Let $\vec { a }$ and $\vec { b }$ be two unit vectors such that the angle between them is $\frac { \pi } { 3 }$. If $\lambda \vec { a } + 2 \vec { b }$ and $3 \vec { a } - \lambda \vec { b }$ are perpendicular to each other, then the number of values of $\lambda$ in $[ - 1,3 ]$ is :
(1) 2
(2) 1
(3) 0
(4) 3

Let the $\operatorname { arc } A C$ of a circle subtend a right angle at the centre $O$. If the point $B$ on the arc $A C$, divides the arc $A C$ such that $\frac { \text { length of } \operatorname { arc } A B } { \text { length of } \operatorname { arc } B C } = \frac { 1 } { 5 }$, and $\overrightarrow { O C } = \alpha \overrightarrow { O A } + \beta \overrightarrow { O B }$, then $\alpha + \sqrt { 2 } ( \sqrt { 3 } - 1 ) \beta$ is equal to
(1) $2 \sqrt { 3 }$
(2) $2 - \sqrt { 3 }$
(3) $5 \sqrt { 3 }$
(4) $2 + \sqrt { 3 }$

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{c}$ be three vectors such that $\vec{c}$ is coplanar with $\vec{a}$ and $\vec{b}$. If the vector $\vec{c}$ is perpendicular to $\vec{b}$ and $\vec{a} \cdot \vec{c} = 5$, then $|\vec{c}|$ is equal to
(1) $\sqrt{\frac{11}{6}}$
(2) $\frac{1}{3\sqrt{2}}$
(3) 16
(4) 18

Q21. For three vectors $\vec { A } = ( - x \hat { i } - 6 \hat { j } - 2 \hat { k } ) , \vec { B } = ( - \hat { i } + 4 \hat { j } + 3 \hat { k } )$ and $\vec { C } = ( - 8 \hat { i } - \hat { j } + 3 \hat { k } )$, if $\vec { A } \cdot ( \vec { B } \times \vec { C } ) = 0$, then value of $x$ is $\_\_\_\_$

Q21. If $\vec { a }$ and $\vec { b }$ makes an angle $\cos ^ { - 1 } \left( \frac { 5 } { 9 } \right)$ with each other, then $| \vec { a } + \vec { b } | = \sqrt { 2 } | \vec { a } - \vec { b } |$ for $| \vec { a } | = n | \vec { b } |$ The integer value of n is $\_\_\_\_$

Q21. The resultant of two vectors $\vec { A }$ and $\vec { B }$ is perpendicular to $\vec { A }$ and its magnitude is half that of $\vec { B }$. The angle between vectors $\vec { A }$ and $\vec { B }$ is $\_\_\_\_$ ○.

Let $\vec { c }$ be the projection vector of $\vec { b } = \lambda \hat { i } + 4 \hat { k } , \lambda > 0$, on the vector $\vec { a } = \hat { i } + 2 \hat { j } + 2 \hat { k }$. If $| \vec { a } + \vec { c } | = 7$, then the area of the parallelogram formed by the vectors $\vec { b }$ and $\vec { c }$ is $\_\_\_\_$

Let $A ( 6,8 ) , B ( 10 \cos \alpha , - 10 \sin \alpha )$ and $C ( - 10 \sin \alpha , 10 \cos \alpha )$, be the vertices of a triangle. If $L ( a , 9 )$ and $G ( h , k )$ be its orthocenter and centroid respectively, then $( 5 a - 3 h + 6 k + 100 \sin 2 \alpha )$ is equal to $\_\_\_\_$

Let $\vec { a } = \hat { \mathrm { i } } + \hat { \mathrm { j } } + \hat { \mathrm { k } } , \overrightarrow { \mathrm { b } } = 2 \hat { \mathrm { i } } + 2 \hat { \mathrm { j } } + \hat { \mathrm { k } }$ and $\overrightarrow { \mathrm { d } } = \vec { a } \times \overrightarrow { \mathrm { b } }$. If $\overrightarrow { \mathrm { c } }$ is a vector such that $\vec { a } \cdot \overrightarrow { \mathrm { c } } = | \overrightarrow { \mathrm { c } } | , | \overrightarrow { \mathrm { c } } - 2 \vec { a } | ^ { 2 } = 8$ and the angle between $\overrightarrow { \mathrm { d } }$ and $\overrightarrow { \mathrm { c } }$ is $\frac { \pi } { 4 }$, then $| 10 - 3 \overrightarrow { \mathrm { b } } \cdot \overrightarrow { \mathrm { c } } | + | \overrightarrow { \mathrm { d } } \times \overrightarrow { \mathrm { c } } | ^ { 2 }$ is equal to

