JEE Main 2024 — Mathematics Vectors & 3D Geometry

$$\begin{aligned} & \vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=(1,8,13) \\ & \vec{a} \times(\vec{a} \times \vec{b})+\vec{a} \times(\vec{a} \times \vec{c})+\vec{a} \times(\vec{b} \times \vec{c}) \\ & =\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\end{aligned}$$​a×b+a×c+b×c=(1,8,13)a×(a×b)+a×(a×c)+a×(b×c)=a×(i^+8j^​+13k^)​

$\begin{aligned} & (\vec{a} \cdot \vec{b}) \vec{a}-a^2 \vec{b}+(\vec{a} \cdot \vec{c}) \vec{a}-a^2 \vec{c}+(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k}) \\ & \Rightarrow-26 \vec{a}-29 \vec{b}+13 \vec{a}-29 \vec{c}+13 \vec{b}+26 \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\end{aligned}$​(a⋅b)a−a2b+(a⋅c)a−a2c+(a⋅c)b−(a⋅b)c=a×(i^+8j^​+13k^)⇒−26a−29b+13a−29c+13b+26c=a×(i^+8j^​+13k^)​ $\begin{aligned}
& \Rightarrow \quad-13 \overrightarrow{\mathrm{a}}-16 \overrightarrow{\mathrm{b}}-3 \overrightarrow{\mathrm{c}}=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}}) \
& \Rightarrow \quad-13 \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}-16 \mathrm{~b}^2-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}={\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})} \cdot \overrightarrow{\mathrm{b}} \
& \Rightarrow \quad(-13)(-26)-16(50)-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=\left|\begin{array}{ccc}
2 & -3 & 4 \
1 & 8 & 13 \
3 & 4 & -5
\end{array}\right| \
& \Rightarrow \quad-462-3 \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-396 \
& \Rightarrow \quad \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-22
\end{aligned}$

Hence $24-\vec{b} \cdot \vec{c}=46$24−b⋅c=46