Vectors - Problems - Geometry Of Cross Product

Vectors - Problems - Geometry of Cross Product

In this lecture, we will solve problems related to the geometry of cross product
The cross product of two vectors gives a vector perpendicular to both vectors
We will use the properties of cross product to solve various geometrical problems

Example 1: Finding the Area of a Triangle

Given two vectors: $\vec{A} = \begin{bmatrix} 3 \ 2 \ 1 \end{bmatrix}$ and $\vec{B} = \begin{bmatrix} 1 \ -1 \ 4 \end{bmatrix}$
To find the area of the triangle formed by these vectors, we can take the magnitude of their cross product
$\vec{A} \times \vec{B} = \begin{vmatrix} i & j & k \ 3 & 2 & 1 \ 1 & -1 & 4 \end{vmatrix} = 10i + 11j + 5k$
The magnitude of $\vec{A} \times \vec{B}$ is $\sqrt{10^2 + 11^2 + 5^2} = \sqrt{246}$
Therefore, the area of the triangle is $\frac{1}{2} \sqrt{246}$

Example 2: Finding the Angle between Two Vectors

Given two vectors: $\vec{A} = \begin{bmatrix} 2 \ 1 \ 3 \end{bmatrix}$ and $\vec{B} = \begin{bmatrix} 1 \ 2 \ -1 \end{bmatrix}$
The angle $\theta$ between the vectors is given by the equation $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{\lVert \vec{A} \rVert \lVert \vec{B} \rVert}$
$\vec{A} \cdot \vec{B} = 2 \cdot 1 + 1 \cdot 2 + 3 \cdot (-1) = -1$
$\lVert \vec{A} \rVert = \sqrt{2^2 + 1^2 + 3^2} = \sqrt{14}$
$\lVert \vec{B} \rVert = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}$
Therefore, $\cos \theta = \frac{-1}{\sqrt{14} \cdot \sqrt{6}}$
$\theta = \arccos \left(\frac{-1}{\sqrt{84}}\right)$

Example 3: Finding the Projection of a Vector

Given two vectors: $\vec{A} = \begin{bmatrix} 3 \ 1 \ -2 \end{bmatrix}$ and $\vec{B} = \begin{bmatrix} 4 \ -2 \ 1 \end{bmatrix}$
The projection of $\vec{A}$ onto $\vec{B}$ is given by the equation $\text{proj}_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{\lVert \vec{B} \rVert^2} \cdot \vec{B}$
$\vec{A} \cdot \vec{B} = 3 \cdot 4 + 1 \cdot (-2) + (-2) \cdot 1 = 10$
$\lVert \vec{B} \rVert^2 = 4^2 + (-2)^2 + 1^2 = 21$
Therefore, $\text{proj}_{\vec{B}} \vec{A} = \frac{10}{21} \cdot \begin{bmatrix} 4 \ -2 \ 1 \end{bmatrix}$

To continue with the lecture, please refer to the next set of slides. Sure! Here are slides 11 to 20:

Slide 11

Example 4: Finding the Equations of Planes
Given three points: $P(1, 2, 3)$, $Q(0, -1, 2)$, $R(-2, 3, 1)$
To find the equation of the plane passing through these points, we can use the cross product of two vectors formed by the points
$\vec{PQ} = \begin{bmatrix} 0-1 \ -1-2 \ 2-3 \end{bmatrix} = \begin{bmatrix} -1 \ -3 \ -1 \end{bmatrix}$
$\vec{PR} = \begin{bmatrix} -2-1 \ 3-2 \ 1-3 \end{bmatrix} = \begin{bmatrix} -3 \ 1 \ -2 \end{bmatrix}$
The equation of the plane is given by the equation $\vec{n} \cdot \vec{r} = \vec{n} \cdot \vec{P}$, where $\vec{n}$ is the normal vector to the plane and $\vec{r}$ is a position vector on the plane
Using $\vec{PQ}$ and $\vec{PR}$, we can find the normal vector $\vec{n} = \vec{PQ} \times \vec{PR}$
Therefore, the equation of the plane is $-x - 2y - z = -9$

