Interactive Vector Geometry Solutions

The scalar product results in a scalar value (not a vector) and has several important properties and applications:

• Geometric Interpretation: The dot product can be interpreted as the product of the magnitude of one vector and the projection of the other vector onto it.
• Testing Perpendicularity: If \(\vec{a} \cdot \vec{b} = 0\), then vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular (orthogonal) to each other.
• Finding Angles: The angle between two vectors can be calculated using \(\cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}\)
• Work in Physics: When a force \(\vec{F}\) moves an object along displacement \(\vec{d}\), the work done is \(W = \vec{F} \cdot \vec{d}\)
• Projection: The scalar projection of \(\vec{a}\) onto \(\vec{b}\) is given by \(\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\)

When we reflect a point P through another point S (the center of reflection), we're essentially finding a point P' such that:

• S is exactly in the middle between P and P'
• The line PP' passes through S
• |PS| = |SP'| (the distances from P to S and from S to P' are equal)

Mathematically, if S is the midpoint of PP', then:

Therefore, the reflection of P(1,-2,4) on point S(-3,2,1) is P'(-7,6,-2).

To prove that the vectors \(\vec{e}\) and \(\vec{f}\) are perpendicular, we'll calculate their dot product and show that it equals zero:

Since the dot product equals zero, the vectors \(\vec{e}\) and \(\vec{f}\) are perpendicular to one another.

To find the fourth vertex D of a parallelogram ABCD given the other three vertices A, B, and C, we can use the properties of a parallelogram:

To find a point P on the y-axis that is equidistant from points A and B:

Therefore, the point on the y-axis that is equidistant from A(2,-2,3) and B(-1,1,2) is P(0,-\(\frac{11}{6}\),0) or approximately P(0,-1.833,0).

Vector \(\overrightarrow{AB} = (3-2, 0-1, 2-0) = (1, -1, 2)\)

Vector \(\overrightarrow{AC} = (0-2, 3-1, -4-0) = (-2, 2, -4)\)

If C is on line g, then \(\overrightarrow{AC} = t \cdot \overrightarrow{AB}\) for some scalar t.

Let's check if such a t exists:

\[ (-2, 2, -4) = t \cdot (1, -1, 2) \]

This gives us three equations:

\[ -2 = t \]

\[ 2 = -t \]

\[ -4 = 2t \]

For the first equation: \(t = -2\)

For the second equation: \(t = -2\)

For the third equation: \(t = -2\)

Since all equations give the same value of t, the point C(0,3,-4) is on line g.

For a right triangle, we need the dot product of two sides to be zero (perpendicular sides).

Vector \(\overrightarrow{AX} = (0-6, 0-2, z-(-2)) = (-6, -2, z+2)\)

Vector \(\overrightarrow{BX} = (0-3, 0-0, z-2) = (-3, 0, z-2)\)

For a right angle at X, we need:

\[ \overrightarrow{AX} \cdot \overrightarrow{BX} = 0 \]

Computing the dot product:

\[ (-6)(-3) + (-2)(0) + (z+2)(z-2) = 0 \]

\[ 18 + 0 + (z^2-4) = 0 \]

\[ z^2 + 14 = 0 \]

\[ z^2 = -14 \]

Since z must be a real number and there's no real solution to \(z^2 = -14\), there is no point X on the z-axis that forms a right triangle with points A and B.

The distance from point P(3,-1) to the origin O(0,0) is given by:

\[ d(P,O) = \sqrt{(3-0)^2 + (-1-0)^2} = \sqrt{9+1} = \sqrt{10} \approx 3.16 \]

a) For a line parallel to vector \(\vec{v} = (3,2)\), the parametric equation is:

\[ \vec{r}(t) = t\vec{v} = (3t, 2t) \]

The Cartesian equation is:

\[ \frac{x}{3} = \frac{y}{2} \]

or

\[ 2x - 3y = 0 \]

b) For a line perpendicular to vector \(\vec{v} = (3,2)\), we need a vector perpendicular to \(\vec{v}\), which is \(\vec{v}_{\perp} = (-2,3)\).

The parametric equation is:

\[ \vec{r}(t) = t\vec{v}_{\perp} = (-2t, 3t) \]

The Cartesian equation is:

\[ \frac{x}{-2} = \frac{y}{3} \]

or

\[ 3x + 2y = 0 \]

The equation of the line through points A(3,0) and B(0,4) is:

\[ \frac{x}{3} + \frac{y}{4} = 1 \]

or

\[ 4x + 3y = 12 \]

The distance from the origin to this line is:

\[ d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}} \]

Where the line is in the form \(ax + by + c = 0\). Rearranging our line equation:

\[ 4x + 3y - 12 = 0 \]

So a=4, b=3, c=-12, and the origin is (0,0):

\[ d = \frac{|4(0) + 3(0) - 12|}{\sqrt{4^2 + 3^2}} = \frac{12}{\sqrt{16 + 9}} = \frac{12}{\sqrt{25}} = \frac{12}{5} = 2.4 \]

