Let c and d be vectors such that | c+ d| = √29 and c (2 i+3 j+4 k) = (2 i+3 j+4 k) d.

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
$$K^2x^2 + (K^2-5K+\lambda_1)xy + \left(3K+\frac{\lambda_2}{2}\right)y^2 - 8x+12y+\lambda_2 = 0$$
represents a circle, for $K$ equal to :

Accepted Answer

From $\vec c\times(2\hat i+3\hat j+4\hat k) = (2\hat i+3\hat j+4\hat k)\times\vec d$, rearranging: $(\vec c+\vec d)\times(2\hat i+3\hat j+4\hat k) = \vec 0$, so $\vec c+\vec d = \lambda(2\hat i+3\hat j+4\hat k)$ for some scalar $\lambda$. Since $|2\hat i+3\hat j+4\hat k|=\sqrt{29}$ and $|\vec c+\vec d|=\sqrt{29}$, we get $\lambda=\pm1$.

$$(\vec c+\vec d)\cdot(-7\hat i+2\hat j+3\hat k) = \pm(-14+6+12) = \pm4$$

So $\lambda_1 = 4,\ \lambda_2=-4$.

For the conic to be a circle: coefficient of $x^2$ = coefficient of $y^2$, and coefficient of $xy$ = 0.
$$K^2 = 3K + \frac{\lambda_2}{2} = 3K - 2 \implies K^2-3K+2=0 \implies K=1,2$$
$$K^2-5K+\lambda_1 = 0 \implies K^2-5K+4=0 \implies K=1,4$$

The common value is $K=1$.

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