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NDA 2015
Exam
NDA
Year
2015
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2015 · Mathematics · Three Dimensional Geometry

The lines 2 x=3 y=-z and 6 x=-y=-4 z

The lines 2x=3y=−z2 x=3 y=-z and 6x=−y=−4z6 x=-y=-4 z

Options

  1. Aare perpendicular
  2. Bare parallel
  3. Cintersect at an angle 45∘45^{{}^{\circ}}
  4. Dintersect at an angle 60∘60^{{}^{\circ}}

Correct answer

A. are perpendicular

Step-by-step solution

2 x=3 y=-z or x3=y2=z−66x=−y=−4z\frac{x}{3}=\frac{y}{2}=\frac{z}{-6}6 x=-y=-4 z or x2=y−12=z−3\frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{-12}=\frac{\mathrm{z}}{-3}cos⁡θ=x1x2+y1y2+z1z2x12+y12+z12⋅x22+y22+z32=(6−24+18)32+32+(−6)2⋅22+(−12)2+(−3)2cos⁡θ=0θ=90∘\begin{aligned} & \cos \theta=\frac{\mathrm{x}_1 \mathrm{x}_2+\mathrm{y}_1 \mathrm{y}_2+\mathrm{z}_1 \mathrm{z}_2}{\sqrt{\mathrm{x}_1^2+\mathrm{y}_1^2+\mathrm{z}_1^2} \cdot \sqrt{\mathrm{x}_2^2+\mathrm{y}_2^2+\mathrm{z}_3^2}} \\ & =\frac{(6-24+18)}{\sqrt{3^2+3^2+(-6)^2} \cdot \sqrt{2^2+(-12)^2+(-3)^2}} \\ & \cos \theta=0 \\ & \theta=90^{{}^{\circ}} \end{aligned}$$ So lines are perpendicular
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