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JEE Physics 2023 Question with Solution

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 B^\hat{B} will be:

  • A

    33

  • B

    5\sqrt{5}

  • C

    33\sqrt{33}

  • D

    6\sqrt{6}

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: A⃗=2i^+3j^+2k^\vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k} and B⃗−A⃗=2j^\vec{B} - \vec{A} = 2\hat{j}.

Find: The magnitude of B⃗\vec{B}.

From the given equation,

B⃗−A⃗=2j^\vec{B} - \vec{A} = 2\hat{j}

Adding A⃗\vec{A} to both sides,

B⃗=2j^+A⃗\vec{B} = 2\hat{j} + \vec{A}

Substituting A⃗=2i^+3j^+2k^\vec{A} = 2\hat{i} + 3\hat{j} + 2\hat{k},

B⃗=2j^+(2i^+3j^+2k^)\vec{B} = 2\hat{j} + (2\hat{i} + 3\hat{j} + 2\hat{k})

So,

B⃗=2i^+5j^+2k^\vec{B} = 2\hat{i} + 5\hat{j} + 2\hat{k}

Now use the magnitude formula,

∣B⃗∣=x2+y2+z2|\vec{B}| = \sqrt{x^2 + y^2 + z^2}

Therefore,

∣B⃗∣=22+52+22|\vec{B}| = \sqrt{2^2 + 5^2 + 2^2} ∣B⃗∣=4+25+4=33|\vec{B}| = \sqrt{4 + 25 + 4} = \sqrt{33}

Therefore, the magnitude of vector B⃗\vec{B} is 33\sqrt{33}. The correct option is C.

Common mistakes

  • Using B⃗−A⃗=2j^\vec{B} - \vec{A} = 2\hat{j} incorrectly as B⃗=A⃗−2j^\vec{B} = \vec{A} - 2\hat{j}. This changes the sign of the j^\hat{j} component. Add A⃗\vec{A} to both sides to get B⃗=2j^+A⃗\vec{B} = 2\hat{j} + \vec{A}.

  • Calculating magnitude by adding components directly, such as 2+5+22 + 5 + 2. Magnitude is not the simple sum of components. Use ∣B⃗∣=x2+y2+z2|\vec{B}| = \sqrt{x^2 + y^2 + z^2} instead.

  • Confusing B^\hat{B} with B⃗\vec{B} in the question statement. The solution works with the vector B⃗\vec{B} and asks for its magnitude. Do not interpret B^\hat{B} as a unit vector here.

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