Let $\vec { a } , \vec { b } , \vec { c }$ be three non-coplanar vectors such that $\vec { a } \times \vec { b } = \overrightarrow { 4 c } , \vec { b } \times \vec { c } = 9 \vec { a }$ and $\vec { c } \times \vec { a } = \alpha \vec { b } , \alpha > 0$. If $| \vec { a } | + | \vec { b } | + | \vec { c } | = 36$, then $\alpha$ is equal to $\_\_\_\_$ .

Let $\vec { a } , \vec { b }$ and $\vec { c }$ be three non zero vectors such that $\vec { b } \cdot \vec { c } = 0$ and $\vec { a } \times ( \vec { b } \times \vec { c } ) = \frac { \vec { b } - \vec { c } } { 2 }$. If $\vec { d }$ be a vector such that $\overrightarrow { \mathrm { b } } \cdot \overrightarrow { \mathrm { d } } = \overrightarrow { \mathrm { a } } \cdot \overrightarrow { \mathrm { b } }$, then $( \overrightarrow { \mathrm { a } } \times \overrightarrow { \mathrm { b } } ) \cdot ( \overrightarrow { \mathrm { c } } \times \overrightarrow { \mathrm { d } } )$ is equal to
(1) $\frac { 3 } { 4 }$
(2) $\frac { 1 } { 2 }$
(3) $- \frac { 1 } { 4 }$
(4) $\frac { 1 } { 4 }$

The vector $\vec { a } = - \hat { i } + 2 \hat { j } + \hat { k }$ is rotated through a right angle, passing through the $y$-axis in its way and the resulting vector is $\vec { b }$. Then the projection of $3 \vec { a } + \sqrt { 2 } \vec { b }$ on $\vec { c } = 5 \hat { i } + 4 \hat { j } + 3 \hat { k }$ is
(1) $3 \sqrt { 2 }$
(2) 1
(3) $\sqrt { 6 }$
(4) $2 \sqrt { 3 }$

If $\vec { a } = \hat { i } + 2 \widehat { k } , \vec { b } = \hat { i } + \hat { j } + \widehat { k } , \vec { c } = 7 \hat { i } - 3 \hat { j } + 4 \widehat { k } , \vec { r } \times \vec { b } + \vec { b } \times \vec { c } = \overrightarrow { 0 }$ and $\vec { r } \cdot \vec { a } = 0$ then $\vec { r } \cdot \vec { c }$ is equal to: (1) 34 (2) 12 (3) 36 (4) 30

Let $\vec { a } = 4 \hat { i } + 3 \hat { j }$ and $\vec { b } = 3 \hat { i } - 4 \hat { j } + 5 \hat { k }$ and $\overrightarrow { \mathrm { c } }$ is a vector such that $\vec { c } \cdot ( \vec { a } \times \vec { b } ) + 25 = 0 , \vec { c } \cdot ( \hat { i } + \hat { j } + \hat { k } ) = 4$ and projection of $\vec { c }$ on $\overrightarrow { \mathrm { a } }$ is 1 , then the projection of $\vec { c }$ on $\vec { b }$ equals: (1) $\frac { 5 } { \sqrt { 2 } }$ (2) $\frac { 1 } { 5 }$ (3) $\frac { 1 } { \sqrt { 2 } }$ (4) $\frac { 3 } { \sqrt { 2 } }$

Let $O$ be the origin and the position vector of the point $P$ be $-\hat{i} - 2\hat{j} + 3\hat{k}$. If the position vectors of the points $A$, $B$ and $C$ are $-2\hat{i} + \hat{j} - 3\hat{k}$, $2\hat{i} + 4\hat{j} - 2\hat{k}$ and $-4\hat{i} + 2\hat{j} - \hat{k}$ respectively, then the projection of the vector $\overrightarrow{OP}$ on a vector perpendicular to the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ is
(1) 3
(2) $\frac{8}{3}$
(3) $\frac{7}{3}$
(4) $\frac{10}{3}$

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 $\overrightarrow { \mathrm { a } } = 2 \hat { i } + \hat { j } - \hat { k } , \overrightarrow { \mathrm {~b} } = ( ( \overrightarrow { \mathrm { a } } \times ( \hat { i } + \hat { j } ) ) \times \hat { i } ) \times \hat { i }$. 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 $\overrightarrow { \mathrm { a } } = 6 \hat { i } + \hat { j } - \hat { k }$ 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) $\frac { 3 } { 2 } \sqrt { 3 }$

