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Worked Examples · Example 4

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

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Collinearity of three points in space is checked by showing that the vectors PQ→\overrightarrow{PQ} and PR→\overrightarrow{PR} are scalar multiples of each other. Here, PR→=3⋅PQ→\overrightarrow{PR} = 3 \cdot \overrightarrow{PQ}, so the points are collinear.

The key idea is simple: three points are collinear if they lie on the same straight line. In vector terms, this means the vector from one point to another is a scalar multiple of the vector from the first point to the third. If PQ→\overrightarrow{PQ} and PR→\overrightarrow{PR} are parallel (i.e., one is a constant times the other), then QQ and RR both lie on the line through PP in the same direction.

Why does this work? Because a line is defined by a point and a direction vector. If both PQ→\overrightarrow{PQ} and PR→\overrightarrow{PR} point in the same (or exactly opposite) direction, then QQ and RR are on the same line through PP. The scalar multiple can be any real number — positive means same direction, negative means opposite direction, but either way the points are collinear.

Let’s apply this to the given points.

  1. Find the vector PQ→\overrightarrow{PQ}. Subtract the coordinates of PP from QQ:

PQ→=(1−(−2),  2−3,  3−5)=(3,  −1,  −2)\overrightarrow{PQ} = (1 - (-2),\; 2 - 3,\; 3 - 5) = (3,\; -1,\; -2)

  1. Find the vector PR→\overrightarrow{PR}. Subtract the coordinates of PP from RR:

PR→=(7−(−2),  0−3,  −1−5)=(9,  −3,  −6)\overrightarrow{PR} = (7 - (-2),\; 0 - 3,\; -1 - 5) = (9,\; -3,\; -6)

  1. Check if PR→\overrightarrow{PR} is a scalar multiple of PQ→\overrightarrow{PQ}. Compare component by component:

93=3,−3−1=3,−6−2=3\frac{9}{3} = 3,\quad \frac{-3}{-1} = 3,\quad \frac{-6}{-2} = 3

All three ratios are equal to 33. Therefore,

PR→=3⋅PQ→\overrightarrow{PR} = 3 \cdot \overrightarrow{PQ}

Watch out

A common mistake is to check only two components and assume the third matches. Always verify all three ratios — if even one differs, the points are not collinear.

Since PR→\overrightarrow{PR} is exactly 33 times PQ→\overrightarrow{PQ}, the two vectors are parallel and share the point PP. This means QQ and RR both lie on the line through PP in the direction of PQ→\overrightarrow{PQ}.

Tip

You could also check collinearity using the area of the triangle formed by the three points. If the area is zero, they are collinear. The area is 12∣PQ→×PR→∣\frac{1}{2} | \overrightarrow{PQ} \times \overrightarrow{PR} |. Here, PQ→×PR→=0\overrightarrow{PQ} \times \overrightarrow{PR} = \mathbf{0}, confirming collinearity — but the scalar multiple method is faster.

✓Final answer

The points P(−2,3,5)P(-2, 3, 5), Q(1,2,3)Q(1, 2, 3) and R(7,0,−1)R(7, 0, -1) are collinear because PR→=3⋅PQ→\overrightarrow{PR} = 3 \cdot \overrightarrow{PQ}.

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