All things are difficult before they are easy, Thomas Fuller.
When you have eliminated the impossible, whatever remains, however improbable, must be the truth, Sherlock Holmes.
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. $\vec{A}·\vec{B} = \sum a_ib_i = ⟨2, 2, -1⟩·⟨5, -3, 2⟩ = a_1b_1 + a_2b_2 + a_3b_3 = 2·5+2·(-3)+(-1)·2 = 10-6-2 = 2.$
The dot product of two vectors 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 projection of $\vec{a}$ onto $\vec{b}$ is $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$.
$\vec{v}$ is defined by its initial point ⟨−1,−3⟩ and terminal point ⟨2,1⟩. We can find $\vec{v}$ by subtracting the coordinates of the initial point from the coordinates of the terminal point: $\vec{v} = ⟨2−(−1),1−(−3)⟩ = ⟨2+1, 1+3⟩ = ⟨3, 4⟩$
The magnitude (length) of v is found using the Pythagorean theorem: $||\vec{v}|| = \sqrt{3^2+4^2}= \sqrt{25} = 5.$
The unit vector in the direction of $\vec{v}$ is obtained by dividing each component of $\vec{v}$ by its magnitude: $\vec{u} = \frac{\vec{v}}{||\vec{v}||} = ⟨\frac{3}{5}, \frac{4}{5}⟩$
For any scalar r and any two vectors $\vec{u} = ⟨x_1, y_1⟩$ and $\vec{v}= ⟨x_2, y_2⟩$
Commutative Addition: $\vec{u}+\vec{v}$=[By the definition of vector addition]$⟨x_1 + x_2, y_1 + y_2⟩$=[Using the commutative property of real numbers (for addition)]$⟨x_2 + x_1, y_2 + y_2⟩ =$[By the definition of vector addition]$ = \vec{v}+\vec{u}$.
Distributive property: $r(\vec{u}+\vec{v})$=[By definition of addition of vectors]$r·⟨x_1 + x_2, y_1 + y_2⟩=$[By the definition of scalar multiplication]$=⟨r(x_1 + x_2), r(y_1 + y_2)⟩$=[Apply the distributive property for real numbers]$⟨rx_1 + rx_2, ry_1 + ry_2⟩=$[We can separate this into two vectors]$=⟨rx_1, ry_1⟩ + ⟨rx_2, ry_2⟩ =$[By the definition of scalar multiplication]$ = r\vec{u}+r\vec{v}$
$\vec{PQ}$ represent the displacement from point P to point Q, then its components can be found by subtracting the components of P from the components of Q = ⟨x_{q} −x_{p}, y_{q} −y_{p}, z_{q} −z_{p}⟩ = (0−1, 1−0, 0−0)= ⟨−1, 1, 0⟩. Another way to reach the same conclusion is as follows: $\vec{OP}+\vec{PQ} = \vec{OQ} ⇒ \vec{PQ} = \vec{OQ} - \vec{OP} = ⟨0, 1, 0⟩ -⟨1, 0, 0⟩ = ⟨-1, 1, 0⟩$.
$\vec{PR} = ⟨0−1,0−0,2−0⟩=⟨−1,0,2⟩$
The dot product of $\vec{PQ}$ and $\vec{PR}$ is given by (where θ is the angle between both vectors):
$\vec{PQ}·\vec{PR} = ||\vec{PQ}||·||\vec{PR}||cos(θ) ⇒ cos(θ) = \frac{\vec{PQ}·\vec{PR}}{||\vec{PQ}||·||\vec{PR}||} =\frac{⟨-1,1,0⟩·⟨-1, 0, 2⟩}{\sqrt{(-1)^2+1^2+0^2}\sqrt{(-1)^2+0^2+2^2}}=\frac{1+0+0}{\sqrt{2}\sqrt{5}} = \frac{1}{\sqrt{10}}$ ⇒To find the angle θ, we take the inverse cosine (arccos) of $\frac{1}{\sqrt{10}}$, θ = cos^{-1}($\frac{1}{\sqrt{10}}$) ≈ 71.5°.
