Let $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ and $\vec{c} = \hat{i} + \hat{j} - \hat{k}$. A vector in the plane of $\vec{a}$ and $\vec{b}$ whose projection on $\vec{c}$ is $\frac{1}{\sqrt{3}}$ is
(1) $4\hat{i} - \hat{j} + 4\hat{k}$
(2) $3\hat{i} + \hat{j} - 3\hat{k}$
(3) $2\hat{i} + \hat{j} - 2\hat{k}$
(4) $4\hat{i} + \hat{j} - 4\hat{k}$

If the four points, whose position vectors are $3 \hat { i } - 4 \hat { j } + 2 \widehat { k } , \hat { i } + 2 \hat { j } - \widehat { k } , - 2 \hat { i } - \hat { j } + 3 \widehat { k }$ and $5 \hat { i } - 2 \alpha \hat { j } + 4 \widehat { k }$ are coplanar, then $\alpha$ is equal to
(1) $\frac { 73 } { 17 }$
(2) $- \frac { 107 } { 17 }$
(3) $- \frac { 73 } { 17 }$
(4) $\frac { 107 } { 17 }$

Let $\vec { a } = - \hat { i } - \hat { j } + \hat { k } , \vec { a } \cdot \vec { b } = 1$ and $\vec { a } \times \vec { b } = \hat { i } - \hat { j }$. Then $\vec { a } - 6 \vec { b }$ is equal to
(1) $3 ( \hat { i } - \hat { j } - \widehat { k } )$
(2) $3 ( \hat { i } + \hat { j } + \hat { k } )$
(3) $3 ( \hat { i } - \hat { j } + \widehat { k } )$
(4) $3 ( \hat { i } + \hat { j } - \widehat { k } )$

If the vectors $\vec { a } = \lambda \hat { i } + \mu \hat { j } + 4 \widehat { k } , \vec { b } = - 2 \hat { i } + 4 \hat { j } - 2 \widehat { k }$ and $\vec { c } = 2 \hat { i } + 3 \hat { j } + \widehat { k }$ are coplanar and the projection of $\vec { a }$ on the vector $\vec { b }$ is $\sqrt { 54 }$ units, then the sum of all possible values of $\lambda + \mu$ is equal to
(1) 0
(2) 6
(3) 24
(4) 18

Let $\vec { a } , \vec { b }$ and $\vec { c }$ be three non-zero non-coplanar vectors. Let the position vectors of four points $A , B , C$ and $D$ be $\overrightarrow { \mathrm { a } } - \overrightarrow { \mathrm { b } } + \overrightarrow { \mathrm { c } } , \lambda \overrightarrow { \mathrm { a } } - 3 \overrightarrow { \mathrm {~b} } + 4 \overrightarrow { \mathrm { c } } , - \vec { a } + 2 \vec { b } - 3 \vec { c }$ and $2 \vec { a } - 4 \vec { b } + 6 \vec { c }$ respectively. If $\overrightarrow { A B } , \overrightarrow { A C }$ and $\overrightarrow { A D }$ are coplanar, then $\lambda$ is :

Let $\lambda \in \mathbb{R}$, $\vec{a} = \lambda\hat{i} + 2\hat{j} - 3\hat{k}$, $\vec{b} = \hat{i} - \lambda\hat{j} + 2\hat{k}$. If $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$, then $|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|^{2}$ is equal to
(1) 140
(2) 132
(3) 144
(4) 136

Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}| = 1$, $|\vec{b}| = 4$ and $\vec{a} \cdot \vec{b} = 2$. If $\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$, then the value of $\vec{b} \cdot \vec{c}$ is
(1) $-24$
(2) $-48$
(3) $-84$
(4) $-60$

Let $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$, and $\vec{b}$ and $\vec{c}$ be two nonzero vectors such that $|\vec{a} + \vec{b} + \vec{c}| = |\vec{a} + \vec{b} - \vec{c}|$ and $\vec{b} \cdot \vec{c} = 0$. Consider the following two statements: $A$: $|\vec{a} + \lambda\vec{c}| \geq |\vec{a}|$ for all $\lambda \in \mathbb{R}$. $B$: $\vec{a}$ and $\vec{c}$ are always parallel.
(1) only (B) is correct
(2) neither (A) nor (B) is correct
(3) only (A) is correct
(4) both (A) and (B) are correct.

