The position vectors of the vertices A, B, C and D of a quadrilateral ABCD are given by 3\hat{i}+4\hat{j}-2\hat{k}, 4\hat{i}-4\hat{j}-3\hat{k}, 2\hat{i}-3\hat{j}+2\hat{k} and 6\hat{i}-2\hat{j}+\hat{k} respectively. What is the angle between the diagonals AC and BD of the quadrilateral?

Explanation

Find the diagonal vectors: \vec{AC} = C - A = (2-3)\hat{i} + (-3-4)\hat{j} + (2 - (-2))\hat{k} = -\hat{i} - 7\hat{j} + 4\hat{k}. Also, \vec{BD} = D - B = (6-4)\hat{i} + (-2 - (-4))\hat{j} + (1 - (-3))\hat{k} = 2\hat{i} + 2\hat{j} + 4\hat{k}. Evaluate their dot product: \vec{AC} \cdot \vec{BD} = (-1)(2) + (-7)(2) + (4)(4) = -2 - 14 + 16 = 0. Since the dot product is 0, the angle between them is 90^\circ.

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