Let A = 2i + 3j, B = pi + 9j, and C = i – j.If points

Solution

Three points A, B, and C are collinear if vectors AB and AC are linearly dependent, i.e., AB is a scalar multiple of AC. Let’s compute: Vector AB = B – A = (p – 2)i + (9 – 3)j = (p – 2)i + 6j Vector AC = C – A = (1 – 2)i + (–1 – 3)j = (–1)i + (–4)j = –i – 4j Now, for AB and AC to be parallel, we must have: AB = λ·AC So compare components: 1.     i-component:  (p – 2) = λ(–1) ⇒ λ = –(p – 2) 2.     j-component:  6 = λ(–4) ⇒ λ = –3/2 Now equate both expressions for λ: –(p – 2) = –3/2 ⇒ p – 2 = 3/2 ⇒ p = 3/2 + 2 = 7/2 Wait — we got 7/2 , but let’s double-check signs. Actually: From j-component: 6 = λ(–4) ⇒ λ = –3/2 Then from i-component: (p – 2) = λ(–1) = –(–3/2) = 3/2 ⇒ p = 2 + 3/2 = 7/2