Slide 12

Example 5: Finding the Volume of a Parallelepiped
Given three vectors: $\vec{A} = \begin{bmatrix} 1 \ 2 \ 3 \end{bmatrix}$, $\vec{B} = \begin{bmatrix} 2 \ 3 \ -1 \end{bmatrix}$, $\vec{C} = \begin{bmatrix} 5 \ 1 \ -2 \end{bmatrix}$
The volume of the parallelepiped formed by these vectors is given by $|\vec{A} \cdot (\vec{B} \times \vec{C})|$
$\vec{B} \times \vec{C} = \begin{vmatrix} i & j & k \ 2 & 3 & -1 \ 5 & 1 & -2 \end{vmatrix} = 5i + 13j + 17k$
$\vec{A} \cdot (\vec{B} \times \vec{C}) = 1 \cdot 5 + 2 \cdot 13 + 3 \cdot 17 = 88$
Therefore, the volume of the parallelepiped is $|88|$

Slide 13

Example 6: Finding the Scalar Triple Product
Given three vectors: $\vec{A} = \begin{bmatrix} -2 \ 1 \ 3 \end{bmatrix}$, $\vec{B} = \begin{bmatrix} 4 \ -1 \ 2 \end{bmatrix}$, $\vec{C} = \begin{bmatrix} 3 \ 2 \ -1 \end{bmatrix}$
The scalar triple product of these vectors is given by $[\vec{A}, \vec{B}, \vec{C}] = \vec{A} \cdot (\vec{B} \times \vec{C})$
$\vec{B} \times \vec{C} = \begin{vmatrix} i & j & k \ 4 & -1 & 2 \ 3 & 2 & -1 \end{vmatrix} = 7i + 14j - 11k$
$\vec{A} \cdot (\vec{B} \times \vec{C}) = -2 \cdot 7 + 1 \cdot 14 + 3 \cdot (-11) = -54$
Therefore, $[\vec{A}, \vec{B}, \vec{C}] = -54$

Slide 14

Example 7: Checking for Coplanarity
Given four points: $P(1, 1, 1)$, $Q(2, 3, -1)$, $R(-1, 4, -3)$, $S(4, 5, -5)$
To check if these points are coplanar, we can find the equation of the plane passing through any three points and check if the fourth point satisfies the equation
Let’s consider points $P$, $Q$, and $R$
Using method similar to Example 4, the equation of the plane passing through $P$, $Q$, and $R$ is $5x - 7y + 6z = -33$
Substituting the coordinates of point $S$ in the equation, we get $5(4) - 7(5) + 6(-5) = -33$
Therefore, these points are coplanar

Slide 15

Example 8: Finding the Angle between a Line and a Plane
Given a line with direction vector $\vec{L} = \begin{bmatrix} 1 \ 2 \ -1 \end{bmatrix}$ and a plane with normal vector $\vec{n} = \begin{bmatrix} 3 \ -1 \ -2 \end{bmatrix}$
The angle $\theta$ between the line and the plane is given by the equation $\cos \theta = \frac{|\vec{n} \cdot \vec{L}|}{\lVert \vec{n} \rVert \lVert \vec{L} \rVert}$
$\vec{n} \cdot \vec{L} = 3 \cdot 1 + (-1) \cdot 2 + (-2) \cdot (-1) = 7$
$\lVert \vec{n} \rVert = \sqrt{3^2 + (-1)^2 + (-2)^2} = \sqrt{14}$
$\lVert \vec{L} \rVert = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{6}$
Therefore, $\cos \theta = \frac{7}{\sqrt{14} \cdot \sqrt{6}}$
$\theta = \arccos \left(\frac{7}{\sqrt{84}}\right)$

Slide 16

Example 9: Finding the Distance between a Point and a Plane
Given a point $P(3, 4, 1)$ and a plane with equation $2x - 3y + z = 7$
The distance between the point and the plane is given by the equation $d = \frac{|\vec{P} \cdot \vec{n}|}{\lVert \vec{n} \rVert}$
$\vec{n} = \begin{bmatrix} 2 \ -3 \ 1 \end{bmatrix}$
$\vec{P} \cdot \vec{n} = 3 \cdot 2 + 4 \cdot (-3) + 1 \cdot 1 = -5$
$\lVert \vec{n} \rVert = \sqrt{2^2 + (-3)^2 + 1^2} = \sqrt{14}$
Therefore, the distance between the point and the plane is $\frac{5}{\sqrt{14}}$

Slide 17

Example 10: Finding the Intersection of a Line and a Plane
Given a line with equation $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-3}{-1}$ and a plane with equation $x - 2y + 2z = 9$
To find the intersection point, we can solve the system of equations formed by the line and the plane
Substituting the values of $x$, $y$, and $z$ from the line equation into the plane equation, we get $\frac{2+2}{3} - 2\left(\frac{-1+1}{2}\right) + 2\left(\frac{3-3}{-1}\right) = 9$
Simplifying the equation, we get $4 = 9$
Since the equation is not true, there is no intersection point

