Q77. Between the following two statements: Statement I : Let $\vec { a } = \hat { i } + 2 \hat { j } - 3 \hat { k }$ and $\vec { b } = 2 \hat { i } + \hat { j } - \hat { k }$. Then the vector $\vec { r }$ satisfying $\vec { a } \times \vec { r } = \vec { a } \times \vec { b }$ and $\vec { a } \cdot \vec { r } = 0$ is of magnitude $\sqrt { 10 }$. Statement II : In a triangle $A B C , \cos 2 A + \cos 2 B + \cos 2 C \geq - \frac { 3 } { 2 }$.
(1) Statement I is incorrect but Statement II is correct.
(2) Both Statement I and Statement II are correct.
(3) Statement I is correct but Statement II is incorrect.
(4) Both Statement I and Statement II are incorrect.

Q78. Let $\vec { a } = 2 \hat { i } + \alpha \hat { j } + \hat { k } , \vec { b } = - \hat { i } + \hat { k } , \vec { c } = \beta \hat { j } - \hat { k }$, where $\alpha$ and $\beta$ are integers and $\alpha \beta = - 6$. Let the values of the ordered pair ( $\alpha , \beta$ ), for which the area of the parallelogram of diagonals $\vec { a } + \vec { b }$ and $\vec { b } + \vec { c }$ is $\frac { \sqrt { 21 } } { 2 }$, be ( $\alpha _ { 1 } , \beta _ { 1 }$ ) and $\left( \alpha _ { 2 } , \beta _ { 2 } \right)$. Then $\alpha _ { 1 } ^ { 2 } + \beta _ { 1 } ^ { 2 } - \alpha _ { 2 } \beta _ { 2 }$ is equal to
(1) 19
(2) 17
(3) 24
(4) 21

If three vectors are given as shown. If angle between vector $\vec{p}$ and $\vec{q}$ is $\theta$ where $\cos\theta = \frac{1}{\sqrt{3}}$ and $|\vec{p}| = 2\sqrt{3}$, $|\vec{q}| = 2$. Then the value of $|\vec{p} \times (\vec{q} - 3\vec{r})|^2 - 3|\vec{r}|^2$ is

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) $\frac { 1 } { 2 } \left( - x ^ { 2 } + x - 1 \right)$

Consider the vectors $\vec { u } = ( - 1,2,3 ) , \quad \vec { v } = ( 2,0 , - 1 )$ and the point $\mathrm { A } ( - 4,4,7 )$. It is requested:
a) (1 point) Determine the vector $\vec { w } _ { 1 }$ that is orthogonal to $\vec { u }$ and $\vec { v }$, unitary, and with negative third coordinate.
b) (0.75 points) Find the non-zero vector $\overrightarrow { w _ { 2 } }$ that is a linear combination of $\vec { u }$ and $\vec { v }$ and orthogonal to $\vec { v }$.
c) (0.75 points) Determine the vertices of the parallelogram whose sides have the directions of vectors $\vec { u }$ and $\vec { v }$ and one of its diagonals is the segment $\overrightarrow { O A }$.

It is known that $P$, $Q$, $R$ are three non-collinear points on the plane $2x - 3y + 5z = \sqrt{7}$ in coordinate space. Let $\overrightarrow{PQ} = (a_{1}, b_{1}, c_{1})$, $\overrightarrow{PR} = (a_{2}, b_{2}, c_{2})$. Select the option in which the absolute value of the determinant is the largest.
(1) $\left|\begin{array}{ccc} -1 & 1 & 1 \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|$
(2) $\left|\begin{array}{ccc} 1 & -1 & 1 \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|$
(3) $\left|\begin{array}{ccc} 1 & 1 & -1 \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|$
(4) $\left|\begin{array}{ccc} -1 & -1 & 1 \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|$
(5) $\left|\begin{array}{ccc} -1 & -1 & -1 \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|$

In coordinate space, there are three mutually perpendicular vectors $\vec { u } , \vec { v } , \vec { w }$. Given that $\vec { u } - \vec { v } = ( 2 , - 1,0 )$ and $\vec { v } - \vec { w } = ( - 1,2,3 )$. What is the volume of the parallelepiped spanned by $\vec { u } , \vec { v } , \vec { w }$?
(1) $2 \sqrt { 5 }$
(2) $5 \sqrt { 2 }$
(3) $2 \sqrt { 10 }$
(4) $4 \sqrt { 5 }$
(5) $4 \sqrt { 10 }$