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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, $\vec{A}·\vec{B}=||\vec{A}||·||\vec{B}||·cos(θ).$
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 (at a right angle) to both of the original vectors and has a magnitude equal to the area of the parallelogram formed by the two input vectors.
The cross product $\vec{A}x\vec{B}$ can be computed using the following formula, $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}) =|\begin{smallmatrix}a_2 & a_3\\ b_2 & b_3\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}a_1 & a_3\\ b_1 & b_3\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}|\vec{k}$, e.g., $\vec{A}x\vec{B} = det(\begin{smallmatrix}i & j & k\\ 1 & 2 & 3\\ 4 & 5 & 6\end{smallmatrix}) =|\begin{smallmatrix}2 & 3\\ 5 & 6\end{smallmatrix}|\vec{i}-|\begin{smallmatrix}1 & 3\\ 4 & 6\end{smallmatrix}|\vec{j}+|\begin{smallmatrix}1 & 2\\ 4 & 5\end{smallmatrix}|\vec{k} = (2⋅6−3⋅5)\vec{i}-(1⋅6−3⋅4)+(1⋅5−2⋅4)\vec{j}\vec{k} = −3\vec{i}+6\vec{j}-3\vec{k} = ⟨-3, 6, -3⟩.$
To find the equation of a plane given three points P_{1}, P_{2}, and P_{3} in the plane, we need to determine a condition that ensures a fourth point P lies in the same plane.
Consider the vectors $\vec{P_1P}, \vec{P_1P_2},and~ \vec{P_1P_3}$. $\vec{P_1P}$ is a vector from point P_{1} to point P.
For point P to be in the planed defined by P_{1}, P_{2}, and P_{3}, the parallelepiped spanned by these vectors $\vec{P_1P}, \vec{P_1P_2},and~ \vec{P_1P_3}$ must be flat. This flatness condition is mathematically represented by the determinant of these vectors being zero: $det(\vec{P_1P}, \vec{P_1P_2},~ \vec{P_1P_3})$ = 0 (Figure vi).
Definition. A normal vector to a plane is a vector that is perpendicular to the plane., A normal vector is perpendicular (orthogonal) to all vectors within the plane.
An alternative way to find the equation of the plane is by using a normal vector. For point P to be in the plane, the vector $\vec{P_1P}$ must be perpendicular to the normal vector $\vec{N}$, $\vec{P_1P}⊥\vec{N}$ where $\vec{N}$ is a normal vector to our plane. This can be expressed using the dot product: $\vec{P_1P}·\vec{N}=0$ [🚀]
A normal vector $\vec{N}$ to the plane can be found using the cross product of the vectors $\vec{P_1P_2}$ and $\vec{P_1P_3}$: $\vec{N}=\vec{P_1P_2}x\vec{P_1P_3}$
Thus, the condition that the point P lies in the plane can be rewritten as:
$\vec{P_1P}·\vec{N}=0 [🚀] ⇒ \vec{P_1P}·(\vec{P_1P_2}x\vec{P_1P_3})=0$ (In the accompanying figure vi, the vectors and the plane can be visualized to better understand the geometric relationships and the conditions derived). This shows that the dot product of $\vec{P_1P}$ with the normal vector $\vec{N}$ must be zero for P to lie in the plane.
Find Vectors in the plane.
To find vectors that lie in the plane, we need to create vectors using the given points. More specifically, those vectors are $\vec{P_1P_2}$ and $\vec{P_1P_3}$, and we just need to subtract the coordinates of one point from the coordinates of the other point.
Let P_{1} has coordinates (x_{1}, y_{1}, z_{1}) and P_{2} has coordinates (x_{2}, y_{2}, z_{2}), then the coordinates of $\vec{P_1P_2}$ are ⟨x_{2}-x_{1}, y_{2}-y_{1}, z_{2}-z_{1}⟩. $\vec{P_1P_2} = ⟨2, 3, 4⟩, \vec{P_1P_3} = ⟨-1, 1, 2⟩$
Calculate the Normal vector to the plane. The normal vector $\vec{N}$ can be found using the cross product of $\vec{P_1P_2}$ and $\vec{P_1P_3}$.
