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Worked Examples · Example 8

Q.Find the angle between the pair of lines x+33=y−15=z+34\dfrac{x+3}{3} = \dfrac{y-1}{5} = \dfrac{z+3}{4} and x+11=y−41=z−52\dfrac{x+1}{1} = \dfrac{y-4}{1} = \dfrac{z-5}{2}.

Punjab PsebTextbookSubjective· 3mImportance★★★★★
Appeared in past exams:COMEDK 2022· Set 2022· 1mreworded
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The angle between two lines in space is found using the dot product of their direction vectors. For the given lines, the direction vectors are (3,5,4)(3,5,4) and (1,1,2)(1,1,2); their dot product is 3+5+8=163+5+8=16, and the cosine of the angle is 16506=16103=853\frac{16}{\sqrt{50}\sqrt{6}} = \frac{16}{10\sqrt{3}} = \frac{8}{5\sqrt{3}}. The angle is cos⁡−1 ⁣(853)\cos^{-1}\!\left(\frac{8}{5\sqrt{3}}\right).

The key idea: two lines in 3D are defined by their direction vectors. The angle between the lines is simply the angle between these vectors — found using the dot product formula. There’s no need to worry about where the lines are located; only their direction matters.


  1. Extract the direction vectors

    Each line is given in symmetric form x−x0a=y−y0b=z−z0c\frac{x-x_0}{a} = \frac{y-y_0}{b} = \frac{z-z_0}{c}, where (a,b,c)(a,b,c) is the direction vector.

    For the first line: x+33=y−15=z+34\frac{x+3}{3} = \frac{y-1}{5} = \frac{z+3}{4} → direction vector d⃗1=(3,5,4)\vec{d}_1 = (3,5,4).

    For the second line: x+11=y−41=z−52\frac{x+1}{1} = \frac{y-4}{1} = \frac{z-5}{2} → direction vector d⃗2=(1,1,2)\vec{d}_2 = (1,1,2).

  2. Recall the formula for the angle between two vectors

    If θ\theta is the angle between d⃗1\vec{d}_1 and d⃗2\vec{d}_2, then

cos⁡θ=d⃗1⋅d⃗2∣d⃗1∣ ∣d⃗2∣.\cos\theta = \frac{\vec{d}_1 \cdot \vec{d}_2}{|\vec{d}_1|\,|\vec{d}_2|}.

  1. Compute the dot product

d⃗1⋅d⃗2=3⋅1+5⋅1+4⋅2=3+5+8=16.\vec{d}_1 \cdot \vec{d}_2 = 3\cdot1 + 5\cdot1 + 4\cdot2 = 3 + 5 + 8 = 16.

  1. Compute the magnitudes

∣d⃗1∣=32+52+42=9+25+16=50=52.|\vec{d}_1| = \sqrt{3^2 + 5^2 + 4^2} = \sqrt{9 + 25 + 16} = \sqrt{50} = 5\sqrt{2}.

∣d⃗2∣=12+12+22=1+1+4=6.|\vec{d}_2| = \sqrt{1^2 + 1^2 + 2^2} = \sqrt{1 + 1 + 4} = \sqrt{6}.

  1. Find cos⁡θ\cos\theta …

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