Vectors - Problems - Finding Vector From Stp Conditio

Slide 1: Vectors - Problems - Finding vector from STP condition

Objective: To learn how to find a vector using the STP condition
STP Condition: A vector that passes through a given point and is parallel to two given vectors
Formula: Given position vectors of two points P and Q, the position vector of any point R on PQ is given by PR = a(PQ) + b(PA), where a and b are scalars "

Slide 2: Example 1

Given:
- Point P(3, 2, -4)
- Point Q(5, 1, 6)
- Position vector of point A(1, 0, -1)
Find the position vector of point R on PQ if PR = 3(PQ) + 2(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (5, 1, 6) - (3, 2, -4) = (2, -1, 10)
- Position vector of PR: PR = 3(PQ) + 2(PA) = 3(2, -1, 10) + 2(1, 0, -1)
- PR = (6, -3, 30) + (2, 0, -2) = (8, -3, 28)
Therefore, the position vector of point R is (8, -3, 28). "

Slide 3: Example 2

Given:
- Point P(2, -3, 1)
- Point Q(4, 5, -2)
- Position vector of point A(1, 0, 3)
Find the position vector of point R on PQ if PR = a(PQ) + b(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (4, 5, -2) - (2, -3, 1) = (2, 8, -3)
- Position vector of PR: PR = a(PQ) + b(PA) = a(2, 8, -3) + b(1, 0, 3)
- PR = (2a + b, 8a, -3a + 3b)
The position vector of point R depends on the values of a and b. "

Slide 4: Finding Vector using STP Condition

Given:
- Point P(x₁, y₁, z₁)
- Point Q(x₂, y₂, z₂)
- Position vector of point A(a₁, b₁, c₁)
To find the position vector of point R on PQ, we use the STP condition:
- PR = a(PQ) + b(PA)
To find the position vector of point R, we need to find the values of a and b.
Let’s consider an example to understand this concept better. "

Slide 5: Example 3

Given:
- Point P(1, 2, -3)
- Point Q(3, -1, 4)
- Position vector of point A(0, -2, 5)
Find the position vector of point R on PQ if PR = 2(PQ) + b(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (3, -1, 4) - (1, 2, -3) = (2, -3, 7)
- Position vector of PR: PR = 2(PQ) + b(PA) = 2(2, -3, 7) + b(0, -2, 5)
- PR = (4, -6, 14) + (0, -2b, 5b) = (4, -2b - 6, 14 + 5b)
Therefore, the position vector of point R is (4, -2b - 6, 14 + 5b). "

Slide 6: Properties of Vector Addition

The position vector of any point on the line joining two points P and Q can be expressed in the form:
- PR = a(PQ) + b(PA), where a and b are scalars
Here are some important properties of vector addition:
1. If a vector is added to itself, the resulting vector is twice the original vector.
2. If a vector is multiplied by a scalar ‘k’, the resulting vector is parallel to the original vector with a magnitude ‘k’ times the magnitude of the original vector.
Let’s see an example to understand these properties. "

Slide 7: Example 4

Given:
- Vector A(2, 3, -4)
Find:
- 2A and A + A
Solution:
- Vector 2A: Scalar multiplication
  - 2A = 2(2, 3, -4) = (4, 6, -8)
- Vector A + A: Vector addition
  - A + A = (2, 3, -4) + (2, 3, -4) = (4, 6, -8)
Therefore, 2A = (4, 6, -8) and A + A = (4, 6, -8). "

Slide 8: Example 5

Given:
- Point P(1, 2, -3)
- Point Q(3, -1, 4)
- Position vector of point A(0, -2, 5)
Find the position vector of point R on PQ if PR = 3(PQ) - 2(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (3, -1, 4) - (1, 2, -3) = (2, -3, 7)
- Position vector of PR: PR = 3(PQ) - 2(PA) = 3(2, -3, 7) - 2(0, -2, 5)
- PR = (6, -9, 21) - (0, 4, -10) = (6, -13, 31)
Therefore, the position vector of point R is (6, -13, 31). "

Slide 11: Vectors - Problems - Finding vector from STP condition

Objective: To learn how to find a vector using the STP condition
STP Condition: A vector that passes through a given point and is parallel to two given vectors
Formula: Given position vectors of two points P and Q, the position vector of any point R on PQ is given by PR = a(PQ) + b(PA), where a and b are scalars

Slide 12: Example 1

Given:
- Point P(3, 2, -4)
- Point Q(5, 1, 6)
- Position vector of point A(1, 0, -1)
Find the position vector of point R on PQ if PR = 3(PQ) + 2(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (5, 1, 6) - (3, 2, -4) = (2, -1, 10)
- Position vector of PR: PR = 3(PQ) + 2(PA) = 3(2, -1, 10) + 2(1, 0, -1)
- PR = (6, -3, 30) + (2, 0, -2) = (8, -3, 28)
Therefore, the position vector of point R is (8, -3, 28).

Slide 13: Example 2

Given:
- Point P(2, -3, 1)
- Point Q(4, 5, -2)
- Position vector of point A(1, 0, 3)
Find the position vector of point R on PQ if PR = a(PQ) + b(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (4, 5, -2) - (2, -3, 1) = (2, 8, -3)
- Position vector of PR: PR = a(PQ) + b(PA) = a(2, 8, -3) + b(1, 0, 3)
- PR = (2a + b, 8a, -3a + 3b)
The position vector of point R depends on the values of a and b.

Slide 14: Finding Vector using STP Condition

Given:
- Point P(x₁, y₁, z₁)
- Point Q(x₂, y₂, z₂)
- Position vector of point A(a₁, b₁, c₁)
To find the position vector of point R on PQ, we use the STP condition:
- PR = a(PQ) + b(PA)
To find the position vector of point R, we need to find the values of a and b.
Let’s consider an example to understand this concept better.

