What minimum number of non-zero non-collinear vectors is required to produce a zero vector?

What minimum number of non-zero non-collinear vectors is required to produce
a zero vector?
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3
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Last Answer : Solution :-

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Last Answer : Let A(x1, y1) ≡ (1, 3), B(x2, y2) ≡ (2, 4), C(x3, y3) ≡ (5, 6) be the vertices of ΔABCArea of ΔABC = \(rac{1}{2}\) |{\(x_1\)(y2 – y3) + \(x_2\)(y3 – y1) + \(x​​_3\)(y1 – y2)}|= \(rac{1}{2}\) |{1(4 – 6) + 2(6 – 3) + 5(3 – 4)}| = \(rac{1}{2}\) |{–2 + 6 – 5}| = \(rac{1}{2}\) sq. units.

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Last Answer : Two lines are parallel if their slopes are equal∴ \(rac{0-(-8)}{3-(-5)}\) = \(rac{a-3}{4-6}\) ⇒ \(rac{8}{8}\) = \(rac{a-3}{-2}\) ⇒ a – 3 = –2 ⇒ a = 1.

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Last Answer : (a) - 2For three points to be collinear, area of the triangle formed by the three points should be equal to zero, i.e.\(rac{1}{2}\) [k(3k - 1) + 2k(1 - 2k) + 3(2k - 3k)] = 0⇒ \(rac{1}{2}\) [3k2 - k + ... = 0 or -2 Neglecting k = 0, as then (k, 2k) and (2k, 3k) will be the same point, we take k = -2.

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Last Answer : (d) 3Let (x, y) be the co-ordinates of the third vertex of the triangle. Then\(rac{0+2+x}{3}\) = 1 and \(rac{0+0+y}{3}\) = 1⇒ 2 + \(x\) = 3 and y = 3 ⇒ \(x\) = 1, y = 3. ∴ Co-ordinates of vertices of the triangle ... - y3) + x2 (y2 - y3) + x3(y1 - y2)]= \(rac{1}{2}\) [0+6+0] = \(rac{6}{2}\) = 3.

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Last Answer : (d) (7, -2)Let the co-ordinates of R be (x, y). As can be easily seen, it is a point of external division Also, PR = 2QR⇒ R divides the join of P and Q externally in the ratio 2:1. ∴ x = \(rac{2 imes2-1 imes-3}{2-1}\), ... }{2-1}\)⇒ x = 4 + 3 = 7 and y = 2 - 4 = -2. ∴ Co-ordinates of R are (7, -2).

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Last Answer : (c) √5 units Area of Δ ABC = 0 for collinearity of A, B, C.⇒ \(rac{1}{2}\)[1(4 – a) + 2(a – 2) + 3(2 – 4)] = 0 ⇒ 4 – a + 2a – 4 + 6 – 12 = 0 ⇒ a – 6 = 0 ⇒ a = 6. ∴ Point C ≡ (3, 6)⇒ BC = \(\sqrt{(3-2)^2+(6-4)^2}\) = \(\sqrt{1+4}\) = √5 units .

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