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Vector Algebra question

2022 · 28 Jun · Shift 2 · Q35
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  5. /2022 · 28 Jun · Shift 2 · Q35

Vector Algebra question

2022 · 28 Jun · Shift 2 · Q35

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a→=αi^+2j^−k^\overrightarrow a = \alpha \widehat i + 2\widehat j - \widehat ka=αi+2j​−k and b→=−2i^+αj^+k^\overrightarrow b = - 2\widehat i + \alpha \widehat j + \widehat kb=−2i+αj​+k, where α∈R\alpha \in Rα∈R. If the area of the parallelogram whose adjacent sides are represented by the vectors a→\overrightarrow aa and b→\overrightarrow bb is 15(α2+4)\sqrt {15({\alpha ^2} + 4)}15(α2+4)​, then the value of 2∣a→∣2+(a→ . b→)∣b→∣22{\left| {\overrightarrow a } \right|^2} + \left( {\overrightarrow a \,.\,\overrightarrow b } \right){\left| {\overrightarrow b } \right|^2}2​a​2+(a.b)​b​2 is equal to :
  1. A
    10
  2. B
    7
  3. C
    9
  4. D
    14
View written solutionFree

Correct answer: D

  1. Given vectors
a⃗=αi^+2j^−k^,b⃗=−2i^+αj^+k^\vec a=\alpha \hat i+2\hat j-\hat k, \qquad \vec b=-2\hat i+\alpha \hat j+\hat ka=αi^+2j^​−k^,b=−2i^+αj^​+k^

The area of the parallelogram formed by a⃗\vec aa and b⃗\vec bb is

∣a⃗×b⃗∣=15(α2+4)|\vec a\times \vec b|=\sqrt{15(\alpha^2+4)}∣a×b∣=15(α2+4)​
  1. Use the identity for area

We know

∣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

First compute the required quantities.

Magnitudes squared

∣a⃗∣2=α2+22+(−1)2=α2+5|\vec a|^2=\alpha^2+2^2+(-1)^2=\alpha^2+5∣a∣2=α2+22+(−1)2=α2+5 ∣b⃗∣2=(−2)2+α2+12=α2+5|\vec b|^2=(-2)^2+\alpha^2+1^2=\alpha^2+5∣b∣2=(−2)2+α2+12=α2+5

So,

∣a⃗∣2=∣b⃗∣2=α2+5|\vec a|^2=|\vec b|^2=\alpha^2+5∣a∣2=∣b∣2=α2+5

Dot product

a⃗⋅b⃗=α(−2)+2(α)+(−1)(1)=−2α+2α−1=−1\vec a\cdot \vec b=\alpha(-2)+2(\alpha)+(-1)(1)=-2\alpha+2\alpha-1=-1a⋅b=α(−2)+2(α)+(−1)(1)=−2α+2α−1=−1

Thus,

a⃗⋅b⃗=−1\vec a\cdot \vec b=-1a⋅b=−1
  1. Apply the area condition

Given

∣a⃗×b⃗∣=15(α2+4)|\vec a\times \vec b|=\sqrt{15(\alpha^2+4)}∣a×b∣=15(α2+4)​

Squaring both sides,

∣a⃗×b⃗∣2=15(α2+4)|\vec a\times \vec b|^2=15(\alpha^2+4)∣a×b∣2=15(α2+4)

But also,

∣a⃗×b⃗∣2=(α2+5)2−(−1)2=(α2+5)2−1|\vec a\times \vec b|^2=(\alpha^2+5)^2-(-1)^2=(\alpha^2+5)^2-1∣a×b∣2=(α2+5)2−(−1)2=(α2+5)2−1

Hence,

(α2+5)2−1=15(α2+4)(\alpha^2+5)^2-1=15(\alpha^2+4)(α2+5)2−1=15(α2+4)

Expand:

α4+10α2+25−1=15α2+60\alpha^4+10\alpha^2+25-1=15\alpha^2+60α4+10α2+25−1=15α2+60 α4+10α2+24=15α2+60\alpha^4+10\alpha^2+24=15\alpha^2+60α4+10α2+24=15α2+60 α4−5α2−36=0\alpha^4-5\alpha^2-36=0α4−5α2−36=0

Let x=α2x=\alpha^2x=α2. Then

x2−5x−36=0x^2-5x-36=0x2−5x−36=0 (x−9)(x+4)=0(x-9)(x+4)=0(x−9)(x+4)=0

Since α2≥0\alpha^2\ge 0α2≥0, we get

α2=9\alpha^2=9α2=9
  1. Evaluate the required expression

We need

2∣a⃗∣2+(a⃗⋅b⃗)∣b⃗∣22|\vec a|^2+(\vec a\cdot \vec b)|\vec b|^22∣a∣2+(a⋅b)∣b∣2

Since ∣a⃗∣2=∣b⃗∣2=α2+5=9+5=14|\vec a|^2=|\vec b|^2=\alpha^2+5=9+5=14∣a∣2=∣b∣2=α2+5=9+5=14 and a⃗⋅b⃗=−1\vec a\cdot \vec b=-1a⋅b=−1,

2∣a⃗∣2+(a⃗⋅b⃗)∣b⃗∣2=2(14)+(−1)(14)=28−14=142|\vec a|^2+(\vec a\cdot \vec b)|\vec b|^2=2(14)+(-1)(14)=28-14=142∣a∣2+(a⋅b)∣b∣2=2(14)+(−1)(14)=28−14=14
  1. Option check
  • A: 101010 ❌
  • B: 777 ❌
  • C: 999 ❌
  • D: 141414 ✅

Therefore, the correct answer is

14\boxed{14}14​
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