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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2017 · Mathematics · Three Dimensional Geometry

The point of intersection of the line joining the points (-3,4,-8)...

The point of intersection of the line joining the points (−3,4,−8)(-3,4,-8) and (5,−6,4)(5,-6,4) with the XY-plane is

Options

  1. A(73,−83,0)\left(\frac{7}{3},-\frac{8}{3}, 0\right)
  2. B(−73,−83,0)\left(-\frac{7}{3},-\frac{8}{3}, 0\right)
  3. C(−73,83,0)\left(-\frac{7}{3}, \frac{8}{3}, 0\right)
  4. D(73,83,0)\left(\frac{7}{3}, \frac{8}{3}, 0\right)

Correct answer

A. (73,−83,0)\left(\frac{7}{3},-\frac{8}{3}, 0\right)

Step-by-step solution

The equation of the line joining the points (−3,4,−8)and(5,−6,4)(-3,4,-8) and (5,-6,4) is x+38=y−4−10=z+812=k( say )⇒x+3=8k;y−4=−10k;z+8=12k⇒x=8k−3;y=−10k+4;z=12k−8\begin{aligned} & \frac{x+3}{8}=\frac{y-4}{-10}=\frac{z+8}{12}=k(\text{ say }) \\ & \Rightarrow x+3=8 k ; y-4=-10 k ; z+8=12 k \\ & \Rightarrow x=8 k-3 ; y=-10 k+4 ; z=12 k-8\end{aligned} Given that this line intersects with xy plane. So,z=0∴12k−8=0⇒12k=8⇒k=812=23.∴x=8(23)−3;y=−10(23)+4;z=12(23)−8⇒x=163−3;y=−203+4;z=243−8\mathrm{So}, \mathrm{z}=0\therefore 12 \mathrm{k}-8=0 \Rightarrow 12 \mathrm{k}=8 \Rightarrow \mathrm{k}=\frac{8}{12}=\frac{2}{3} .\therefore x=8\left(\frac{2}{3}\right)-3 ; y=-10\left(\frac{2}{3}\right)+4 ; z=12\left(\frac{2}{3}\right)-8\Rightarrow \mathrm{x}=\frac{16}{3}-3 ; \mathrm{y}=\frac{-20}{3}+4 ; \mathrm{z}=\frac{24}{3}-8⇒x=73;y=−83;z=0∴(x,y,z)=(73,−83,0)\begin{aligned} & \Rightarrow \mathrm{x}=\frac{7}{3} ; \mathrm{y}=\frac{-8}{3} ; \mathrm{z}=0 \\ & \therefore(\mathrm{x}, \mathrm{y}, \mathrm{z})=\left(\frac{7}{3}, \frac{-8}{3}, 0\right)\end{aligned}$
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