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Exercise 11.2 · Q2

Q.Show that the points (−2,3,5)(-2, 3, 5), (1,2,3)(1, 2, 3) and (7,0,−1)(7, 0, -1) are collinear.

Uttarakhand UbseTextbookSubjective· 3mImportance★★★★★est
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✓ Free question

Three points are collinear if the vectors formed by them are parallel — i.e., one is a scalar multiple of the other. Here, AB→=(3,−1,−2)\overrightarrow{AB} = (3, -1, -2) and BC→=(6,−2,−4)\overrightarrow{BC} = (6, -2, -4), and since BC→=2⋅AB→\overrightarrow{BC} = 2 \cdot \overrightarrow{AB}, the points are collinear.

The idea is simple: three points AA, BB, CC lie on a straight line if the direction from AA to BB is the same as the direction from BB to CC (or from AA to CC). In vector terms, this means the vectors AB→\overrightarrow{AB} and BC→\overrightarrow{BC} are parallel — one is a scalar multiple of the other.

Why does this work? Because if AA, BB, CC are collinear, then BB lies somewhere on the line through AA and CC. So the displacement from AA to BB and from BB to CC must point along the same line. Checking this scalar multiple condition is the cleanest method — no need to find equations of lines or compute distances.

Let’s apply it step by step.

  1. Label the points

    Let A=(−2,3,5)A = (-2, 3, 5), B=(1,2,3)B = (1, 2, 3), C=(7,0,−1)C = (7, 0, -1).

  2. Find vector AB→\overrightarrow{AB}

    AB→=B−A=(1−(−2),  2−3,  3−5)=(3,−1,−2)\overrightarrow{AB} = B - A = (1 - (-2),\; 2 - 3,\; 3 - 5) = (3, -1, -2).

  3. Find vector BC→\overrightarrow{BC}

    BC→=C−B=(7−1,  0−2,  −1−3)=(6,−2,−4)\overrightarrow{BC} = C - B = (7 - 1,\; 0 - 2,\; -1 - 3) = (6, -2, -4).

  4. Check if BC→\overrightarrow{BC} is a scalar multiple of AB→\overrightarrow{AB}

    Compare component by component:

    63=2\frac{6}{3} = 2, −2−1=2\frac{-2}{-1} = 2, −4−2=2\frac{-4}{-2} = 2.

    All ratios are equal to 22. Hence BC→=2⋅AB→\overrightarrow{BC} = 2 \cdot \overrightarrow{AB}.

Tip

You could also check AC→\overrightarrow{AC} against AB→\overrightarrow{AB}: AC→=(9,−3,−6)=3⋅(3,−1,−2)\overrightarrow{AC} = (9, -3, -6) = 3 \cdot (3, -1, -2). Any pair works — the key is that all three vectors are parallel.

Since BC→\overrightarrow{BC} is exactly twice AB→\overrightarrow{AB}, the two vectors are parallel and share point BB. That means AA, BB, CC lie on the same straight line.

Watch out

A common mistake is to check only two pairs of coordinates and assume collinearity. Always verify all three components — a single mismatch (e.g., ratios not equal) means the points are not collinear.

✓Final answer

The points (−2,3,5)(-2, 3, 5), (1,2,3)(1, 2, 3) and (7,0,−1)(7, 0, -1) are collinear because BC→=2⋅AB→\overrightarrow{BC} = 2 \cdot \overrightarrow{AB}.

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