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Notely Maths 12
Class 12 Maths
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7.1 Q-5

Question Statement

Find the vector from the point A to the origin, given that AB=4i^−2j^\mathbf{A B} = 4\hat{i} - 2\hat{j} and point B is (−2,5)(-2, 5).

Background and Explanation

In this problem, we are asked to find the vector from point A to the origin. We are given the vector A B, which represents the vector from point A to point B, and the coordinates of point B. The general approach to solve this problem involves using the vector addition and subtraction rules.

  • Vector Subtraction: The vector from point A to the origin is calculated by subtracting the vector from A to B from the position vector of point B.
  • Vector from Origin to Point: The vector from the origin to any point is simply the negative of the position vector of that point relative to the origin.

Solution

Step 1: Find the position vector of A

We are given that the vector AB⃗A \vec{B} is 4i^−2j^4\hat{i} - 2\hat{j} and the coordinates of point B are (−2,5)(-2, 5).

The vector from point A to point B is given by: AB⃗=Position vector of B−Position vector of AA \vec{B} = \text{Position vector of B} - \text{Position vector of A}

Rearranging the equation: Position vector of A=Position vector of B−AB⃗\text{Position vector of A} = \text{Position vector of B} - A \vec{B}

Substitute the values: Position vector of A=(−2,5)−(4i^−2j^)\text{Position vector of A} = (-2, 5) - (4\hat{i} - 2\hat{j})

This gives: Position vector of A=−2i^+5j^−4i^+2j^\text{Position vector of A} = -2\hat{i} + 5\hat{j} - 4\hat{i} + 2\hat{j}
Position vector of A=−6i^+7j^\text{Position vector of A} = -6\hat{i} + 7\hat{j}

So, the position vector of point A is −6i^+7j^-6\hat{i} + 7\hat{j}.

Step 2: Find the vector from A to the origin

The vector from the origin to point A is the negative of the position vector of A.

So, the vector from the origin to A is: HO⃗=(0,0)−(−6,7)H \vec{O} = (0, 0) - (-6, 7)

Simplifying: HO⃗=6i^−7j^H \vec{O} = 6\hat{i} - 7\hat{j}

Thus, the vector from the origin to point A is: HO⃗=6i^−7j^H \vec{O} = 6\hat{i} - 7\hat{j}

Key Formulas or Methods Used

  • Vector Subtraction:
    AB⃗=Position vector of B−Position vector of AA \vec{B} = \text{Position vector of B} - \text{Position vector of A}

  • Vector from Origin to Point:
    The vector from the origin to point AA is simply the negative of the position vector of AA:
    HO⃗=−(Position vector of A)H \vec{O} = -( \text{Position vector of A} )

Summary of Steps

  1. Use the formula for vector subtraction to find the position vector of point A:
    Position vector of A=Position vector of B−AB⃗\text{Position vector of A} = \text{Position vector of B} - A \vec{B}

  2. Find the vector from the origin to A by negating the position vector of A:
    HO⃗=−Position vector of AH \vec{O} = - \text{Position vector of A}

  3. Simplify the resulting expression to obtain the final vector from the origin to A.