Sometimes people don’t want to hear the truth because they don’t want their illusions destroyed, the truth is too hard to bear, they don’t have the time or just don’t care or some unfathomable combination of the above, JustToThePoint, Anawim, #justtothepoint.
Definition. A vector $\vec{AB}$ is a geometric object or quantity that has both magnitude (or length) and direction. Vectors in an n-dimensional Euclidean space can be represented as coordinates vectors in a Cartesian coordinate system.
Definition. The magnitude of the vector $\vec{A}$, also known as length or norm, is given by the square root of the sum of its components squared, $|\vec{A}|~ or~ ||\vec{A}|| = \sqrt{a_1^2+a_2^2+a_3^2}$, e.g., $||< 3, 2, 1 >|| = \sqrt{3^2+2^2+1^2}=\sqrt{14}$, $||< 3, -4, 5 >|| = \sqrt{3^2+(-4)^2+5^2}=\sqrt{50}=5\sqrt{2}$, or $||< 1, 0, 0 >|| = \sqrt{1^2+0^2+0^2}=\sqrt{1}=1$.
The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a_{1}+b_{1}), (a_{2}+b_{2}), (a_{3}+b_{3}) >.
The subtraction of two vectors is similar to addition and is also done component-wise, it is given by simply subtracting their corresponding components (x, y, and z in 3D), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a_{1}-b_{1}), (a_{2}-b_{2}), (a_{3}-b_{3}) >.
Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction. It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = < ca_1, ca_2, ca_3>$.
The dot or scalar product is a fundamental operation between two vectors. It produces a scalar quantity that represents the projection of one vector onto another. The dot product is the sum of the products of their corresponding components: $\vec{A}·\vec{B} = \sum a_ib_i = a_1b_1 + a_2b_2 + a_3b_3.$, e.g. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$
It is the product of their magnitudes multiplied by the cosine of the angle between them, $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$
Let $\vec{A}, \vec{B},~\text{and} \vec{C}$ be three vectors in ℝ^{3} with components,
$\vec{A} = (\begin{smallmatrix}a_1\\a_2\\a_3\end{smallmatrix}), \vec{B} = (\begin{smallmatrix}b_1\\b_2\\b_3\end{smallmatrix}), \vec{C} = (\begin{smallmatrix}c_1\\c_2\\c_3\end{smallmatrix})$.
The determinant of the matrix formed by these three vectors, denoted as $det(\vec{A}, \vec{B}, \vec{C})$, is given by
$det(\vec{A}, \vec{B}, \vec{C}) = |\begin{smallmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{smallmatrix}| = a_1det(\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix})-a_2det(\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix})+a_3det(\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix})$
Where the 2 × 2 determinants (also called minors) are calculated as follows:
$det(\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}) = b_2c_3-b_3c_2, det(\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}) = b_1c_3 -b_3c_1, det(\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix})=b_1c_2-b_2c_1$
In conclusion, $det(\vec{A}, \vec{B}, \vec{C}) = a_1(b_2c_3-b_3c_2) -a_2(b_1c_3 -b_3c_1) + a_3(b_1c_2-b_2c_1)$
Imagine each vector $\vec{A}, \vec{B},\text{and} \vec{C}$ originating from the same point. They extend out in space, forming the edges of a three-dimensional shape called a parallelepiped (a three-dimensional figure with parallelogram-like faces).
Geometrically, the determinant $det(\vec{A}, \vec{B}, \vec{C})$ represents the volume of the parallelepiped spanned or formed by the vectors $\vec{A}, \vec{B},and~ \vec{C}$ (Figure ii).
The sign of the determinant indicates the orientation of the three vectors:
Definition. The cross product, denoted by $\vec{A}x\vec{B}$, is a binary operation on two vectors in three-dimensional space. It results in a vector that is perpendicular to both of the input vectors, has a magnitude equal to the area of the parallelogram formed by the two input vectors, and has a direction determined by the right-hand rule.
This operation is particularly useful in maths, physics, and engineering for finding a vector that is orthogonal (normal or perpendicular) to a plane defined by two other vectors, among other applications.