The angle between two vectors can be calculated using the dot product formula:

\[ \cos(\theta) = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} \]

Computing the dot product:

\[ \vec{a} \cdot \vec{b} = 1 \cdot (-2) + 1 \cdot 3 = -2 + 3 = 1 \]

Computing the magnitudes:

\[ |\vec{a}| = \sqrt{1^2 + 1^2} = \sqrt{2} \]

\[ |\vec{b}| = \sqrt{(-2)^2 + 3^2} = \sqrt{4 + 9} = \sqrt{13} \]

Substituting into the formula:

\[ \cos(\theta) = \frac{1}{\sqrt{2} \cdot \sqrt{13}} = \frac{1}{\sqrt{26}} \]

Therefore:

\[ \theta = \arccos\left(\frac{1}{\sqrt{26}}\right) \approx \arccos(0.196) \approx 78.7° \]

For an isosceles right triangle with the right angle at vertex A, we need:

|AC| = |AB| (isosceles property)

\(\overrightarrow{AC} \cdot \overrightarrow{AB} = 0\) (right angle at A)

Vector \(\overrightarrow{AB} = (6-5, -2-(-4), 3-3) = (1, 2, 0)\)

Let's denote C as (x, y, z). Then:

Vector \(\overrightarrow{AC} = (x-5, y-(-4), z-3) = (x-5, y+4, z-3)\)

For condition 2 (right angle at A):

\[ \overrightarrow{AC} \cdot \overrightarrow{AB} = 0 \]

\[ (x-5)(1) + (y+4)(2) + (z-3)(0) = 0 \]

\[ x-5 + 2y+8 = 0 \]

\[ x + 2y = -3 \]

For condition 1 (isosceles property):

\[ |\overrightarrow{AC}|^2 = |\overrightarrow{AB}|^2 \]

\[ (x-5)^2 + (y+4)^2 + (z-3)^2 = 1^2 + 2^2 + 0^2 \]

\[ (x-5)^2 + (y+4)^2 + (z-3)^2 = 5 \]

We have two equations and three unknowns, which means there are multiple solutions.

To find the equation of line L, I'll use the fact that L is the perpendicular bisector of segment PP'.

The length of a vector \(\vec{v}\) in R³ is calculated using the Euclidean norm:

\[ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]

Proof:

This formula is derived from the Pythagorean theorem. Consider a vector \(\vec{v} = (v_1, v_2, v_3)\) in three-dimensional space:

This formula generalizes to n-dimensional space where: \[ |\vec{v}| = \sqrt{\sum_{i=1}^{n} v_i^2} \]

Two lines in 3D space can relate to each other in four possible ways:

1. Intersecting lines: The lines meet at exactly one point.
2. Parallel lines: The lines have the same direction but never meet.
3. Skew lines: The lines are neither parallel nor intersecting. They exist in different planes and never meet.
4. Coincident lines: The lines are identical, sharing all their points.

Mathematical determination:

Consider two lines L₁ and L₂ with parametric equations:

L₁: \(\vec{r}_1(t) = \vec{p}_1 + t\vec{v}_1\)

L₂: \(\vec{r}_2(s) = \vec{p}_2 + s\vec{v}_2\)

Where \(\vec{p}_1\) and \(\vec{p}_2\) are points on the lines, and \(\vec{v}_1\) and \(\vec{v}_2\) are their direction vectors.

• The lines are parallel if \(\vec{v}_1 = \lambda \vec{v}_2\) for some scalar λ, but there's no solution to \(\vec{p}_1 + t\vec{v}_1 = \vec{p}_2 + s\vec{v}_2\).
• The lines intersect if there exist parameters t and s such that \(\vec{p}_1 + t\vec{v}_1 = \vec{p}_2 + s\vec{v}_2\).
• The lines are skew if \(\vec{v}_1\) and \(\vec{v}_2\) are not parallel and there's no solution to \(\vec{p}_1 + t\vec{v}_1 = \vec{p}_2 + s\vec{v}_2\).
• The lines are coincident if \(\vec{v}_1 = \lambda \vec{v}_2\) and \(\vec{p}_1\) lies on L₂ (or vice versa).

A parametric equation represents a line through a point with a specific direction:

Example: For a line through point (1, 2, 3) with direction vector (2, -1, 4):

\[ \vec{r}(t) = (1, 2, 3) + t(2, -1, 4) = (1+2t, 2-t, 3+4t) \]

Notes:

• As t varies through all real numbers, \(\vec{r}(t)\) traces the entire line.
• When t = 0, we get the initial point P₀.
• The direction vector can be found from two points on the line: \(\vec{v} = \vec{P_2} - \vec{P_1}\).

In 2D Space:

Consider two lines with direction vectors \(\vec{v}_1 = (a_1, b_1)\) and \(\vec{v}_2 = (a_2, b_2)\) or with equations \(y = m_1x + c_1\) and \(y = m_2x + c_2\).

