Vectors. The Dot Product and Projections.

All those who seem stupid, they are, and also so are half of those who do not, Quevedo

Recall: Vector Basics

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., $\vert{}\vert{}\langle 3, 2, 1 \rangle\vert{}\vert{} = \sqrt{3^2+2^2+1^2} = \sqrt{14}$, $\vert{}\vert{}\langle 3, -4, 5 \rangle\vert{}\vert{} = \sqrt{3^2+(-4)^2+5^2} = \sqrt{50} = 5\sqrt{2}$, or $\vert{}\vert{}\langle 1, 0, 0 \rangle\vert{}\vert{} = \sqrt{1^2+0^2+0^2} = \sqrt{1} = 1$.

Basic Vector Operations

• Vector Addition: The sum of two vectors is the sum of their components, $\vec{A}+ \vec{B} = (a_1+b_1)\vec{i}+(a_2+b_2)\vec{j}+(a_3+b_3)\vec{k} = \langle a_1+b_1, a_2+b_2, a_3+b_3 \rangle$.

• Vector Subtraction: 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), $\vec{A} - \vec{B} = (a_1-b_1)\vec{i}+(a_2-b_2)\vec{j}+(a_3-b_3)\vec{k} = \langle a_1-b_1, a_2-b_2, a_3-b_3 \rangle$.

• Scalar multiplication is the multiplication of a vector by a scalar, a real number, changing its magnitude without altering its direction (unless the scalar is negative, which reverses the direction). It is effectively multiplying each component of the vector by the scalar value, $c\vec{A} = (ca_1)\vec{i}+(ca_2)\vec{j}+(ca_3)\vec{k} = \langle ca_1, ca_2, ca_3 \rangle$.

• 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. $\vec{A}·\vec{B} = \sum a_ib_i = \langle 2, 2, -1 \rangle \cdot \langle 5, -3, 2 \rangle = (2 \cdot 5) + (2 \cdot -3) + (-1 \cdot 2) = 10 - 6 - 2 = 2.$

Geometrically, it is the product of their magnitudes multiplied by the cosine of the angle $\theta$ between them, $\boxed{\vec{A} \cdot \vec{B} = \vert{}\vert{}\vec{A}\vert{}\vert{} \cdot \vert{}\vert{}\vec{B}\vert{}\vert{} \cos(\theta)}$

Geometric Definition of the Dot Product

Theorem. The dot product of two vectors $\vec{A}$ and $\vec{B}$ is the product of their magnitudes multiplied by the cosine of the angle $\theta$ between them: $\vec{A} \cdot \vec{B} = \Vert{}\vec{A}\Vert{} \Vert{}\vec{B}\Vert{} \cos(\theta)$

Proof

1. Self Dot Product: The dot product of a vector with itself yields the square of its magnitude: $\vec{A} \cdot \vec{A} = \sum_{i=1}^{n} a_i^2 = a_1^2 + a_2^2 + a_3^2 = \Vert{}\vec{A}\Vert{}^2$
   This matches the geometric formula since $\cos(0) = 1$, giving $\Vert{}\vec{A}\Vert{} \Vert{}\vec{A}\Vert{} \cos(0) = \Vert{}\vec{A}\Vert{}^2$.

2. The Law of cosine relates the lengths of the sides of a triangle to the cosine of one of its angle. For a triangle with sides a, b, and c, $c^2 = a^2 + b^2 -2ab\cos(\theta)$ where $\theta$ is the angle opposite to the side c.
   Consider the triangle formed by vectors $\vec{A}$, $\vec{B}$, and $\vec{C} = \vec{A} - \vec{B}$ (Figure v).
   By the Law of Cosines: $\Vert{}\vec{C}\Vert{}^2 = \Vert{}\vec{A}\Vert{}^2 + \Vert{}\vec{B}\Vert{}^2 - 2\Vert{}\vec{A}\Vert{} \Vert{}\vec{B}\Vert{} \cos(\theta) \quad \text{--- (Equation 1)}$