Direction and Right-Hand Rule. The direction of the resulting vector $\vec{A}x\vec{B}$ is determined by the right-hand rule. If you curl the fingers of your right hand from $\vec{A}$ to $\vec{B}$ (first or index finger points in the direction of $\vec{A}$, second or middle finger points to $\vec{B}$), your thumb, when extended, points in the direction of the cross product $\vec{A} \times \vec{B}$ (Figure iii).
The cross product $\vec{A}x\vec{B}$ can be computed using the following formula involving determinants:
$\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$
Where the 2×2 determinants (minors) are calculated as follows:
$|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}| = a_2b_3-a_3b_2, |\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}| = a_1b_3-a_3b_1, |\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}| = a_1b_2-a_2b_1.$
Putting it all together, we get: $\vec{A}x\vec{B} = (a_2b_3-a_3b_2)\vec{i}-(a_1b_3-a_3b_1)\vec{j}+(a_1b_2-a_2b_1)\vec{k}$
To better understand the concept of the cross product, let’s go through a few examples in detail.
Cross Product of Unit Vectors. Find the cross product $\vec{i}x\vec{j}$. $\vec{i}x\vec{j} = det(\begin{smallmatrix}i & j & k\\ 1 & 0 & 0\\ 0 & 1 & 0\end{smallmatrix}) = (0·0-0·1)\vec{i}-(1·0-0·0)\vec{j}+(1·1-0·0)\vec{k} = \vec{k}$
Cross product of two vectors. Find the cross product $⟨1,0,2⟩x⟨−1,−2,4⟩$. $⟨1,0,2⟩x⟨−1,−2,4⟩ = det(\begin{smallmatrix}i & j & k\\ 1 & 0 & 2 \\ -1 & -2 & 4\end{smallmatrix})=(0⋅4−2⋅−2)\vec{i}−(1⋅4−2⋅−1)\vec{j}+(1⋅−2−0⋅−1)\vec{k}=4\vec{i}-6\vec{j}+-2\vec{k}=⟨4, -6, -2⟩$.
$⟨−2,1,3⟩x⟨5,2,−1⟩ = det(\begin{smallmatrix}i & j & k\\ -2 & 1 & 3 \\ 5 & 2 & -1\end{smallmatrix})=-7\vec{i}+13\vec{j}+-9\vec{k}=⟨-7, 13, -9⟩$.
To calculate the volume of a parallelepiped spanned by the vectors $\vec{A}, \vec{B},and~ \vec{C}$ (Figure v), we can use the base times heigh formula. This can be expressed as:
Volume = $|\vec{B}x\vec{C}||\vec{A}·\vec{n}|$ where $\vec{n}$ is a unit vector perpendicular to the parallelogram formed by $\vec{B} \text{and} \vec{C}$. First, we need to find $\vec{n}$:
$\vec{n} = \frac{\vec{B}x\vec{C}}{|\vec{B}x\vec{C}|}$
Volume = $|\vec{B}x\vec{C}||\vec{A}·\frac{\vec{B}x\vec{C}}{|\vec{B}x\vec{C}|}| = |\vec{A}·(\vec{B}x\vec{C})| =[🚀] |det(\vec{A},\vec{B},\vec{C})|$
This shows that the volume of the parallelepiped can also be found using the determinant of the matrix formed by the vectors $\vec{A}, \vec{B},and~ \vec{C}$.
To verify the last step [🚀], let’s look at the determinant form:
$det(\vec{A}, \vec{B}, \vec{C}) = |\begin{smallmatrix}a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\\ c_1 & c_2 & c_3\end{smallmatrix}| = a_1|\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}|-a_2|\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}|+a_3|\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix}|$
Expanding $\vec{A}·(\vec{B}x\vec{C})$
$\vec{A}·(\vec{B}x\vec{C}) =\vec{A}·(|\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix}|\vec{k}) = a_1|\begin{smallmatrix}b_2 & b_3\\ c_2 & c_3\end{smallmatrix}|-a_2|\begin{smallmatrix}b_1 & b_3\\ c_1 & c_3\end{smallmatrix}|+a_3|\begin{smallmatrix}b_1 & b_2\\ c_1 & c_2\end{smallmatrix}|$ ∎
The cross product, a fundamental operation in vector algebra, has several important properties. Let $\vec{u}, \vec{v}$, and $\vec{w}$ be vectors in space, and let c be a scalar, here are the key properties of the cross product:
First, we need to find the vectors representing two sides of the triangle. Let’s use the vectors $\vec{PQ}$ and $\vec{PR}$.
