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AIIMS 201827 May
Exam
AIIMS
Year
2018
Date
27 May
Subject
Physics
Chapter
Mathematics in Physics

AIIMS 2018 · 27 May · Physics · Mathematics in Physics

Two vectors are given by A =(3 i + j +3 k ) and B =(3 i +5 j -2 k )...

Two vectors are given by A⃗=(3i^+j^+3k^)\vec{A}=(3 \hat{i}+\hat{j}+3 \hat{k}) and B⃗=(3i^+5j^−2k^)\vec{B}=(3 \hat{i}+5 \hat{j}-2 \hat{k}). Find the third vector C⃗\vec{C} if A⃗+3B⃗−C⃗=0\vec{A}+3 \vec{B}-\vec{C}=0.

Options

  1. A(12i^+14j^+12k^)(12 \hat{i}+14 \hat{j}+12 \hat{k})
  2. B(13i^+17j^+12k^)(13 \hat{i}+17 \hat{j}+12 \hat{k})
  3. C(12i^+16j^−3k^)(12 \hat{i}+16 \hat{j}-3 \hat{k})
  4. D(15i^+13j^+4k^)(15 \hat{i}+13 \hat{j}+4 \hat{k})

Correct answer

C. (12i^+16j^−3k^)(12 \hat{i}+16 \hat{j}-3 \hat{k})

Step-by-step solution

 A⃗+3B⃗−C⃗=03i^+j^+3k^+3(3i^+5j^−2k^)−C⃗=012i^+16j^−3k^−C⃗=0,C⃗=12i^+16j^−3k^\begin{aligned} & \text{ } \vec{A}+3 \vec{B}-\vec{C}=0 \\ & 3 \hat{i}+\hat{j}+3 \hat{k}+3(3 \hat{i}+5 \hat{j}-2 \hat{k})-\vec{C}=0 \\ & 12 \hat{i}+16 \hat{j}-3 \hat{k}-\vec{C}=0, \vec{C}=12 \hat{i}+16 \hat{j}-3 \hat{k}\end{aligned}
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