Q77. Consider three vectors $\vec { a } , \vec { b } , \vec { c }$. Let $| \vec { a } | = 2 , | \vec { b } | = 3$ and $\vec { a } = \vec { b } \times \vec { c }$. If $\alpha \in \left[ 0 , \frac { \pi } { 3 } \right]$ is the angle between the vectors $\vec { b }$ and $\vec { c }$, then the minimum value of $27 | \vec { c } - \vec { a } | ^ { 2 }$ is equal to:
(1) 110
(2) 124
(3) 121
(4) 105

Q78. Let $\overrightarrow { \mathrm { a } } = 2 \hat { i } + 5 \hat { j } - \hat { k } , \overrightarrow { \mathrm {~b} } = 2 \hat { i } - 2 \hat { j } + 2 \hat { k }$ and $\overrightarrow { \mathrm { c } }$ be three vectors such that $( \vec { c } + \hat { i } ) \times ( \vec { a } + \vec { b } + \hat { i } ) = \vec { a } \times ( \vec { c } + \hat { i } )$. If $\vec { a } \cdot \vec { c } = - 29$, then $\vec { c } \cdot ( - 2 \hat { i } + \hat { j } + \hat { k } )$ is equal to:
(1) 15
(2) 12
(3) 10
(4) 5

Let $| \vec { a } | = 1 , | \vec { b } | = 4 \& | \vec { c } | = 2$. If $\left( \vec { a } \times \vec { b } = 2 ( \vec { a } \times \vec { c } ) \right.$ and $\vec { b } ^ { \wedge } \vec { c } = \frac { \pi } { 3 }$ then find $\left. \left. | \vec { a } \cdot \vec { c } | \right| ^ { 2 } \right| ^ { 2 } - 2 \vec { c } \left| = | \lambda \vec { a } | ^ { 2 } \right.$

Let $| \vec { a } | = | \vec { b } | = | \vec { c } | = 1 \cdot | \vec { a } - \vec { b } | ^ { 2 } + | \vec { b } - \vec { c } | ^ { 2 } + | \vec { c } - \vec { a } | ^ { 2 } = 9$ and $| 2 \vec { a } + k \vec { b } + k \vec { c } | = 9$ then positive value of $k$ is

For given vectors $\vec { a } = - \hat { i } + \hat { j } + 2 \hat { k }$ and $\vec { b } = 2 \hat { i } - \hat { j } + \hat { k }$ where $\vec { c } = \vec { a } \times \vec { b }$ and $\vec { d } = \vec { c } \times \vec { b }$. Then the value of $( \vec { a } - \vec { b } ) \cdot \vec { d }$ is:\ (A) 35\ (B) -35\ (C) 30\ (D) -30

Consider two straight lines
$$y=1, \quad y=-1$$
and the point $\mathrm{A}(0,3)$ in the $xy$-plane. Take a point P on the straight line $y=1$ and a point Q on the straight line $y=-1$ such that
$$\angle\mathrm{PAQ}=90^\circ.$$
Let the two points P and Q move preserving the above conditions. We are to find the minimum value of the length of the line segment PQ.
First, denote the coordinates of P by $(\alpha,1)$ and the coordinates of Q by $(\beta,-1)$. Then the condition $\angle\mathrm{PAQ}=90^\circ$ is reduced to the conditions $\alpha\neq 0$, $\beta\neq 0$ and
$$\alpha\beta = \mathbf{AB}.$$
Since we know that $\alpha$ and $\beta$ have opposite signs, let us assume that $\alpha<0<\beta$.
Then we have
$$\begin{aligned} \mathrm{PQ}^2 &= (\beta-\alpha)^2+\mathbf{C} \\ &= \alpha^2+\beta^2+\mathbf{DE} \\ &\geqq 2|\alpha\beta|+\mathbf{DE} = \mathbf{FG}. \end{aligned}$$
So we have
$$\mathrm{PQ}\geqq\mathbf{H}.$$
Hence, when
$$\alpha=\mathbf{IJ}\sqrt{\mathbf{K}} \text{ and } \beta=\mathbf{L}\sqrt{\mathbf{M}},$$
PQ takes the minimum value $\mathbf{H}$.

