Scalar and Vector Resolute✅

Scalar and Vector Resolute, Dot Product Applications

Scalar Resolute

• Component of vector $b$ in the direction of vector $a$.
• Given vectors $a$ and $b$ forming an angle $\theta$, the component of $b$ in direction $a$ is $|b| \cos(\theta)$.
• Vector $b$ decomposes into two components:
  • Component in direction $a$: $|b| \cos(\theta)$
  • Component perpendicular to $a$: $|b| \sin(\theta)$
• Component of $b$ in direction $a$ is $|b| \cos(\theta)$.
• Recall dot product: $a \cdot b = |a| |b| \cos(\theta)$.
• Solving for $\cos(\theta)$: $\cos(\theta) = \frac{a \cdot b}{|a| |b|}$.
• Substitute $\cos(\theta)$ into the component formula.

Component of B in Direction A

• Component of $b$ in direction $a$: $\frac{a \cdot b}{|a|}$.
• Rewrite as: $\frac{a}{|a|} \cdot b$.
• $\frac{a}{|a|}$ is the unit vector $\hat{a}$ in direction $a$.
• The component of $b$ in direction $a$ is $\hat{a} \cdot b$.

Examples

• Example 1:
  • $a = i$, $b = 2i + 3j$
  • Component of $b$ in direction $a$: $\frac{a \cdot b}{|a|} = \frac{i \cdot (2i + 3j)}{|i|} = \frac{2}{1} = 2$
• Example 2:
  • $a = -3i$, $b = 2i + 3j$
  • Component of $b$ in direction $a$: $\frac{a \cdot b}{|a|} = \frac{-3i \cdot (2i + 3j)}{|-3i|} = \frac{-6}{3} = -2$

Interpretation of Component Sign

• Positive Component:
  • Vector $a$ and $b$ have an angle less than $\frac{\pi}{2}$.
  • The component of $b$ in direction $a$ is positive and points in the same direction as $a$.
• Negative Component:
  • Vector $a$ and $b$ have an angle greater than $\frac{\pi}{2}$.
  • The component of $b$ in the direction of $a$ is negative and points in the opposite direction of $a$.

Vector Resolute (Vector Projection)

• Vector projection of $b$ onto $a$ is denoted as $proj_{a}b$.
• Defined as: $proj<em>{a}b = (comp</em>{a}b) \hat{a}$ where $\hat{a}$ is the unit vector in the direction of a.

Intuition

• Scalar component (comp_a b) is the length $|b| \cos(\theta)$.
• $\hat{a} = \frac{a}{|a|}$ is the unit vector with magnitude 1.
• Multiply the scalar component by the unit vector to get the vector projection.
• $proj_{a}b$ is a vector in the direction of $a$ with magnitude equal to the scalar component.

Formula Expansion

• $proj<em>{a}b = (comp</em>{a}b) \hat{a} = (\frac{a \cdot b}{|a|}) \frac{a}{|a|} = \frac{a \cdot b}{|a|^2} a$

Example

• Given $a = 2i$ and $b = 2i + 3j$, find $proj_{a}b$.
• $comp_{a}b = \frac{a \cdot b}{|a|} = \frac{(2i) \cdot (2i + 3j)}{|2i|} = \frac{4}{2} = 2$
• $\hat{a} = \frac{a}{|a|} = \frac{2i}{2} = i$
• $proj<em>{a}b = (comp</em>{a}b) \hat{a} = 2i$

Exercise

• Find the projection of $b$ onto $a$ with $a = -3i$ and $b = 2i + 3j$.

Vector Rejection

• When projecting vector b onto vector a, the difference between the original vector b and the projection is perpendicular to a.
• This new vector is the vector rejection.
• $b = b<em>{parallel} + b</em>{perpendicular}$
• $b_{parallel}$ is the projection of $b$ onto $a$ (parallel component).
• $b_{perpendicular}$ is the vector rejection (perpendicular component).
• $b<em>{perpendicular} = b - proj</em>{a}b$

Applications of the Dot Product

• Work: $W = F \cdot d$ where F is force and d is displacement.
• Angle between two vectors: $\cos(\theta) = \frac{a \cdot b}{|a||b|}$

Vector Resolution

• Given vectors A and B, their sum is C = A + B.
• Projecting A onto C yields Aparallel and Aperpendicular.
• Projecting B onto C yields Bparallel and Bperpendicular.
• Aparallel and Bparallel point in the same direction as C.
• C = Aparallel + Bparallel.
• Aperpendicular + Bperpendicular = 0. Any vector can be decomposed into parallel and perpendicular components with regards to another vector. The perpendicular component is zero in the case of vector resolution. This is useful in solving questions involving equilibrium or balanced forces.

Example: Work Done by Multiple Forces

• Three forces act on a particle on a planar surface.
  • $F<em>1, F</em>2, F_3$ in Newtons
• Particle moves from point (1,8,1) to (2,2) in meters.
  • Total force: $F = F<em>1 + F</em>2 + F_3 = 4i + 2j$ N
  • Displacement vector: $d = (2 - 1)i + (2 - 8)j + (2 - 1)k = i -6j + k$ m
  • Work done: $W = F \cdot d = (4i + 2j) \cdot (i - 6j + k) = 4 - 12 + 0 = -8$ Joules

Example: Angle Between Vectors in a Cube

• Cube centered at the origin with side length 2d.
• Find the angle between vectors from the origin to points (-d, d, d) and (d, d, d).
  • Vector $X_1 = -di + dj + dk$
  • Vector $X_2 = di + dj + dk$
  • $\cos(\theta) = \frac{X<em>1 \cdot X</em>2}{|X<em>1||X</em>2|} = \frac{(-di + dj + dk) \cdot (di + dj + dk)}{\sqrt{3d^2} \sqrt{3d^2}} = \frac{-d^2 + d^2 + d^2}{3d^2} = \frac{d^2}{3d^2} = \frac{1}{3}$
  • $\theta = \arccos(\frac{1}{3})$

Exercise

• Given the same cube configuration, find the angle between different vectors.

Example: Forces on a Pole

• A pole is supported by three ropes with tension T1, T2, and T3.
• Ropes are attached to the ground at (-6,0,0), (0,-6,0), and (d,d,0).
• Pole is at (0,0,10).
• Tension in each rope is 100 N.
• Need to find d such that the pole is vertical.
• $T_1 = 100 \frac{(-6,0,-10)}{\sqrt{(-6)^2 + 0^2 + (-10)^2}}$
• $T_2 = 100 \frac{(0,-6,-10)}{\sqrt{0^2 + (-6)^2 + (-10)^2}}$
• $T_3 = 100 \frac{(d,d,-10)}{\sqrt{d^2 + d^2 + (-10)^2}}$
• Let A be the vector (d,d,0). To keep the pole vertical, the component of T in the direction of A must be zero (forces balance out).
• $proj<em>{A}T = \frac{A \cdot T}{|A|^2} A = \frac{A \cdot (T</em>1 + T<em>2 + T</em>3)}{|A|^2} A$. Since vector A is not zero, then:
• $A \cdot (T<em>1 + T</em>2 + T_3) = 0$. By working out the details, we find that d = 7.5 meters.