Parallel lines:

• Direction vectors are scalar multiples: \(\vec{v}_1 = \lambda \vec{v}_2\) for some scalar λ
• Using slopes: \(m_1 = m_2\)
• Using direction vectors: \(\frac{a_1}{a_2} = \frac{b_1}{b_2}\)

Perpendicular lines:

• Dot product of direction vectors equals zero: \(\vec{v}_1 \cdot \vec{v}_2 = a_1a_2 + b_1b_2 = 0\)
• Using slopes: \(m_1 \cdot m_2 = -1\) (product of slopes equals -1)

In 3D Space:

Consider two lines with direction vectors \(\vec{v}_1 = (a_1, b_1, c_1)\) and \(\vec{v}_2 = (a_2, b_2, c_2)\).

Parallel lines:

• Direction vectors are scalar multiples: \(\vec{v}_1 = \lambda \vec{v}_2\) for some scalar λ
• Proportional components: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
• Cross product equals zero: \(\vec{v}_1 \times \vec{v}_2 = \vec{0}\)

Perpendicular lines:

• Dot product of direction vectors equals zero: \(\vec{v}_1 \cdot \vec{v}_2 = a_1a_2 + b_1b_2 + c_1c_2 = 0\)

Note: In 3D, two perpendicular lines may not intersect (they could be skew), so perpendicularity alone doesn't guarantee intersection.

Two planes in 3D space can relate to each other in three possible ways:

1. Parallel planes: The planes have parallel normal vectors and never intersect.
2. Coincident planes: The planes are identical, sharing all their points.
3. Intersecting planes: The planes intersect along a line.

Mathematical determination:

Consider two planes with equations:

Plane 1: \(a_1x + b_1y + c_1z + d_1 = 0\) with normal vector \(\vec{n}_1 = (a_1, b_1, c_1)\)

Plane 2: \(a_2x + b_2y + c_2z + d_2 = 0\) with normal vector \(\vec{n}_2 = (a_2, b_2, c_2)\)

• Parallel planes: Normal vectors are parallel but planes are not coincident
  • \(\vec{n}_1 = \lambda \vec{n}_2\) for some scalar λ ≠ 0
  • \(\frac{d_1}{d_2} \neq \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
• Coincident planes: Normal vectors are parallel and planes are identical
  • \(\vec{n}_1 = \lambda \vec{n}_2\) for some scalar λ ≠ 0
  • \(\frac{d_1}{d_2} = \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}\)
• Intersecting planes: Normal vectors are not parallel
  • \(\vec{n}_1\) is not a scalar multiple of \(\vec{n}_2\)
  • The planes intersect along a line with direction vector \(\vec{v} = \vec{n}_1 \times \vec{n}_2\)
  • The dihedral angle between the planes can be found from: \(\cos(\theta) = \frac{\vec{n}_1 \cdot \vec{n}_2}{|\vec{n}_1||\vec{n}_2|}\)

The midpoint of two points is the point that lies exactly halfway between them.

Formula:

For two points P₁(x₁, y₁, z₁) and P₂(x₂, y₂, z₂), the midpoint M is given by:

\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) \]

Proof:

The midpoint is found by traveling half the distance from P₁ to P₂:

\[ M = P_1 + \frac{1}{2}(P_2 - P_1) = P_1 + \frac{1}{2}P_2 - \frac{1}{2}P_1 = \frac{P_1 + P_2}{2} \]

In component form:

\[ M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) \]

Example:

The midpoint of P₁(3, -2, 5) and P₂(-1, 4, 7) is:

\[ M = \left(\frac{3 + (-1)}{2}, \frac{-2 + 4}{2}, \frac{5 + 7}{2}\right) = (1, 1, 6) \]

Applications:

• Finding the center of a line segment
• Constructing perpendicular bisectors
• Finding centroids or center points in geometric figures

A vector is a mathematical object that has both magnitude (size) and direction.

Key characteristics:

• Magnitude: The length or size of the vector, denoted as |v| or ||v||
• Direction: The orientation of the vector in space
• No position: Vectors are not fixed to a specific location in space (free vectors)

Representations:

1. Geometric: As directed line segments (arrows) in a coordinate system
2. Algebraic: As ordered tuples of numbers, e.g., v = (v₁, v₂, v₃) in 3D space
3. Component form: As a linear combination of unit vectors, e.g., v = v₁i + v₂j + v₃k

Vector operations:

• Addition: (a₁, a₂, a₃) + (b₁, b₂, b₃) = (a₁+b₁, a₂+b₂, a₃+b₃)
• Scalar multiplication: λ(a₁, a₂, a₃) = (λa₁, λa₂, λa₃)
• Dot product: a·b = a₁b₁ + a₂b₂ + a₃b₃ (results in a scalar)
• Cross product: a×b = (a₂b₃-a₃b₂, a₃b₁-a₁b₃, a₁b₂-a₂b₁) (results in a vector)

Applications:

• Physics: Representing forces, velocities, accelerations, etc.
• Computer graphics: Describing positions, directions, colors, etc.
• Engineering: Analyzing structures, fluid flow, electrical fields, etc.
• Machine learning: Feature vectors in various algorithms