We fix $\boldsymbol{x}, \boldsymbol{y} \in \mathscr{E}_{d}^{n}(\mathbb{R})$ and we introduce for all $(\tau, R) \in \operatorname{Dep}(\mathbb{R}^{d})$ $$J(\tau, R) = \sum_{i=1}^{n} |\boldsymbol{y}_{i} - (R\boldsymbol{x}_{i} + \tau)|^{2} = \|\boldsymbol{y} - g \cdot \boldsymbol{x}\|^{2}$$ where $g = (\tau, R)$. We denote $\overline{\boldsymbol{x}} = \frac{1}{n} \sum_{i=1}^{n} \boldsymbol{x}_{i}$ and $\overline{\boldsymbol{y}} = \frac{1}{n} \sum_{i=1}^{n} \boldsymbol{y}_{i}$.

• [(a)] Show that $J(\tau, R) = \left(\sum_{i=1}^{n} |\boldsymbol{y}_{i} - \overline{\boldsymbol{y}} - R(\boldsymbol{x}_{i} - \overline{\boldsymbol{x}})|^{2}\right) + n|\overline{\boldsymbol{y}} - R\overline{\boldsymbol{x}} - \tau|^{2}$.
• [(b)] Deduce that for all $R \in \mathrm{SO}_{d}(\mathbb{R})$, the map $\tau \mapsto J(\tau, R)$ from $\mathbb{R}^{d}$ to $\mathbb{R}$ has a unique minimum, denoted $\tau(R)$, which we will express explicitly.

Two particles of mass $m$ each are tied at the ends of a light string of length $2a$. The whole system is kept on a frictionless horizontal surface with the string held tight so that each mass is at a distance '$a$' from the center P (as shown in the figure). Now, the mid-point of the string is pulled vertically upwards with a small but constant force $F$. As a result, the particles move towards each other on the surface. The magnitude of acceleration, when the separation between them becomes $2x$, is
(A) $\frac{F}{2m}\frac{a}{\sqrt{a^2-x^2}}$
(B) $\frac{F}{2m}\frac{x}{\sqrt{a^2-x^2}}$
(C) $\frac{F}{2m}\frac{x}{a}$
(D) $\frac{F}{2m}\frac{\sqrt{a^2-x^2}}{x}$

Two forces are such that the sum of their magnitudes is 18 N and their resultant is 12 N which is perpendicular to the smaller force. Then the magnitudes of the forces are
(1) $12 \mathrm{~N} , 6 \mathrm{~N}$
(2) $13 \mathrm{~N} , 5 \mathrm{~N}$
(3) $10 \mathrm{~N} , 8 \mathrm{~N}$
(4) $16 \mathrm{~N} , 2 \mathrm{~N}$

A uniform sphere of weight $W$ and radius 5 cm is being held by a string as shown in the figure. The tension in the string will be :
(1) $12 \frac { \mathrm {~W} } { 5 }$
(2) $5 \frac { \mathrm {~W} } { 12 }$
(3) $13 \frac { \mathrm {~W} } { 5 }$
(4) $13 \frac { \mathrm {~W} } { 12 }$

Two vectors $\vec { A }$ and $\vec { B }$ have equal magnitudes. The magnitude of $( \vec { A } + \vec { B } )$ is ' $n$ ' times the magnitude of $( \vec { A } - \vec { B } )$. The angle between $\vec { A }$ and $\vec { B }$ is:
(1) $\cos ^ { - 1 } \left[ \frac { n ^ { 2 } - 1 } { n ^ { 2 } + 1 } \right]$
(2) $\sin ^ { - 1 } \left[ \frac { n - 1 } { n + 1 } \right]$
(3) $\cos ^ { - 1 } \left[ \frac { n - 1 } { n + 1 } \right]$
(4) $\sin ^ { - 1 } \left[ \frac { n ^ { 2 } - 1 } { n ^ { 2 } + 1 } \right]$

Two forces $P$ and $Q$, of magnitude $2F$ and $3F$, respectively, are at an angle $\theta$ with each other. If the force $Q$ is doubled, then their resultant also gets doubled. Then, the angle $\theta$ is:
(1) $120 ^ { \circ }$
(2) $60 ^ { \circ }$
(3) $30 ^ { \circ }$
(4) $90 ^ { \circ }$

The sum of two forces $\overrightarrow { \mathrm { P } }$ and $\overrightarrow { \mathrm { Q } }$ is $\overrightarrow { \mathrm { R } }$ such that $| \overrightarrow { \mathrm { R } } | = | \overrightarrow { \mathrm { P } } |$. Find the angle between resultant of $2 \overrightarrow { \mathrm { P } }$ and $\overrightarrow { \mathrm { Q } }$ and $\overrightarrow { \mathrm { Q } }$.

The resultant of these forces $\overrightarrow{OP}, \overrightarrow{OQ}, \overrightarrow{OR}, \overrightarrow{OS}$ and $\overrightarrow{OT}$ is approximately $\_\_\_\_$ N. [Take $\sqrt{3} = 1.7, \sqrt{2} = 1.4$ Given $\hat{\mathrm{i}}$ and $\hat{\mathrm{j}}$ unit vectors along $x, y$ axis]
(1) $-1.5\hat{\mathrm{i}} - 15.5\hat{\mathrm{j}}$
(2) $9.25\hat{i} + 5\hat{j}$
(3) $3\hat{i} + 15\hat{j}$
(4) $2.5\hat{\mathrm{i}} - 14.5\hat{\mathrm{j}}$

Two forces $\bar { F } _ { 1 }$ and $\bar { F } _ { 2 }$ are acting on a body. One force has magnitude thrice that of the other force and the resultant of the two forces is equal to the force of larger magnitude. The angle between $\vec { F } _ { 1 }$ and $\vec { F } _ { 2 }$ is $\cos ^ { - 1 } \left( \frac { 1 } { n } \right)$. The value of $| n |$ is $\_\_\_\_$.