Match Statements on Vectors Integrals and Complex Numbers

To match the statements in Column I with the correct values in Column II, we analyze each part step-by-step:

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(A)

We are given the vectors:
$\vec{a} = \hat{j} + \sqrt{3}\,\hat{k}, \quad \vec{b} = -\hat{j} + \sqrt{3}\,\hat{k}, \quad \vec{c} = 2\sqrt{3}\,\hat{k}$

Notice that:
$\vec{a} + \vec{b} = (\hat{j} + \sqrt{3}\,\hat{k}) + (-\hat{j} + \sqrt{3}\,\hat{k}) = 2\sqrt{3}\,\hat{k} = \vec{c}$

Let the vertices of the triangle be $A, B,$ and $C$ such that:
$\vec{AB} = \vec{a}, \quad \vec{BC} = \vec{b}, \quad \text{and} \quad \vec{AC} = \vec{a} + \vec{b} = \vec{c}$

The two sides of the triangle meeting at vertex $B$ are along the vectors $\vec{BA} = -\vec{a}$ and $\vec{BC} = \vec{b}$.

The internal angle $\theta$ between sides $\vec{a}$ and $\vec{b}$ at vertex $B$ is given by:
$\cos\theta = \frac{\vec{BA} \cdot \vec{BC}}{|\vec{BA}||\vec{BC}|} = \frac{(-\vec{a}) \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

First, compute the magnitudes and the dot product:
$|\vec{a}| = \sqrt{0^2 + 1^2 + (\sqrt{3})^2} = \sqrt{4} = 2$
$|\vec{b}| = \sqrt{0^2 + (-1)^2 + (\sqrt{3})^2} = \sqrt{4} = 2$
$\vec{a} \cdot \vec{b} = (0)(0) + (1)(-1) + (\sqrt{3})(\sqrt{3}) = -1 + 3 = 2$

Thus:
$\cos\theta = -\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} = -\frac{2}{2 \times 2} = -\frac{1}{2}$

Since $\theta \in (0, \pi)$, we have:
$\theta = \frac{2\pi}{3}$

Therefore, (A) matches with (q).

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(B)

We are given:
$\int_{a}^{b} (f(x) - 3x)\,dx = a^2 - b^2$

Rewriting the right-hand side in integral form:
$a^2 - b^2 = -(b^2 - a^2) = -\int_{a}^{b} 2x\,dx$

Thus:
$\int_{a}^{b} (f(x) - 3x)\,dx = \int_{a}^{b} (-2x)\,dx$
$\int_{a}^{b} (f(x) - x)\,dx = 0$

Assuming this relation holds identically for all real limits $a$ and $b$, differentiating both sides with respect to $b$ gives:
$f(b) - b = 0 \implies f(x) = x$

Evaluating at $x = \frac{\pi}{6}$:
$f\left(\frac{\pi}{6}\right) = \frac{\pi}{6}$

(Note: If the relation is only given for specific constants $a$ and $b$, $f\left(\frac{\pi}{6}\right)$ is not uniquely determined, leading to any value being possible. Hence, officially (p) or all options (p, q, r, s, t) are accepted).

Therefore, (B) matches with (p) [or (p, q, r, s, t)].

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(C)

We evaluate the integral:
$I = \int_{-1/6}^{1/6} \sec(\pi x)\,dx$

Using the substitution $u = \pi x$, so $du = \pi\,dx$:
$I = \frac{1}{\pi}\int_{-\pi/6}^{\pi/6} \sec u\,du$

Since $\sec u$ is an even function:
$I = \frac{2}{\pi}\int_{0}^{\pi/6} \sec u\,du = \frac{2}{\pi}\left[ \ln|\sec u + \tan u| \right]_{0}^{\pi/6}$

Evaluating at the limits:
$\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}, \quad \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}$
$\sec\left(\frac{\pi}{6}\right) + \tan\left(\frac{\pi}{6}\right) = \frac{3}{\sqrt{3}} = \sqrt{3}$
$\sec(0) + \tan(0) = 1$

Thus:
$I = \frac{2}{\pi}\ln(\sqrt{3}) = \frac{2}{\pi}\left(\frac{1}{2}\ln 3\right) = \frac{\ln 3}{\pi}$

Now, substituting $I$ into the given expression:
$\frac{\pi^2}{\ln 3}\int_{-1/6}^{1/6} \sec(\pi x)\,dx = \frac{\pi^2}{\ln 3} \times \frac{\ln 3}{\pi} = \pi$

Therefore, (C) matches with (s).

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(D)

We are given $|z| = 1$ with $z \neq 1$. Let $z = e^{i\theta}$ where $\theta \in (0, 2\pi)$.

Then:
$1 - z = 1 - e^{i\theta} = 1 - \cos\theta - i\sin\theta = 2\sin^2\left(\frac{\theta}{2}\right) - 2i\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)$
$1 - z = -2i\sin\left(\frac{\theta}{2}\right)\left(\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right) = 2\sin\left(\frac{\theta}{2}\right)e^{-i\pi/2}e^{i\theta/2} = 2\sin\left(\frac{\theta}{2}\right)e^{i\left(\frac{\theta - \pi}{2}\right)}$

Taking the reciprocal:
$\frac{1}{1-z} = \frac{1}{2\sin\left(\frac{\theta}{2}\right)} e^{i\left(\frac{\pi - \theta}{2}\right)}$

Since $\theta \in (0, 2\pi)$, we have $\frac{\theta}{2} \in (0, \pi)$, which implies $\sin\left(\frac{\theta}{2}\right) > 0$. Thus:
$\operatorname{Arg}\left(\frac{1}{1-z}\right) = \frac{\pi - \theta}{2}$

As $\theta \to 0^+$, $\operatorname{Arg}\left(\frac{1}{1-z}\right) \to \frac{\pi}{2}$.
As $\theta \to 2\pi^-$, $\operatorname{Arg}\left(\frac{1}{1-z}\right) \to -\frac{\pi}{2}$.

Hence:
$\left|\operatorname{Arg}\left(\frac{1}{1-z}\right)\right| < \frac{\pi}{2}$

The maximum (supremum) value attained is:
$\frac{\pi}{2}$

Therefore, (D) matches with (t).

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Final Matching:

• A $\to$ q
• B $\to$ p (or p, q, r, s, t)
• C $\to$ s
• D $\to$ t