Consider a triangle OAB. We denote by M the point dividing the side OA internally in the ratio 3:1 and denote by N the point dividing the side OB internally in the ratio 1:2. Also we denote by P the point of intersection of the line segment AN and the line segment BM.
(1) When the vectors $\overrightarrow{\mathrm{OA}}$ and $\overrightarrow{\mathrm{OB}}$ are denoted by $\vec{a}$ and $\vec{b}$ respectively, we are to express the vector $\overrightarrow{\mathrm{OP}}$ in terms of $\vec{a}$ and $\vec{b}$.
When we set
$$\begin{array}{ll} \mathrm{AP}:\mathrm{PN} = s:(1-s) & (0we have
$$\overrightarrow{\mathrm{OP}} = (\mathbf{A} - t)\vec{b} = \frac{\mathbf{D}}{\mathbf{E}}t\vec{a}+\frac{\mathbf{B}}{\mathbf{E}}s\vec{b},$$
from which we obtain
$$s=\frac{\mathbf{G}}{\mathbf{H}}, \quad t=\frac{\square}{\mathbf{I}}.$$
Hence, $\overrightarrow{\mathrm{OP}}$ can be expressed in terms of $\vec{a}$ and $\vec{b}$ as
$$\overrightarrow{\mathrm{OP}} = \frac{\mathbf{K}}{\mathbf{L}}\vec{a}+\frac{\mathbf{M}}{\mathbf{N}}\vec{b}.$$
(2) We are to look at the relation between the length of the line segment OP and the size of $\angle\mathrm{AOB}$, when $\mathrm{OA}=6$ and $\mathrm{OB}=9$.
Let us denote the length of OP by $\ell$. When we express $\ell^2$ in terms of $\vec{a}\cdot\vec{b}$, we have
$$\ell^2 = \frac{\mathbf{Q}}{\mathbf{PQ}}\vec{a}\cdot\vec{b}+\mathbf{RS}$$
where $\vec{a}\cdot\vec{b}$ represents the inner product of $\vec{a}$ and $\vec{b}$.
Hence, for instance, if $\ell=4$, then we have
$$\cos\angle\mathrm{AOB}=\frac{\mathbf{TU}}{\mathbf{V}}$$
When the size of $\angle\mathrm{AOB}$ varies, the range of the values which $\ell$ can take is
$$\mathbf{W}<\ell<\mathbf{X}.$$

Take four points
$$\mathrm { A } ( 1,0 ) , \quad \mathrm { B } ( 0,1 ) , \quad \mathrm { C } ( 3,0 ) , \quad \mathrm { D } ( 0,2 )$$
on a plane with the coordinate system having the origin O. Take two points P and Q on segments AB and CD respectively, such that
$$\mathrm { AP } : \mathrm { PB } = \mathrm { CQ } : \mathrm { QD } = k : 2 .$$
We are to find the minimum possible length of the segment PQ.
(1) First, let us find the value of $x + 2 y$, where $\overrightarrow { \mathrm { PQ } } = ( x , y )$.
Since
$$\overrightarrow { \mathrm { OP } } = \frac { \square \mathbf { A } \overrightarrow { \mathrm { OA } } + k \overrightarrow { \mathrm { OB } } } { k + \mathbf { B } } , \quad \overrightarrow { \mathrm { OQ } } = \frac { \square \mathbf { C } \overrightarrow { \mathrm { OC } } + k \overrightarrow { \mathrm { OD } } } { k + \mathbf { D } } ,$$
we have
$$( x , y ) = \frac { 1 } { k + \mathbf { E } } ( \mathbf { F } , k ) ,$$
which gives $x + 2 y = \mathbf { G }$.
(2) When we represent $\mathrm { PQ } ^ { 2 }$ in terms of $y$, we have
$$\mathrm { PQ } ^ { 2 } = \mathbf { H } y ^ { 2 } - \mathbf { I } y + \mathbf { J } .$$
Hence PQ takes the minimum at $y = \frac { \mathbf { K } } { \mathbf { L } }$ and its value there is $\mathrm { PQ } = \frac { \square \mathbf { M } \sqrt { \mathbf { N } } } { \mathbf { O } }$. In this case, the value of $k$ is $k = \square \mathbf { P }$.

