Match Column Statements for Lines Vectors Functions and Integrals

To solve the given matching problem, we analyze each column statement step-by-step.

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Part A

The given lines are:
$L_1: \frac{x-2}{1} = \frac{y-1}{-2} = \frac{z+1}{1} = \lambda$
$L_2: \frac{x-\frac{8}{3}}{2} = \frac{y+3}{-1} = \frac{z-1}{1} = \mu$

Any general point $P$ on $L_1$ and $Q$ on $L_2$ can be represented as:
$P = (\lambda + 2, -2\lambda + 1, \lambda - 1)$
$Q = \left(2\mu + \frac{8}{3}, -\mu - 3, \mu + 1\right)$

Since $P$, $Q$, and the origin $O(0,0,0)$ lie on the same line, the position vectors $\vec{OP}$ and $\vec{OQ}$ are collinear. Thus, their corresponding coordinates must be proportional:
$\frac{\lambda + 2}{2\mu + \frac{8}{3}} = \frac{-2\lambda + 1}{-\mu - 3} = \frac{\lambda - 1}{\mu + 1}$

Equating the second and third ratios:
$\frac{2\lambda - 1}{\mu + 3} = \frac{\lambda - 1}{\mu + 1}$
$(2\lambda - 1)(\mu + 1) = (\lambda - 1)(\mu + 3)$
$2\lambda\mu + 2\lambda - \mu - 1 = \lambda\mu + 3\lambda - \mu - 3$
$\lambda\mu - \lambda + 2 = 0 \implies \lambda\mu = \lambda - 2 \quad \text{--- (1)}$

Equating the first and third ratios:
$(\lambda + 2)(\mu + 1) = (\lambda - 1)\left(2\mu + \frac{8}{3}\right)$
$\lambda\mu + \lambda + 2\mu + 2 = 2\lambda\mu + \frac{8}{3}\lambda - 2\mu - \frac{8}{3}$
$\lambda\mu + \frac{5}{3}\lambda - 4\mu - \frac{14}{3} = 0 \quad \text{--- (2)}$

Substituting $\lambda\mu = \lambda - 2$ into (2):
$(\lambda - 2) + \frac{5}{3}\lambda - 4\mu - \frac{14}{3} = 0$
$\frac{8}{3}\lambda - 4\mu - \frac{20}{3} = 0 \implies \mu = \frac{2\lambda - 5}{3} \quad \text{--- (3)}$

Substitute (3) back into (1):
$\lambda \left(\frac{2\lambda - 5}{3}\right) = \lambda - 2$
$2\lambda^2 - 8\lambda + 6 = 0 \implies \lambda^2 - 4\lambda + 3 = 0 \implies \lambda = 1 \text{ or } \lambda = 3$

• For $\lambda = 1$, $\mu = -1$, the point $P = (3, -1, 0)$ and $Q = \left(\frac{2}{3}, -2, 0\right)$ are not collinear with $O(0,0,0)$ because $\frac{3}{2/3} \neq \frac{-1}{-2}$.
• For $\lambda = 3$, $\mu = \frac{1}{3}$, the points are:
  $P = (5, -5, 2)$
  $Q = \left(\frac{10}{3}, -\frac{10}{3}, \frac{4}{3}\right) = \frac{2}{3} P$

Since $Q = \frac{2}{3}P$, $O, P, Q$ are collinear.
The vector $\vec{PQ} = \vec{OQ} - \vec{OP} = -\frac{1}{3} \vec{OP}$.
Hence, the distance $d$ is:
$d^2 = |\vec{PQ}|^2 = \frac{1}{9} |\vec{OP}|^2 = \frac{1}{9} (5^2 + (-5)^2 + 2^2) = \frac{54}{9} = 6$

Matching for A: t

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Part B

Given equation:
$\tan^{-1}(x+3) - \tan^{-1}(x-3) = \sin^{-1}\left(\frac{3}{5}\right)$

Taking tangent on both sides (using $\tan(\sin^{-1}(3/5)) = \frac{3}{4}$):
$\tan\left(\tan^{-1}(x+3) - \tan^{-1}(x-3)\right) = \frac{3}{4}$
$\frac{(x+3) - (x-3)}{1 + (x+3)(x-3)} = \frac{3}{4}$
$\frac{6}{1 + x^2 - 9} = \frac{3}{4} \implies \frac{6}{x^2 - 8} = \frac{3}{4}$
$3(x^2 - 8) = 24 \implies x^2 - 8 = 8 \implies x^2 = 16 \implies x = \pm 4$

Checking both values:

• For $x = 4$: $\tan^{-1}(7) - \tan^{-1}(1) = \tan^{-1}\left(\frac{7-1}{1+7}\right) = \tan^{-1}\left(\frac{3}{4}\right) = \sin^{-1}\left(\frac{3}{5}\right)$ (Valid).
• For $x = -4$: $\tan^{-1}(-1) - \tan^{-1}(-7) = \tan^{-1}(7) - \tan^{-1}(1) = \sin^{-1}\left(\frac{3}{5}\right)$ (Valid).