Let $\vec{a}$ and $\vec{b}$ be two vectors such that $|\vec{a}| = \sqrt{14}$, $|\vec{b}| = \sqrt{6}$ and $|\vec{a} \times \vec{b}| = \sqrt{48}$. Then $(\vec{a} \cdot \vec{b})^2$ is equal to $\underline{\hspace{1cm}}$.

Let $\vec { a } = 6 \hat { i } + 9 \hat { j } + 12 \hat { k } , \vec { b } = \alpha \hat { i } + 11 \hat { j } - 2 \hat { k }$ and $\vec { c }$ be vectors such that $\vec { a } \times \vec { c } = \vec { a } \times \vec { b }$. If $\vec { a } \cdot \vec { c } = - 12$, and $\vec { c } \cdot ( \hat { i } - 2 \hat { j } + \hat { k } ) = 5$ then $\vec { c } \cdot ( \hat { i } + \hat { j } + \hat { k } )$ is equal to $\_\_\_\_$

Let a unit vector $\widehat { u } = x \hat { i } + y \hat { j } + z \widehat { k }$ make angles $\frac { \pi } { 2 } , \frac { \pi } { 3 }$ and $\frac { 2 \pi } { 3 }$ with the vectors $\frac { 1 } { \sqrt { 2 } } \hat { i } + \frac { 1 } { \sqrt { 2 } } \widehat { k } , \frac { 1 } { \sqrt { 2 } } \hat { j } + \frac { 1 } { \sqrt { 2 } } \widehat { k }$ and $\frac { 1 } { \sqrt { 2 } } \hat { i } + \frac { 1 } { \sqrt { 2 } } \hat { j }$ respectively. If $\vec { v } = \frac { 1 } { \sqrt { 2 } } \hat { i } + \frac { 1 } { \sqrt { 2 } } \hat { j } + \frac { 1 } { \sqrt { 2 } } \hat { k }$, then $| \hat { u } - \vec { v } | ^ { 2 }$ is equal to
(1) $\frac { 11 } { 2 }$
(2) $\frac { 5 } { 2 }$
(3) 9
(4) 7

Let $\mathrm { P } ( 3,2,3 ) , \mathrm { Q } ( 4,6,2 )$ and $\mathrm { R } ( 7,3,2 )$ be the vertices of $\triangle \mathrm { PQR }$. Then, the angle $\angle \mathrm { QPR }$ is
(1) $\frac { \pi } { 6 }$
(2) $\cos ^ { - 1 } \left( \frac { 7 } { 18 } \right)$
(3) $\cos ^ { - 1 } \left( \frac { 1 } { 18 } \right)$
(4) $\frac { \pi } { 3 }$

Let $\overrightarrow { \mathrm { a } } = \hat { i } + 2 \hat { j } + 3 \hat { k } , \overrightarrow { \mathrm {~b} } = 2 \hat { i } + 3 \hat { j } - 5 \hat { k }$ and $\overrightarrow { \mathrm { c } } = 3 \hat { i } - \hat { j } + \lambda \hat { k }$ be three vectors. Let $\overrightarrow { \mathrm { r } }$ be a unit vector along $\vec { b } + \vec { c }$. If $\vec { r } \cdot \vec { a } = 3$, then $3 \lambda$ is equal to: (1) 21 (2) 30 (3) 25 (4) 27

