QJEEM24Ma406f: JEE Main 2024 Physics - Class 11 JEEM24Ma406f Solution: Let and . If is the unit vector in the direction of such that , …

$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}, \vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in R$

$\text { also, } \vec{b}+\vec{c}=(x+2) \hat{i}+6 \hat{j}-2 \hat{k}$

$\vec{d} \text { is the unit vector in the direction of } \vec{b}+\vec{c}$

$\begin{aligned}& |\vec{b}+\vec{c}|=\sqrt{(x+2)^2+6^2+2^2} \\& =\sqrt{40+(x+2)^2} \\& \vec{d}=\frac{x+2}{\sqrt{40+(x+2)^2}} \hat{i}+\frac{6}{\sqrt{40+(x+2)^2}} \hat{j} -\frac{2}{\sqrt{40+(x+2)^2}} \hat{k} \\\end{aligned}$

$\begin{aligned}& \vec{a} \cdot \vec{d}=1 \\& \frac{x+2+6-2}{\sqrt{40+(x+2)^2}}=1 \\& x+6=\sqrt{40+(x+2)^2}\end{aligned}$

$\begin{aligned}&(x+6)^2=40+(x+2)^2 \\& x^2+36+ 12 x=40+x^2+4+4 x \\& 8 x=8 \\& \Rightarrow x=1 \\&(\vec{a} \times \vec{b}) \cdot \vec{c}=[\vec{a} \vec{b} \vec{c}] \\&=\left[\begin{array}{ccc}1 & 1 & 1 \\2 & 4 & -5 \\1 & 2 & 3\end{array}\right] \\&=1(12+10)-1(6+5)+1(4-4) \\&=22-11=11\end{aligned}$