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Vector Algebra question

2019 · 8 Apr · Shift 2 · Q54
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Vector Algebra question

2019 · 8 Apr · Shift 2 · Q54

JEE MainPhysicsVector AlgebraMCQ+4 / −1
Let ∣A1→∣=3\left| {\mathop {{A_1}}\limits^ \to } \right| = 3​A1​→​​=3, ∣A2→∣=5\left| {\mathop {{A_2}}\limits^ \to } \right| = 5​A2​→​​=5 and ∣A1→+A2→∣=5\left| {\mathop {{A_1}}\limits^ \to + \mathop {{A_2}}\limits^ \to } \right| = 5​A1​→​+A2​→​​=5. The value of (2A1→+3A2→)(3A1→−2A2→)\left( {2\mathop {{A_1}}\limits^ \to + 3\mathop {{A_2}}\limits^ \to } \right)\left( {3\mathop {{A_1}}\limits^ \to - \mathop {2{A_2}}\limits^ \to } \right)(2A1​→​+3A2​→​)(3A1​→​−2A2​→​) is :-
  1. A
    –118.5
  2. B
    –112.5
  3. C
    –99.5
  4. D
    –106.5
View written solutionFree

Correct answer: A

  1. Interpret the given expression as a dot product:
(2A⃗1+3A⃗2)⋅(3A⃗1−2A⃗2)(2\vec A_1+3\vec A_2)\cdot(3\vec A_1-2\vec A_2)(2A1​+3A2​)⋅(3A1​−2A2​)

We are given:

  1. First find A⃗1⋅A⃗2\vec A_1\cdot \vec A_2A1​⋅A2​ using
∣A⃗1+A⃗2∣2=∣A⃗1∣2+∣A⃗2∣2+2A⃗1⋅A⃗2|\vec A_1+\vec A_2|^2=|\vec A_1|^2+|\vec A_2|^2+2\vec A_1\cdot \vec A_2∣A1​+A2​∣2=∣A1​∣2+∣A2​∣2+2A1​⋅A2​

Substitute values:

52=32+52+2A⃗1⋅A⃗25^2=3^2+5^2+2\vec A_1\cdot \vec A_252=32+52+2A1​⋅A2​ 25=9+25+2A⃗1⋅A⃗225=9+25+2\vec A_1\cdot \vec A_225=9+25+2A1​⋅A2​ 25=34+2A⃗1⋅A⃗225=34+2\vec A_1\cdot \vec A_225=34+2A1​⋅A2​ 2A⃗1⋅A⃗2=−92\vec A_1\cdot \vec A_2=-92A1​⋅A2​=−9 A⃗1⋅A⃗2=−92\vec A_1\cdot \vec A_2=-\frac{9}{2}A1​⋅A2​=−29​
  1. Now expand the required dot product:
(2A⃗1+3A⃗2)⋅(3A⃗1−2A⃗2)(2\vec A_1+3\vec A_2)\cdot(3\vec A_1-2\vec A_2)(2A1​+3A2​)⋅(3A1​−2A2​)

Using distributive property,

=2A⃗1⋅3A⃗1+2A⃗1⋅(−2A⃗2)+3A⃗2⋅3A⃗1+3A⃗2⋅(−2A⃗2)=2\vec A_1\cdot 3\vec A_1+2\vec A_1\cdot(-2\vec A_2)+3\vec A_2\cdot 3\vec A_1+3\vec A_2\cdot(-2\vec A_2)=2A1​⋅3A1​+2A1​⋅(−2A2​)+3A2​⋅3A1​+3A2​⋅(−2A2​) =6∣A⃗1∣2−4(A⃗1⋅A⃗2)+9(A⃗2⋅A⃗1)−6∣A⃗2∣2=6|\vec A_1|^2-4(\vec A_1\cdot\vec A_2)+9(\vec A_2\cdot\vec A_1)-6|\vec A_2|^2=6∣A1​∣2−4(A1​⋅A2​)+9(A2​⋅A1​)−6∣A2​∣2

Since A⃗1⋅A⃗2=A⃗2⋅A⃗1\vec A_1\cdot\vec A_2=\vec A_2\cdot\vec A_1A1​⋅A2​=A2​⋅A1​,

=6∣A⃗1∣2+5(A⃗1⋅A⃗2)−6∣A⃗2∣2=6|\vec A_1|^2+5(\vec A_1\cdot\vec A_2)-6|\vec A_2|^2=6∣A1​∣2+5(A1​⋅A2​)−6∣A2​∣2
  1. Substitute the values:
=6(32)+5(−92)−6(52)=6(3^2)+5\left(-\frac{9}{2}\right)-6(5^2)=6(32)+5(−29​)−6(52) =6(9)−452−6(25)=6(9)-\frac{45}{2}-6(25)=6(9)−245​−6(25) =54−22.5−150=54-22.5-150=54−22.5−150 =−118.5=-118.5=−118.5
  1. Therefore, the correct option is:
A: −118.5\boxed{\text{A: }-118.5}A: −118.5​
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