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

Let $\vec{\mathrm{a}} = 3\hat{i} - \hat{j} + 2\hat{k}$, $\vec{\mathrm{b}} = \vec{\mathrm{a}} \times (\hat{i} - 2\hat{k})$ and $\vec{\mathrm{c}} = \vec{\mathrm{b}} \times \hat{k}$. Then the projection of $\vec{\mathrm{c}} - 2\hat{j}$ on $\vec{a}$ is:
(1) $2\sqrt{14}$
(2) $\sqrt{14}$
(3) $3\sqrt{7}$
(4) $2\sqrt{7}$

Let $\hat { a }$ be a unit vector perpendicular to the vector $\overrightarrow { \mathrm { b } } = \hat { i } - 2 \hat { j } + 3 \hat { k }$ and $\overrightarrow { \mathrm { c } } = 2 \hat { i } + 3 \hat { j } - \hat { k }$, and makes an angle of $\cos ^ { - 1 } \left( - \frac { 1 } { 3 } \right)$ with the vector $\hat { i } + \hat { j } + \hat { k }$. If $\hat { a }$ makes an angle of $\frac { \pi } { 3 }$ with the vector $\hat { i } + \alpha \hat { j } + \hat { k }$, then the value of $\alpha$ is :
(1) $\sqrt { 6 }$
(2) $- \sqrt { 6 }$
(3) $- \sqrt { 3 }$
(4) $\sqrt { 3 }$

Given a sphere $S$ whose center is at O and whose radius is 1 , take three points A , B and C on $S$ such that
$$\overrightarrow { \mathrm { OA } } \cdot \overrightarrow { \mathrm { OB } } = \overrightarrow { \mathrm { OB } } \cdot \overrightarrow { \mathrm { OC } } = \overrightarrow { \mathrm { OC } } \cdot \overrightarrow { \mathrm { OA } } = 0 .$$
Note that $\overrightarrow { \mathrm { OA } } \cdot \overrightarrow { \mathrm { OB } }$, etc., refers to the inner product of the two vectors.
(1) It follows that $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { AC } } = \square \mathbf { A } , | \overrightarrow { \mathrm { AB } } | = \sqrt { \mathbf { B } } , \cos \angle \mathrm { BAC } = \frac { \square \mathbf { C } } { \square }$ and the area of the triangle ABC is $\frac { \sqrt { \mathbf { E } } } { \mathbf { F } }$.
(2) Let G be the center of gravity of triangle ABC and P be the intersection point of the ray (half line) OG and sphere $S$.
Since $\overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { G } } ( \overrightarrow { \mathrm { OA } } + \overrightarrow { \mathrm { OB } } + \overrightarrow { \mathrm { OC } } )$, we have
$$\begin{gathered} | \overrightarrow { \mathrm { OG } } | = \frac { \sqrt { \mathbf { I } } } { \sqrt { \mathbf { J } } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \square \mathbf { K } - \sqrt { \square \mathbf { L } } } { \square \mathbf { M } } \\ \overrightarrow { \mathrm { AG } } \cdot \overrightarrow { \mathrm { PG } } = \square \mathbf { N } . \end{gathered}$$
Hence the volume of the tetrahedron PABC is $\frac { \sqrt { \mathbf { Q } } - \mathbf { P } } { \mathbf { Q } }$.