Slide 18

Example 11: Finding the Angle between Two Planes
Given two planes with equations $2x + y - 3z = 4$ and $x + 2y + z = 1$
The angle $\theta$ between the planes is given by the equation $\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{\lVert \vec{n_1} \rVert \lVert \vec{n_2} \rVert}$
$\vec{n_1} = \begin{bmatrix} 2 \ -1 \ -3 \end{bmatrix}$
$\vec{n_2} = \begin{bmatrix} 1 \ 2 \ 1 \end{bmatrix}$
$\vec{n_1} \cdot \vec{n_2} = 2 \cdot 1 + (-1) \cdot 2 + (-3) \cdot 1 = -1$
$\lVert \vec{n_1} \rVert = \sqrt{2^2 + (-1)^2 + (-3)^2} = \sqrt{14}$
$\lVert \vec{n_2} \rVert = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}$
Therefore, $\cos \theta = \frac{1}{\sqrt{84}}$
$\theta = \arccos \left(\frac{1}{\sqrt{84}}\right)$

Slide 19

Recap:
- The cross product of two vectors gives a vector perpendicular to both vectors.
- The area of a triangle formed by two vectors $\vec{A}$ and $\vec{B}$ can be found using the magnitude of their cross product: $\frac{1}{2} \lVert \vec{A} \times \vec{B} \rVert$.
- The angle between two vectors $\vec{A}$ and $\vec{B}$ can be found using $\cos \theta = \frac{\vec{A} \cdot \vec{B}}{\lVert \vec{A} \rVert \lVert \vec{B} \rVert}$.
- The projection of a vector $\vec{A}$ onto another vector $\vec{B}$ can be found using $\text{proj}_{\vec{B}} \vec{A} = \frac{\vec{A} \cdot \vec{B}}{\lVert \vec{B} \rVert^2} \cdot \vec{B}$.
- The volume of a parallelepiped formed by three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ can be found using $|\vec{A} \cdot (\vec{B} \times \vec{C})|$.
- The scalar triple product of three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ is given by $[\vec{A}, \vec{B}, \vec{C}] = \vec{A} \cdot (\vec{B} \times \vec{C})$.

Keep following for more examples and practice problems! Sure! Here are slides 21 to 30 in markdown format: ``

Slide 21

Example 12: Finding the Direction Ratios of a Line
Given the equation of a line $\frac{x-2}{3} = \frac{y+1}{2} = \frac{z-3}{-1}$
The direction ratios of the line are the coefficients of $x$, $y$, and $z$ in the equation
Therefore, the direction ratios are $3$, $2$, and $-1$

Slide 22

Example 13: Finding the Angle between a Line and a Plane
Given a line with direction ratios $3$, $1$, $-2$ and a plane with equation $x - 2y + 2z = 9$
To find the angle $\theta$ between the line and the plane, we can use the formula $\cos \theta = \frac{|\vec{d} \cdot \vec{n}|}{\lVert \vec{d} \rVert \lVert \vec{n} \rVert}$
$\vec{d} = \begin{bmatrix} 3 \ 1 \ -2 \end{bmatrix}$ (direction ratios of the line)
$\vec{n} = \begin{bmatrix} 1 \ -2 \ 2 \end{bmatrix}$ (normal vector of the plane)
$\vec{d} \cdot \vec{n} = 3 \cdot 1 + 1 \cdot (-2) + (-2) \cdot 2 = -5$
$\lVert \vec{d} \rVert = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{14}$
$\lVert \vec{n} \rVert = \sqrt{1^2 + (-2)^2 + 2^2} = \sqrt{9}$
Therefore, $\cos \theta = \frac{5}{\sqrt{126}}$

Slide 23

Example 14: Finding the Shortest Distance between Two Skew Lines
Given two lines with equations:
- Line 1: $\frac{x-1}{3} = \frac{y+2}{-2} = \frac{z-3}{1}$
- Line 2: $\frac{x+2}{1} = \frac{y-1}{-3} = \frac{z+4}{2}$
To find the shortest distance between the lines, we can use the formula $d = \frac{|\vec{P_1P_2} \cdot \vec{n}|}{\lVert \vec{n} \rVert}$
$\vec{P_1P_2}$ is