Question ID: #648
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 $\lambda_1, \lambda_2 (\lambda_1 \gt \lambda_2)$ are the possible values of $(\vec{c}+\vec{d}).(-7\hat{i}+2\hat{j}+3\hat{k})$ then the equation
$$K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+(3K+\frac{\lambda_{2}}{2})y^{2}-8x+12y+\lambda_{2}=0$$
represents a circle, for K equal to:
- (1) 4
- (2) 1
- (3) -1
- (4) 2
Solution:
Let $\vec{a} = 2\hat{i} + 3\hat{j} + 4\hat{k}$.
The given condition is $\vec{c} \times \vec{a} = \vec{a} \times \vec{d}$.
$\vec{c} \times \vec{a} = -(\vec{d} \times \vec{a})$
$\vec{c} \times \vec{a} + \vec{d} \times \vec{a} = 0$
$(\vec{c} + \vec{d}) \times \vec{a} = 0$
This implies vectors $(\vec{c} + \vec{d})$ and $\vec{a}$ are parallel.
$\vec{c} + \vec{d} = \lambda \vec{a} = \lambda(2\hat{i} + 3\hat{j} + 4\hat{k})$
Given $|\vec{c} + \vec{d}| = \sqrt{29}$:
$|\lambda|\sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$
$|\lambda|\sqrt{4 + 9 + 16} = \sqrt{29}$
$|\lambda|\sqrt{29} = \sqrt{29} \Rightarrow \lambda = \pm 1$.
We need to find values of $(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$:
$= \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$
$= \lambda(-14 + 6 + 12) = 4\lambda$.
Possible values are $4(1) = 4$ and $4(-1) = -4$.
Let $\lambda_1 = 4$ and $\lambda_2 = -4$.
The equation of the curve is:
$K^2 x^2 + (K^2 – 5K + 4)xy + (3K + \frac{-4}{2})y^2 – 8x + 12y – 4 = 0$
For this to represent a circle:
1. Coefficient of $xy$ term must be zero:
$K^2 – 5K + 4 = 0 \Rightarrow (K-1)(K-4) = 0 \Rightarrow K = 1, 4$.
2. Coefficient of $x^2$ must equal Coefficient of $y^2$:
$K^2 = 3K – 2 \Rightarrow K^2 – 3K + 2 = 0 \Rightarrow (K-1)(K-2) = 0 \Rightarrow K = 1, 2$.
The common value satisfying both conditions is $K = 1$.
Ans. (2)
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