Question

(a) Find the equations of the line through the points (4,-1,2) and (3,1,4) (b) Find the equation of the plane through the points (0,0,0),(1,2,3) and (2,1,1) (c) Find the distance from the point (1,1,1) to the plane \(3 x-2 y+6 z=12\) (d) Find the distance from the point (1,0,2) to the line \(\mathbf{r}=2 \mathbf{i}+\mathbf{j}-\mathbf{k}+(\mathbf{i}-2 \mathbf{j}+2 \mathbf{k}) t\) (e) Find the angle between the plane in (c) and the line in (d).

Short answer

Line equation is \( x = 4 - t \), \( y = -1 + 2t \), \( z = 2 + 2t \).Plane equation: \( -x + 5y - 3z = 0\).Distance from (1,1,1) to plane: \(\frac{5}{7}\). Distance from (1,0,2) to line: \(\frac{\sqrt{59}}{3}\). \(\theta = \frac{4\sqrt{5}}{13}\).

Step-by-step solution

Find vector through the given points for the line

To find the equation of the line through points (4,-1,2) and (3,1,4), first find the direction vector. Subtract coordinates of the points: \[ \textbf{v} = (3-4, 1+1, 4-2) = (-1, 2, 2). \]

Write the parametric equations of the line

Using point (4,-1,2) and direction vector (-1,2,2), the parametric equations of the line are: \[ \textbf{r}(t) = (4, -1, 2) + t(-1, 2, 2) \] \[ x = 4 - t \] \[ y = -1 + 2t \] \[ z = 2 + 2t. \]

Find two vectors in the plane using three points

For the plane through points (0,0,0), (1,2,3), (2,1,1), find vectors \[ \textbf{v}_1 = (1-0, 2-0, 3-0) = (1, 2, 3) \] \[ \textbf{v}_2 = (2-0, 1-0, 1-0) = (2, 1, 1). \]

Find normal vector to the plane

Find the cross product of \( \textbf{v}_1 \) and \( \textbf{v}_2 \): \[ \textbf{n} = \textbf{v}_1 \times \textbf{v}_2 = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ 1 & 2 & 3 \ 2 & 1 & 1 \ \textbf{i}(2 \times 1 - 3 \times 1) - \textbf{j}(1 \times 1 - 3 \times 2) + \textbf{k}(1 \times 1 - 2 \times 2) \ = -\textbf{i} + 5\textbf{j} -3\textbf{k} \] Thus, \( \textbf{n} = (-1, 5, -3) \).

Write the equation of the plane

Using point (0,0,0) and normal vector (-1, 5, -3), the plane equation is: \[ -1(x-0) + 5(y-0) - 3(z-0) = 0 \] \[ -x + 5y - 3z = 0. \]

Use the point-to-plane distance formula

To find the distance from point (1,1,1) to the plane \( 3x - 2y + 6z = 12 \), use: \[ d = \frac{|3(1) - 2(1) + 6(1) - 12|}{\sqrt{3^2 + (-2)^2 + 6^2}} \] \[ = \frac{|3 - 2 + 6 - 12|}{\sqrt{9 + 4 + 36}} \] \[ = \frac{| -5 |}{\sqrt{49}} \] \[ = \frac{5}{7}. \]

Use the point-to-line distance formula

For finding the distance from (1,0,2) to the line \( \textbf{r} = 2 \textbf{i} + \textbf{j} - \textbf{k} + t(\textbf{i} - 2 \textbf{j} + 2 \textbf{k}) \), use: \[ \textbf{r}_0 = (2, 1, -1), \textbf{v} = (1, -2, 2), \] point \( P = (1,0,2) \): \[ d = \frac{|(P - \textbf{r}_0) \times \textbf{v}|}{|\textbf{v}|} \] \[ = \frac{|(1-2, 0-1, 2+1) \times (1, -2, 2)|}{\sqrt{1^2 + (-2)^2 + 2^2}} \] \[ = \frac{|(-1, -1, 3) \times (1, -2, 2)|}{3} \] \[ = \frac{|(-1-6, 2-3, 1+2)|}{3} \] \[ = \frac{|(-7, -1, 3)|}{3} \] \[ = \frac{\sqrt{7^2 + (-1)^2 + 3^2}}{3} \] \[ = \frac{\sqrt{49 + 1 + 9}}{3} = \frac{\sqrt{59}}{3}. \]

Find angle between line and plane

To find the angle \( \theta \) between the plane \( 3x - 2y + 6z = 12 \) and line \( \textbf{r} = 2\textbf{i} + \textbf{j} -\textbf{k} + t(\textbf{i} - 2\textbf{j} + 2\textbf{k}) \), use: \[ \tan(\theta) = \frac{| \textbf{n} \times \textbf{d}|}{|\textbf{n} \bullet \textbf{d}|}\] \(\textbf{n} = (3, -2, 6) \) and \( \textbf{d} = (1,-2,2) \): \[ = \frac{|(3, -2, 6) \times (1, -2, 2)|}{| (3, -2, 6) \cdot (1, -2, 2)|} \] \[ = \frac{|(8, 0, -4)|}{13} = \frac{\sqrt{64 + 16}}{13} = \frac{4\sqrt{5}}{13}. \]

