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Vector Algebra question

2022 · 25 Jun · Shift 1 · Q40
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  5. /2022 · 25 Jun · Shift 1 · Q40

Vector Algebra question

2022 · 25 Jun · Shift 1 · Q40

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let θ\thetaθ be the angle between the vectors a→\overrightarrow aa and b→\overrightarrow bb, where ∣a→∣=4,∣b→∣=3|\overrightarrow a | = 4,|\overrightarrow b | = 3∣a∣=4,∣b∣=3 and θ∈(π4,π3)\theta \in \left( {{\pi \over 4},{\pi \over 3}} \right)θ∈(4π​,3π​). Then ∣(a→−b→)×(a→+b→)∣2+4(a→ . b→)2{\left| {\left( {\overrightarrow a - \overrightarrow b } \right) \times \left( {\overrightarrow a + \overrightarrow b } \right)} \right|^2} + 4{\left( {\overrightarrow a \,.\,\overrightarrow b } \right)^2}​(a−b)×(a+b)​2+4(a.b)2 is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 576

  1. We need to evaluate
∣ (a⃗−b⃗)×(a⃗+b⃗) ∣2+4(a⃗⋅b⃗)2\left|\, (\vec a-\vec b)\times(\vec a+\vec b)\,\right|^2+4(\vec a\cdot\vec b)^2​(a−b)×(a+b)​2+4(a⋅b)2

given ∣a⃗∣=4,∣b⃗∣=3,θ∈(π4,π3).|\vec a|=4,\quad |\vec b|=3,\quad \theta\in\left(\frac\pi4,\frac\pi3\right).∣a∣=4,∣b∣=3,θ∈(4π​,3π​).

  1. First simplify the cross product:
(a⃗−b⃗)×(a⃗+b⃗)=a⃗×a⃗+a⃗×b⃗−b⃗×a⃗−b⃗×b⃗.(\vec a-\vec b)\times(\vec a+\vec b) =\vec a\times\vec a+\vec a\times\vec b-\vec b\times\vec a-\vec b\times\vec b.(a−b)×(a+b)=a×a+a×b−b×a−b×b.

Now,

a⃗×a⃗=0⃗,b⃗×b⃗=0⃗,\vec a\times\vec a=\vec 0,\qquad \vec b\times\vec b=\vec 0,a×a=0,b×b=0,

and

b⃗×a⃗=−(a⃗×b⃗).\vec b\times\vec a=-(\vec a\times\vec b).b×a=−(a×b).

So,

(a⃗−b⃗)×(a⃗+b⃗)=a⃗×b⃗−(−(a⃗×b⃗))=2(a⃗×b⃗).(\vec a-\vec b)\times(\vec a+\vec b) =\vec a\times\vec b-\bigl(-(\vec a\times\vec b)\bigr) =2(\vec a\times\vec b).(a−b)×(a+b)=a×b−(−(a×b))=2(a×b).

Hence,

∣(a⃗−b⃗)×(a⃗+b⃗)∣2=∣2(a⃗×b⃗)∣2=4∣a⃗×b⃗∣2.\left|(\vec a-\vec b)\times(\vec a+\vec b)\right|^2 =|2(\vec a\times\vec b)|^2 =4|\vec a\times\vec b|^2.​(a−b)×(a+b)​2=∣2(a×b)∣2=4∣a×b∣2.
  1. Therefore the required expression becomes
4∣a⃗×b⃗∣2+4(a⃗⋅b⃗)2=4(∣a⃗×b⃗∣2+(a⃗⋅b⃗)2).4|\vec a\times\vec b|^2+4(\vec a\cdot\vec b)^2 =4\left(|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2\right).4∣a×b∣2+4(a⋅b)2=4(∣a×b∣2+(a⋅b)2).
  1. Use the identity
∣a⃗×b⃗∣2+(a⃗⋅b⃗)2=∣a⃗∣2∣b⃗∣2.|\vec a\times\vec b|^2+(\vec a\cdot\vec b)^2=|\vec a|^2|\vec b|^2.∣a×b∣2+(a⋅b)2=∣a∣2∣b∣2.

So,

4(∣a⃗∣2∣b⃗∣2)=4(42)(32).4\left(|\vec a|^2|\vec b|^2\right)=4(4^2)(3^2).4(∣a∣2∣b∣2)=4(42)(32).
  1. Compute:
4⋅16⋅9=576.4\cdot 16\cdot 9=576.4⋅16⋅9=576.
  1. Thus the required integer is
576.\boxed{576}.576​.
  1. Comparison with stored answer: Stored correct answer = 576576576, which matches our derived result.
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