3. Algebraic Expansion: Expanding $\Vert{}\vec{C}\Vert{}^2 = (\vec{A} - \vec{B}) \cdot (\vec{A} - \vec{B})$ using the distributive and commutative properties ($\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$):
   $\Vert{}\vec{C}\Vert{}^2 = \vec{A} \cdot \vec{A} - 2(\vec{A} \cdot \vec{B}) + \vec{B} \cdot \vec{B} = \Vert{}\vec{A}\Vert{}^2 + \Vert{}\vec{B}\Vert{}^2 - 2(\vec{A} \cdot \vec{B}) \quad \text{--- (Equation 2)}$

4. Equating Both Expressions: Setting Equation 1 equal to Equation 2: $\Vert{}\vec{A}\Vert{}^2 + \Vert{}\vec{B}\Vert{}^2 - 2\Vert{}\vec{A}\Vert{} \Vert{}\vec{B}\Vert{} \cos(\theta) = \Vert{}\vec{A}\Vert{}^2 + \Vert{}\vec{B}\Vert{}^2 - 2(\vec{A} \cdot \vec{B})$.
   Finally, canceling $\Vert{}\vec{A}\Vert{}^2 + \Vert{}\vec{B}\Vert{}^2$ from both sides and dividing by $-2$: $\vec{A} \cdot \vec{B} = \Vert{}\vec{A}\Vert{} \Vert{}\vec{B}\Vert{} \cos(\theta) \quad \blacksquare$

Vector Projections and Decomposition

The dot product allows us to resolve any vector into components aligned with or perpendicular to a given direction.

• Unit Direction Vector: The unit vector pointing along $\vec{b}$ is $\vec{u} = \frac{\vec{b}}{\Vert{}\vec{b}\Vert{}}$, where $\Vert{}\vec{u}\Vert{} = 1$.
  $||\vec{u}|| = \sqrt{\frac{b_1^2}{||\vec{b}||^2}+\frac{b_2^2}{||\vec{b}||^2}+\frac{b_3^2}{||\vec{b}||^2}} = \sqrt{\frac{b_1^2+b_2^2+b_3^2}{||\vec{b}||^2}} = \frac{\sqrt{b_1^2+b_2^2+b_3^2}}{||\vec{b}||} = \frac{||\vec{b}||}{||\vec{b}||} = 1$, i.e., $\vec{u}$ is an unit vector.

• Scalar/Vector Projection (Component):
  The dot product of $\vec{a}$ with the unit vector $\vec{u}$, denoted as $\vec{a}·\vec{u}$, is the projection of $\vec{a}$ in the direction of $\vec{u}$, or the amount that $\vec{a}$ is pointing (or aligned) in the same direction as the unit vector $\vec{u}$. By forming a right triangle with the vector $\vec{a}$ and this shadow, you can calculate $\cos(\theta)=\frac{c_1}{\Vert{}\Vert{}\vec{a}\Vert{}\Vert{}} \implies c_1 = \Vert{}\Vert{}\vec{a}\Vert{}\Vert{}\cdot \cos(\theta)$.
  $\vec{a}·\vec{u} =[\text{Previous theorem}] ||\vec{a}|| \cdot 1 \cdot \cos(\theta) = c_1$ (Figure B).
  If $\vec{a}$ and $\vec{u}$ were to be perpendicular vectors, then there would be no shadow and that corresponds to the case when $\cos(\theta) = \cos(\frac{\pi}{2}) = 0 \implies \vec{a}\cdot \vec{u} = 0.$ Generalizing the previous result (where we are not dealing with a unit vector, but an arbitrary vector $\vec{b}$), $\vec{a}\cdot \vec{b} =||\vec{a}|| \cdot ||\vec{b}|| \cdot \cos(\theta)$ ⇒[Considering $\vec{a} \cdot \vec{b} = \vec{a}\cdot ||\vec{b}||\cdot \vec{u}$ and that the projection of $\vec{a}$ onto $\vec{b}$ is independent of the magnitude of $\vec{b}$, but rather depends solely on its direction] $|proj_{\vec{b}}\vec{a}| = ||\vec{a}||\cdot \cos(\theta) = \frac{||\vec{a}||\cdot||\vec{b}||\cdot \cos(\theta)}{||\vec{b}||} = \frac{\vec{a}\cdot\vec{b}}{||\vec{b}||},$ the projection of $\vec{a}$ onto $\vec{b}$ is $proj_{\vec{b}}\vec{a} = \frac{\vec{a}\cdot\vec{b}}{||\vec{b}||^2}\vec{b}$ (Figure C.).
  Similarly, the projection of $\vec{b}$ onto $\vec{a}$ is $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}$.

