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Vector Algebra question

2023 · 24 Jan · Shift 2 · Q26
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  5. /2023 · 24 Jan · Shift 2 · Q26

Vector Algebra question

2023 · 24 Jan · Shift 2 · Q26

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let α→=4i^+3j^+5k^\overrightarrow \alpha = 4\widehat i + 3\widehat j + 5\widehat kα=4i+3j​+5k and β→=i^+2j^−4k^\overrightarrow \beta = \widehat i + 2\widehat j - 4\widehat kβ​=i+2j​−4k. Let β→1{\overrightarrow \beta _1}β​1​ be parallel to α→\overrightarrow \alphaα and β→2{\overrightarrow \beta _2}β​2​ be perpendicular to α→\overrightarrow \alphaα. If β→=β→1+β→2\overrightarrow \beta = {\overrightarrow \beta _1} + {\overrightarrow \beta _2}β​=β​1​+β​2​, then the value of 5β→2 .(i^+j^+k^)5{\overrightarrow \beta _2}\,.\left( {\widehat i + \widehat j + \widehat k} \right)5β​2​.(i+j​+k) is :
  1. A
    9
  2. B
    7
  3. C
    6
  4. D
    11
View written solutionFree

Correct answer: B

  1. We decompose β⃗\vec\betaβ​ into components parallel and perpendicular to α⃗\vec\alphaα:
β⃗=β⃗1+β⃗2\vec\beta = \vec\beta_1 + \vec\beta_2β​=β​1​+β​2​

where β⃗1∥α⃗\vec\beta_1 \parallel \vec\alphaβ​1​∥α and β⃗2⊥α⃗\vec\beta_2 \perp \vec\alphaβ​2​⊥α.

So,

β⃗1=proj⁡α⃗β⃗=β⃗⋅α⃗∣α⃗∣2 α⃗\vec\beta_1 = \operatorname{proj}_{\vec\alpha} \vec\beta = \frac{\vec\beta\cdot\vec\alpha}{|\vec\alpha|^2}\,\vec\alphaβ​1​=projα​β​=∣α∣2β​⋅α​α
  1. Compute the required dot products:

Given

α⃗=4i^+3j^+5k^,β⃗=i^+2j^−4k^\vec\alpha = 4\hat i+3\hat j+5\hat k, \qquad \vec\beta = \hat i+2\hat j-4\hat kα=4i^+3j^​+5k^,β​=i^+2j^​−4k^

First,

β⃗⋅α⃗=(1)(4)+(2)(3)+(−4)(5)=4+6−20=−10\vec\beta\cdot\vec\alpha = (1)(4)+(2)(3)+(-4)(5)=4+6-20=-10β​⋅α=(1)(4)+(2)(3)+(−4)(5)=4+6−20=−10

And,

∣α⃗∣2=42+32+52=16+9+25=50|\vec\alpha|^2 = 4^2+3^2+5^2=16+9+25=50∣α∣2=42+32+52=16+9+25=50

Hence,

β⃗1=−1050α⃗=−15α⃗\vec\beta_1 = \frac{-10}{50}\vec\alpha = -\frac15\vec\alphaβ​1​=50−10​α=−51​α

So,

β⃗1=−15(4i^+3j^+5k^)=−45i^−35j^−k^\vec\beta_1 = -\frac15(4\hat i+3\hat j+5\hat k) = -\frac45\hat i-\frac35\hat j-\hat kβ​1​=−51​(4i^+3j^​+5k^)=−54​i^−53​j^​−k^
  1. Now find β⃗2\vec\beta_2β​2​:
β⃗2=β⃗−β⃗1\vec\beta_2 = \vec\beta-\vec\beta_1β​2​=β​−β​1​ β⃗2=(i^+2j^−4k^)−(−45i^−35j^−k^)\vec\beta_2 = \left(\hat i+2\hat j-4\hat k\right)-\left(-\frac45\hat i-\frac35\hat j-\hat k\right)β​2​=(i^+2j^​−4k^)−(−54​i^−53​j^​−k^) β⃗2=95i^+135j^−3k^\vec\beta_2 = \frac95\hat i+\frac{13}{5}\hat j-3\hat kβ​2​=59​i^+513​j^​−3k^
  1. Compute β⃗2⋅(i^+j^+k^)\vec\beta_2\cdot(\hat i+\hat j+\hat k)β​2​⋅(i^+j^​+k^):
β⃗2⋅(i^+j^+k^)=95+135−3\vec\beta_2\cdot(\hat i+\hat j+\hat k) = \frac95+\frac{13}{5}-3β​2​⋅(i^+j^​+k^)=59​+513​−3 =225−3=225−155=75= \frac{22}{5}-3 = \frac{22}{5}-\frac{15}{5}=\frac75=522​−3=522​−515​=57​

Therefore,

5 β⃗2⋅(i^+j^+k^)=5⋅75=75\,\vec\beta_2\cdot(\hat i+\hat j+\hat k)=5\cdot\frac75=75β​2​⋅(i^+j^​+k^)=5⋅57​=7
  1. Hence the correct option is:
7\boxed{7}7​

So, option B\boxed{\text{B}}B​ is correct.

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