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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2018 · Mathematics · Three Dimensional Geometry

Let the coordinates of the points A, B, C be (1,8,4) , (0,-11,4)...

Let the coordinates of the points A,B,C\mathrm{A}, \mathrm{B}, \mathrm{C} be (1,8,4)(1,8,4), (0,−11,4)(0,-11,4) and (2,−3,1)(2,-3,1) respectively. What are the coordinates of the point DD which is the foot of the perpendicular from A on BC ?

Options

  1. A(3,4,−2)(3,4,-2)
  2. B(4,−2,5)(4,-2,5)
  3. C(4,5,−2)(4,5,-2)
  4. D(2,4,5)(2,4,5)

Correct answer

C. (4,5,−2)(4,5,-2)

Step-by-step solution

A(1,8,4),B(0,−11,4),C(2,−3,1)Let D=(α,β,γ)\begin{aligned} & \mathrm{A}(1,8,4), \mathrm{B}(0,-11,4), \mathrm{C}(2,-3,1) \\ & \text{Let } \mathrm{D}=(\alpha, \beta, \gamma)\end{aligned} Direction ratios of BC=(x2−x1,y2−y1,z2−z1)Let,\mathrm{BC}= \left(\mathrm{x}_2- \mathrm{x}_1, \mathrm{y}_2- \mathrm{y}_1, \mathrm{z}_2- \mathrm{z}_1\right) Let,( a,b,c)=(2,8,−3)\mathrm{a}, \mathrm{b}, \mathrm{c})=(2,8,-3) Direction ratios of A D=(x2−x1,y2−y1,z2−z1)Let⁡(a′,b′,c′)=(α−1,β−8,γ−4)D=\left(\mathrm{x}_2- \mathrm{x}_1, \mathrm{y}_2- \mathrm{y}_1, \mathrm{z}_2- \mathrm{z}_1\right) \operatorname{Let}\left( a^{\prime}, b^{\prime}, c^{\prime}\right)=(\alpha-1, \beta-8, \gamma-4) Since, A Disperpendiculartois perpendicular toB C,,aa′+bb′+cc′=0⇒2(\aalpha−1)+8(\bbeta−8)−3(γ−4)=0⇒2α−2+8β−64−3γ+12=0⇒2α+8β−3γ−54=0...(1)\begin{aligned}& \mathrm{aa}^{\prime}+\mathrm{bb}^{\prime}+\mathrm{cc}^{\prime}=0 \\ & \Rightarrow 2(\a alpha-1)+8(\bbeta-8)-3(\gamma-4)=0 \\ & \Rightarrow 2 \alpha-2+8 \beta-64-3 \gamma+12=0 \\ & \Rightarrow 2 \alpha+8 \beta-3 \gamma-54=0...(1) \end{aligned}On substituting the options, we find option (c) is correct. when $(\alpha, \beta, \gamma)=(4,5,$-2)\begin{aligned}& (1) \Rightarrow 2(4)+8(5)-3(-2)-54=0 \\ & \Rightarrow 8+40+6-54=0 \\ & \Rightarrow 0=0 \end{aligned}$$
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