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Question: Show that the points (2, 3, 4), (-1, -2, 1) and (5, 8, 7) are collinear....

Show that the points (2, 3, 4), (-1, -2, 1) and (5, 8, 7) are collinear.

Explanation

Solution

Hint: To solve this problem, a distance formula between two points will be used. The distance formula between two points (x1,y1,z1),(x2,y2,z2)\left( {{x_1},{y_1},{z_1}} \right),\left( {{x_2},{y_2},{z_2}} \right) is-
d=(x2−x1)2+(y2−y1)2+(z2−z1)2\mathrm d=\sqrt{\left({\mathrm x}_2-{\mathrm x}_1\right)^2+\left({\mathrm y}_2-{\mathrm y}_1\right)^2+\left({\mathrm z}_2-{\mathrm z}_1\right)^2}
Complete step by step answer:
Let the points be A(2, 3, 4), B(-1, -2, 1) and C(5, 8, 7). For these points to be collinear, it is sufficient to prove that the sum of distances between two pairs of points is equal to the distance between the third pair. By applying distance formula in these three points-
\mathrm{AB}=\sqrt{\left(2-\left(-1\right)\right)^2+\left(3-\left(-2\right)\right)^2+\left(4-1\right)^2}\\\=\sqrt{3^2+5^2+3^2}=\sqrt{43}\\\\\mathrm{CA}=\sqrt{\left(5-2\right)^2+\left(8-3\right)^2+\left(7-4\right)^2}\\\=\sqrt{3^2+5^2+3^2}=\sqrt{43}\\\\\mathrm{BC}=\sqrt{\left(-1-5\right)^2+\left(-2-8\right)^2+\left(1-7\right)^2}\\\=\sqrt{\left(-6\right)^2+\left(-10\right)^2+\left(-6\right)^2}=\sqrt{172}=\sqrt{4\times43}=2\sqrt{43}\\\\\\\\\mathrm{Clearly}\;\mathrm{we}\;\mathrm{can}\;\mathrm{see}\;\mathrm{that}\;\mathrm{AB}+\mathrm{CA}=\sqrt{43}+\sqrt{43}=2\sqrt{43}=\mathrm{BC}
Hence, the three points A, B, C are collinear. Hence, proved.

Note: We can also solve this problem by using vector algebra. First we find the two vectors AB→  and  BC→\overrightarrow {AB} \;and\;\overrightarrow {BC} . Then we can show that the angle between them is zero.
AB→=(2−(−1))i^+(3−(−2))j^+(4−1)k^ AB→=3i^+5j^+3k^ Similarly;, BC→=(−5−1)i^+(−2−8)j^+(1−7)k^ BC→=−6i^−10j^−6k^ \begin{aligned} &\overrightarrow {AB} = \left( {2 - \left( { - 1} \right)} \right)\hat i + \left( {3 - \left( { - 2} \right)} \right)\hat j + \left( {4 - 1} \right)\hat k \\\ &\overrightarrow {AB} = 3\hat i + 5\hat j + 3\hat k \\\ &Similarly;, \\\ &\overrightarrow {BC} = \left( { - 5 - 1} \right)\hat i + \left( { - 2 - 8} \right)\hat j + \left( {1 - 7} \right)\hat k \\\ &\overrightarrow {BC} = - 6\hat i - 10\hat j - 6\hat k \\\ \end{aligned}

Clearly, it is visible that the direction ratios of AB→  and  BC→\overrightarrow {AB} \;and\;\overrightarrow {BC} are in proportion, hence they have an angle between them as 0. Hence, proved.