• Orthogonal Component: The orthogonal projection of $\vec{a}$ with respect to $\vec{b}$ is $ort_{\vec{b}}\vec{a}=\vec{SR}=\vec{PR}-\vec{PS}=\vec{a}-proj_{\vec{b}}\vec{a}$. This represents the component of $\vec{a}$ that is orthogonal (perpendicular) to $\vec{b}$ (Figure 4).

• Vector Projection: The actual vector shadow lying along the line of $\vec{b}$: $\text{proj}_{\vec{b}}\vec{a} = \left(\text{comp}_{\vec{b}}\vec{a}\right) \hat{u} = \frac{\vec{a} \cdot \vec{b}}{\Vert{}\vec{b}\Vert{}^2} \vec{b}$

• Examples:

1. Projection of $\vec{v} = ⟨3, 4⟩$ onto $\vec{u} = ⟨1, 0⟩, proj_{\vec{u}}\vec{v} = \frac{\vec{v}·\vec{u}}{||\vec{u}||^2}\vec{u} = \frac{3⋅1+4⋅0}{1^2+0^2}⟨1, 0⟩ = \frac{3}{1}⟨1, 0⟩ = ⟨3, 0⟩$.
2. The projection of $\vec{b}$ = ⟨2, -1, 3⟩ onto $\vec{a}$ = ⟨1, 1, 1⟩. $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{1⋅2+1⋅(−1)+1⋅3}{1^2+1^2+1^2}⟨1, 1, 1⟩=\frac{4}{3}⟨1, 1, 1⟩=⟨\frac{4}{3}, \frac{4}{3}, \frac{4}{3}⟩$.
3. The projection of $\vec{b}$ = ⟨-1, 2, -3⟩ onto $\vec{a}$ = ⟨1, -1, 1⟩. $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{-6}{3}⟨1, -1, 1⟩=⟨-2, 2, -2⟩$.
4. The projection of $\vec{b}$ = ⟨2, 1, -1⟩ onto $\vec{a}$ = ⟨1, 0, -2⟩. $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{4}{5}⟨1, 0, -2⟩=⟨\frac{4}{5}, 0, \frac{-8}{5}⟩$.
5. The projection of $\vec{b}$ = ⟨3, 3, 4⟩ onto $\vec{a}$ = ⟨-1, -2, 2⟩. $proj_{\vec{a}}\vec{b} = \frac{\vec{a}·\vec{b}}{||\vec{a}||^2}\vec{a}=\frac{3·(-1)+3·(-2)+2·4}{(-1)^2+(-2)^2+(-2)^2}⟨-1, -2, 2⟩=\frac{-1}{9}⟨-1, -2, 2⟩=⟨\frac{1}{9}, \frac{2}{9}, \frac{-2}{9}⟩$.
6. $ort_{\vec{a}}\vec{b}=\vec{b}-proj_{\vec{a}}\vec{b} = ⟨3, 3, 4⟩-⟨\frac{1}{9}, \frac{2}{9}, \frac{-2}{9}⟩ = ⟨\frac{26}{9}, \frac{25}{9}, \frac{38}{9}⟩$.

Solved exercises

Scaled Parallel Vector

• Given $\vec{a} = -3\hat{\mathbf{i}} + 4\hat{\mathbf{j}}$ and $\vec{b} = \hat{\mathbf{i}} + 2\hat{\mathbf{j}}$, find a vector $\vec{v}$ such that $\Vert{}\vec{v}\Vert{} = 3$ and $\vec{v} \parallel (2\vec{a} - 3\vec{b})$.

