Animated Solution for Mathematics - Three Dimensional Geometry: Match the following :

A Journey Through Complexity

Welcome, future engineer. Today, we are not just solving a 'Match the Following' problem; we are conducting an orchestra of mathematical concepts. We will traverse coordinate geometry, vector algebra, integral calculus, and trigonometric identities.

Take a deep breath, clear your workspace, and let us dive into the elegance of these solutions.

The Geometry of Intersection

We begin with Part A. We are presented with two lines: $x+y=|a|$ and $ax-y=1$. The problem asks us to find the condition for $a$ such that these lines intersect in the first quadrant.

Imagine these lines on a Cartesian plane. The first line, $x+y=|a|$, is a line with a negative slope, shifting its distance from the origin based on the parameter $a$. The second line, $ax-y=1$, is a line with a positive slope $a$.

To find the intersection, we add the two equations:

This yields:

Substituting this back, we find:

Now, here is the crucial realization: the 'first quadrant' constraint is not a suggestion; it is a strict boundary. We require $x>0$ and $y>0$.

For $x>0$, we need $a>-1$. But for $y>0$, we need $a|a|-1>0$, which simplifies to $a^2>1$, or $a>1$. Thus, the intersection is only guaranteed in the first quadrant when $a>1$. The threshold $a_0=1$ is our first victory.

The Elegance of Vectors

Next, we move to Part B. We are given $\hat{k}\times (\hat{k}\times \vec{a})=0$. This looks intimidating, but it is a classic application of the vector triple product identity:

Applying this to our equation, where $\vec{A}=\hat{k}$, $\vec{B}=\hat{k}$, and $\vec{C}=\vec{a}$, we get:

Since $\hat{k}\cdot \vec{a}=\gamma$ and $\hat{k}\cdot \hat{k}=1$, the equation simplifies to $\gamma \hat{k}-\vec{a}=0$. Substituting $\vec{a}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}$, we see the $\gamma \hat{k}$ terms vanish, leaving $-\alpha \hat{i}-\beta \hat{j}=0$. This forces $\alpha =0$ and $\beta =0$.

Since the point $(\alpha ,\beta ,\gamma )$ lies on the plane $x+y+z=2$, we substitute $0+0+\gamma =2$, giving us $\gamma =2$. The geometry here is beautiful: the vector $\vec{a}$ is forced to lie entirely along the z-axis, resulting in $\vec{a}=2\hat{k}$.

The Beauty of Integrals

Part C invites us to evaluate:

Many students panic at the sight of integrals, but look closely at the limits. The second integral is flipped! By the fundamental property ${\int }_a^{b}f(x)dx=-{\int }_b^{a}f(x)dx$, we can rewrite the second integral as $-{\int }_0^1(y^2-1)dy$, which is ${\int }_0^1(1-y^2)dy$.

Now, we have:

Integrating $1-y^2$ gives $y-\frac{y^3}{3}$. Evaluating from $0$ to $1$, we get:

This matches perfectly with the integral ${\int }_0^1\sqrt{1-x}dx+{\int }_{-1}^0\sqrt{1+x}dx$, which also evaluates to $4/3$. It is a perfect symmetry.

The Trigonometric Trap

Finally, Part D. The equation $\sin A\sin B\sin C+\cos A\cos B=1$ seems unsolvable at first glance. But let us rearrange it:

We know that $\sin C\le 1$. If we replace $\sin C$ with its maximum value of $1$, the inequality becomes $\sin A\sin B\ge 1-\cos A\cos B$. Rearranging gives:

This is the expansion of $\cos (A-B)\ge 1$. Since the cosine function can never exceed $1$, the only way this inequality holds is if $\cos (A-B)=1$, which implies $A=B$.

Substituting $A=B$ back into the original equation, we get ${\sin }^{2}A\sin C+{\cos }^{2}A=1$, which simplifies to ${\sin }^{2}A\sin C={\sin }^{2}A$. Assuming $\sin A eq 0$, we get $\sin C=1$.

You have successfully navigated through coordinate geometry, vector identities, integral properties, and trigonometric bounding. This is the essence of JEE Advanced—not just calculation, but insight. Keep this momentum going!