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BITSAT 2012
Exam
BITSAT
Year
2012
Subject
Mathematics
Chapter
Vector Algebra

BITSAT 2012 · Mathematics · Vector Algebra

The unit vector perpendicular to the vectors 6 i +2 j +3 k and 3 i...

The unit vector perpendicular to the vectors 6i^+2j^+3k^6 \hat{i}+2 \hat{j}+3 \hat{k} and 3i^−6j^−2k^3 \hat{i}-6 \hat{j}-2 \hat{k} is −-

Options

  1. A2i^−3j^+6k^7\frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7}
  2. B2i^−3j^−6k^7\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}
  3. C2i^+3j^−6k^7\frac{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}
  4. D2i^+3j^+6k^7\frac{2 \hat{i}+3 \hat{j}+6 \hat{k}}{7}

Correct answer

C. 2i^+3j^−6k^7\frac{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}

Step-by-step solution

Unit vector perpendicular to both the given vectors is, (6i^+2j^+3k^)×(3i^−6j^−2k^)∣(6i^+2j^+3k^)×(3i^−6j^−2k^)∣=2i^+3j^−6k^7\frac{(6 \hat{i}+2 \hat{j}+3 \hat{k}) \times(3 \hat{i}-6 \hat{j}-2 \hat{k})}{|(6 \hat{i}+2 \hat{j}+3 \hat{k}) \times(3 \hat{i}-6 \hat{j}-2 \hat{k})|}=\frac{2 \hat{i}+3 \hat{j}-6 \hat{k}}{7}
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