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QJEEM23Ma555a: JEE Main 2023 Physics - Class 11 JEEM23Ma555a Solution: If the points with position vectors are collinear, then is…

If the points with position vectors αi^+10j^+13k^,6i^+11j^+11k^,92i^+βj^−8k^\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}, \frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k} are collinear, then (19α−6β)2(19 \alpha-6 \beta)^{2} is equal to :

Multiple Choice Options

  • A 16
  • B 49
  • C 36
  • D 25

Step-by-Step Verified Solution

Given : Points with position vectors

αi^+10j^+13k^,6i^+11j^+11k^\alpha \hat{i}+10 \hat{j}+13 \hat{k}, 6 \hat{i}+11 \hat{j}+11 \hat{k}

and 92i^+βj^−8k^\frac{9}{2} \hat{i}+\beta \hat{j}-8 \hat{k} are collinear.

So, α−66−92=10−1111−β=13−1111+8\frac{\alpha-6}{6-\frac{9}{2}}=\frac{10-11}{11-\beta}=\frac{13-11}{11+8}

amp;⇒2(α−6)3=−111−β=219amp;⇒23(α−6)=219\begin{aligned}& \Rightarrow \frac{2(\alpha-6)}{3}=\frac{-1}{11-\beta}=\frac{2}{19} \\\\& \Rightarrow \frac{2}{3}(\alpha-6)=\frac{2}{19}\end{aligned}

amp;⇒19α−114=3⇒19α=117amp;⇒α=11719\begin{aligned}& \Rightarrow 19 \alpha-114=3 \Rightarrow 19 \alpha=117 \\\\& \Rightarrow \alpha=\frac{117}{19}\end{aligned}

And, −111−β=219\frac{-1}{11-\beta}=\frac{2}{19}

amp;⇒−19=22−2βamp;⇒2β=41amp;⇒β=412\begin{aligned}& \Rightarrow-19=22-2 \beta \\\\& \Rightarrow 2 \beta=41 \\\\& \Rightarrow \beta=\frac{41}{2}\end{aligned}

amp;∴(19α−6β)2=(19×11719−6×412)2amp;=(117−123)2=36\begin{aligned}& \therefore(19 \alpha-6 \beta)^2=\left(19 \times \frac{117}{19}-\frac{6 \times 41}{2}\right)^2 \\\\& =(117-123)^2=36\end{aligned}

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