In a regular tetrahedron OABC, each side of which has the length 1, let L denote the point which divides segment OA internally in the ratio $3:1$, let M denote the midpoint of side BC, and let P denote the point which divides segment LM internally in the ratio $t:(1-t)$, where $0 < t < 1$.
(1) When we set $\overrightarrow{\mathrm{OA}} = \vec{a}$, $\overrightarrow{\mathrm{OB}} = \vec{b}$, $\overrightarrow{\mathrm{OC}} = \vec{c}$ and express $\overrightarrow{\mathrm{OP}}$ in terms of $\vec{a}, \vec{b}, \vec{c}$, we have
$$\overrightarrow{\mathrm{OP}} = \frac{\mathbf{A}}{\mathbf{B}}(\mathbf{C} - t)\vec{a} + \frac{\mathbf{D}}{\mathbf{E}}t(\vec{b} + \vec{c}).$$
Since $\vec{a} \cdot (\vec{b} + \vec{c}) = \mathbf{F}$ and $|\vec{b} + \vec{c}|^2 = \mathbf{G}$, we have
$$|\overrightarrow{\mathrm{OP}}| = \frac{1}{\mathbf{H}}\sqrt{\mathbf{I}t^2 - \mathbf{JJ}t + \mathbf{K}},$$
where $\vec{a} \cdot (\vec{b} + \vec{c})$ means the inner product of the vectors $\vec{a}$ and $(\vec{b} + \vec{c})$.
(2) The value of $t$ at which $|\overrightarrow{\mathrm{OP}}|$ is minimized is
$$t = \frac{\mathbf{L}}{\mathbf{L}}$$
and the minimum value of $|\overrightarrow{\mathrm{OP}}|$ is $\frac{\sqrt{\mathbf{N}}}{\mathbf{O}}$.
(3) When $|\overrightarrow{\mathrm{OP}}|$ is minimized as in (2), we have $\cos\angle\mathrm{AOP} = \frac{\square\mathbf{P}\sqrt{\mathbf{Q}}}{\square\mathbf{R}}$.

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,
$$k = \frac { \mathbf { J } } { \mathbf { K } } .$$

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 $\frac { \mathbf { J } \sqrt { \mathbf { K } } } { \mathbf { L } }$.

For $\mathbf { C } , \mathbf { D } , \mathbf { E } , \mathbf { F } , \mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below. For the other $\square$, enter the correct number.
Consider a regular tetrahedron OABC with sides of length 1. Let $x$ be a number satisfying $0 < x < 1$, and let P be the point that divides side AB by the ratio $x : ( 1 - x )$ and Q be the point that divides side BC by the ratio $x : ( 1 - x )$. Also, let $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$. We are to find the range of values of $\cos \angle \mathrm { POQ }$.
The vectors $\vec { a } , \vec { b }$ and $\vec { c }$ satisfy
$$\vec { a } \cdot \vec { b } = \vec { b } \cdot \vec { c } = \vec { c } \cdot \vec { a } = \frac { \mathbf { A } } { \mathbf { B } }$$
Next, since we can express $\overrightarrow { \mathrm { OP } }$ and $\overrightarrow { \mathrm { OQ } }$ as $\overrightarrow { \mathrm { OP } } = \square \mathbf { C }$ and $\overrightarrow { \mathrm { OQ } } = \mathbf { D }$, we have
$$| \overrightarrow { \mathrm { OP } } | = | \overrightarrow { \mathrm { OQ } } | = \sqrt { \vec { E } } , \quad \overrightarrow { \mathrm { OP } } \cdot \overrightarrow { \mathrm { OQ } } = \square \mathbf { F } .$$
Hence we obtain
$$\cos \angle \mathrm { POQ } = \frac { 1 } { \mathbf { G } } - \frac { \mathbf { H } } { \mathbf { I } } .$$
From this we finally obtain
$$\frac { \square \mathbf { J } } { \mathbf { K } } < \cos \angle \mathrm { POQ } \leqq \frac { \mathbf { L } } { \mathbf { M } } .$$
(0) $( 1 - x ) \vec { a } + x \vec { b }$
(1) $x \vec { a } + ( 1 - x ) \vec { b }$
(2) $( 1 - x ) \vec { b } + x \vec { c }$
(3) $x \vec { b } + ( 1 - x ) \vec { c }$
(4) $x ^ { 2 } + x + 1$
(5) $x ^ { 2 } - x + 1$ (6) $x ^ { 2 } - x - 1$ (7) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x + 1 \right)$ (8) $\frac { 1 } { 2 } \left( - x ^ { 2 } - x + 1 \right)$ (9) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x - 1 \right)$