Verified Solution • 2025–26 Session moderate

QJEEM23Mab81a: JEE Main 2023 Physics - Class 11 JEEM23Mab81a Solution: Let and . Let be parallel to and be perpendicular to . If ,…

Let α→=4i^+3j^+5k^\overrightarrow \alpha = 4\widehat i + 3\widehat j + 5\widehat k and β→=i^+2j^−4k^\overrightarrow \beta = \widehat i + 2\widehat j - 4\widehat k . Let β→1{\overrightarrow \beta _1} be parallel to α→\overrightarrow \alpha and β→2{\overrightarrow \beta _2} be perpendicular to α→\overrightarrow \alpha . If β→=β→1+β→2\overrightarrow \beta = {\overrightarrow \beta _1} + {\overrightarrow \beta _2} , then the value of 5β→2 .(i^+j^+k^)5{\overrightarrow \beta _2}\,.\left( {\widehat i + \widehat j + \widehat k} \right) is :

Multiple Choice Options

  • A 9
  • B 7
  • C 6
  • D 11

Step-by-Step Verified Solution

Let β⃗1=λα⃗\vec{\beta}_1=\lambda \vec{\alpha}

Now β⃗2=β⃗−β⃗1\vec{\beta}_2=\vec{\beta}-\vec{\beta}_1

amp;=(i^+2j^−4k^)−λ(4i^+3j^+5k^)amp;=(1−4λ)i^+(2−3λ)j^−(5λ+4)k^amp;β⃗2⋅α⃗=0amp;⇒4(1−4λ)+3(2−3λ)−5(5λ+4)=0amp;⇒4−16α+6−9λ−25λ−20=0amp;⇒50λ=−10amp;⇒λ=−15amp;β⃗2=(1+45)i^+(2+35)j^−(−1+4)k^amp;β⃗2=95i^+135j^−3k^amp;5β⃗2=9i^+13j^−15k^amp;5β⃗2⋅(i^+j^+k^)=9+13−15=7\begin{aligned}& =(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-4 \hat{\mathrm{k}})-\lambda(4 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+5 \hat{\mathrm{k}}) \\\\& =(1-4 \lambda) \hat{\mathrm{i}}+(2-3 \lambda) \hat{\mathrm{j}}-(5 \lambda+4) \hat{\mathrm{k}} \\\\& \vec{\beta}_2 \cdot \vec{\alpha}=0 \\\\& \Rightarrow 4(1-4 \lambda)+3(2-3 \lambda)-5(5 \lambda+4)=0 \\\\& \Rightarrow 4-16 \alpha+6-9 \lambda-25 \lambda-20=0 \\\\& \Rightarrow 50 \lambda=-10 \\\\& \Rightarrow \lambda=\frac{-1}{5} \\\\& \vec{\beta}_2=\left(1+\frac{4}{5}\right) \hat{\mathrm{i}}+\left(2+\frac{3}{5}\right) \hat{\mathrm{j}}-(-1+4) \hat{\mathrm{k}} \\\\& \vec{\beta}_2=\frac{9}{5} \hat{\mathrm{i}}+\frac{13}{5} \hat{\mathrm{j}}-3 \hat{\mathrm{k}} \\\\& 5 \vec{\beta}_2=9 \hat{\mathrm{i}}+13 \hat{\mathrm{j}}-15 \hat{\mathrm{k}} \\\\& 5 \vec{\beta}_2 \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})=9+13-15=7\end{aligned}

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