Vector Algebra
Cross Product Condition & Angle Between Vectors
nta_pyq_2024_jan
Grade 12
Question:
Let $\vec{a}=\hat{i}+\hat{j}+\hat{k}$, $\vec{b}=-\hat{i}-8\hat{j}+2\hat{k}$ and $\vec{c}=4\hat{i}+c_2\hat{j}+c_3\hat{k}$ be three vectors. If $\vec{b}\times\vec{a}=\vec{c}\times\vec{a}$, and 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:
Step-by-Step Solution
Key Concept: $\vec{b}\times\vec{a}=\vec{c}\times\vec{a}\Rightarrow(\vec{b}-\vec{c})\times\vec{a}=\vec{0}\Rightarrow\vec{b}-\vec{c}=\lambda\vec{a}$. Use this to find $c_2$ and $c_3$. Then compute $\cos\theta=\dfrac{\vec{c}\cdot(3\hat{i}+4\hat{j}+\hat{k})}{|\vec{c}||3\hat{i}+4\hat{j}+\hat{k}|}$ and $\tan^2\theta$.
$\vec{b}-\vec{c}=\lambda\vec{a}\Rightarrow(-5,c_2-(-8+c_2+c_3...))$: $\lambda+4=-1\Rightarrow\lambda=-5$; $c_2=-3$; $c_3=7$. $\vec{c}\cdot(3\hat{i}+4\hat{j}+\hat{k})=12-12+7=7$. $\cos^2\theta=\dfrac{49}{74\cdot26}$. $\tan^2\theta=\dfrac{74\cdot26-49}{49}=\dfrac{1875}{49}\approx38.26$. $\lfloor\tan^2\theta\rfloor=38$.
Correct Answer: 38