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Question: Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear....

Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear.

Explanation

Solution

Hint: Here to show that the 3 points are collinear we have to find the vectors of the given points and calculate its magnitude. If the points are collinear it means all points lie in a straight line.

Complete step-by-step answer:
As you know in question, we have to prove three points are collinear. First of all, you have to know the condition for collinearity.

Three points A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear
If and only if ∣AB→∣+∣BC→∣=∣AC→∣\left| {\overrightarrow {AB} } \right| + \left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {AC} } \right|

First find the vectors from AB→,BC→,AC→\overrightarrow {AB} ,\overrightarrow {BC} ,\overrightarrow {AC}
AB→=(2−1)i^+(6−2)j^+(3−7)k^\overrightarrow {AB} = \left( {2 - 1} \right)\widehat i + \left( {6 - 2} \right)\widehat j + \left( {3 - 7} \right)\widehat k
AB→=i^+4j^−4k^\overrightarrow {AB} = \widehat i + 4\widehat j - 4\widehat k
BC→=(3−2)i^+(10−6)j^+(−1−3)k^\overrightarrow {BC} = \left( {3 - 2} \right)\widehat i + \left( {10 - 6} \right)\widehat j + \left( { - 1 - 3} \right)\widehat k
BC→=i^+4j^−4k^\overrightarrow {BC} = \widehat i + 4\widehat j - 4\widehat k
AC→=(3−1)i^+(10−2)j^+(−1−7)k^\overrightarrow {AC} = \left( {3 - 1} \right)\widehat i + \left( {10 - 2} \right)\widehat j + \left( { - 1 - 7} \right)\widehat k
AC→=2i^+8j^−8k^\overrightarrow {AC} = 2\widehat i + 8\widehat j - 8\widehat k
Now we have to calculate magnitude of these vectors AB→,BC→,AC→\overrightarrow {AB} ,\overrightarrow {BC} , \overrightarrow {AC}
Magnitude of ∣AB→∣=12+42+(−4)2\left| {\overrightarrow {AB} } \right| = \sqrt {{1^2} + {4^2} + {{\left( { - 4} \right)}^2}}
∣AB→∣=1+16+16=33\left| {\overrightarrow {AB} } \right| = \sqrt {1 + 16 + 16} = \sqrt {33}
Magnitude of ∣BC→∣=12+42+(−4)2\left| {\overrightarrow {BC} } \right| = \sqrt {{1^2} + {4^2} + {{\left( { - 4} \right)}^2}}
∣BC→∣=1+16+16=33\left| {\overrightarrow {BC} } \right| = \sqrt {1 + 16 + 16} = \sqrt {33}
Magnitude of ∣AC→∣=22+82+(−8)2\left| {\overrightarrow {AC} } \right| = \sqrt {{2^2} + {8^2} + {{\left( { - 8} \right)}^2}}
∣AC→∣=4+64+64=132=4×33\left| {\overrightarrow {AC} } \right| = \sqrt {4 + 64 + 64} = \sqrt {132} = \sqrt {4 \times 33}
∣AC→∣=233\left| {\overrightarrow {AC} } \right| = 2\sqrt {33}

Now put the magnitude of these vectors In condition of collinearity.
∣AB→∣+∣BC→∣=33+33=233\left| {\overrightarrow {AB} } \right| + \left| {\overrightarrow {BC} } \right| = \sqrt {33} + \sqrt {33} = 2\sqrt {33}
∣AC→∣=233\left| {\overrightarrow {AC} } \right| = 2\sqrt {33}

Now you can easily see condition of collinearity satisfy
∣AB→∣+∣BC→∣=∣AC→∣=233\left| {\overrightarrow {AB} } \right| + \left| {\overrightarrow {BC} } \right| = \left| {\overrightarrow {AC} } \right| = 2\sqrt {33}

Hence proved three point A(1, 2, 7), B(2, 6, 3) and C(3, 10, -1) are collinear

Note: Whenever you come to this type of problem, always apply the condition of collinearity. If some points are collinear it means all points lie in a straight line. It’s the geometrical application of collinearity.