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Vector Algebra question

2023 · 11 Apr · Shift 2 · Q57
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  5. /2023 · 11 Apr · Shift 2 · Q57

Vector Algebra question

2023 · 11 Apr · Shift 2 · Q57

JEE MainPhysicsVector AlgebraMCQ+4 / −1
When vector A⃗=2i^+3j^+2k^\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k}A=2i^+3j^​+2k^ is subtracted from vector B→\overrightarrow{\mathrm{B}}B, it gives a vector equal to 2j^2 \hat{j}2j^​. Then the magnitude of vector B→\overrightarrow{\mathrm{B}}B will be :
  1. A
    3
  2. B
    33\sqrt{33}33​
  3. C
    6\sqrt66​
  4. D
    5\sqrt55​
View written solutionFree

Correct answer: B

  1. Interpret the statement

    “Vector A⃗\vec AA is subtracted from vector B⃗\vec BB and it gives 2j^2\hat j2j^​” means B⃗−A⃗=2j^\vec B - \vec A = 2\hat jB−A=2j^​

  2. Write the given vector

    A⃗=2i^+3j^+2k^\vec A = 2\hat i + 3\hat j + 2\hat kA=2i^+3j^​+2k^

    Also, 2j^=0i^+2j^+0k^2\hat j = 0\hat i + 2\hat j + 0\hat k2j^​=0i^+2j^​+0k^

  3. Find B⃗\vec BB

    From B⃗−A⃗=2j^\vec B - \vec A = 2\hat jB−A=2j^​ we get B⃗=A⃗+2j^\vec B = \vec A + 2\hat jB=A+2j^​

    Substitute A⃗\vec AA: B⃗=(2i^+3j^+2k^)+2j^\vec B = (2\hat i + 3\hat j + 2\hat k) + 2\hat jB=(2i^+3j^​+2k^)+2j^​ B⃗=2i^+5j^+2k^\vec B = 2\hat i + 5\hat j + 2\hat kB=2i^+5j^​+2k^

  4. Calculate the magnitude of B⃗\vec BB

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

  5. Check options

    • A: 333
    • B: 33\sqrt{33}33​ ✅
    • C: 6\sqrt66​
    • D: 5\sqrt55​

Therefore, the correct option is B.

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