$\vec{N}=\vec{P_1P_2}x\vec{P_1P_3} = [\begin{smallmatrix}i & j & k\\ a_1 & a_2 & a_3\\ b_1 & b_2 & b_3\end{smallmatrix}] = [\begin{smallmatrix}i & j & k\\ 2 & 3 & 4\\ -1 & 1 & 2\end{smallmatrix}] = (3⋅2−4⋅1)\vec{i}-((2⋅2−4⋅(−1)))\vec{j}+(2⋅1−3⋅(−1))\vec{k} = 2\vec{i}-8\vec{j}+5\vec{k}$
Form the Plane equation.
A point P with coordinates ⟨x, y, z⟩ is in the plane if and only if the dot product of the normal vector $\vec{N}$ with any vector $\vec{P_1P}$ from point P_{1} to a point P must be zero, that is, $\vec{P_1P}⊥\vec{N} ↭ \vec{P_1P}·\vec{N}=⟨x-1, y+2, z-0⟩·(2\vec{i}-8\vec{j}+5\vec{k}) = (x-1)2-8(y+2)+5z = 0 ↭ 2x -2 -8y -16+5z = 0 ↭ 2x -8y +5z = 18$, and this is the equation of the plane.
Conclusion: The equation of the plane containing the points P_{1} = ⟨1, −2, 0⟩, P_{2} = ⟨3, 1, 4⟩, and P_{3} = ⟨0, −1, 2⟩ is: 2x −8y +5z = 18.
Parallel planes have the same normal vectors. The normal vector of a plane ax + by + cz = d is $\vec{N}$ =⟨a,b,c⟩, hence any parallel to 2x +3y +z = 5, has a normal vector $\vec{N} = ⟨2, 3, 1⟩$.
A point P with coordinates (x, y, z) is in the plane ↭ $\vec{QP}·\vec{N} = 0 ↭ (x-1, y-1, z-1)·⟨2, 3, 1⟩ = 0 ↭ 2(x-1) +3(y-1)+z-1 = 0 ↭ 2x -2 + 3y -3 + z -1 = 0 ↭ 2x + 3y + z = 6.$
Therefore, the equation of the plane passing through the point Q (1, 1, 1) and parallel to the plane 2x +3y +z = 5 is 2x + 3y + z = 6.
Another way. A plane with normal vector $\vec{N} = ⟨2, 3, 1⟩$ has the form 2x + 3y + z = c, where c is a constant. Q(1, 1, 1) is in the plane. Substitute Q into the plane equation to find c: 2·1 + 3·1 + 1 = c ⇒ c = 6, so the plane is 2x + 3y + z = 6.
Find the Direction Vector. The direction vector $\vec{d}$ (a vector parallel to the line) can be found by subtracting the coordinates of point A from the coordinates of point B: $\vec{d}$ = ⟨−3−1, 5−4, 0+2⟩ = ⟨−4, 1, 2⟩.
Parametric Equations. Parametric equations express the coordinates (x, y, z) of any point on the line as functions of a parameter t.
The general form for the parametric equations is:
$ \begin{cases} x = x_0 + t·d_x \\ y = y_0 + t·d_y \\ z = z_0 + t·d_z \end{cases}$
where (x_{0}, y_{0},z_{0}) is a point on the line (we can use point A) and $\vec{d} = ⟨d_x, d_y, d_z⟩$ is the direction vector.
Substituting A = ⟨1, 4, -2⟩ and $\vec{d} = ⟨-4, 1, 2⟩$, the parametric equations of the line are:
$ \begin{cases} x = 1 - 4t \\ y = 4 + t \\ z = -2 + 2t \end{cases}$
Symmetric equations eliminate the parameter t and relate x, y, and z directly. To find the symmetric equations, we solve each of the parametric equations for t: $t = \frac{x-1}{-4}, t = y - 4, t = \frac{x + 2}{2}$.
Setting these expressions equal to each other: $\frac{x-1}{-4} = y - 4 = \frac{x + 2}{2}$ (symmetric equation of the line).
