QJEEM24Ma4de5: JEE Main 2024 Physics - Class 11 JEEM24Ma4de5 Solution: Let and be three vectors. Let be a unit vector along . If ,…

$\begin{aligned}& \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\& \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k} \\& \vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k} \\& \vec{b}+\vec{c}=5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}\end{aligned}$

$\vec{r}$ is a unit vector along $\vec{b}+\vec{c}$

$\therefore \quad \vec{r}=\frac{5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}}{\sqrt{25+4+(\lambda-5)^2}}$

Now, $\vec{r} \cdot \vec{a}=3$

$\frac{1}{\sqrt{29+(\lambda-5)^2}}[5+4+3(\lambda-5)]=3$

Squaring both sides

$\begin{aligned}& \Rightarrow \quad \frac{1}{29+(\lambda-5)^2}\left[9+3(\lambda-5)^2\right]=9 \\& \Rightarrow \quad[3+(\lambda-5)]^2=29+(\lambda-5)^2 \\& \Rightarrow \quad 9+(\lambda-5)^2+6(\lambda-5)=29+(\lambda-5)^2 \\& \Rightarrow \quad 9+6(\lambda-5)=29 \\& \Rightarrow \quad \lambda=\frac{20}{6}+5=\frac{25}{3} \\& \therefore \quad 3 \lambda=3 \times \frac{25}{3}=25\end{aligned}$