Vector Algebra Question 3
Question 3 - 2024 (01 Feb Shift 2)
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}, \quad \vec{b}=-\hat{i}-8 \hat{j}+2 \hat{k} \quad$ and $\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}+\mathrm{c}{2} \hat{\mathrm{j}}+\mathrm{c}{3} \hat{\mathrm{k}}$ be three vectors such that $\vec{b} \times \vec{a}=\vec{c} \times \vec{a}$. If the angle between the vector $\vec{c}$ and the vector $3 \hat{i}+4 \hat{j}+\hat{k}$ is $\theta$, then the greatest integer less than or equal to $\tan ^{2} \theta$ is :
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Answer (38)
Solution
$\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+\hat{\mathrm{j}}+\mathrm{k}$
$\overrightarrow{\mathrm{b}}=\hat{\mathrm{i}}+8 \hat{\mathrm{j}}+2 \mathrm{k}$
$\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}+\mathrm{c}{2} \hat{\mathrm{j}}+\mathrm{c}{3} \mathrm{k}$
$\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}=\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}}$
$(\vec{b}-\vec{c}) \times \vec{a}=0$
$\overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}}=\lambda \vec{\alpha}$
$\overrightarrow{\mathrm{b}}=\overrightarrow{\mathrm{c}}+\lambda \vec{\alpha}$
$-\hat{i}-8 \hat{j}+2 k=\left(4 \hat{i}+c_{2} \hat{j}+c_{3} \mathrm{k}\right)+\lambda(\hat{i}+\hat{j}+\mathrm{k})$
$\lambda+4=-1 \Rightarrow \lambda=-5$
$\lambda+c_{2}=-8 \Rightarrow c_{2}=-3$
$\lambda+c_{3}=2 \Rightarrow c_{3}=7$
$\overrightarrow{\mathrm{c}}=4 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}+7 \mathrm{k}$
$\cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}}$
$\tan ^{2} \theta=\frac{625 \times 3}{49}$
$\left[\tan ^{2} \theta\right]=38$