Find the Direction Vector $\vec{AB}$. It is obtained by subtracting the coordinates of point A from the coordinates of point B: $\vec{AB} = (3-1)\hat{\mathbf{i}+(6-3)\hat{\mathbf{j}}}-\hat{\mathbf{k}} = 2\hat{\mathbf{i}}+3\hat{\mathbf{j}}+\hat{\mathbf{k}}$
Find the vector equation of the line L. The vector equation of the line passing through A with direction vector $\vec{AB}$ is: $\vec{r} = \vec{a}+t\vec{AB} = \hat{\mathbf{i}}+3\hat{\mathbf{j}}+\hat{\mathbf{k}} +t(2\hat{\mathbf{i}}+3\hat{\mathbf{j}}-\hat{\mathbf{k}}) = (2t+1)\hat{\mathbf{i}} + (3t + 3)\hat{\mathbf{j}} + (t-1)\hat{\mathbf{k}}$
To find the shortest distance from point C to the line, we need to determine where the perpendicular from C meets the line L. Let’s call this point D, which lies on the line L.
$\vec{CD} = ⟨2t+1+2, 3t+3-4, t-1+3⟩ = ⟨2t+3, 3t-1, t +2⟩$
Find t by ensuring that $\vec{CD}$ is perpendicular to $\vec{AB}$, their dot product must be zero: $⟨2t+3, 3t-1, t +2⟩·⟨2, 3, 1⟩ = 0 ↭ 2(2t+3)+3(3t-1)+1·(t+2)=0 ↭ 4t+6+9t-3+t+2=0 ↭ 14t + 5 = 0 ↭ t = \frac{-5}{14}$.
$\vec{CD} = ⟨2·\frac{-5}{14}+3, 3(\frac{-5}{14})-1, \frac{-5}{14}+2⟩ = ⟨\frac{16}{7}, \frac{-29}{14}, \frac{23}{14}⟩$ The required distance is $|\vec{CD}| = \sqrt{(\frac{32}{14})^2 + (\frac{-29}{14})^2+ (\frac{23}{14})^2} = \sqrt{\frac{2394}{196}}≈ 3.5$
The normal vector of a plane can be directly obtained from the coefficients of x, y, and z in the plane’s equation. For the plane 3x + 4y+ 6z = 12, the normal vector $\vec{n}$ is: $\vec{n}=⟨3, 4, 6⟩$.
To graph the plane, we need to find the intercepts where the plane intersects the x-, y-, and z-axes.
x-intercept: Set y = 0 and z = 0 in the plane equation: x = 4. So, the x-intercept is (4, 0, 0). Analogously, the y-intercept is (0, 3, 0) and the z-intercept is (0, 0, 2). To graph the plane, we plot these points on a 3D coordinate system. Then, we draw a triangular plane that passes through these points. This triangular plane represents the portion of the plane 3x + 4y + 6z =12 that intersects the positive x-, y-, and z-axes.
In addition to finding the intercepts, we can also graph the traces of the plane in the xy-, xz-, and yz-planes.
xy-trace: Set z = 0: 3x + 4y = 12. This is the equation of a line in the xy-plane. xz-trace: 3x + 6z = 12, the equation of a line in the xz-plane. yz-trace: 4y + 6z = 12, the equation of a line in the yz-plane.
From the equations, we can extract a point and a direction vector for the line: (Figure D)
Therefore, $\vec{PQ}=⟨−1−(−1),−3−0,2−2⟩ = ⟨0, -3, 0⟩$ is in the plane, too.
The normal vector to the plane can be found by taking the cross product of the direction vector $\vec{v}$ and the vector $\vec{PQ}$: $\vec{n} = \vec{v}x\vec{PQ} = (\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ -2 & 4 & 1\\ 0 & -3 & 0\end{smallmatrix}) = (4⋅0−1⋅(−3))\hat{\mathbf{i}}-((−2)⋅0−1⋅0)\hat{\mathbf{j}}+((−2)⋅(-3)-0·1)\hat{\mathbf{j}} = ⟨3, 0, 6⟩$.
The equation of a plane with normal vector $\vec{n} = ⟨3, 0, 6⟩$ and a point Q(-1, -3, 2) in the plane is: 3x +6z = c, 3·(-1)+6·2 = c ⇒ c = 9 ⇒3x + 6z = 9 ⇒[Divide the entire equation by 3 to simplify:] x + 2z = 3 is the equation of the plane (Figure E).
Alternatively, 3(x + 1) +0(y + 3) + 6(z − 2) = 0 ⇒3x+6z=9 ⇒ x + 2z = 3.