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Physics Question on Vectors

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=0⃗\vec{A}+3 \vec{B}-\vec{C}=\vec{0}.

A

12i^+14j^+12k^12\hat{i}+14\hat{j}+12\hat{k}

B

13i^+17j^+12k^13\hat{i}+17\hat{j}+12\hat{k}

C

12i^+16j^−3k^12\hat{i}+16\hat{j}-3\hat{k}

D

15i^+13j^+4k^15\hat{i}+13\hat{j}+4\hat{k}

Answer

12i^+16j^−3k^12\hat{i}+16\hat{j}-3\hat{k}

Explanation

Solution

The correct answer is C:12i^+16j^−3k^12\hat{i}+16\hat{j}-3\hat{k}
Given, A⃗=3i^+j^+3k^\vec{A}=3\hat{i}+\hat{j}+3\hat{k}
B⃗=3i^+5j^−2k^\vec{B}=3\hat{i}+5\hat{j}-2\hat{k}
Here, A⃗+3B⃗=C⃗\vec{A}+3\vec{B}=\vec{C}
(3i^+j^+3k^)+3(3i^+5j^−2k^)=C⃗(3\hat{i}+\hat{j}+3\hat{k})+3(3\hat{i}+5\hat{j}-2\hat{k})=\vec{C}
⇒C⃗=12i^+16j^−3k^\Rightarrow \vec{C}=12\hat{i}+16\hat{j}-3\hat{k}