Suppose that a triangle ABC which is inscribed in a circle O of radius 2 satisfies
$$3\overrightarrow{\mathrm{OA}} + 4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}} = \vec{0}.$$
Let D denote the point of intersection of the straight line AO and the segment BC. We are to find the lengths of the segments AD and BD.
(1) When we set $\overrightarrow{\mathrm{OD}} = k\overrightarrow{\mathrm{OA}}$ where $k$ is a real number, we have
$$\overrightarrow{\mathrm{OD}} = -\frac{\mathbf{A}}{\mathbf{B}}k\overrightarrow{\mathrm{OB}} - \frac{\mathbf{C}}{\mathbf{D}}k\overrightarrow{\mathrm{OC}}.$$
As the three points B, C and D are located on a straight line, we obtain $k = \frac{\mathbf{EF}}{\mathbf{GH}}$. From this we derive that $\mathrm{OD} = \mathbf{H}$ and finally obtain
$$\mathrm{AD} = 1.$$
(2) From (1) we see that $\mathrm{BD} = \frac{\mathbf{J}}{\mathbf{K}}\mathrm{BC}$. So in order to find the length of the segment BD, we should find the length of the segment BC.
First we note that
$$\mathrm{BC}^2 = \mathbf{L} - \mathbf{M}\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$$
where $\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}}$ represents the inner product of $\overrightarrow{\mathrm{OB}}$ and $\overrightarrow{\mathrm{OC}}$. Since we know from (1) that $|4\overrightarrow{\mathrm{OB}} + 2\overrightarrow{\mathrm{OC}}|^2 = \mathbf{NO}$, we have
$$\overrightarrow{\mathrm{OB}} \cdot \overrightarrow{\mathrm{OC}} = \frac{\mathbf{PQR}}{\mathbf{S}}.$$
Hence we obtain $\mathrm{BC} = \frac{\square\sqrt{\mathbf{U}}}{\square\mathbf{V}}$ and finally from that
$$\mathrm{BD} = \frac{\sqrt{\mathrm{W}}}{\mathrm{X}}.$$

Let $S$ be a circle with its center at point O and a radius of 1. Let $\triangle \mathrm { ABC }$ be a triangle such that all its vertices are on $S$ and $\mathrm { AB } : \mathrm { AC } = 3 : 2$. As shown in the figure, let D be a point on the extension of side BC and $k$ be the number where
$$\mathrm { BC } : \mathrm { CD } = 2 : k .$$
Moreover, set
$$\overrightarrow { \mathrm { OA } } = \vec { a } , \quad \overrightarrow { \mathrm { OB } } = \vec { b } , \quad \overrightarrow { \mathrm { OC } } = \vec { c }$$
Answer the following questions.
(1) When we express $\overrightarrow { \mathrm { OD } }$ in terms of $\vec { b } , \vec { c }$ and $k$, we have
$$\overrightarrow { \mathrm { OD } } = \left( \frac { k } { \mathbf { A } } + \mathbf { B } \right) \vec { c } - \frac { k } { \mathbf { C } } \vec { b }$$
(2) Since the equality
$$| \vec { b } - \vec { a } | = \frac { \mathbf { D } } { \mathbf { E } } | \vec { c } - \vec { a } |$$
holds, by expressing the inner product $\vec { a } \cdot \vec { b }$ in terms of the inner product $\vec { a } \cdot \vec { c }$, we have
$$\vec { a } \cdot \vec { b } = \frac { \mathbf { F } } { \mathbf { G } } \vec { a } \cdot \vec { c } - \frac { \mathbf { H } } { \mathbf { I } }$$
(3) It follows that when the tangent to $S$ at the point A passes through the point D,
$$k = \frac { \mathbf { J } } { \mathbf { K } } .$$

