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7.1 Q-4

Question Statement

Find the sum of vectors AB⃗A \vec{B} and CD\mathbf{C D}, given the four points:

  • A(1,βˆ’1)A(1, -1)
  • B(2,0)B(2, 0)
  • C(βˆ’1,3)C(-1, 3)
  • D(βˆ’2,2)D(-2, 2)

Background and Explanation

To find the sum of vectors ABβƒ—A \vec{B} and CD\mathbf{C D}, we first need to calculate the individual vectors. The vector from point AA to point BB is written as ABβƒ—A \vec{B}, and the vector from point CC to point DD is written as CDβƒ—C \vec{D}. The vector between two points P(x1,y1)P(x_1, y_1) and Q(x2,y2)Q(x_2, y_2) is calculated using the formula: PQ=(x2βˆ’x1)i^+(y2βˆ’y1)j^\mathbf{P Q} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}

Once we have these two vectors, we add them together by combining their i and j components separately.


Solution

Step 1: Calculate AB⃗A \vec{B}

  1. Find the components of AB⃗A \vec{B}:
    The coordinates of A(1,βˆ’1)A(1, -1) and B(2,0)B(2, 0) give us: ABβƒ—=(2,0)βˆ’(1,βˆ’1)A \vec{B} = (2, 0) - (1, -1)

  2. Subtract the corresponding components:
    ABβƒ—=(2βˆ’1,0+1)=i^+j^A \vec{B} = (2 - 1, 0 + 1) = \hat{i} + \hat{j}

So, the vector AB⃗=i^+j^A \vec{B} = \hat{i} + \hat{j} (Equation 1).


Step 2: Calculate CD⃗C \vec{D}

  1. Find the components of CD⃗C \vec{D}:
    The coordinates of C(βˆ’1,3)C(-1, 3) and D(βˆ’2,2)D(-2, 2) give us: CDβƒ—=(βˆ’2,2)βˆ’(βˆ’1,3)C \vec{D} = (-2, 2) - (-1, 3)

  2. Subtract the corresponding components:
    CDβƒ—=(βˆ’2+1,2βˆ’3)=βˆ’i^βˆ’j^C \vec{D} = (-2 + 1, 2 - 3) = -\hat{i} - \hat{j}

So, the vector CDβƒ—=βˆ’i^βˆ’j^C \vec{D} = -\hat{i} - \hat{j} (Equation 2).


Step 3: Add AB⃗A \vec{B} and CD⃗C \vec{D}

  1. Add the two vectors:
    ABβƒ—+CDβƒ—=(i^+j^)+(βˆ’i^βˆ’j^)A \vec{B} + C \vec{D} = (\hat{i} + \hat{j}) + (-\hat{i} - \hat{j})

  2. Simplify by combining the i and j components:
    ABβƒ—+CDβƒ—=i^βˆ’i^+j^βˆ’j^A \vec{B} + C \vec{D} = \hat{i} - \hat{i} + \hat{j} - \hat{j}
    AB⃗+CD⃗=0A \vec{B} + C \vec{D} = 0

Thus, the sum of the vectors is the null vector:
AB⃗+CD⃗=0A \vec{B} + C \vec{D} = \mathbf{0}


Key Formulas or Methods Used

  • Vector between two points:
    PQ=(x2βˆ’x1)i^+(y2βˆ’y1)j^\mathbf{P Q} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j}

  • Vector addition:
    To add two vectors, combine their i and j components:
    A+B=(x1+x2)i^+(y1+y2)j^\mathbf{A} + \mathbf{B} = (x_1 + x_2)\hat{i} + (y_1 + y_2)\hat{j}


Summary of Steps

  1. Find the vector AB⃗A \vec{B} by subtracting the coordinates of AA from BB.
  2. Find the vector CD⃗C \vec{D} by subtracting the coordinates of CC from DD.
  3. Add the two vectors by combining their i and j components.
  4. Simplify to find the sum of the vectors, which in this case is the null vector 0\mathbf{0}.