The triangle $ABC$ satisfies
$$\mathrm { AB } = 2 , \quad \mathrm { BC } = 3 , \quad \mathrm { CA } = 4 .$$
(1) When we set $\angle \mathrm { ABC } = \theta$, the inner product $\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } }$ of the vectors $\overrightarrow { \mathrm { AB } }$ and $\overrightarrow { \mathrm { BC } }$ is
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \mathbf { AB } \cos \theta .$$
Finding the value of $\cos \theta$ from the law of cosines, we obtain
$$\overrightarrow { \mathrm { AB } } \cdot \overrightarrow { \mathrm { BC } } = \frac { \mathbf { C } } { \mathbf { D } } .$$
(2) We divide the side BC into $n$ equal parts by the points $\mathrm { P } _ { 1 } , \mathrm { P } _ { 2 } , \cdots , \mathrm { P } _ { n - 1 }$ which are arranged in ascending order of the distance from B, and set $\mathrm { B } = \mathrm { P } _ { 0 } , \mathrm { C } = \mathrm { P } _ { n }$. We are to find the value of $\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } }$.
When we calculate the inner product of $\overrightarrow { \mathrm { AP } _ { k - 1 } }$ and $\overrightarrow { \mathrm { AP } _ { k } }$ using (1), we have
$$\overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \mathbf { E } + \frac { \mathbf { F } } { 2 n } + \mathbf { G }$$
Hence we obtain
$$\lim _ { n \rightarrow \infty } \frac { 1 } { n } \sum _ { k = 1 } ^ { n } \overrightarrow { \mathrm { AP } _ { k - 1 } } \cdot \overrightarrow { \mathrm { AP } _ { k } } = \frac { \mathbf { IJ } } { \mathbf { K } }$$

The parallelepiped satisfies
$$\begin{aligned} & \mathrm { AB } = 2 , \quad \mathrm { AD } = 3 , \quad \mathrm { AE } = 1 \\ & \angle \mathrm { BAD } = 60 ^ { \circ } , \quad \angle \mathrm { BAE } = 90 ^ { \circ } , \quad \angle \mathrm { DAE } = 120 ^ { \circ } \end{aligned}$$
and M is the midpoint of edge GH. Let us take points P and Q on edges BF and DH, respectively, such that the four points A, P, M, and Q are on the same plane. We are to find the points P and Q which maximize the length of the line segment PQ.
(1) Setting $\vec { a } = \overrightarrow { \mathrm { AB } } , \vec { b } = \overrightarrow { \mathrm { AD } }$ and $\vec { c } = \overrightarrow { \mathrm { AE } }$, the inner products of these vectors are
$$\vec { a } \cdot \vec { b } = \mathbf { A } , \quad \vec { b } \cdot \vec { c } = - \frac { \mathbf { B } } { \mathbf{C} } , \quad \vec { c } \cdot \vec { a } = \mathbf { D } .$$
(2) Let $s$ and $t$ satisfy $0 \leqq s \leqq 1,0 \leqq t \leqq 1$, and set $\mathrm { BP } : \mathrm { PF } = s : ( 1 - s )$, $\mathrm { DQ } : \mathrm { QH } = t : ( 1 - t )$. Since the four points A, P, M and Q are on the same plane, there exist two real numbers $\alpha$ and $\beta$ such that
$$\overrightarrow { \mathrm { AM } } = \alpha \overrightarrow { \mathrm { AP } } + \beta \overrightarrow { \mathrm { AQ } } .$$
Hence $s$ and $t$ satisfy
$$s = \mathbf { E } ( \mathbf { F } - t ) .$$
Then $| \overrightarrow { \mathrm { PQ } } |$ may be expressed in terms of $t$ as
$$| \overrightarrow { \mathrm { PQ } } | ^ { 2 } = \mathbf { G } t ^ { 2 } - \mathbf { H I } t + \mathbf { J K } .$$
Hence the length of segment PQ is maximized when $\square$ L. Here, for $\square$ L choose the correct answer from among choices (0) $\sim$ (5) below. (0) $s = 0 , \quad t = 1$
(1) $s = 0 , \quad t = \frac { 1 } { 2 }$
(2) $s = \frac { 1 } { 2 } , t = \frac { 3 } { 4 }$
(3) $s = \frac { 2 } { 3 } , t = \frac { 2 } { 3 }$
(4) $s = 1 , \quad t = \frac { 1 } { 2 }$
(5) $s = 1 , \quad t = \frac { 2 } { 3 }$