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = 1$, $|\vec{b}| = 2$ and the angle formed by $\vec{a}$ and $\vec{b}$ is $60^\circ$. Set $\vec{u} = x\vec{a} + \vec{b}$ and $\vec{v} = x\vec{a} - \vec{b}$ for a real number $x$. When $x > 1$, we are to find the value of $x$ such that the angle formed by $\vec{u}$ and $\vec{v}$ is $30^\circ$. In the following, $\vec{u} \cdot \vec{v}$ denotes the inner product of $\vec{u}$ and $\vec{v}$, and $\vec{a} \cdot \vec{b}$ denotes the inner product of $\vec{a}$ and $\vec{b}$.
First of all, since the angle formed by $\vec{u}$ and $\vec{v}$ is $30^\circ$, we obtain $$(\vec{u} \cdot \vec{v})^2 = \frac{\mathbf{A}}{\mathbf{B}}|\vec{u}|^2|\vec{v}|^2.$$
When we express this equation in terms of $x$, noting $\vec{a} \cdot \vec{b} = \mathbf{C}$, we have $$x^4 - \mathbf{DE}x^2 + \mathbf{FG} = 0.$$
By transforming this, we also have $$\left(x^2 - \mathbf{H}\right)^2 = (\mathbf{I}x)^2.$$
When this is solved for $x$, we obtain $$x = \mathbf{J} + \sqrt{\mathbf{KL}},$$ noting $x > 1$.

In the figure to the right, let $\angle \mathrm { XOY } = 60 ^ { \circ }$, and let OZ be the half-line (ray) which bisects $\angle \mathrm { XOY }$. In addition, the points A and B on the half-lines OX and OY satisfy $\mathrm { OA } = \mathrm { OB } = 1$.
Let $\mathrm { P } , \mathrm { Q }$ and R be moving points on $\mathrm { OX } , \mathrm { OZ }$ and OY that start simultaneously from $\mathrm { A } , \mathrm { O }$ and B, moving in the direction away from point O at the speeds of $1 , \sqrt { 3 }$ and 2 units per second.
We are to find the moment at which the three points $\mathrm { P } , \mathrm { Q }$ and R are arranged on a straight line by considering the area of the triangle PQR.
First, the lengths of $\mathrm { OP } , \mathrm { OQ }$ and OR at $t$ seconds after the start are
$$\mathrm { OP } = t + \mathbf { A } , \quad \mathrm { OQ } = \sqrt { \mathbf { B } } t , \quad \mathrm { OR } = \mathbf { C } t + \mathbf { D } .$$
At this time the areas of the triangles are
$$\begin{aligned} \triangle \mathrm { OPQ } & = \frac { \sqrt { \mathbf { E } } t ( t + \mathbf { F } ) } { 4 } , \\ \triangle \mathrm { ORQ } & = \frac { \sqrt { \mathbf { G } } t ( \mathbf { H } t + \mathbf { I } ) } { 4 } , \\ \triangle \mathrm { OPR } & = \frac { \sqrt { \mathbf { G } } ( t + \mathbf { K } ) ( \mathbf { L } t + \mathbf { M } ) } { 4 } \end{aligned}$$
Hence we obtain
$$\triangle \mathrm { PQR } = \frac { \sqrt { \mathbf { N } } } { 4 } \left| - t ^ { 2 } + t + \mathbf { O } \right| .$$
So, to find the moment such that the three points $\mathrm { P } , \mathrm { Q }$ and R are arranged on a straight line, we should find the case where
$$t ^ { 2 } - t - \mathbf { O } = \mathbf { P } .$$
Thus the time required is
$$t = \frac { \mathbf { Q } + \sqrt { \mathbf { R } } } { \mathbf { S } } \text{ (seconds).}$$

