Related Formula
u × v = -( v × u)$$\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$$
General equation of a circle requires coefficient of x²$x^2$ = coefficient of y²$y^2$ and coefficient of xy$xy$ = 0$0$.
Core Logic
Given c × (2 i + 3 j + 4 k) = (2 i + 3 j + 4 k) × d$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}$
Rearranging:
c × (2 i + 3 j + 4 k) + d × (2 i + 3 j + 4 k) = 0$$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) + \vec{d} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$
( c + d) × (2 i + 3 j + 4 k) = 0$$(\vec{c} + \vec{d}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$
This implies c + d$\vec{c} + \vec{d}$ is parallel to 2 i + 3 j + 4 k$2\hat{i} + 3\hat{j} + 4\hat{k}$.
Let c + d = λ (2 i + 3 j + 4 k)$\vec{c} + \vec{d} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})$.
Step 1: Find the scaling factor
Given | c + d| = √(29)$|\vec{c} + \vec{d}| = \sqrt{29}$
|λ| √(2² + 3² + 4²) = √(29)$$|\lambda| \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$$
|λ| √(4 + 9 + 16) = √(29)$$|\lambda| \sqrt{4 + 9 + 16} = \sqrt{29}$$
|λ| √(29) = √(29) ⇒ λ = ± 1$$|\lambda| \sqrt{29} = \sqrt{29} \Rightarrow \lambda = \pm 1$$
Therefore, c + d = ±(2 i + 3 j + 4 k)$\vec{c} + \vec{d} = \pm(2\hat{i} + 3\hat{j} + 4\hat{k})$.
Step 2: Evaluate the dot product limits
We need values of ( c + d) · (-7 i + 2 j + 3 k)$(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$.
Dot product = λ(2(-7) + 3(2) + 4(3)) = λ(-14 + 6 + 12) = 4λ$= \lambda(2(-7) + 3(2) + 4(3)) = \lambda(-14 + 6 + 12) = 4\lambda$.
Since λ = ± 1$\lambda = \pm 1$, the possible values are 4$4$ and -4$-4$.
Given λ₁ > λ₂$\lambda_1 > \lambda_2$, we have λ₁ = 4$\lambda_1 = 4$ and λ₂ = -4$\lambda_2 = -4$.
Step 3: Apply circle constraints
Substitute λ₁$\lambda_1$ and λ₂$\lambda_2$ into the conic equation:
K² x² + (K² - 5K + 4)xy + (3K - 2)y² - 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:
- Coefficient of xy$xy$ must be 0$0$:
K² - 5K + 4 = 0 ⇒ K = 1, 4$$K^2 - 5K + 4 = 0 \Rightarrow K = 1, 4$$
- Coefficient of x²$x^2$ must equal coefficient of y²$y^2$:
K² = 3K - 2 ⇒ K² - 3K + 2 = 0 ⇒ 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 × A = -Y × A$X \times A = -Y \times A$ perfectly factor out to (X+Y) × A = 0$(X+Y) \times A = 0$, guaranteeing (X+Y)$(X+Y)$ is a scalar multiple of A$A$. This immediately unlocks vector magnitudes.
Chapter Mix
Class 12 Maths: Vector Algebra
Class 11 Maths: Conic Sections
