Circles
Vectors + Circle — Finding K
nta_pyq_2026_jan
Grade 11
Question:
Let $\vec{c}$ and $\vec{d}$ be vectors such that $|\vec{c}+\vec{d}|=\sqrt{29}$ and $\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$. If $\lambda_1,\lambda_2$ ($\lambda_1>\lambda_2$) are the possible values of $(\vec{c}+\vec{d})\cdot(-7\hat{i}+2\hat{j}+3\hat{k})$, then the equation $K^2x^2+(K^2-5K+\lambda_1)xy+\left(3K+\dfrac{\lambda_2}{2}\right)y^2-8x+12y+\lambda_2=0$ represents a circle, for K equal to:
Step-by-Step Solution
Key Concept: $\vec{c}\times\vec{a}=\vec{a}\times\vec{d}$ (where $\vec{a}=2\hat{i}+3\hat{j}+4\hat{k}$) $\Rightarrow(\vec{c}+\vec{d})\times\vec{a}=0\Rightarrow\vec{c}+\vec{d}=t\vec{a}$. $|t|\sqrt{29}=\sqrt{29}\Rightarrow t=\pm1$. $\lambda_1=4,\lambda_2=-4$.
$\lambda_1=4,\lambda_2=-4$. $K=1$.
Correct Answer: 1