Question

Let \( \vec{a}=2 \hat{i}-3 \hat{j}+4 \hat{k}, \vec{b}=3 \hat{i}+4 \hat{j}-5 \hat{k} \) and a vector \( \vec{c} \) be such that \( \vec{a} \times(\vec{b}+\vec{c})+\vec{b} \times \vec{c}=\hat{i}+8 \hat{j}+13 \hat{k} \). If \( \vec{a} \cdot \vec{c}=13 \), then \( (24-\vec{b} \cdot \vec{c}) \) is equal to \( \qquad \)

Answer 1

\(\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}=(1,8,13)\)

\(\vec{a} \times(\vec{a} \times \vec{b})+\vec{a} \times(\vec{a} \times \vec{c})+\vec{a} \times(\vec{b} \times \vec{c})\)

\(=\overrightarrow{\mathrm{a}} \times(\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+13 \hat{\mathrm{k}})\)

\((\vec{a} \cdot \vec{b}) \vec{a}-a^{2} \vec{b}+(\vec{a} \cdot \vec{c}) \vec{a}-a^{2} \vec{c}+(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\)

\(\Rightarrow-26 \vec{a}-29 \vec{b}+13 \vec{a}-29 \vec{c}+13 \vec{b}+26 \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\)

\(\Rightarrow -13 \vec{a}-16 \vec{b}-3 \vec{c}=\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\)

\(\Rightarrow -13 \vec{a} \cdot \vec{b}-16 b^{2}-3 \vec{b} \cdot \vec{c}=\{\vec{a} \times(\hat{i}+8 \hat{j}+13 \hat{k})\} \cdot \vec{b}\)

\(\Rightarrow (-13)(-26)-16(50)-3 \vec{b} \cdot \vec{c}=\left|\begin{array}{ccc}2 & -3 & 4 \\ 1 & 8 & 13 \\ 3 & 4 & -5\end{array}\right|\)

\(\Rightarrow -462-3 \vec{b} \cdot \vec{c}=-396\)

\(\Rightarrow \overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-22\)

Hence \(24-\vec{b} \cdot \vec{c}=46\)