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NDA 2019
Exam
NDA
Year
2019
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2019 · Mathematics · Three Dimensional Geometry

If the position vectors of points A and B are 3 i -2 j + k and 2 i...

If the position vectors of points A and B are 3i^−2j^+k^3 \hat{i}-2 \hat{j}+\hat{k} and 2i^+4j^−3k^2 \hat{i}+4 \hat{j}-3 \hat{k} respectively, then what is the length of AB→\overrightarrow{A B} ?

Options

  1. A14\sqrt{14}
  2. B29\sqrt{29}
  3. C43\sqrt{43}
  4. D53\sqrt{53}

Correct answer

D. 53\sqrt{53}

Step-by-step solution

Given, OA→=3i^−2j^+k^\overrightarrow{O A}=3 \hat{i}-2 \hat{j}+\hat{k}OB→=2i^+4j^−3k^AB→=OB→−OA→=(2i^+4j^−3k^)−(3i^−2j^+k^)=i^(2−3)+j^(4+2)+k^(−3−1)=−i^+6j^−4k^\begin{aligned} & \overrightarrow{O B}=2 \hat{i}+4 \hat{j}-3 \hat{k} \\ & \overrightarrow{A B}=\overrightarrow{O B}-\overrightarrow{O A} \\ & =(2 \hat{i}+4 \hat{j}-3 \hat{k})-(3 \hat{i}-2 \hat{j}+\hat{k}) \\ & =\hat{i}(2-3)+\hat{j}(4+2)+\hat{k}(-3-1) \\ & =-\hat{i}+6 \hat{j}-4 \hat{k}\end{aligned}$\begin{aligned} & \text{ Length of } A B=\sqrt{(-1)^2+(6)^2+(-4)^2} \\ & =\sqrt{1+36+16}=\sqrt{53}\end{aligned}
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