Let a = 3i + 2j + k, b = 2i - j + 3k and c be a vector such that (a + b) x c = 2(a x b) + 24j - 6k

Question

Let \(\vec{a}=3 \hat{i}+2 \hat{j}+\hat{k}, \vec{b}=2 \hat{i}-\hat{j}+3 \hat{k}\) and \(\vec{c}\) be a vector such that \((\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}\) and \((\vec{\mathrm{a}}-\vec{\mathrm{b}}+\hat{\mathrm{i}}) \cdot \vec{\mathrm{c}}=-3\). Then \(|\vec{\mathrm{c}}|^{2}\) is equal to ____.

Answer 1

Correct answer: 38

\((\vec{a}+\vec{b}) \times \vec{c}=2(\vec{a} \times \vec{b})+24 \hat{j}-6 \hat{k}\)

\((5 \hat{i}+\hat{j}+4 \hat{k}) \times \vec{c}=2(7 \hat{i}-7 \hat{j}-7 \hat{k})+24 \hat{j}-6 \hat{k}\)

\(\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 5 & 1 & 4 \\ x & y & z\end{array}\right|=14 \hat{i}+10 \hat{j}-20 \hat{k}\)

\(\Rightarrow \hat{\mathrm{i}}(\mathrm{z}-4 \mathrm{y})-\hat{\mathrm{j}}(5 \mathrm{z}-4 \mathrm{x})+\hat{\mathrm{k}}(5 \mathrm{y}-\mathrm{x})=14 \hat{\mathrm{i}}+10 \hat{\mathrm{j}}-20 \hat{\mathrm{k}}\)

\(z-4 y=14,4 x-5 z=10,5 y-x=-20\)

\((\mathrm{a}-\mathrm{b}+\mathrm{i}) \cdot \vec{\mathrm{c}}=-3\)

\((2 \hat{i}+3 \hat{j}-2 \hat{k}) \cdot \vec{c}=-3\)

\(2 \mathrm{x}+3 \mathrm{y}-2 \mathrm{z}=-3\)

\(\therefore \mathrm{x}=5, \mathrm{y}=-3, \mathrm{z}=2\)

\(|\vec{\mathrm{c}}|^{2}=25+9+4=38\)