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Given points A(1, 2, 0) and B(2, -1, 3), find the vector AB.

Mathematics

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Answer

(c)i^−3j^+3k^\text{(c)} \hat{i} - 3\hat{j} + 3\hat{k}

Step-by-step solution

4 steps
  1. Identify the coordinates of points A and B.

    Point A is (1,2,0)(1, 2, 0).
    Point B is (2,−1,3)(2, -1, 3).

  2. Express the position vectors of points A and B.

    The position vector of A is OA⃗=1i^+2j^+0k^\vec{OA} = 1\hat{i} + 2\hat{j} + 0\hat{k}.
    The position vector of B is OB⃗=2i^−1j^+3k^\vec{OB} = 2\hat{i} - 1\hat{j} + 3\hat{k}.

  3. Calculate the vector AB⃗\vec{AB}.

    The vector AB⃗\vec{AB} is found by subtracting the position vector of A from the position vector of B:

    AB⃗=OB⃗−OA⃗\vec{AB} = \vec{OB} - \vec{OA} AB⃗=(2i^−j^+3k^)−(i^+2j^+0k^)\vec{AB} = (2\hat{i} - \hat{j} + 3\hat{k}) - (\hat{i} + 2\hat{j} + 0\hat{k}) AB⃗=(2−1)i^+(−1−2)j^+(3−0)k^\vec{AB} = (2-1)\hat{i} + (-1-2)\hat{j} + (3-0)\hat{k} AB⃗=1i^−3j^+3k^\vec{AB} = 1\hat{i} - 3\hat{j} + 3\hat{k} AB⃗=i^−3j^+3k^\vec{AB} = \hat{i} - 3\hat{j} + 3\hat{k}
  4. Compare the result with the given options.

    The calculated vector AB⃗=i^−3j^+3k^\vec{AB} = \hat{i} - 3\hat{j} + 3\hat{k} matches option (c).

    The final answer is (c)i^−3j^+3k^\text{(c)} \hat{i} - 3\hat{j} + 3\hat{k}.

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