Given two points $\mathrm { A } ( 1 , - 1,0 )$ and $\mathrm { B } ( - 2,1,2 )$ in a coordinate space with the origin O, let us set $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$.
(1) First, we are to find the value of $t$ at which $| \vec { a } + t \vec { b } |$ is minimized. Since
$$| \vec { a } + t \vec { b } | ^ { 2 } = \mathbf { A } t ^ { 2 } - \mathbf { B } t + \mathbf { C }$$
$| \vec { a } + t \vec { b } |$ is minimized at $t = \frac { \mathbf { D } } { \mathbf { E } }$, and its minimum value is $\mathbf { F }$.
(2) Next, the vectors $\vec { c }$ which are orthogonal to the vectors $\vec { a }$ and $\vec { b }$ can be represented as
$$\vec { c } = s ( \mathbf { G } , \mathbf { H } , 1 )$$
where $s$ is a non-zero real number. Now, let C and D be the points such that $\overrightarrow { \mathrm { OC } } = ( \mathbf { G } , \mathbf { H } , 1 )$ and $\overrightarrow { \mathrm { OD } } = 3 \vec { a } + \vec { b }$. Since $\angle \mathrm { CBD } = \frac { \pi } { \mathbf { I } }$, the area of the triangle BCD is $\frac { \mathbf { J } \sqrt { \mathbf { K } } } { \mathbf { L } }$.

A triangle OAB and a triangle OAC share one side OA. Further, they satisfy the following two conditions:
(i) $\overrightarrow { \mathrm { OC } } = x \overrightarrow { \mathrm { OA } } + \frac { 1 } { 2 } \overrightarrow { \mathrm { OB } }$;
(ii) the center of gravity G of the triangle OAC is on the segment AB.
We are to find the value of $x$, and express $\overrightarrow { \mathrm { OG } }$ in terms of $\overrightarrow { \mathrm { OA } }$ and $\overrightarrow { \mathrm { OB } }$.
Let us denote the point of intersection of the segment OC and the segment AB by D. Then we have
$$\overrightarrow { \mathrm { OD } } = \frac { x } { \mathbf { A } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { B } } { \mathbf { A } } \overrightarrow { \mathrm { OB } } .$$
Since D is on the segment AB, we have $x = \frac { \mathbf { D } } { \mathbf { E } }$. Hence we obtain
$$\overrightarrow { \mathrm { OG } } = \frac { \mathbf { F } } { \mathbf { G } } \overrightarrow { \mathrm { OA } } + \frac { \mathbf { H } } { \mathbf { I } } \overrightarrow { \mathrm { OB } } .$$
In particular, when $\mathrm { OA } = 1 , \mathrm { OB } = 2$ and $\angle \mathrm { AOB } = 60 ^ { \circ }$, we have $\mathrm { OG } = \frac { \sqrt { \mathbf { J K } } } { \mathbf { L } }$.

5. For ALL APPLICANTS.

A particular robot has three commands: F: Move forward a unit distance; L: Turn left $90 ^ { \circ }$; R: Turn right $90 ^ { \circ }$.
A program is a sequence of commands. We consider particular programs $P _ { n }$ (for $n \geqslant 0$ ) in this question. The basic program $P _ { 0 }$ just instructs the robot to move forward:
$$P _ { 0 } = \mathbf { F } .$$
The program $P _ { n + 1 }$ (for $n \geqslant 0$ ) involves performing $P _ { n }$, turning left, performing $P _ { n }$ again, then turning right:
$$P _ { n + 1 } = P _ { n } \mathbf { L } P _ { n } \mathbf { R }$$
So, for example, $P _ { 1 } = \mathbf { F L F R }$.
(i) Write down the program $P _ { 2 }$.
(ii) How far does the robot travel during the program $P _ { n }$ ? In other words, how many $\mathbf { F }$ commands does it perform?
(iii) Let $l _ { n }$ be the total number of commands in $P _ { n }$; so, for example, $l _ { 0 } = 1$ and $l _ { 1 } = 4$.
Write down an equation relating $l _ { n + 1 }$ to $l _ { n }$. Hence write down a formula for $l _ { n }$ in terms of $n$. No proof is required. Hint: consider $l _ { n } + 2$.
(iv) The robot starts at the origin, facing along the positive $x$-axis. What direction is the robot facing after performing the program $P _ { n }$ ?
(v) The left-hand diagram on the opposite page shows the path the robot takes when it performs the program $P _ { 1 }$. On the right-hand diagram opposite, draw the path it takes when it performs the program $P _ { 4 }$.
(vi) Let $\left( x _ { n } , y _ { n } \right)$ be the position of the robot after performing the program $P _ { n }$, so $\left( x _ { 0 } , y _ { 0 } \right) = ( 1,0 )$ and $\left( x _ { 1 } , y _ { 1 } \right) = ( 1,1 )$. Give an equation relating $\left( x _ { n + 1 } , y _ { n + 1 } \right)$ to $\left( x _ { n } , y _ { n } \right)$.
What is $\left( x _ { 8 } , y _ { 8 } \right)$ ? What is $\left( x _ { 8 k } , y _ { 8 k } \right)$ ? [Figure] [Figure]