$\vec{PQ} =⟨0−1,1−0,0−0⟩=⟨−1,1,0⟩, \vec{PR} = ⟨0−1,0−0,1−0⟩=⟨−1,0,1⟩.$
The area of the parallelogram with adjacent sides $\vec{PQ}$ and $\vec{PR}$ is given by the magnitude of their cross product $|\vec{PQ}x\vec{PR}| = \Biggl \vert\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ -1 & 1 & 0\\ -1 & 0 & 1\end{smallmatrix}\Biggr \vert$
This determinant is calculated as follows: $\vec{PQ}x\vec{PR} = (1·1-0·0)\hat{\mathbf{i}}-(-1·1-0·(-1))\hat{\mathbf{j}}+(-1·0-(-1)·1) = \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Now, we find the magnitude of the resulting vector $\vec{PQ}x\vec{PR} = \sqrt{1^2+1^2+1^2} = \sqrt{3}$
The area the triangle with vertices P = (1, 0, 0), Q = (0, 1, 0), and R = (0, 0, 1) is half the area of the parallelogram with adjacent sides $\vec{PQ}$ and $\vec{PR}$, that is, $\frac{\sqrt{3}}{2}$
✴️ Fundamental idea ✴️ If the points in space P(1, 0, 1), Q(2, 4, 6), R(3, -1, 2) and S(6, 2, 8) are coplanar, the volume of the parallelepiped spanned by the vector $\vec{PQ}, \vec{PR}$ and $\vec{PS}$ will be zero (Figure 1).
This volume can be calculated as follows:
Volume = $|det(\vec{PQ}, \vec{PR}, \vec{PS})| = |\vec{PQ}·(\vec{PR} x \vec{PS})|$
First, we need to find the vectors $\vec{PQ}, \vec{PR}$ and $\vec{PS}$.
$\vec{PQ} = ⟨Q−P⟩ = ⟨2−1,4−0,6−1⟩ = ⟨1, 4, 5⟩, \vec{PR} = ⟨R−P⟩=⟨3−1,−1−0,2−1⟩ = ⟨2, -1, 1⟩, \vec{PS} = ⟨S−P⟩ = ⟨6−1,2−0,8−1⟩ = ⟨5, 2, 7⟩$
Compute the Cross Product $\vec{PR} x \vec{PS}$:
$\vec{PR} x \vec{PS} = \Biggl \vert\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ 2 & -1 & 1\\ 5 & 2 & 7\end{smallmatrix}\Biggr \vert = (-7-2)\hat{\mathbf{i}}-(14-5)\hat{\mathbf{j}}+(4+5)\hat{\mathbf{j}} = ⟨-9, -9, 9⟩$
Volume = $|\vec{PQ}·(\vec{PR} x \vec{PS})| = |⟨1, 4, 5⟩·⟨-9, -9, 9⟩| = |-9-36+45| = 0$
Only a single plane can pass through any set of three non-colinear points. The plane must contain the vectors $\vec{PQ}$ and $\vec{PR}$.
First, we need to find the vectors $\vec{PQ}$ and $\vec{PS}$ using the coordinates of points P, Q, and R.
$\vec{PQ} = Q−P = ⟨2−1,4−0,5−(−1)⟩ = ⟨1, 4, 6⟩, \vec{PR} = R−P = ⟨3−1,1−0,7−(−1)⟩ = ⟨2, 1, 8⟩$.
The cross product $\vec{PQ}x\vec{PR}$ will give us a vector orthogonal (normal) to both vectors, and thus to the plane containing P, Q, and R.
$\vec{PQ}x\vec{PR} = \Biggl \vert\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ 1 & 4 & 6\\ 2 & 1 & 8\end{smallmatrix}\Biggr \vert = (4⋅8−6⋅1)\hat{\mathbf{i}}-(1⋅8−6⋅2)\hat{\mathbf{j}}+(1⋅1−4⋅2)\hat{\mathbf{k}} = ⟨26, 4, -7⟩$
The area of the triangle formed by the points P, Q, and R is half the area of the parallelogram formed by $\vec{PQ}x\vec{PR}$. The magnitude of the cross product $|\vec{PQ}x\vec{PR}|$ gives the area of this parallelogram.