Key concepts

Direction Vector

The direction vector is crucial when dealing with lines in 3-dimensional space. It's essentially a vector that points in the direction the line is heading. To find it, you simply subtract the coordinates of two given points on the line. For example, if we have points \(A(4, -1, 2)\) and \(B(3, 1, 4)\), the direction vector \(\textbf{v}\) is obtained by:
\[ \textbf{v} = (3-4, 1+1, 4-2) = (-1, 2, 2) \] This vector helps define the orientation of the line.

Parametric Equations

Parametric equations provide a way to express the coordinates of points on a line using a parameter \(t\). They are very useful because they allow you to easily describe the line in terms of a variable. Given a point \(P(4, -1, 2)\) and a direction vector \(\textbf{v} = (-1, 2, 2)\), the parametric equations of the line can be written as:
\[ \textbf{r}(t) = (4, -1, 2) + t(-1, 2, 2) \] This gives us:
\[ x = 4 - t \]
\[ y = -1 + 2t \]
\[ z = 2 + 2t \]
These equations allow us to generate any point on the line by simply varying \(t\).

Cross Product

The cross product is a vector operation that helps find a vector perpendicular to two given vectors. This is particularly useful when finding the normal vector to a plane. For instance, given vectors \(\textbf{v}_1 = (1, 2, 3)\) and \(\textbf{v}_2 = (2, 1, 1)\), the cross product \(\textbf{n} = \textbf{v}_1 \times \textbf{v}_2\) is:
\[ \textbf{n} = \textbf{i}(2 \cdot 1 - 3 \cdot 1) - \textbf{j}(1 \cdot 1 - 3 \cdot 2) + \textbf{k}(1 \cdot 1 - 2 \cdot 2) \] This simplifies to:
\[ \textbf{n} = -\textbf{i} + 5\textbf{j} - 3\textbf{k} \] or \[ \textbf{n} = (-1, 5, -3) \] This vector is perpendicular to both \(\textbf{v}_1\) and \(\textbf{v}_2\), making it a normal vector to the plane.

Normal Vector

A normal vector to a plane is crucial for defining the orientation of the plane. It is a vector that is perpendicular to every vector lying on the plane. Once we have the normal vector, we can write the equation of the plane. For example, for a normal vector \(\textbf{n} = (-1, 5, -3)\) passing through point \(P(0, 0, 0)\), the equation of the plane is:
\[ -1(x-0) + 5(y-0) - 3(z-0) = 0 \] This simplifies to:
\[ -x + 5y - 3z = 0 \] This equation represents the plane in 3-dimensional space.

Point-to-Plane Distance

To find the distance from a point to a plane, you can use a specific formula. If the point is \(P(1, 1, 1)\) and the plane's equation is \(3x - 2y + 6z = 12\), the distance \(d\) is given by:
\[ d = \frac{|3(1) - 2(1) + 6(1) - 12|}{\sqrt{3^2 + (-2)^2 + 6^2}} \] This simplifies to:
\[ d = \frac{|3 - 2 + 6 - 12|}{\sqrt{9 + 4 + 36}} = \frac{5}{7} \] This formula measures the shortest distance from the point to the plane.

Point-to-Line Distance

The shortest distance from a point to a line can be calculated using a different formula. Given a line described by \(\textbf{r} = 2\textbf{i} + \textbf{j} - \textbf{k} + t(\textbf{i} - 2\textbf{j} + 2\textbf{k})\) and a point \(P = (1, 0, 2)\), the distance \(d\) is:
\[ \textbf{r}_0 = (2, 1, -1), \textbf{v} = (1, -2, 2) \]
\[ d = \frac{|(1-2, 0-1, 2+1) \times (1, -2, 2)|}{|\textbf{v}|} \] Simplifying, we get:
\[ d = \frac{|-7, -1, 3|}{3} = \frac{\sqrt{59}}{3} \] This gives the shortest distance between the point and the line.

Angle Between Line and Plane

The angle between a line and a plane is found using the dot and cross products. Given a line with direction vector \(\textbf{d} = (1, -2, 2)\) and a plane with normal vector \(\textbf{n} = (3, -2, 6)\), the angle \(\theta\) between them is:
\[ \tan(\theta) = \frac{|\textbf{n} \times \textbf{d}|}{|\textbf{n} \cdot \textbf{d}|} \] Calculating, we have:
\[ |\textbf{n} \times \textbf{d}| = |(8, 0, -4)| = \sqrt{64 + 16} = 4\sqrt{5} \]
\[ |\textbf{n} \cdot \textbf{d}| = |13| \] Thus:
\[ \tan(\theta) = \frac{4\sqrt{5}}{13} \] This gives us the angle between the line and the plane.