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 $\overrightarrow{c}=4 \hat{i}+c _2 \hat{j}+c _3 \hat{k}$ 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
$\overrightarrow{a}=\hat{i}+\hat{j}+k$
$\overrightarrow{b}=\hat{i}+8 \hat{j}+2 k$
$\overrightarrow{c}=4 \hat{i}+c _2 \hat{j}+c _3 k$
$\overrightarrow{b} \times \overrightarrow{a}=\overrightarrow{c} \times \overrightarrow{a}$
$(\vec{b}-\vec{c}) \times \vec{a}=0$
$\overrightarrow{b}-\overrightarrow{c}=\lambda \vec{\alpha}$
$\overrightarrow{b}=\overrightarrow{c}+\lambda \vec{\alpha}$
$-\hat{i}-8 \hat{j}+2 k=\left(4 \hat{i}+c _2 \hat{j}+c _3 k\right)+\lambda(\hat{i}+\hat{j}+k)$
$\lambda+4=-1 \Rightarrow \lambda=-5$
$\lambda+c _2=-8 \Rightarrow c _2=-3$
$\lambda+c _3=2 \Rightarrow c _3=7$
$\overrightarrow{c}=4 \hat{i}-3 \hat{j}+7 k$
$\cos \theta=\frac{12-12+7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{\sqrt{26} \cdot \sqrt{74}}=\frac{7}{2 \sqrt{481}}$
$\tan ^{2} \theta=\frac{625 \times 3}{49}$
$\left[\tan ^{2} \theta\right]=38$
