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We are given the position vectors of three points: A = 3i + 2j B = pi + 9j C = i + j To determine the value of p such that points A, B, and C are collinear, we use the condition that vectors AB and AC must be parallel, i.e., their cross product must be zero: AB × AC = 0 AB = B – A = (p – 3)i + (9 – 2)j = (p – 3)i + 7j AC = C – A = (1 – 3)i + (1 – 2)j = –2i – j Take the cross product in 2D (which is a scalar): In 2D, the cross product of vectors u = ai + bj and v = ci + dj is: u × v = ad – bc So: AB × AC = (p – 3)(–1) – (7)(–2) = –(p – 3) + 14 = –p + 3 + 14 = –p + 17 Set this equal to zero: –p + 17 = 0 ⇒ p = 17
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