In this question we will write $( x , y )$ for a vector instead of the usual $\binom { x } { y }$ notation. So, for example, $3 ( 2,1 ) = ( 6,3 )$.
This question is about vectors $( x , y )$ where $x$ and $y$ are whole numbers, and whether or not such vectors can be written in the form $a ( 5,0 ) + b ( 0,7 ) + c ( 2,1 )$ where $a$ and $b$ and $c$ are whole numbers each greater than or equal to zero.
We will consider the set $S$ of vectors $( p , q )$ with $0 \leq p \leq 4$ and $0 \leq q \leq 6$, with $p$ and $q$ whole numbers.
Then, by considering the vectors $( x , y ) , ( x , y ) - ( 2,1 ) , ( x , y ) - 2 ( 2,1 ) , \ldots , ( x , y ) - 34 ( 2,1 )$, we will find conditions on $x$ and $y$ that imply that $( x , y )$ can be written in the form $a ( 5,0 ) + b ( 0,7 ) + c ( 2,1 )$ where $a$ and $b$ and $c$ are whole numbers each greater than or equal to zero.
(i) Consider the set $S$ of vectors $( p , q )$ with $0 \leq p \leq 4$ and $0 \leq q \leq 6$, with $p$ and $q$ whole numbers.
(a) How many vectors are in $S$ ?
(b) Explain why for any vector $( x , y )$ with $x$ and $y$ whole numbers, we can find whole numbers $a$ and $b$, and a vector $\mathbf { v }$ in $S$ such that
$$( x , y ) = a ( 5,0 ) + b ( 0,7 ) + \mathbf { v }$$
In the rest of this question, we'll call such a vector $\mathbf { v }$ the residue of ( $x , y$ ), and we will assume that the residue is uniquely defined for each vector $( x , y )$.
(ii) Consider vectors $( x , y ) + k ( 2,1 )$ and $( x , y ) + m ( 2,1 )$ where $k$ and $m$ are whole numbers.
(a) Prove that if these vectors have the same residue (as defined in the previous part of the question), then $( k - m )$ is a multiple of 35 .
(b) Explain why the vectors $( x , y ) , ( x , y ) - ( 2,1 ) , ( x , y ) - 2 ( 2,1 ) , \ldots , ( x , y ) - 34 ( 2,1 )$ all have different residues.
(iii) Hence show that if $x$ is at least 68 and $y$ is at least 34 , with $x$ and $y$ whole numbers, then the vector $( x , y )$ can be written in the form $a ( 5,0 ) + b ( 0,7 ) + c ( 2,1 )$ where $a$ and $b$ and $c$ are whole numbers each greater than or equal to zero.
You do not need to find the values of $a$ and $b$ and $c$.
(iv) A student claims: "if $x$ and $y$ are whole numbers with $x > 0$ and $y > 0$ and $x + y$ at least 102 then $( x , y )$ can be written in the form $a ( 5,0 ) + b ( 0,7 ) + c ( 2,1 )$ where $a$ and $b$ and $c$ are whole numbers each greater than or equal to zero."
Is the student's claim correct? Justify your answer.