Vectors III. Planes and Areas

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.

Recall

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, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k}$ = < (a₁+b₁), (a₂+b₂), (a₃+b₃) >.

• 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), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k}$ = < (a₁-b₁), (a₂-b₂), (a₃-b₃) >.

• 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 $proj_{\vec{b}}\vec{a} = \frac{\vec{a}·\vec{b}}{||\vec{b}||^2}\vec{b}$.

Vectors III. Exercises

• A vector $\vec{v}$ has initial point ⟨−1,−3⟩ and terminal point ⟨2, 1⟩. Find the unit vector in the direction of $\vec{v}$.

$\vec{v}$ 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: $\vec{u} = \frac{\vec{v}}{||\vec{v}||} = ⟨\frac{3}{5}, \frac{4}{5}⟩$

• Proof of Commutative and distributive property.

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}$

• Let’s P = (1, 0, 0), Q = (0, 1, 0), and R = (0, 0, 2). They form a triangle and we want to calculate the angle θ formed by two of their sides (Figure vi).

$\vec{PQ}$ 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°.

• Martha’s car is stuck in the mud. A truck pulls it with a horizontal force of 300 lb, while Martha and her friend push at a 15° upward angle with a force of 150 lb. Calculate the resultant force’s magnitude and direction angle from the positive x-axis (Figure A).

Represent the forces as vectors:

1. The truck’s force can be represented as a horizontal vector $\vec{F_1}$ with a magnitude of 300 lb. Since it’s horizontal, $\vec{F_1} = ⟨300, 0⟩$.
2. Martha and her friend’s pushing force can be represented as a vector $\vec{F_2}$ with a magnitude of 150 lb at an angle of 15° to the truck’s force. $\vec{F_2} =$[Using basic trigonometry] ⟨150cos(15°), 150sin(15°)⟩ = ⟨150×0.9659, 150×0.2588⟩ ≈ ⟨144.9, 38.8⟩.

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°.

• What is the set of points defined by the equation x + 2y + 3z = 0.

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).

• Calculate or express the Area of a Triangle using Vectors (Figure D)

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 $\vec{B}$.

Using the Determinant for Calculation

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, ⟨-a₂, a₁⟩. 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.

• Find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).

1. Find the vectors $\vec{AB}$ and $\vec{AC}$: $\vec{AB}=(2−1,−1−2,4−3)=(1,−3,1),\vec{AC} = (4−1,5−2,−1−3)=(3,3,−4)$
2. Compute the cross product $\vec{AB}x\vec{AC}$: $\vec{AB}x\vec{AC} = (\begin{smallmatrix}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}}\\ 1 & -3 & 1\\3 & 3 & -4\end{smallmatrix}) = \hat{\mathbf{i}}((−3)(−4)−(1)(3))−\hat{\mathbf{j}}((1)(−4)−(1)(3))+\hat{\mathbf{k}}((1)(3)−(−3)(3)) = \hat{\mathbf{i}}(12−3)−\hat{\mathbf{j}}(−4−3)+\hat{\mathbf{k}}(3+9)=⟨9,7,12⟩$
3. Compute the Area of the triangle, Area = $\frac{1}{2}||\vec{AB}x\vec{AC}|| = \frac{1}{2}\sqrt{9^2+7^2+12^2}=\frac{1}{2}\sqrt{274}=\frac{\sqrt{274}}{2}.$

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