Q.Find the shortest distance between the lines given by $\vec{r} = (8 + 3\lambda)\hat{i} - (9 + 16\lambda)\hat{j} + (10 + 7\lambda)\hat{k}$ and $\vec{r} = 15\hat{i} + 29\hat{j} + 5\hat{k} + \mu(3\hat{i} + 8\hat{j} - 5\hat{k})$.

Assertion (A) — do the lines intersect?

The lines are $\vec{r} = \vec{a} + t\,\vec{b}$ and $\vec{r} = \vec{b} + s\,\vec{a}$. For a common point set them equal:

For non-parallel $\vec{a},\vec{b}$ this forces $s=1$ and $t=1$, giving the common point

So the two lines do intersect — (A) is true.

Reason (R) — is the shortest-distance statement correct?

For $\vec{r}=\vec{p}+t\,\vec{q}$ and $\vec{r}=\vec{c}+s\,\vec{d}$, the shortest distance is

which is exactly the length of the projection of $(\vec{p}-\vec{c})$ onto $\vec{q}\times\vec{d}$. (R) is true.

Does (R) explain (A)? …

We have two curves:

• Hyperbola: $\dfrac{x^2}{16} - \dfrac{y^2}{9} = 1$ (center at origin, transverse axis along x-axis, $a=4$, $b=3$).
• Circle: $x^2 + y^2 = 16$ (center at origin, radius $R=4$).

The key idea: Common tangents exist only if the curves are positioned so that a line can touch both. For a circle and a hyperbola centered at the same point, the number of common tangents depends on whether the circle lies inside, touches, or crosses the hyperbola’s asymptotes or branches.

Intuition: The hyperbola opens left and right, with asymptotes $y = \pm \frac{3}{4}x$. The circle of radius 4 is centered at the same origin. Since the hyperbola’s vertices are at $(\pm 4, 0)$, the circle passes exactly through the vertices. But the hyperbola’s branches curve away from the center, so the circle is actually inside the hyperbola’s “bow” near the vertices, but outside near the asymptotes? Let’s check carefully.

1. Understand the shapes and boundaries

   • Hyperbola: vertices at $(\pm 4, 0)$. For any point on the hyperbola, $x^2/16 - y^2/9 = 1$ implies $x^2 = 16 + (16/9)y^2 \ge 16$. So $|x| \ge 4$.
   • Circle: $x^2 + y^2 = 16$ implies $|x| \le 4$ and $|y| \le 4$.
   • At $x = \pm 4$, the hyperbola gives $y=0$; the circle also gives $y=0$. So the circle and hyperbola intersect only at the two vertices? Let’s check: For any other point on the circle, $|x| < 4$, but hyperbola requires $|x| \ge 4$. So indeed, the only intersection points are $(\pm 4, 0)$.

2. Relative position: Is the circle inside or outside the hyperbola?

   • For a fixed $x$ with $|x| < 4$, the hyperbola has no real $y$ (since $y^2 = (9/16)(x^2 - 16) < 0$). So the hyperbola’s branches exist only for $|x| \ge 4$.
   • The circle exists for all $|x| \le 4$. So for $|x| < 4$, the circle’s points are between the two branches of the hyperbola. But the hyperbola’s branches are at $|x| \ge 4$, so the circle is entirely inside the region bounded by the two branches? Actually, the hyperbola’s branches are separate; the circle is a closed curve around the origin. Since the circle’s radius is 4, it touches the hyperbola at the vertices but otherwise lies inside the “gap” between the two branches? Let’s visualize: The hyperbola’s left branch is at $x \le -4$, right branch at $x \ge 4$. The circle is centered at origin with radius 4, so its leftmost point is $(-4,0)$ and rightmost $(4,0)$. So the circle is entirely contained in the vertical strip $-4 \le x \le 4$. The hyperbola’s branches are outside that strip except at the vertices. So the circle lies between the two branches, touching them only at the vertices.

3. Implication for common tangents

   • A common tangent must touch both curves. Since the circle is inside the region between the hyperbola’s branches, any line that touches the circle will either cross the hyperbola’s branches or miss them entirely.
   • At the vertices, the circle and hyperbola share a point, but the tangent to the circle at $(4,0)$ is vertical line $x=4$. Does that also touch the hyperbola? The hyperbola’s tangent at $(4,0)$ is also vertical (since derivative gives slope infinite). So $x=4$ is a common tangent? Wait: At $(4,0)$, the circle’s tangent is vertical. The hyperbola’s tangent at its vertex is also vertical. So the line $x=4$ touches both curves at that same point. That is a common tangent? Actually, a common tangent is a line that touches both curves (possibly at different points). Here it touches both at the same point, so it is a common tangent. Similarly, $x=-4$ is another. So that gives 2 common tangents? But careful: Are these considered tangents? Yes, they are. So we have at least 2. …

We have two curves:

• Hyperbola: $\frac{x^2}{16} - \frac{y^2}{9} = 1$ (center at origin, transverse axis along x-axis, $a=4$, $b=3$).
• Circle: $x^2 + y^2 = 16$ (center at origin, radius $R=4$).

The key idea: The number of common tangents between two curves depends on whether they intersect, touch, or are separate. For conics, common tangents are lines tangent to both. If the curves intersect, they share exactly two common tangents (the "external" ones). If one lies completely inside the other without touching, there are none. If they touch externally, there are three. If they are separate, there are four.

Let’s determine the relative position.

1. Check if the curves intersect. Substitute the circle equation into the hyperbola: From $x^2 + y^2 = 16$, we have $y^2 = 16 - x^2$. Plug into hyperbola:

Multiply by 144:

So $x = \pm 4$. Then $y^2 = 16 - 16 = 0$, so $y = 0$.

Thus the curves intersect at exactly two points: $(4,0)$ and $(-4,0)$.

2. Interpret the intersection.

   The circle of radius 4 passes through the vertices of the hyperbola (since the hyperbola’s vertices are at $(\pm 4, 0)$). So the curves cross at those two points. They are not tangent there (the circle’s slope is vertical at those points; the hyperbola’s slope is also vertical? Let’s check quickly: For the hyperbola, implicit differentiation gives $\frac{x}{8} - \frac{2y}{9}y' = 0$ → at $(4,0)$, $y'$ is undefined (vertical tangent). For the circle, $2x + 2y y' = 0$ → at $(4,0)$, $y'$ is also undefined. So they actually share the same vertical tangent at each intersection? That would mean the curves are tangent at those points, not crossing. Let’s verify carefully.)

   Watch out

   A common mistake: assuming intersection always means crossing. Here the curves actually touch at the vertices because both have vertical tangents at $(\pm 4,0)$. So they are tangent to each other at two points. That changes the count of common tangents.

3. Re-evaluate the tangency.

   At $(4,0)$:

   • Circle: $x^2 + y^2 = 16$ → derivative: $2x + 2y y' = 0$ → at $(4,0)$, $8 + 0 = 0$ is impossible, so the tangent is vertical (infinite slope). …