Let a = i + j + k, b = -i - 8j + 2k and c = 4i + c2j + c3k be three vectors

Question

Let \(\vec{\mathrm a}=\hat{\mathrm i}+\hat{\mathrm j}+\hat{\mathrm k}, \vec{\mathrm b}=-\hat{\mathrm i}-8 \hat{\mathrm j}+2 \hat{\mathrm k}\) and \(\vec{\mathrm{c}}=4 \hat{\mathrm{i}}+\mathrm{c}_{2} \hat{\mathrm{j}}+\mathrm{c}_{3} \hat{\mathrm{k}}\) be three vectors such that \(\vec{\mathrm b} \times \vec{\mathrm a}=\vec{\mathrm c} \times \vec{\mathrm a}\). If the angle between the vector \(\vec{\mathrm{c}}\) and the vector \(3 \hat{\mathrm{i}}+4 \hat{\mathrm{j}}+\hat{\mathrm{k}}\) is \(\theta\), then the greatest integer less than or equal to \(\tan ^{2} \theta\) is ______.

Answer 1

Correct answer: 38

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

\(\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+2 \mathrm{k}\)

\(\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}+\mathrm{c}_{2} \hat{\mathrm{j}}+\mathrm{c}_{3} \mathrm{k}\)

\(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}\)

\((\overrightarrow{\mathrm b}-\overrightarrow{\mathrm c}) \times \overrightarrow{\mathrm a}=0\)

\(\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\alpha}\)

\(\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\alpha}\)

\(-\hat{\mathrm i}-8 \hat{\mathrm j}+2 \mathrm k=\left(4 \hat{\mathrm i}+\mathrm c_{2} \hat{\mathrm j}+\mathrm c_{3} \mathrm k\right)+\lambda(\hat{\mathrm i}+\hat{\mathrm j}+\mathrm k)\)

\(\lambda+4=-1 \Rightarrow \lambda=-5\)

\(\lambda+\mathrm{c}_{2}=-8 \Rightarrow \mathrm{c}_{2}=-3\)

\(\lambda+\mathrm c_{3}=2 \Rightarrow \mathrm c_{3}=7\)

\(\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \mathrm{k}\)

\(\cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}}\)

\(\tan ^{2} \theta=\frac{625 \times 3}{49}\)

\(\left[\tan ^{2} \theta\right]=38\)