Verified Solution • 2025–26 Session moderate

QJEEM23Ma8cdb: JEE Main 2023 Physics - Class 11 JEEM23Ma8cdb Solution: If the vectors , and are coplanar and the projection of on the…

If the vectors a→=λi^+μj^+4k^\overrightarrow a = \lambda \widehat i + \mu \widehat j + 4\widehat k , b→=−2i^+4j^−2k^\overrightarrow b = - 2\widehat i + 4\widehat j - 2\widehat k and c→=2i^+3j^+k^\overrightarrow c = 2\widehat i + 3\widehat j + \widehat k are coplanar and the projection of a→\overrightarrow a on the vector b→\overrightarrow b is 54\sqrt {54} units, then the sum of all possible values of λ+μ\lambda + \mu is equal to :

Multiple Choice Options

  • A 24
  • B 0
  • C 18
  • D 6

Step-by-Step Verified Solution

a⃗=λi^+μj^+4k^,b⃗=−2i^+4j^−2k^,c⃗=2i^+3j^+k^\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}, \vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}

Now, a⃗⋅b⃗=54⇒−2λ+4μ−824=54\vec{a} \cdot \vec{b}=\sqrt{54} \Rightarrow \frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}

⇒−2λ+4μ−8=36\Rightarrow-2 \lambda+4 \mu-8=36

⇒2μ−λ=22...(i)\Rightarrow 2 \mu-\lambda=22\quad...(i)

and ∣λamp;μamp;4−2amp;4amp;−22amp;3amp;1∣=0\left|\begin{array}{ccc}\lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{array}\right|=0

10λ−2μ−56=0...(ii)10 \lambda-2 \mu-56=0 \quad...(ii)

By (i) & (ii) λ=789,μ=1389\lambda=\frac{78}{9}, \mu=\frac{138}{9}

∴μ+λ=24\therefore \mu+\lambda=24

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