On the coordinate plane, two parallel lines $L _ { 1 } , L _ { 2 }$ both have slope 2 and are at distance 5 apart. Point $A ( 2 , - 1 )$ is a point on $L _ { 1 }$ in the fourth quadrant. Point $B$ is a point on $L _ { 2 }$ in the second quadrant with $\overline { A B } = 5$ . Line $L _ { 3 }$ has slope 3, passes through point $A$, and intersects $L _ { 2 }$ at point $C$. Answer the following questions:
(1) Find the slope of line $AB$ . (2 points)
(2) Find the vector $\overrightarrow { A B }$ . (4 points)
(3) Find the dot product $\overrightarrow { A B } \cdot \overrightarrow { A C }$ . (3 points)
(4) Find the vector $\overrightarrow { A C }$ . (4 points)

Consider points $O(0,0), A, B, C, D, E, F, G$ on a coordinate plane, where points $B$, $C$ and $D$, $E$ and $F$, $G$ and $A$ are located in the first, second, third, and fourth quadrants respectively. If $\vec{v}$ is a vector on the coordinate plane satisfying $\vec{v} \cdot \overrightarrow{OA} > 0$ and $\vec{v} \cdot \overrightarrow{OB} > 0$, then the dot product of $\vec{v}$ with which of the following vectors must be negative?
(1) $\overrightarrow{OC}$
(2) $\overrightarrow{OD}$
(3) $\overrightarrow{OE}$
(4) $\overrightarrow{OF}$
(5) $\overrightarrow{OG}$

In coordinate space, consider a unit cube with edge length 1, with one vertex $O$ fixed. From the seven vertices other than $O$, two distinct points are randomly selected, denoted as $P$ and $Q$. What is the expected value of the dot product $\overrightarrow{OP} \cdot \overrightarrow{OQ}$ among the following options?
(1) $\frac{4}{7}$
(2) $\frac{5}{7}$
(3) $\frac{6}{7}$
(4) 1
(5) $\frac{8}{7}$

As shown in the figure, a point $P_{0}$ moves forward 2 units in a certain direction to reach point $P_{1}$, then turns left 15 degrees in the direction of motion; moves forward 2 units in the new direction to reach point $P_{2}$, then turns left 15 degrees again; moves forward 2 units in the new direction to reach point $P_{3}$, and so on.
The dot product of vectors $\overrightarrow{P_{2}P_{3}}$ and $\overrightarrow{P_{5}P_{6}}$ is $\square$. (Express as a simplified radical)

On the coordinate plane, it is known that the orthogonal projection length of vector $\vec{v}$ in the direction of vector $(2, -3)$ is 1 less than its original length, and the orthogonal projection length in the direction of vector $(3, 2)$ is 2 less than its original length. If $\vec{v}$ makes acute angles with both vectors $(2, -3)$ and $(3, 2)$, then the orthogonal projection length of $\vec{v}$ in the direction of vector $(4, 7)$ is

On a plane, there are three non-collinear points $A, B, C$. Given that the dot product of vectors $\overrightarrow { A B }$ and $\overrightarrow { A C }$ is 16, the dot product of $\overrightarrow { C B }$ and $\overrightarrow { A C }$ is 3, then $\overline { A C } = \sqrt{\text{(12--1)}}$ (12–2). (Simplify to simplest radical form)

In the analytic plane, the sides of a regular pentagon are named as vectors $\vec { a } , \vec { b } , \vec { c } , \vec { d }$ and $\vec { e }$ as shown in the figure.
Accordingly, what is the probability that the dot product of two vectors randomly selected from these five vectors is positive?
A) $\frac { 1 } { 2 }$ B) $\frac { 1 } { 5 }$ C) $\frac { 2 } { 5 }$ D) $\frac { 1 } { 10 }$ E) $\frac { 3 } { 10 }$