf{b} |}{| \mathbf{b} |} \] where \( \mathbf{a} = 2 \mathbf{i} + 1 \mathbf{j} + 0 \mathbf{k} \) and \( \mathbf{b} = 2 \mathbf{i} + \mathbf{j} - 2\mathbf{k} \). \[ = \frac{ \sqrt{|-2m(-2)|^2 + |-2 |-m^2}}}{\sqrt{2a^3 + b^2 + c^2}} = 2 \]

- Equation of the plane perpendicular to the line through point (2,1,0)

The direction vector of the line is \( \mathbf{d} = 2 \mathbf{i} + \mathbf{j} - 2 \mathbf{k} \). This direction vector is also the normal to the plane. The point (2,1,0) is on the plane. The equation of the plane is hence:\[ 2(x - 2) + (y - 1) - 2(z - 0) = 0 \] \that simplifies to \[ 2x + y - 2z = 5 \]

- Equation of the line of intersection of (y=2z) and plane from (Step 5)

Substituting \( y = 2z \) into the equation \( 2x + y - 2z = 5 \): \( 2x + 2z - 2z = 5 \) \( 2x = 5 \) So, \( x = \frac{5}{2} \).So, the Intersection line is: {\(x,z\) = (\frac{5}{2} + t)} where t is the direction

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Plane Equations

A plane in 3D space can be described using its equation. The standard form of a plane equation is given by:

\(ax + by + cz = d\)

Here, \(a\), \(b\), and \(c\) are the components of the normal vector to the plane, and \(d\) is a constant. The normal vector \((a, b, c)\) determines the orientation of the plane. For example, if the normal vector is \((2, 1, -2)\), the plane equation is: \(2x + y - 2z = d\). To find \(d\), you can use a known point on the plane by substituting its coordinates into the equation. For instance, for a point \((2, 1, 0)\), solving the equation \(2(2) + 1(1) - 2(0) = 4 + 1 - 0 \) gives \(d = 5\). Thus, the specific plane equation is \(2x + y - 2z = 5\).

Direction Vectors

Direction vectors are essential in understanding lines and planes in 3D geometry. They define the direction and magnitude along a line. For a line given by the parametric equations:

\(x = x_0 + at\)
\(y = y_0 + bt\)
\(z = z_0 + ct\)

The vector \((a, b, c)\) is the direction vector. For instance, the line \(\mathbf{r} = (2 + 2t)\mathbf{i} + t\mathbf{j} - 2t\mathbf{k}\) has a direction vector \(\mathbf{d} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\). This means that as \(t\) changes, the point moves along the direction specified by \(\mathbf{d}\).

Perpendicular Distance

To find the perpendicular distance from a point to a plane or between geometric objects, you often use projection formulas. For a point \((x_1, y_1, z_1)\) and a plane \(ax + by + cz = d\), the distance \(D\) can be computed as:

\(D = \frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}\)

This formula comes from projecting the point onto the normal vector of the plane, which measures the shortest distance. For example, if you need to find the perpendicular distance from the line \(\mathbf{r}\) to the \(x\)-axis, you first find a direction vector \(\mathbf{b} = 2\mathbf{i} + \mathbf{j} - 2\mathbf{k}\) and a point vector \(\mathbf{a} = (2, 1, 0)\). Then use the cross product and magnitude formulas to compute the perpendicular distance.

Cross Product

The cross product of two vectors results in a third vector that is perpendicular to the plane containing the original two vectors. For vectors \(\mathbf{u} = (u_1, u_2, u_3)\) and \(\mathbf{v} = (v_1, v_2, v_3)\), their cross product is:

\(\mathbf{u} \times \mathbf{v} = (u_2v_3 - u_3v_2)\mathbf{i} + (u_3v_1 - u_1v_3)\mathbf{j} + (u_1v_2 - u_2v_1)\mathbf{k}\)

This is useful in finding normals to planes. For instance, for vectors \(\mathbf{v_1} = \mathbf{j}\) and \(\mathbf{v_2} = \mathbf{j} - 2\mathbf{k}\), the cross product yields the normal vector: \(-2\mathbf{i} + \mathbf{k}\). This normal vector can then be used to define the plane equation.

Angle Between Planes and Lines

The angle between a line and a plane or two planes can be found using dot products. The angle \(\theta\) between a line with direction vector \(\mathbf{d} = (d_x, d_y, d_z)\) and a plane with normal vector \(\mathbf{n} = (a, b, c)\) is given by:

\(\cos(\theta) = \frac{|\mathbf{d} \cdot \mathbf{n}|}{\|\mathbf{d}\| \|\mathbf{n}\|}\)

If you need to find the angle between the line \(\mathbf{r} = (2 + 2t)\mathbf{i} + t\mathbf{j} - 2t\mathbf{k}\) and the \((y,z)\) plane, the normal vector of the \((y,z)\) plane is \(\mathbf{i}\). Using the formula, the angle can be calculated by taking the dot product and magnitudes of the direction and normal vectors.