1. Calculate the combined direction vector $\vec{w}$: $\vec{w} = 2\vec{a} - 3\vec{b} = 2(-3\hat{\mathbf{i}} + 4\hat{\mathbf{j}}) - 3(\hat{\mathbf{i}} + 2\hat{\mathbf{j}}) = -6\hat{\mathbf{i}} + 8\hat{\mathbf{j}} - 3\hat{\mathbf{i}} - 6\hat{\mathbf{j}} = -9\hat{\mathbf{i}} + 2\hat{\mathbf{j}}$
2. Normalize $\vec{w}$ to a unit vector $\vec{u}$: $\Vert{}\vec{w}\Vert{} = \sqrt{(-9)^2 + 2^2} = \sqrt{81 + 4} = \sqrt{85}$
   $\vec{u} = \frac{\vec{w}}{\Vert{}\vec{w}\Vert{}} = \frac{1}{\sqrt{85}}(-9\hat{\mathbf{i}} + 2\hat{\mathbf{j}})$
3. Scale by the desired magnitude $\Vert{}\vec{v}\Vert{} = 3$: $\vec{v} = \pm 3 \vec{u} = \pm \frac{3\sqrt{85}}{85}(-9\hat{\mathbf{i}} + 2\hat{\mathbf{j}}) = \pm \left( -\frac{27\sqrt{85}}{85}\hat{\mathbf{i}} + \frac{6\sqrt{85}}{85}\hat{\mathbf{j}} \right)$

Direction Cosines & Angles of $\vec{v} = \langle 2, 1, -4 \rangle$

1. In three-dimensional space, any vector $\vec{v}$ can be represented as a linear combination of its components along the three coordinate axes (x, y, and z axes in Cartesian coordinate system). These angles $\alpha, \beta, \gamma$ are called direction angles and the cosines of these angles are called direction cosines, and their formulas are: $cos(α)=\frac{\vec{v}·\vec{i}}{||\vec{v}||} = \frac{v_x}{||\vec{v}||}, cos(β)=\frac{\vec{v}·\vec{j}}{||\vec{v}||} = \frac{v_y}{||\vec{v}||}, cos(γ)=\frac{\vec{v}·\vec{k}}{||\vec{v}||} = \frac{v_z}{||\vec{v}||}$ (Figure i).
2. Calculate magnitude: $\Vert{}\vec{v}\Vert{} = \sqrt{2^2 + 1^2 + (-4)^2} = \sqrt{4 + 1 + 16} = \sqrt{21}$
3. Determine Direction Cosines: $\cos(\alpha) = \frac{v_x}{\Vert{}\vec{v}\Vert{}} = \frac{2}{\sqrt{21}}, \quad \cos(\beta) = \frac{v_y}{\Vert{}\vec{v}\Vert{}} = \frac{1}{\sqrt{21}}, \quad \cos(\gamma) = \frac{v_z}{\Vert{}\vec{v}\Vert{}} = \frac{-4}{\sqrt{21}}$
4. Compute Direction Angles:
   $\alpha = \arccos\left(\frac{2}{\sqrt{21}}\right) \approx 1.119 \text{ rad} \approx 64.12^\circ$
   $\beta = \arccos\left(\frac{1}{\sqrt{21}}\right) \approx 1.353 \text{ rad} \approx 77.40^\circ$
   $\gamma = \arccos\left(\frac{-4}{\sqrt{21}}\right) \approx 2.632 \text{ rad} \approx 150.79^\circ$
5. Direction angles from direction cosines:
   $\alpha = \arccos(\frac{2}{\sqrt{21}}) = 1.119\text{ radians}$. Convert to degrees: $1.119\text{radians} = 1.119\cdot \frac{180}{\pi}= 64.11^\circ$
   $\beta = \arccos(\frac{1}{\sqrt{21}}) = 1.3531\text{ radians}$. Convert to degrees: $1.3531\text{radians}= 1.3531\cdot \frac{180}{\pi}= 77.396^\circ$
   $\gamma = \arccos(\frac{-4}{\sqrt{21}}) = 2.632\text{ radians}$. Convert to degrees: $2.632\text{radians}= 2.632\cdot \frac{180}{\pi}= 150.794^\circ$.

