Let vecc$\vec{c}$ and vecd$\vec{d}$ be vectors such that |vecc+vecd|=sqrt29$|\vec{c}+\vec{d}|=\sqrt{29}$ and vecctimes(2hati+3hatj+4hatk)=(2hati+3hatj+4hatk)timesvecd$\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$ . If lambda_1,lambda_2(lambda_1>lambda_2)$\lambda_{1},\lambda_{2}(\lambda_{1}>\lambda_{2})$ are the possible values of (vecc+vecd).(-7hati+2hatj+3hatk)$(\vec{c}+\vec{d}).(-7\hat{i}+2\hat{j}+3\hat{k})$ , then the equation K^2x^2+(K^2-5K+lambda_1)xy+left(3K+fraclambda_22right)y^2-8x+12y+lambda_2=0$K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+\left(3K+\frac{\lambda_{2}}{2}\right)y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for k equal to:

### Related Formula vecu times vecv = -(vecv times vecu)$$\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$$ General equation of a circle requires coefficient of x^2$x^2$ = coefficient of y^2$y^2$ and coefficient of xy$xy$ = 0$0$. ### Core Logic Given vecc times (2hati + 3hatj + 4hatk) = (2hati + 3hatj + 4hatk) times vecd$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}$ Rearranging: vecc times (2hati + 3hatj + 4hatk) + vecd times (2hati + 3hatj + 4hatk) = vec0$$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) + \vec{d} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$ (vecc + vecd) times (2hati + 3hatj + 4hatk) = vec0$$(\vec{c} + \vec{d}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$ This implies vecc + vecd$\vec{c} + \vec{d}$ is parallel to 2hati + 3hatj + 4hatk$2\hat{i} + 3\hat{j} + 4\hat{k}$. Let vecc + vecd = lambda (2hati + 3hatj + 4hatk)$\vec{c} + \vec{d} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})$. ### Step 1: Find the scaling factor Given |vecc + vecd| = sqrt29$|\vec{c} + \vec{d}| = \sqrt{29}$ |lambda| sqrt2^2 + 3^2 + 4^2 = sqrt29$$|\lambda| \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$$ |lambda| sqrt4 + 9 + 16 = sqrt29$$|\lambda| \sqrt{4 + 9 + 16} = \sqrt{29}$$ |lambda| sqrt29 = sqrt29 Rightarrow lambda = pm 1$$|\lambda| \sqrt{29} = \sqrt{29} \Rightarrow \lambda = \pm 1$$ Therefore, vecc + vecd = pm(2hati + 3hatj + 4hatk)$\vec{c} + \vec{d} = \pm(2\hat{i} + 3\hat{j} + 4\hat{k})$. ### Step 2: Evaluate the dot product limits We need values of (vecc + vecd) cdot (-7hati + 2hatj + 3hatk)$(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$. Dot product = lambda(2(-7) + 3(2) + 4(3)) = lambda(-14 + 6 + 12) = 4lambda$= \lambda(2(-7) + 3(2) + 4(3)) = \lambda(-14 + 6 + 12) = 4\lambda$. Since lambda = pm 1$\lambda = \pm 1$, the possible values are 4$4$ and -4$-4$. Given lambda_1 > lambda_2$\lambda_1 > \lambda_2$, we have lambda_1 = 4$\lambda_1 = 4$ and lambda_2 = -4$\lambda_2 = -4$. ### Step 3: Apply circle constraints Substitute lambda_1$\lambda_1$ and lambda_2$\lambda_2$ into the conic equation: K^2 x^2 + (K^2 - 5K + 4)xy + left(3K - 2right)y^2 - 8x + 12y - 4 = 0$$K^2 x^2 + (K^2 - 5K + 4)xy + \left(3K - 2\right)y^2 - 8x + 12y - 4 = 0$$ For this to represent a circle: 1) Coefficient of xy$xy$ must be 0$0$: K^2 - 5K + 4 = 0 Rightarrow K = 1, 4$$K^2 - 5K + 4 = 0 \Rightarrow K = 1, 4$$ 2) Coefficient of x^2$x^2$ must equal coefficient of y^2$y^2$: K^2 = 3K - 2 Rightarrow K^2 - 3K + 2 = 0 Rightarrow K = 1, 2$$K^2 = 3K - 2 \Rightarrow K^2 - 3K + 2 = 0 \Rightarrow K = 1, 2$$ The common value satisfying both conditions is K = 1$K = 1$. ### Pattern Recognition Cross-product equations mapping to X times A = -Y times A$X \times A = -Y \times A$ perfectly factor out to (X+Y) times A = 0$(X+Y) \times A = 0$, guaranteeing (X+Y)$(X+Y)$ is a scalar multiple of A$A$. This immediately unlocks vector magnitudes. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Vector Algebra Class 11 Maths: Conic Sections