Matching for B: p, r

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Part C

Given conditions:

1. $\vec{a} \cdot \vec{b} = 0$
2. $(\vec{b}-\vec{a}) \cdot (\vec{b}+\vec{c}) = 0 \implies |\vec{b}|^2 + \vec{b}\cdot\vec{c} - \vec{a}\cdot\vec{c} = 0$
3. $2|\vec{b}+\vec{c}| = |\vec{b}-\vec{a}| \implies 4|\vec{b}+\vec{c}|^2 = |\vec{b}-\vec{a}|^2$
   $4(|\vec{b}|^2 + |\vec{c}|^2 + 2\vec{b}\cdot\vec{c}) = |\vec{b}|^2 + |\vec{a}|^2$
   $3|\vec{b}|^2 + 4|\vec{c}|^2 + 8\vec{b}\cdot\vec{c} = |\vec{a}|^2 \quad \text{--- (1)}$

Given $\vec{a} = \mu\vec{b} + 4\vec{c}$, take dot product with $\vec{b}$:
$\vec{a}\cdot\vec{b} = \mu|\vec{b}|^2 + 4\vec{b}\cdot\vec{c} = 0 \implies \vec{b}\cdot\vec{c} = -\frac{\mu}{4}|\vec{b}|^2$

From condition (2):
$\vec{a}\cdot\vec{c} = |\vec{b}|^2 + \vec{b}\cdot\vec{c} = |\vec{b}|^2 \left(1 - \frac{\mu}{4}\right)$

Now, compute $|\vec{a}|^2$:
$|\vec{a}|^2 = \vec{a}\cdot(\mu\vec{b} + 4\vec{c}) = 4(\vec{a}\cdot\vec{c}) = 4|\vec{b}|^2 \left(1 - \frac{\mu}{4}\right) = |\vec{b}|^2 (4 - \mu)$

Also, from $\vec{a} = \mu\vec{b} + 4\vec{c}$:
$|\vec{a}|^2 = \mu^2|\vec{b}|^2 + 16|\vec{c}|^2 + 8\mu(\vec{b}\cdot\vec{c}) = \mu^2|\vec{b}|^2 + 16|\vec{c}|^2 - 2\mu^2|\vec{b}|^2 = 16|\vec{c}|^2 - \mu^2|\vec{b}|^2$

Equating the two expressions for $|\vec{a}|^2$:
$|\vec{b}|^2(4 - \mu) = 16|\vec{c}|^2 - \mu^2|\vec{b}|^2 \implies 16|\vec{c}|^2 = |\vec{b}|^2(\mu^2 - \mu + 4)$

Substitute $|\vec{c}|^2$, $\vec{b}\cdot\vec{c}$, and $|\vec{a}|^2$ into (1):
$3|\vec{b}|^2 + 4\left(\frac{|\vec{b}|^2(\mu^2 - \mu + 4)}{16}\right) + 8\left(-\frac{\mu}{4}|\vec{b}|^2\right) = |\vec{b}|^2(4 - \mu)$

Dividing by $|\vec{b}|^2 \neq 0$:
$3 + \frac{\mu^2 - \mu + 4}{4} - 2\mu = 4 - \mu$
$12 + \mu^2 - \mu + 4 - 8\mu = 16 - 4\mu$
$\mu^2 - 5\mu = 0 \implies \mu = 0 \text{ or } \mu = 5$

Matching for C: q, s (or q)

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Part D

Given function:
$f(x) = \frac{\sin\left(\frac{9x}{2}\right)}{\sin\left(\frac{x}{2}\right)} \quad \text{for } x \neq 0, \text{ and } f(0) = 9$

Using the trigonometric identity:
$\frac{\sin\left((2n+1)\frac{x}{2}\right)}{\sin\left(\frac{x}{2}\right)} = 1 + 2\sum_{k=1}^{n} \cos(kx)$

For $n = 4$:
$f(x) = 1 + 2\cos(x) + 2\cos(2x) + 2\cos(3x) + 2\cos(4x)$

Since $f(x)$ is an even function:
$\int_{-\pi}^{\pi} f(x) \, dx = 2 \int_{0}^{\pi} \left(1 + 2\cos(x) + 2\cos(2x) + 2\cos(3x) + 2\cos(4x)\right) \, dx$
$= 2 \left[ x + 2\sin(x) + \sin(2x) + \frac{2}{3}\sin(3x) + \frac{1}{2}\sin(4x) \right]_{0}^{\pi} = 2\pi$

Therefore:
$\frac{2}{\pi} \int_{-\pi}^{\pi} f(x) \, dx = \frac{2}{\pi} \cdot 2\pi = 4$

Matching for D: r

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Final Summary of Matches:

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