Let $\overrightarrow{c}$ and $\overrightarrow{d}$ be vectors such that $|\overrightarrow{c} + \overr

Question

Let $\overrightarrow{c}$ and $\overrightarrow{d}$ be vectors such that $|\overrightarrow{c} + \overrightarrow{d}| = \sqrt{29}$ and $\overrightarrow{c} \times (2\widehat{i} + 3\widehat{j} + 4\widehat{k}) = (2\widehat{i} + 3\widehat{j} + 4\widehat{k}) \times \overrightarrow{d}$. If $\lambda_{1},\lambda_{2}\left( \lambda_{1} > \lambda_{2} \right)$ are the possible values of $(\overrightarrow{c} + \overrightarrow{d}) \cdot ( - 7\widehat{i} + 2\widehat{j} + 3\widehat{k})$, then the equation

$K^{2}x^{2} + \left( K^{2} - 5\text{ }K + \lambda_{1} \right)xy + \left( 3\text{ }K + \frac{\lambda_{2}}{2} \right)y^{2} - 8x + 12y + \lambda_{2} = 0$ represents a circle, for k equal to :

Accepted Answer

$|\overrightarrow{c} + \overrightarrow{d}| = \sqrt{29}$

$$\begin{matrix} & \overrightarrow{c} + \overrightarrow{d} = \lambda((2\widehat{i} + 3\widehat{j} + 4\widehat{k}) \\ & \lambda = \pm 1 \\ & \lambda( - 14 + 6 + 12) = 4\lambda,\lambda_{1} = 4,\lambda_{2} = - 4 \\ & k^{2}x^{2} + \left( k^{2} - 5k + 4 \right)xy + (3k - 2)y^{2} - 8x + 12y - 4 \\ & \ = 0 \end{matrix}$$

is circle

$$\begin{matrix} & k^{2} - 5k + 4 = 0 \Rightarrow k = 1,4 \\ & k^{2} = 3k - 2 \Rightarrow k = 1,2 \\ & k = 1 \end{matrix}$$

Question

Options

Solution

More Circle Questions