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JEE Main 202311 AprilEvening Shift
Exam
JEE Main
Year
2023
Date
11 April
Shift
Evening Shift
Subject
Physics
Chapter
Mathematics in Physics

JEE Main 2023 · 11 April · Evening Shift · Physics · Mathematics in Physics

When vector → A = 2 ^ i + 3 ^ j + 2 ^ k is subtracted from vector →...

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

Options

  1. A5\sqrt{5}
  2. B33
  3. C6\sqrt{6}
  4. D33\sqrt{33}

Correct answer

D. 33\sqrt{33}

Step-by-step solution

It is given that \overset{\to}{A} = 2\overset{^}{i} + 3\overset{^}{j} + 2\overset{^}{k}.

By the given condition,

\overset{\to}{B} - ( 2\overset{^}{i} + 3\overset{^}{j} + 2\overset{^}{k} ) = 2\overset{^}{j}

⇒ \overset{\to}{B} = 2\overset{^}{i} + 5\overset{^}{j} + 2\overset{^}{k}

The magnitude is given by the formula,\left|\overset{\to}{B}\right| = \sqrt{{B}^{2}x + {B}^{2}y + {B}^{2}z}<br/><br />⇒ | \overset{\to}{B} | = \sqrt{4 + 25 + 4} = \sqrt{33}

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