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NDA 2020
Exam
NDA
Year
2020
Subject
Mathematics
Chapter
Three Dimensional Geometry

NDA 2020 · Mathematics · Three Dimensional Geometry

Let A be a point in space such that | O A |=12 , where O is the...

Let AA be a point in space such that ∣OA→∣=12|\overrightarrow{O A}|=12, where OO is the origin. If OA→\overrightarrow{O A} is inclined at angles 45∘45^{{}^{\circ}} and 60∘60^{{}^{\circ}} with xx axis and yy-axis respectively, then what is OA→\overrightarrow{O A} equal to ?

Options

  1. A6i^+6j^±2k^6 \hat{i}+6 \hat{j} \pm \sqrt{2} \hat{k}
  2. B6i^+62j^±6k^6 \hat{i}+6 \sqrt{2} \hat{j} \pm 6 \hat{k}
  3. C62i^+6j^±6k^6 \sqrt{2} \hat{i}+6 \hat{j} \pm 6 \hat{k}
  4. D32i^+3j^±6k^3 \sqrt{2} \hat{i}+3 \hat{j} \pm 6 \hat{k}

Correct answer

C. 62i^+6j^±6k^6 \sqrt{2} \hat{i}+6 \hat{j} \pm 6 \hat{k}

Step-by-step solution

Let A=(x,y,z)A=(x, y, z) Then, ∣OA→∣=x2+y2+z2=12|\overrightarrow{O A}|=\sqrt{x^2+y^2+z^2}=12 Also, x=∣OA→∣⋅cos⁡45∘=12⋅12=62i^x=|\overrightarrow{O A}| \cdot \cos 45^{{}^{\circ}}=12 \cdot \frac{1}{\sqrt{2}}=6 \sqrt{2} \hat{i}. y=∣OA∣⋅cos⁡60∘=12⋅12=6j^y=|O A| \cdot \cos 60^{{}^{\circ}}=12 \cdot \frac{1}{2}=6 \hat{j} Hence, OA→=62i^+6j^±6k^\overrightarrow{O A}=6 \sqrt{2} \hat{i}+6 \hat{j} \pm 6 \hat{k}
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