Three non-collinear points in space form a triangle. The position vector of each vertex is given by $\vec{A}=(6,7,1)$, $\vec{B}=(2,−3,1)$ and $\vec{C}=(4,−5,0)$. You can calculate the normal vector at each vertex, lets take vertex $\vec{A}$, for example: $$ \vec{N} = (\vec{B} - \vec{A}) \times (\vec{C} - \vec{A}) $$ If the absolute value of the normal is zero, then the three points are aligned in space. You can convince yourself of this by looking at the expression for the module of a vector: $$ |\vec{N}| = |(\vec{B} - \vec{A}) \times (\vec{C} - \vec{A})| = |(\vec{B} - \vec{A})| \cdot |(\vec{C} - \vec{A})| \cdot sin{\space\theta} $$ where $\theta$ is the angle between the vectors $(\vec{B} - \vec{A})$ and $(\vec{C} - \vec{A})$, i.e. the angle of vertex $\vec{A}$). This expression will be zero if $\theta = 0, \pi$. If, lets say, your points are aligned in the following order: $B - A - C$, you can see that the vertex angle at A is $\pi$ (180). If you calculated the normal at B or C then the angle would be zero. Either way, the normal module is zero.
We will discuss here how to prove the conditions of collinearity of three points. Collinear points: Three points A, B and C are said to be collinear if they lie on the same straight line. There points A, B and C will be collinear if AB + BC = AC as is clear from the adjoining figure. In general, three points A, B and C are collinear if the sum of the lengths of any two line segments among AB, BC and CA is equal to the length of the remaining line segment, that is, either AB + BC = AC or AC +CB = AB or BA + AC = BC. In other words, There points A, B and C are collinear iff: (i) AB + BC = AC i.e., Or, (ii) AB + AC = BC i.e. , Or, AC + BC = AB i.e., Solved examples to prove the collinearity of three points: 1. Prove that the points A (1, 1), B (-2, 7) and (3, -3) are collinear. Solution: Let A (1, 1), B (-2, 7) and C (3, -3) be the given points. Then, AB = \(\sqrt{(-2 - 1)^{2} + (7 - 1)^{2}}\) = \(\sqrt{(-3)^{2} + 6^{2}}\) = \(\sqrt{9 + 36}\) = \(\sqrt{45}\) = 3\(\sqrt{5}\) units. BC = \(\sqrt{(3 + 2)^{2} + (-3 - 7)^{2}}\) = \(\sqrt{5^{2} + (-10)^{2}}\) = \(\sqrt{25 + 100}\) = \(\sqrt{125}\) = 5\(\sqrt{5}\) units. AC = \(\sqrt{(3 - 1)^{2} + (-3 - 1)^{2}}\) = \(\sqrt{2^{2} + (-4)^{2}}\) = \(\sqrt{4 + 16}\) = \(\sqrt{20}\) = 2\(\sqrt{5}\) units. Therefore, AB + AC = 3\(\sqrt{5}\) + 2\(\sqrt{5}\) units = 5\(\sqrt{5}\) = BC Thus, AB + AC = BC Hence, the given points A, B, C are collinear. 2. Use the distance formula to show the points (1, -1), (6, 4) and (4, 2) are collinear. Solution: Let the points be A (1, -1), B (6, 4) and C (4, 2). Then, AB = \(\sqrt{(6 - 1)^{2} + (4 + 1)^{2}}\) = \(\sqrt{5^{2} + 5^{2}}\) = \(\sqrt{25 + 25}\) = \(\sqrt{50}\) = 5\(\sqrt{2}\) BC = \(\sqrt{(4 - 6)^{2} + (2 - 4)^{2}}\) = \(\sqrt{(-2)^{2} + (-2)^{2}}\) = \(\sqrt{4 + 4}\) = \(\sqrt{8}\) = 2\(\sqrt{2}\) and AC = \(\sqrt{(4 - 1)^{2} + (2 + 1)^{2}}\) = \(\sqrt{3^{2} + 3^{2}}\) = \(\sqrt{9 + 9}\) = \(\sqrt{18}\) = 3\(\sqrt{2}\) ⟹ BC + AC = 2\(\sqrt{2}\) + 3\(\sqrt{2}\) = 5\(\sqrt{2}\) = AB So, the points A, B and C are collinear with C lying between A and B. 3. Use the distance formula to show the points (2, 3), (8, 11) and (-1, -1) are collinear. Solution: Let the points be A (2, 3), B (8, 11) and C (-1, -1). Then, AB = \(\sqrt{(2 - 8)^{2} + (3 - 11)^{2}}\) = \(\sqrt{6^{2} + (-8)^{2}}\) = \(\sqrt{36 + 64}\) = \(\sqrt{100}\) = 10 BC = \(\sqrt{(8 - (-1))^{2} + (11 - (-1))^{2}}\) = \(\sqrt{9^{2} + 12^{2}}\) = \(\sqrt{81 + 144}\) = \(\sqrt{225}\) = 15 and CA = \(\sqrt{((-1) - 2)^{2} + ((-1) + 3)^{2}}\) = \(\sqrt{(-3)^{2} + (-4)^{2}}\) = \(\sqrt{9 + 16}\) = \(\sqrt{25}\) = 5 ⟹ AB + CA = 10 + 5 = 15 = BC Hence, the given points A, B, C are collinear. ● Distance and Section Formulae 10th Grade Math From Conditions of Collinearity of Three Points to HOME PAGE
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Determine whether the following point is collinear. A(–1, –1), B(0, 1), C(1, 3) Slope of BC = \[\frac{3 - 1}{1 - 0} = \frac{2}{1} = 2\] Slope of AB = Slope of BC = 2 Is there an error in this question or solution? Page 2Determine whether the following point is collinear. D(–2, –3), E(1, 0), F(2, 1) D(–2, –3), E(1, 0), F(2, 1) \[\text { Slope of DE } = \frac{0 - \left( - 3 \right)}{1 - \left( - 2 \right)} = \frac{3}{3} = 1\] \[\text { Slope of EF } = \frac{1 - 0}{2 - 1} = \frac{1}{1} = 1\] Slope of DE = Slope of EF = 1 Is there an error in this question or solution? Page 3Determine whether the following point is collinear. L(2, 5), M(3, 3), N(5, 1) L(2, 5), M(3, 3), N(5, 1) \[\text { Slope of LM } = \frac{3 - 5}{3 - 2} = \frac{- 2}{1} = - 2\] \[\text { Slope of MN} = \frac{1 - 3}{5 - 3} = \frac{- 2}{2} = - 1\] Slope of LM not equal to slope of MN. Thus, the given points are not collinear. Is there an error in this question or solution? Page 4Determine whether the following point is collinear. P(2, –5), Q(1, –3), R(–2, 3) P(2, –5), Q(1, –3), R(–2, 3) \[\text { Slope of PQ } = \frac{- 3 - \left( - 5 \right)}{1 - 2} = \frac{2}{- 1} = - 2\] \[\text{ Slope of QR } = \frac{3 - \left( - 3 \right)}{- 2 - 1} = \frac{6}{- 3} = - 2\] Slope of PQ = Slope of QR Is there an error in this question or solution? Page 5Determine whether the following point is collinear. R(1, –4), S(–2, 2), T(–3, 4) R(1, –4), S(–2, 2), T(–3, 4) \[\text { Slope of RS } = \frac{2 - \left( - 4 \right)}{- 2 - 1} = \frac{6}{- 3} = - 2\] \[\text {Slope of ST} = \frac{4 - 2}{- 3 - \left( - 2 \right)} = \frac{2}{- 1} = - 2\] Slope of RS = Slope of ST Is there an error in this question or solution? Page 6A(–4, 4), \[K\left( - 2, \frac{5}{2} \right),\] N (4, –2) \[\text { Slope of AK } = \frac{\frac{5}{2} - 4}{- 2 - \left( - 4 \right)} = \frac{\frac{- 3}{2}}{2} = \frac{- 3}{4}\] \[\text { Slope of KN } = \frac{- 2 - \frac{5}{2}}{4 - \left( - 2 \right)} = \frac{- 3}{4}\] Slope of AK=Slope of KN |
Check whether the points A(1,3), B(0,0) and C 1,3 are collinear or not
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