Slide 15: Example 3

Given:
- Point P(1, 2, -3)
- Point Q(3, -1, 4)
- Position vector of point A(0, -2, 5)
Find the position vector of point R on PQ if PR = 2(PQ) + b(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (3, -1, 4) - (1, 2, -3) = (2, -3, 7)
- Position vector of PR: PR = 2(PQ) + b(PA) = 2(2, -3, 7) + b(0, -2, 5)
- PR = (4, -6, 14) + (0, -2b, 5b) = (4, -2b - 6, 14 + 5b)
Therefore, the position vector of point R is (4, -2b - 6, 14 + 5b).

Slide 16: Properties of Vector Addition

The position vector of any point on the line joining two points P and Q can be expressed in the form:
- PR = a(PQ) + b(PA), where a and b are scalars
Here are some important properties of vector addition:
1. If a vector is added to itself, the resulting vector is twice the original vector.
2. If a vector is multiplied by a scalar ‘k’, the resulting vector is parallel to the original vector with a magnitude ‘k’ times the magnitude of the original vector.
Let’s see an example to understand these properties.

Slide 17: Example 4

Given:
- Vector A(2, 3, -4)
Find:
- 2A and A + A
Solution:
- Vector 2A: Scalar multiplication
  - 2A = 2(2, 3, -4) = (4, 6, -8)
- Vector A + A: Vector addition
  - A + A = (2, 3, -4) + (2, 3, -4) = (4, 6, -8)
Therefore, 2A = (4, 6, -8) and A + A = (4, 6, -8).

Slide 18: Example 5

Given:
- Point P(1, 2, -3)
- Point Q(3, -1, 4)
- Position vector of point A(0, -2, 5)
Find the position vector of point R on PQ if PR = 3(PQ) - 2(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (3, -1, 4) - (1, 2, -3) = (2, -3, 7)
- Position vector of PR: PR = 3(PQ) - 2(PA) = 3(2, -3, 7) - 2(0, -2, 5)
- PR = (6, -9, 21) - (0, 4, -10) = (6, -13, 31)
Therefore, the position vector of point R is (6, -13, 31).

Example 6:
- Given:
  - Point P(1, 2, 3)
  - Point Q(4, -2, 5)
  - Position vector of point A(-1, 0, 1)
Find the position vector of point R on PQ if PR = a(PQ) + b(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (4, -2, 5) - (1, 2, 3) = (3, -4, 2)
- Position vector of PR: PR = a(PQ) + b(PA) = a(3, -4, 2) + b(-1, 0, 1)
- PR = (3a - b, -4a, 2a + b)
The position vector of point R depends on the values of a and b.

Example 7:
- Given:
  - Point P(-2, 5, 1)
  - Point Q(3, -1, 4)
  - Position vector of point A(0, -3, 2)
Find the position vector of point R on PQ if PR = 2(PQ) - 3(PA).
Solution:
- Position vector of PQ: PQ = Q - P = (3, -1, 4) - (-2, 5, 1) = (5, -6, 3)
- Position vector of PR: PR = 2(PQ) - 3(PA) = 2(5, -6, 3) - 3(0, -3, 2)
- PR = (10, -12, 6) - (0, -9, 6) = (10, -3, 0)
Therefore, the position vector of point R is (10, -3, 0).

Use of Vectors in Real-Life Applications:
1. Navigation and GPS systems use vectors to calculate distances and directions.
2. Physics, especially in mechanics and electromagnetism, utilizes vectors to describe quantities like velocity, force, or electric field.
3. Computer graphics and animation use vectors to define the position, orientation, and movement of objects.
4. Engineering and architecture use vectors to analyze and design structures, calculate forces, and simulate fluid flow.
5. Vectors are extensively used in robotics, including robot motion planning and control algorithms.

Properties of Vector Multiplication:
1. Dot Product: The dot product of two vectors A and B, denoted as A · B, is a scalar quantity given by the formula A · B = |A| |B| cosθ, where |A| and |B| are the magnitudes of vectors A and B, and θ is the angle between them.
2. Cross Product: The cross product of two vectors A and B, denoted as A × B, is a vector quantity perpendicular to both A and B, and its magnitude is given by |A × B| = |A| |B| sinθ, where θ is the angle between them. The direction of A × B can be determined using the right-hand rule.

Example 8:
- Given:
  - Vector A = (2, -1, 3)
  - Vector B = (4, 0, -2)
Find:
- A · B and A × B
Solution:
- Dot Product:
  - A · B = (2)(4) + (-1)(0) + (3)(-2) = 8 - 6 = 2
- Cross Product:
  - A × B = [(2)(-2) - (-1)(0), (-1)(4) - (2)(-2), (2)(0) - (-1)(4)]
  - A × B = (-4, -6, 4)
Therefore, A · B = 2 and A × B = (-4, -6, 4).

Example 9:
- Given:
  - Vector A = (1, -3, 2)
  - Vector B = (-2, 4, -1)
Find:
- A · B and A × B
Solution:
- Dot Product:
  - A · B = (1)(-2) + (-3)(4) + (2)(-1) = -2 - 12 - 2 = -16
- Cross Product:
  - A × B = [(1)(-1) - (-3)(4), (-3)(-2) - (1)(-1), (1)(4) - (-3)(-2)]
  - A × B = (-13, 5, -10)
Therefore, A · B = -16 and A × B = (-13, 5, -10).

Applications of Vectors in Physics:
1. Motion: Vectors are used to describe the displacement, velocity, and acceleration of objects in various types of motion, such as linear, projectile, and circular motion.
2. Forces: Vectors are used to represent forces acting on objects and to calculate their resultant force and direction.
3. Electric and Magnetic Fields: Vectors are used to describe