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 $\_\_\_\_$

Let $\overrightarrow { \mathrm { OA } } = \overrightarrow { \mathrm { a } } , \overrightarrow { \mathrm { OB } } = 12 \overrightarrow { \mathrm { a } } + 4 \overrightarrow { \mathrm {~b} }$ and $\overrightarrow { \mathrm { OC } } = \overrightarrow { \mathrm { b } }$, where O is the origin. If $S$ is the parallelogram with adjacent sides OA and OC, then $\frac { \text { area of the quadrilateral } \mathrm { OABC } } { \text { area of } \mathrm { S } }$ is equal to
(1) 6
(2) 10
(3) 7
(4) 8

Let $\vec { a } , \vec { b }$ and $\vec { c }$ be three non-zero vectors such that $\vec { b }$ and $\vec { c }$ are non-collinear if $\vec { a } + 5 \vec { b }$ is collinear with $\overrightarrow { c , b } + 6 \overrightarrow { c c }$ is collinear with $\vec { a }$ and $\vec { a } + \alpha \vec { b } + \beta \vec { c } = \overrightarrow { 0 }$, then $\alpha + \beta$ is equal to
(1) 35
(2) 30
(3) - 30
(4) - 25

Let $\vec{a} = \hat{i} + \alpha\hat{j} + \beta\hat{k}$, $\alpha, \beta \in R$. Let a vector $\vec{b}$ be such that the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{4}$ and $|\vec{b}|^2 = 6$. If $\vec{a} \cdot \vec{b} = 3\sqrt{2}$, then the value of $(\alpha^2 + \beta^2)|\vec{a} \times \vec{b}|^2$ is equal to
(1) 90
(2) 75
(3) 95
(4) 85

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

The set of all $\alpha$, for which the vectors $\vec { a } = \alpha t \hat { i } + 6 \hat { j } - 3 \hat { k }$ and $\vec { b } = t \hat { i } - 2 \hat { j } - 2 \alpha t \hat { k }$ are inclined at an obtuse angle for all $t \in \mathbb { R }$, is
(1) $\left( - \frac { 4 } { 3 } , 1 \right)$
(2) $[ 0,1 )$
(3) $\left( - \frac { 4 } { 3 } , 0 \right]$
(4) $( - 2,0 ]$

Let $\overrightarrow { \mathrm { a } } = 2 \hat { i } + \hat { j } - \hat { k } , \overrightarrow { \mathrm {~b} } = ( ( \overrightarrow { \mathrm { a } } \times ( \hat { i } + \hat { j } ) ) \times \hat { i } ) \times \hat { i }$. Then the square of the projection of $\overrightarrow { \mathrm { a } }$ on $\overrightarrow { \mathrm { b } }$ is:
(1) $\frac { 1 } { 3 }$
(2) $\frac { 2 } { 3 }$
(3) 2
(4) $\frac { 1 } { 5 }$

Let $O$ be the origin and the position vector of $A$ and $B$ be $2 \hat { i } + 2 \hat { j } + \widehat { k }$ and $2 \hat { i } + 4 \hat { j } + 4 \widehat { k }$ respectively. If the internal bisector of $\angle A O B$ meets the line $A B$ at $C$, then the length of $O C$ is
(1) $\frac { 2 } { 3 } \sqrt { 31 }$
(2) $\frac { 2 } { 3 } \sqrt { 34 }$
(3) $\frac { 3 } { 4 } \sqrt { 34 }$
(4) $\frac { 3 } { 2 } \sqrt { 31 }$

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{b}| = 1$ and $|\vec{b} \times \vec{a}| = 2$. Then $|(\vec{b} \times \vec{a}) - \vec{b}|^2$ is equal to
(1) 3
(2) 5
(3) 1
(4) 4

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 $\overrightarrow { \mathrm { a } } = 6 \hat { i } + \hat { j } - \hat { k }$ and $\overrightarrow { \mathrm { b } } = \hat { i } + \hat { j }$. If $\overrightarrow { \mathrm { c } }$ is a vector such that $| \overrightarrow { \mathrm { c } } | \geq 6 , \overrightarrow { \mathrm { a } } \cdot \overrightarrow { \mathrm { c } } = 6 | \overrightarrow { \mathrm { c } } | , | \overrightarrow { \mathrm { c } } - \overrightarrow { \mathrm { a } } | = 2 \sqrt { 2 }$ and the angle between $\vec { a } \times \vec { b }$ and $\vec { c }$ is $60 ^ { \circ }$, then $| ( \vec { a } \times \vec { b } ) \times \vec { c } |$ is equal to:
(1) $\frac { 9 } { 2 } ( 6 - \sqrt { 6 } )$
(2) $\frac { 3 } { 2 } \sqrt { 6 }$
(3) $\frac { 9 } { 2 } ( 6 + \sqrt { 6 } )$
(4) $\frac { 3 } { 2 } \sqrt { 3 }$