Represent the forces as vectors:
Add the vectors to find the resultant force: $\vec{R}= \vec{F_1}+\vec{F_2} = ⟨300, 0⟩ + ⟨144.9,38.8⟩ = ⟨300+144.9,0+38.8⟩=⟨444.9,38.8⟩$ with magnitude $||\vec{R}||=\sqrt{444.9^2+38.8^2}=\sqrt{200,441.45}≈447.7$ lb, and direction θ = $tan^{-1}(\frac{38.8}{444.9}) = tan^{-1}(0.0873)≈5°$. Hence, the resultant force acting on the car is approximately 447.7 lb, and the direction angle of this force from the positive x-axis is approximately 5°.
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).
Let $\vec{A}$ and $\vec{B}$ be two vectors forming the sides of the triangle. These vectors can be represented in component form: $\vec{A} = ⟨a_1, a_2⟩, \vec{B} = ⟨b_1, b_2⟩$
The area of a triangle can be expressed as: Area_{Triangle} = $\frac{basexheight}{2}$.
If $\vec{A}$ is considered as the base, the height can be found using the sine of the angle θ between $\vec{A}$ and $\vec{B}: sin(θ)=\frac{h}{||\vec{B}||}⇒ h = sin(θ)x||\vec{B}||$.
Therefore, the area becomes: Area_{Triangle} = $\frac{1}{2}x||\vec{A}||x||\vec{B}||xsin(θ)$ where θ is the angle between the two vectors $\vec{A}$ and $\vec{B}$.
We define a new vector $\vec{A’}$ by rotating $\vec{A}$ 90 degrees. This rotation can be achieved by switching the components of $\vec{A}$ and changing the sign of one of them, ⟨-a_{2}, a_{1}⟩. Let $\vec{A} = ⟨a_1, a_2⟩$ and rotate it 90° anti-clockwise ⇒ $\vec{A’} = ⟨-a_2, a_1⟩$. If we rotate $\vec{A} = ⟨a_1, a_2⟩$ 90° clockwise ⇒ $\vec{A’} = ⟨a_2, -a_1⟩$
Let’s rotate $\vec{A}$ 90° anti-clockwise, the angle θ’ between $\vec{A’}$ and $\vec{B}$ satisfies: $θ’=\frac{π}{2}-θ$ ↭ θ and θ’ are complementary angles, cos(θ’) = sin(θ).
Area_{Triangle} = $\frac{1}{2}x||\vec{A}||x||\vec{B}||xsin(θ) = \frac{1}{2}x||\vec{A’}||x||\vec{B}||xcos(θ’) = \frac{1}{2}\vec{A’}·\vec{B} =$[🚀]
$\vec{A’}·\vec{B} = ⟨-a_2, a_1⟩·⟨b_1, b_2⟩ = a_1b_2-a_2b_1 =det(\vec{A}, \vec{B}) = [\begin{smallmatrix}a_1 & a_2\\ b_1 & b_2\end{smallmatrix}]$. In words, the dot product of the rotated vector $\vec{A’}$ and vector $\vec{B}$ gives the signed area of the parallelogram formed by $\vec{A}and \vec{B}$ (Figure ii).
The area of a parallelogram is the product of its base $||\vec{A}|| = ||\vec{A’}||$ and the corresponding altitude, $sin(θ)||\vec{B}||=cos(θ’)||\vec{B}||$
Area_{Triangle} = $\frac{1}{2}·||\vec{A}||·||\vec{B}||·sin(θ) = \frac{1}{2}·|det(\vec{A}, \vec{B})| = \frac{1}{2}·|\begin{smallmatrix}a_1 & a_2\\\ b_1 & b_2\end{smallmatrix}|$ where $det(\vec{A}, \vec{B})$ represents the determinant of the matrix formed by the components of vectors $\vec{A}$ and $\vec{B}$ respectively.