Mathematics in Physics

JEE Main level, single correct, medium

13 of 121

Question 13 of 121

When vector A⃗=2i^+3j^+2k^\vec{A}=2\hat{i}+3\hat{j}+2\hat{k} is subtracted from vector B⃗\vec{B}, it gives a vector equal to 2j^2\hat{j}. Then the magnitude of vector B⃗\vec{B} will be:

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Answer: (D) 33\sqrt{33}

It is given that A⃗=2i^+3j^+2k^\vec{A}=2\hat{i}+3\hat{j}+2\hat{k}.
By the given condition,
B⃗−(2i^+3j^+2k^)=2j^\vec{B}-(2\hat{i}+3\hat{j}+2\hat{k})=2\hat{j}
⇒B⃗=2i^+5j^+2k^\Rightarrow \vec{B}=2\hat{i}+5\hat{j}+2\hat{k}
The magnitude is given by the formula,∣B⃗∣=B2x+B2y+B2z\left|\vec{B}\right|=\sqrt{{{B}^{2}}_{x}+{{B}^{2}}_{y}+{{B}^{2}}_{z}}
⇒∣B⃗∣=4+25+4=33\Rightarrow |\vec{B}|=\sqrt{4+25+4}=\sqrt{33}

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