  1.119 radiant ----> x degrees
  π ----> 180 ⇒ x degrees = (1.119·180)/π

Direction Angles of $\vec{v} = 3\hat{\mathbf{i}} + 5\hat{\mathbf{j}} - 4\hat{\mathbf{k}}$

1. Calculate magnitude:
   $\Vert{}\vec{v}\Vert{} = \sqrt{3^2 + 5^2 + (-4)^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}$
2. Calculate angles:
   $\cos(\alpha)=\frac{\vec{v} \cdot \vec{i}}{||\vec{v}||} = \frac{v_x}{||\vec{v}||} = \frac{3}{5\sqrt{2}}$
   $\cos(\beta)=\frac{\vec{v} \cdot \vec{j}}{||\vec{v}||} = \frac{v_y}{||\vec{v}||} = \frac{5}{5\sqrt{2}} = \frac{1}{\sqrt{2}}$
   $\cos(\gamma)=\frac{\vec{v} \cdot \vec{k}}{||\vec{v}||} = \frac{v_z}{||\vec{v}||} = \frac{-4}{5\sqrt{2}}$
   $\alpha = \arccos\left(\frac{3}{5\sqrt{2}}\right) \approx \arccos(0.4243) \approx 64.89^\circ$
   $\beta = \arccos\left(\frac{5}{5\sqrt{2}}\right) = \arccos\left(\frac{1}{\sqrt{2}}\right) = 45.00^\circ$
   $\gamma = \arccos\left(\frac{-4}{5\sqrt{2}}\right) \approx \arccos(-0.5657) \approx 124.45^\circ$

Angle Classification Between Two Vectors

• Find the angle $\theta$ between $\vec{a} = \langle 3, -1 \rangle$ and $\vec{b} = \langle -4, 0 \rangle$. Determine whether $\theta$ is acute or obtuse.

1. Compute Dot Product and Magnitudes:
   $\vec{a} \cdot \vec{b} = (3)(-4) + (-1)(0) = -12$
   $\Vert{}\vec{a}\Vert{} = \sqrt{3^2 + (-1)^2} = \sqrt{10}, \quad \Vert{}\vec{b}\Vert{} = \sqrt{(-4)^2 + 0^2} = 4$
2. Calculate $\cos(\theta)$ and $\theta$:
   $\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{\Vert{}\vec{a}\Vert{} \Vert{}\vec{b}\Vert{}} = \frac{-12}{4\sqrt{10}} = \frac{-3}{\sqrt{10}} \approx -0.9487$
   $\theta = \arccos\left(-\frac{3}{\sqrt{10}}\right) \approx 2.82 \text{ rad} \approx 2.82 \times \frac{180}{\pi} = 161.57^\circ$
3. Classification:
   Since $\vec{a} \cdot \vec{b} = -12 < 0$ and $\cos(\theta) < 0$, the angle lies in the interval $\frac{\pi}{2} < \theta < \pi$.
   Therefore, $\theta$ is an obtuse angle.

The sign of the dot product $\vec{a} \cdot \vec{b} = ||\vec{a}|| \cdot ||\vec{b}|| \cdot \cos(\theta) = -12$ depends on the cosine of the angle between the vectors.

1. If the angle between the vectors is acute ($0 \lt \theta \lt \frac{\pi}{2}$), the dot product is positive.
2. If the angle between the vectors is obtuse ($\frac{\pi}{2} \lt \theta \lt \pi$), the dot product is negative. Since $\theta \approx 161.57^\circ$ and the dot product is -12 < 0, our angle $\theta$ is obtuse.
3. If the angle between the vectors is exactly 90 degrees ($\theta = \frac{\pi}{2}$ is a right angle), the dot product is zero (Figure iii).

• A man exerts a force of 8 lb on the handle of a wagon at an angle of 55°. If the man pulls the wagon a distance of 50 ft, what is the work done by the force? (Figure ii)

Understanding Work as a Vector Dot Product

Work is the measure of energy transfer that occurs when a force acts upon an object to cause its displacement. Mathematically, work is defined as the dot product of the force vector $\vec{F}$ and the displacement vector $\vec{PQ}$. When a force is applied at an angle to the direction of motion, only the component of the force acting along the path of motion contributes to the work done.

$W = \vec{F} \cdot \vec{PQ} = \Vert{}\vec{F}\Vert{} \Vert{}\vec{PQ}\Vert{} \cos(\theta)$

Imagine pulling an object with a constant force of 8 lb over a distance of 50 ft. If the force is applied at an upward angle of 55° relative to the horizontal displacement, we can calculate the work done:

• Force magnitude: $\Vert{}\vec{F}\Vert{} = 8$ lb
• Displacement magnitude: $\Vert{}\vec{PQ}\Vert{} = 50$ ft
• Angle: $\theta = 55^\circ$

$W = 8 \cdot 50 \cdot \cos(55^\circ) = 400 \cdot 0.5736 \approx 229.44 \text{ ft}\cdot\text{lb}$. The work done by the force is approximately 229.44 foot-pounds.