For $\lambda > 0$, let $\theta$ be the angle between the vectors $\vec { a } = \hat { i } + \lambda \hat { j } - 3 \hat { k }$ and $\vec { b } = 3 \hat { i } - \hat { j } + 2 \hat { k }$. If the vectors $\vec { a } + \vec { b }$ and $\vec { a } - \vec { b }$ are mutually perpendicular, then the value of $( 14 \cos \theta ) ^ { 2 }$ is equal to
(1) 50
(2) 40
(3) 25
(4) 20

Let $P ( x , y , z )$ be a point in the first octant, whose projection in the $x y$-plane is the point $Q$. Let $O P = \gamma$; the angle between $O Q$ and the positive $x$-axis be $\theta$; and the angle between $O P$ and the positive $z$-axis be $\phi$, where $O$ is the origin. Then the distance of $P$ from the $x$-axis is
(1) $\gamma \sqrt { 1 - \sin ^ { 2 } \phi \cos ^ { 2 } \theta }$
(2) $\gamma \sqrt { 1 - \sin ^ { 2 } \theta \cos ^ { 2 } \phi }$
(3) $\gamma \sqrt { 1 + \cos ^ { 2 } \phi \sin ^ { 2 } \theta }$
(4) $\gamma \sqrt { 1 + \cos ^ { 2 } \theta \sin ^ { 2 } \phi }$

The least positive integral value of $\alpha$, for which the angle between the vectors $\alpha \hat { \mathrm { i } } - 2 \hat { \mathrm { j } } + 2 \widehat { \mathrm { k } }$ and $\alpha \hat { \mathrm { i } } + 2 \alpha \hat { \mathrm { j } } - 2 \widehat { \mathrm { k } }$ is acute, is $\_\_\_\_$.

Let $\vec { a } = 9 \hat { i } - 13 \hat { j } + 25 \hat { k } , \vec { b } = 3 \hat { i } + 7 \hat { j } - 13 \hat { k }$ and $\vec { c } = 17 \hat { i } - 2 \hat { j } + \hat { k }$ be three given vectors. If $\vec { r }$ is a vector such that $\vec { r } \times \vec { a } = ( \vec { b } + \vec { c } ) \times \vec { a }$ and $\vec { r } \cdot ( \vec { b } - \vec { c } ) = 0$, then $\frac { | 593 \vec { r } + 67 \vec { a } | ^ { 2 } } { ( 593 ) ^ { 2 } }$ is equal to $\_\_\_\_$

If the components of $\overrightarrow { \mathrm { a } } = \alpha \hat { i } + \beta \hat { j } + \gamma \hat { k }$ along and perpendicular to $\overrightarrow { \mathrm { b } } = 3 \hat { i } + \hat { j } - \hat { k }$ respectively, are $\frac { 16 } { 11 } ( 3 \hat { i } + \hat { j } - \hat { k } )$ and $\frac { 1 } { 11 } ( - 4 \hat { i } - 5 \hat { j } - 17 \hat { k } )$, then $\alpha ^ { 2 } + \beta ^ { 2 } + \gamma ^ { 2 }$ is equal to :
(1) 26
(2) 18
(3) 23
(4) 16

Let the point A divide the line segment joining the points $P ( - 1 , - 1,2 )$ and $Q ( 5,5,10 )$ internally in the ratio $\mathrm { r } : 1 ( \mathrm { r } > 0 )$. If O is the origin and $( \overrightarrow { \mathrm { OQ } } \cdot \overrightarrow { \mathrm { OA } } ) - \frac { 1 } { 5 } | \overrightarrow { \mathrm { OP } } \times \overrightarrow { \mathrm { OA } } | ^ { 2 } = 10$, then the value of r is :
(1) $\sqrt { 7 }$
(2) 14
(3) 3
(4) 7

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 $\vec { c }$ 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 $\vec { a } = \hat { \mathrm { i } } + \hat { \mathrm { j } } + \hat { \mathrm { k } } , \overrightarrow { \mathrm { b } } = 2 \hat { \mathrm { i } } + 2 \hat { \mathrm { j } } + \hat { \mathrm { k } }$ 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

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 $k = \square \mathbf { P }$.

Let D, E and F be the three points which divide internally the three sides AB, BC and CA, respectively, of a triangle ABC in the ratio of $k:(1-k)$, where $0 < k \leq \frac{1}{2}$.
(1) When $k = \frac{1}{3}$, we are to find how many times greater the area of triangle ABC is than the area of triangle DEF. Since
$$\triangle\mathrm{ADF} = \triangle\mathrm{BED} = \triangle\mathrm{CFE} = \frac{\mathbf{M}}{\mathbf{N}} \triangle\mathrm{ABC},$$
it follows that
$$\triangle\mathrm{ABC} = \mathbf{O}\, \triangle\mathrm{DEF}.$$
(2) The area of the triangle DEF is half of the area of the triangle ABC when
$$k(1-k) = \frac{\mathbf{P}}{\mathbf{Q}},$$
that is, when
$$k = \frac{\mathbf{R} - \sqrt{\mathbf{S}}}{\mathbf{T}}.$$

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 $x > 1$.