Area_{TrianglePQR} = $\frac{1}{2}|\vec{PQ}x\vec{PR}| = \frac{1}{2}\sqrt{26^2+4^2+(-7)^2} = \frac{\sqrt{741}}{2}≈ 13.61$
The general form of a plane in 3D space is given by: Ax + By + Cz = D. In our case, the equation is x +2y+3z=0, which can be written in the form: 1⋅x + 2⋅y + 3⋅z=0
A plane can be defined by a point on the plane and a normal vector (a vector perpendicular to the plane). The coefficients of x, y, and z in the equation x + 2y + 3z= 0 give us the normal vector to the plane. Therefore, the normal vector $\vec{A} = ⟨1, 2, 3⟩$.
Any point P = ⟨x, y, z⟩ that satisfies the equation x + 2y + 3z = 0 lies on the plane. For example, the origin O= ⟨0, 0, 0⟩ satisfies the equation: x +2y +3z = 0. Hence, the plane passes through the origin.
Let $\vec{OP} = ⟨x, y, z⟩$ be the position vector of a point P = ⟨x, y, z⟩ on the plane. Then, the dot product of $\vec{OP}$ and the normal vector $\vec{A}$ must be zero, $\vec{OP}·\vec{A} = 0 ↭ x·1 + y·2 + z·3 = 0 ↭ x + 2y +3z = 0 ↭ \vec{OP}⟂\vec{A}$ (Figure A).
Find the line of intersection of the two given planes. The line of intersection of the planes x − z = 1 and y + 2z = 3 can be found by parameterizing the solution to the system of equations. Let’s solve for x and y in terms of z.
From the first plane (i), x = z + 1. From the second plane (ii), y = 3 - 2z. We can let z = t, where t is a parameter ⇒ the parametric form of the line of intersection is r(t) = ⟨t + 1, 3 -2t, t⟩.
Calculate the Normal Vector to the given Plane. The equation of the given plane is x + y -2z = 1.
Recall that the general form of the equation of a plane in ℝ^{3} is ax + by + cz + d = 0 where a, b, and c, are the components of the normal vector $\vec{N} = ⟨a, b, c⟩$, which is perpendicular to the plane. For our given plane, the normal vector is: $\vec{N} = ⟨1, 1, -2⟩$
Determine a plane perpendicular to the given plane x + y - 2z = 1. A plane perpendicular to the plane x + y -2x = 1 must have a normal vector $\vec{M} = ⟨a, b, c⟩$ that is perpendicular to $\vec{N} ⇒ \vec{M}·\vec{N} = 0$ ↭ a + b -2c = 0 ⇒[Solving for a in terms of b and c:] a = 2c -b$.
Form the Equation of the Third Plane. A point P with coordinates (x, y, z) is in the plane ↭ $\vec{P_1P}·\vec{M} = 0$ where P_{1} is a point in this third plane containing the line, r(t) = ⟨t + 1, 3 -2t, t⟩.
Choose t = 0, we have P_{1} = ⟨1, 3, 0⟩, therefore the equation of the plane is: $\vec{P_1P}·\vec{M} = ⟨x-1, y-3, z⟩·⟨a, b, c⟩ = (x-1)a+(y-3)b+zc = 0.$
Let’s consider another point on the line r(t) = ⟨t + 1, 3 -2t, t⟩. Choose t = 1, P_{2} = ⟨2, 1, 1⟩. Substituting these coordinates in the plane equation: (2−1)a + (1−3)b +1c = 0⇒ a -2b + c = 0 ⇒[Using the relation a = 2c−b] 2c -b -2b +c = 0 ⇒ 3c -3b = 0 ⇒c =b ⇒ a = 2c -b = c ⇒ a = b = c. If we take a = 1⇒ b = c = 1, and the equation of the plane becomes (x -1) + (y -3) + z = 0 ⇒ x + y + z = 4.
Using the point P_{1} = ⟨1, 3, 0⟩ and the normal vector $\vec{M}=⟨1, 1, 1⟩$, the equation of the third plane is x + y + z = 4.