Forces on an Inclined Plane

When an object rests on a tilted surface, gravity exerts a force straight down. Because the ramp is angled, this single downward gravitational pull splits into two distinct, perpendicular components:

1. Parallel Force ($\vec{F_{\parallel}}$): The component pulling the object down the ramp.
2. Perpendicular Force ($\vec{F_{\perp}}$): The component pressing the object into the ramp (counteracted by the surface’s normal force).

To keep an object from sliding, we only need to apply an opposing force equal to the parallel component $\vec{F_{\parallel}}$.

• A 20-pound object sits on a ramp at an angle of 45∘ from the horizontal. Find the force pulling the object down the ramp and the force pulling the object directly into the ramp.

To solve for the forces acting on a 20-pound object sitting on a ramp at an angle of 45° from the horizontal, we need to resolve the weight of the object into two components using standard trigonometry:

1. The force pulling the object down the ramp is the parallel component of the weight (this is the part of the gravitational force that makes the object slide down the ramp), $F_{\parallel} = W \sin(\theta) = 20 \sin(45^\circ) = 20 \left(\frac{\sqrt{2}}{2}\right) = 10\sqrt{2} \approx 14.14 \text{ lb}$
2. The force pressing the object into the ramp is the perpendicular component of the weight. This is calculated using the cosine function: $F_{\perp} = W \cos(\theta) = 20 \cos(45^\circ) = 20 \left(\frac{\sqrt{2}}{2}\right) = 10\sqrt{2} \approx 14.14 \text{ lb}$
3. 🔍 Why both components are equal at 45? At $\theta =45^\circ: \sin(45)=\cos(45) = \frac{\sqrt{2}}{2}$. So the parallel and perpendicular components are identical. This symmetry disappears at any other angle.

Preventing Motion Using Vector Projections

• A 200 pound cart sits on a ramp inclined at 30°. What force is required to keep the cart from rolling down the ramp? (Figure iv)

When an object is on an inclined plane, gravity exerts a force straight down. The gravitational force acting on the cart is vertical and downwards, we can represent this force by the vector $\vec{F}=-200\vec{j}$ where $\vec{j}$ is the unit vector in the downward direction.

Gravity tries to pull the cart straight down, but because the ramp is tilted, this downward pull can be thought of as being split into two parts:

1. One part tries to make the cart slide down the ramp.
2. The other part presses the cart into the ramp. To keep the cart from sliding, we only need to counteract the part of the force pulling it down the ramp.

We define a unit vector $\vec{v}$ that points up along the incline of the ramp, $\vec{v} = cos(30)\vec{i} + sen(30)\vec{j} = \frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j}$, where $\vec{i}$ and $\vec{j}$ are unit vectors in the horizontal and vertical directions, respectively.

To find the force required to prevent the cart from rolling down the ramp, we need to project the gravitation force $\vec{F}$ onto the unit vector representing (in the direction of) the ramp. This projection gives us the component of the gravitational force acting along the ramp $\vec{v}$. This projection, $w_1 = proj_{\vec{v}}\vec{F} = (\frac{\vec{F}·\vec{v}}{||\vec{v}||^2})·\vec{v} =$[Since $\vec{v}$ is a unit vector, $||\vec{v}|| = 1$] $(\vec{F}·\vec{v})·\vec{v} = (-200\vec{j})·(\frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j})·\vec{v} = -100·\vec{v} = -100(\frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j})$

The magnitude of this force (projection vector) is $||(-100\vec{j})(\frac{\sqrt{3}}{2}\vec{i} + \frac{1}{2}\vec{j})|| = 100(\frac{3}{4}+\frac{1}{4}) = 100·1 = 100$ ⇒ Therefore, a force of 100 pounds is required to keek the cart stationary, that is, to prevent the cart from rolling down the ramp.

Bibliography

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