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NDA 2018 Maths · Mathematics · Vector Algebra · hard

A vector vec a satisfies vec a·hat i=vec a·(hat i+hat j)=vec a·(hat i+hat j+hat k)=1. Then vec a equals

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A vector a⃗\vec a satisfies a⃗⋅i^=a⃗⋅(i^+j^)=a⃗⋅(i^+j^+k^)=1\vec a\cdot\hat i=\vec a\cdot(\hat i+\hat j)=\vec a\cdot(\hat i+\hat j+\hat k)=1. Then a⃗\vec a equals
  1. A. i^+j^+k^\hat i+\hat j+\hat k
  2. B. i^\hat i
  3. C. j^+k^\hat j+\hat k
  4. D. i^+j^\hat i+\hat j
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Answer: (B) i^\hat i

Write a⃗=a1i^+a2j^+a3k^\vec a=a_1\hat i+a_2\hat j+a_3\hat k. The first condition gives a1=1a_1=1, the second gives a1+a2=1a_1+a_2=1 so a2=0a_2=0, and the third gives a1+a2+a3=1a_1+a_2+a_3=1 so a3=0a_3=0. Thus a⃗=i^\vec a=\hat i. The vector i^+j^+k^\hat i+\hat j+\hat k satisfies only the last condition.
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