3D Geometry
Direction Cosines — Cosine of Acute Angle
nta_pyq_2026_jan
Grade None
Question:
Let the direction cosines of two lines satisfy the equations: $4l+m-n=0$ and $2mn+10nl+3lm=0$. Then the cosine of the acute angle between these lines is:
\dfrac{20}{3\sqrt{38}}
\dfrac{10}{3\sqrt{38}}
\dfrac{10}{7\sqrt{38}}
\dfrac{10}{\sqrt{38}}
Step-by-Step Solution
Key Concept: From $n=4l+m$; substitute in second: $2m(4l+m)+10l(4l+m)+3lm=0\Rightarrow40l^2+21lm+2m^2=0$. Let $t=l/m$: $40t^2+21t+2=0$.
$\cos\theta=\dfrac{10}{3\sqrt{38}}$.
Correct Answer: 2