Question

Let $L_1$ and $L_2$ denote the lines $\vec{r_1} = -9\hat{i}+7\hat{j}+6\hat{k}+\lambda(\hat{i}-\hat{j}), \lambda \in \mathbb{R}$ and $\vec{r_2} = 7\hat{i}+3\hat{j}+4\hat{k}+\mu(\hat{i}+3\hat{k}), \mu \in \mathbb{R}$

Let $\mathcal{F}$ be the family of all circles that has a point on $L_1$ as well as a point on $L_2$. Let C be the circle with radius r that has the smallest radius among the circles in $\mathcal{F}$, then the value of $r^2$ is _______.

Answer 1

19

Answer 2

Let $L_1$ be the line $\vec{r_1} = \vec{a_1} + \lambda\vec{b_1}$ and $L_2$ be the line $\vec{r_2} = \vec{a_2} + \mu\vec{b_2}$. From the given equations: $\vec{a_1} = -9\hat{i}+7\hat{j}+6\hat{k}$ and $\vec{b_1} = \hat{i}-\hat{j}$. $\vec{a_2} = 7\hat{i}+3\hat{j}+4\hat{k}$ and $\vec{b_2} = \hat{i}+3\hat{k}$.

Let $C$ be a circle in the family $\mathcal{F}$. By definition, $C$ must contain at least one point from $L_1$ and at least one point from $L_2$. Let $P \in C \cap L_1$ and $Q \in C \cap L_2$. The distance between any two points on a circle is less than or equal to its diameter. If $r$ is the radius of $C$, its diameter is $2r$. Thus, $|PQ| \le 2r$. Since $P \in L_1$ and $Q \in L_2$, the distance $|PQ|$ is greater than or equal to the shortest distance between the lines $L_1$ and $L_2$. Let $d_{min}$ be the shortest distance between $L_1$ and $L_2$. So, $d_{min} \le |PQ|$. Combining the inequalities, we have $d_{min} \le |PQ| \le 2r$, which implies $d_{min} \le 2r$, or $r \ge d_{min}/2$. This shows that the radius of any circle in $\mathcal{F}$ must be at least $d_{min}/2$. The smallest possible radius for a circle in $\mathcal{F}$ is $d_{min}/2$.

To confirm that a circle with this radius exists in $\mathcal{F}$, let $P_0$ be the point on $L_1$ and $Q_0$ be the point on $L_2$ such that $|P_0Q_0| = d_{min}$. Consider the circle with $P_0Q_0$ as its diameter. The center of this circle is the midpoint of $P_0Q_0$, and its radius is $|P_0Q_0|/2 = d_{min}/2$. This circle passes through $P_0 \in L_1$ and $Q_0 \in L_2$, so it belongs to the family $\mathcal{F}$. Its radius is $d_{min}/2$. Thus, the circle with the smallest radius in $\mathcal{F}$ has radius $r = d_{min}/2$.

We need to calculate $d_{min}$, the shortest distance between the skew lines $L_1$ and $L_2$. The formula for the shortest distance is $d_{min} = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{||\vec{b_1} \times \vec{b_2}||}$.

First, calculate $\vec{a_2} - \vec{a_1}$: $\vec{a_2} - \vec{a_1} = (7\hat{i}+3\hat{j}+4\hat{k}) - (-9\hat{i}+7\hat{j}+6\hat{k}) = (7 - (-9))\hat{i} + (3 - 7)\hat{j} + (4 - 6)\hat{k} = 16\hat{i} - 4\hat{j} - 2\hat{k}$.

Next, calculate $\vec{b_1} \times \vec{b_2}$: $\vec{b_1} \times \vec{b_2} = (\hat{i}-\hat{j}) \times (\hat{i}+3\hat{k}) = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\1 & -1 & 0 \\1 & 0 & 3\end{vmatrix} = \hat{i}((-1)(3) - (0)(0)) - \hat{j}((1)(3) - (0)(1)) + \hat{k}((1)(0) - (-1)(1))$ $= -3\hat{i} - 3\hat{j} + \hat{k}$.

Now, calculate the dot product $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})$: $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (16\hat{i} - 4\hat{j} - 2\hat{k}) \cdot (-3\hat{i} - 3\hat{j} + \hat{k}) = (16)(-3) + (-4)(-3) + (-2)(1) = -48 + 12 - 2 = -38$.

Calculate the magnitude of $\vec{b_1} \times \vec{b_2}$: $||\vec{b_1} \times \vec{b_2}|| = ||-3\hat{i} - 3\hat{j} + \hat{k}|| = \sqrt{(-3)^2 + (-3)^2 + (1)^2} = \sqrt{9 + 9 + 1} = \sqrt{19}$.

Now, calculate the shortest distance $d_{min}$: $d_{min} = \frac{|-38|}{\sqrt{19}} = \frac{38}{\sqrt{19}} = \frac{2 \times 19}{\sqrt{19}} = 2\sqrt{19}$.

The radius of the smallest circle is $r = d_{min}/2$: $r = \frac{2\sqrt{19}}{2} = \sqrt{19}$.

The question asks for the value of $r^2$. $r^2 = (\sqrt{19})^2 = 19$.

The final answer is 19.

Explanation of the solution:

The smallest circle having a point on line $L_1$ and a point on line $L_2$ must have its diameter equal to the shortest distance between the two lines. This is because for any circle in the family, there exist points $P \in C \cap L_1$ and $Q \in C \cap L_2$. The distance $|PQ|$ is less than or equal to the diameter of the circle, $2r$. Since $P \in L_1$ and $Q \in L_2$, $|PQ|$ is always greater than or equal to the shortest distance between the lines, $d_{min}$. Thus, $d_{min} \le |PQ| \le 2r$, which implies $r \ge d_{min}/2$. The minimum radius is $d_{min}/2$. We calculate the shortest distance $d_{min}$ between the skew lines $L_1$ and $L_2$ using the formula $d_{min} = \frac{|(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})|}{||\vec{b_1} \times \vec{b_2}||}$. We find $\vec{a_1} = -9\hat{i}+7\hat{j}+6\hat{k}$, $\vec{b_1} = \hat{i}-\hat{j}$, $\vec{a_2} = 7\hat{i}+3\hat{j}+4\hat{k}$, $\vec{b_2} = \hat{i}+3\hat{k}$. $\vec{a_2} - \vec{a_1} = 16\hat{i} - 4\hat{j} - 2\hat{k}$. $\vec{b_1} \times \vec{b_2} = -3\hat{i} - 3\hat{j} + \hat{k}$. $(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (16)(-3) + (-4)(-3) + (-2)(1) = -48 + 12 - 2 = -38$. $||\vec{b_1} \times \vec{b_2}|| = \sqrt{(-3)^2 + (-3)^2 + 1^2} = \sqrt{9+9+1} = \sqrt{19}$. $d_{min} = \frac{|-38|}{\sqrt{19}} = \frac{38}{\sqrt{19}} = 2\sqrt{19}$. The radius of the smallest circle is $r = d_{min}/2 = \frac{2\sqrt{19}}{2} = \sqrt{19}$. The value of $r^2$ is $(\sqrt{19})^2 = 19$.