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Vector Algebra question

2021 · 25 Feb · Shift 2 · Q46
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Vector Algebra question

2021 · 25 Feb · Shift 2 · Q46

JEE MainMathematicsVector AlgebraNumerical+4 / −1
Let a→=i^+αj^+3k^\overrightarrow a = \widehat i + \alpha \widehat j + 3\widehat ka=i+αj​+3k and b→=3i^−αj^+k^\overrightarrow b = 3\widehat i - \alpha \widehat j + \widehat kb=3i−αj​+k. If the area of the parallelogram whose adjacent sides are represented by the vectors a→\overrightarrow aa and b→\overrightarrow bb is 838\sqrt 383​ square units, then a→\overrightarrow aa. b→\overrightarrow bb is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. The area of the parallelogram formed by vectors a⃗\vec aa and b⃗\vec bb is
∣a⃗×b⃗∣=83.|\vec a \times \vec b| = 8\sqrt{3}.∣a×b∣=83​.

Given

a⃗=i^+αj^+3k^=(1,α,3),\vec a = \hat i + \alpha \hat j + 3\hat k = (1,\alpha,3),a=i^+αj^​+3k^=(1,α,3), b⃗=3i^−αj^+k^=(3,−α,1).\vec b = 3\hat i - \alpha \hat j + \hat k = (3,-\alpha,1).b=3i^−αj^​+k^=(3,−α,1).
  1. First compute the dot product:
a⃗⋅b⃗=(1)(3)+(α)(−α)+(3)(1)=3−α2+3=6−α2.\vec a \cdot \vec b = (1)(3) + (\alpha)(-\alpha) + (3)(1) = 3 - \alpha^2 + 3 = 6 - \alpha^2.a⋅b=(1)(3)+(α)(−α)+(3)(1)=3−α2+3=6−α2.
  1. Compute the magnitudes squared:
∣a⃗∣2=12+α2+32=α2+10,|\vec a|^2 = 1^2 + \alpha^2 + 3^2 = \alpha^2 + 10,∣a∣2=12+α2+32=α2+10, ∣b⃗∣2=32+(−α)2+12=α2+10.|\vec b|^2 = 3^2 + (-\alpha)^2 + 1^2 = \alpha^2 + 10.∣b∣2=32+(−α)2+12=α2+10.

So,

∣a⃗∣2∣b⃗∣2=(α2+10)2.|\vec a|^2|\vec b|^2 = (\alpha^2+10)^2.∣a∣2∣b∣2=(α2+10)2.
  1. Use the identity
∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2.|\vec a \times \vec b|^2 = |\vec a|^2|\vec b|^2 - (\vec a\cdot \vec b)^2.∣a×b∣2=∣a∣2∣b∣2−(a⋅b)2.

Since ∣a⃗×b⃗∣=83|\vec a \times \vec b|=8\sqrt3∣a×b∣=83​,

(83)2=(α2+10)2−(6−α2)2.(8\sqrt3)^2 = (\alpha^2+10)^2 - (6-\alpha^2)^2.(83​)2=(α2+10)2−(6−α2)2. 192=(α2+10)2−(α2−6)2.192 = (\alpha^2+10)^2 - (\alpha^2-6)^2.192=(α2+10)2−(α2−6)2.
  1. Apply difference of squares:
A2−B2=(A−B)(A+B),A^2-B^2=(A-B)(A+B),A2−B2=(A−B)(A+B),

with

A=α2+10,B=α2−6.A=\alpha^2+10, \quad B=\alpha^2-6.A=α2+10,B=α2−6.

Then

192=[(α2+10)−(α2−6)] [(α2+10)+(α2−6)].192 = [(\alpha^2+10)-(\alpha^2-6)]\,[(\alpha^2+10)+(\alpha^2-6)].192=[(α2+10)−(α2−6)][(α2+10)+(α2−6)]. 192=(16)(2α2+4).192 = (16)(2\alpha^2+4).192=(16)(2α2+4). 192=32(α2+2).192 = 32(\alpha^2+2).192=32(α2+2). α2+2=6\alpha^2+2=6α2+2=6 α2=4.\alpha^2=4.α2=4.
  1. Now find a⃗⋅b⃗\vec a\cdot\vec ba⋅b:
a⃗⋅b⃗=6−α2=6−4=2.\vec a\cdot\vec b = 6-\alpha^2 = 6-4 = 2.a⋅b=6−α2=6−4=2.

Therefore,

a⃗⋅b⃗=2.\boxed{\vec a\cdot\vec b = 2}.a⋅b=2​.
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