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JEE Mains · Maths · STD 12 - 10. vector algebra

જો સદિશો \(\vec{a}=\lambda \hat{i}+\mu \hat{j}+4 \hat{k}, \vec{b}=-2 \hat{i}+4 \hat{j}-2 \hat{k}\) અને \(\vec{c}=2 \hat{i}+3 \hat{j}+\hat{k}\) સમતલી હોય અને \(\vec{a}\) નો સદિશ \(\vec{b}\) પરનો પ્રક્ષેપ \(\sqrt{54}\) એકમ હોય, તો \(\lambda+\mu\) ની શક્ય તમામ કિંમતોનો સરવાળો \(............\) છે.

  1. A \(0\)
  2. B \(6\)
  3. C \(24\)
  4. D \(18\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(24\)

Step-by-step Solution

Detailed explanation

\(\left|\begin{array}{ccc}\lambda & \mu & 4 \\ -2 & 4 & -2 \\ 2 & 3 & 1\end{array}\right|=0\) \(\lambda(10)=\mu(2)+4(-14)=0\) \(10 \lambda-2 \mu=56\) \(5 \lambda-\mu=28\) \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}=\sqrt{54}\) \(\frac{-2 \lambda+4 \mu-8}{\sqrt{24}}=\sqrt{54}\)…
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