Q80. Let P be the point of intersection of the lines $\frac { x - 2 } { 1 } = \frac { y - 4 } { 5 } = \frac { z - 2 } { 1 }$ and $\frac { x - 3 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 3 } { 2 }$. Then, the shortest distance of P from the line $4 x = 2 y = z$ is
(1) $\frac { 5 \sqrt { 14 } } { 7 }$
(2) $\frac { 3 \sqrt { 14 } } { 7 }$
(3) $\frac { \sqrt { 14 } } { 7 }$
(4) $\frac { 6 \sqrt { 14 } } { 7 }$

Q89. Consider a line L passing through the points $\mathrm { P } ( 1,2,1 )$ and $\mathrm { Q } ( 2,1 , - 1 )$. If the mirror image of the point $\mathrm { A } ( 2,2,2 )$ in the line L is $( \alpha , \beta , \gamma )$, then $\alpha + \beta + 6 \gamma$ is equal to $\_\_\_\_$

Q89. Let $\vec { a } = 2 \hat { i } - 3 \hat { j } + 4 \hat { k } , \vec { b } = 3 \hat { i } + 4 \hat { j } - 5 \hat { k }$ and a vector $\vec { c }$ be such that $\vec { a } \times ( \vec { b } + \vec { c } ) + \vec { b } \times \vec { c } = \hat { i } + 8 \hat { j } + 13 \hat { k }$. If $\vec { a } \cdot \vec { c } = 13$, then $( 24 - \vec { b } \cdot \vec { c } )$ is equal to $\_\_\_\_$

Q89. 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 $\_\_\_\_$

Q90. Let $P$ be the point $( 10 , - 2 , - 1 )$ and $Q$ be the foot of the perpendicular drawn from the point $R ( 1,7,6 )$ on the line passing through the points $( 2 , - 5,11 )$ and $( - 6,7 , - 5 )$. Then the length of the line segment $P Q$ is equal to

ANSWER KEYS

\begin{tabular}{|l|l|l|l|l|l|l|l|} \hline 1. (1) & 2. (3) & 3. (4) & 4. (2) & 5. (1) & 6. (2) & 7. (3) & 8. (4) \hline 9. (2) & 10. (3) & 11. (2) & 12. (4) & 13. (1) & 14. (3) & 15. (1) & 16. (3) \hline 17. (3) & 18. (1) & 19. (1) & 20. (3) & 21. (4) & 22. (13) & 23. (1) & 24. (12) \hline 25. (2) & 26. (16) & 27. (5) & 28. (250) & 29. (60) & 30. (16) & 31. (4) & 32. (2) \hline 33. (4) & 34. (2) & 35. (3) & 36. (2) & 37. (2) & 38. (2) & 39. (2) & 40. (1) \hline 41. (4) & 42. (3) & 43. (3) & 44. (2) & 45. (4) & 46. (1) & 47. (1) & 48. (4) \hline 49. (1) & 50. (2) & 51. (661) & 52. (5) & 53. (274) & 54. (2) & 55. (76) & 56. (3) \hline 57. (877) & 58. (6) & 59. (8) & 60. (20) & 61. (2) & 62. (4) & 63. (3) & 64. (2) \hline 65. (4) & 66. (1) & 67. (4) & 68. (4) & 69. (3) & 70. (4) & 71. (2) & 72. (4) \hline 73. (1) & 74. (3) & 75. (1) & 76. (2) & 77. (2) & 78. (1) & 79. (2) & 80. (1) \hline

Q90. The square of the distance of the image of the point $( 6,1,5 )$ in the line $\frac { x - 1 } { 3 } = \frac { y } { 2 } = \frac { z - 2 } { 4 }$, from the origin is

ANSWER KEYS

\begin{tabular}{|l|l|l|l|} \hline 1. (1) & 2. (2) & 3. (1) & 4. (4) \hline 9. (1) & 10. (1) & 11. (1) & 12. (2) \hline 17. (4) & 18. (1) & 19. (2) & 20. (2) \hline 25. (22) & 26. (16) & 27. (2500) & 28. (28) \hline 33. (1) & 34. (2) & 35. (2) & 36. (4) \hline 41. (4) & 42. (1) & 43. (1) & 44. (3) \hline

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 $\cos\angle\mathrm{AOP} = \frac{\square\mathbf{P}\sqrt{\mathbf{Q}}}{\square\mathbf{R}}$.

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) $s = 1 , \quad t = \frac { 2 } { 3 }$

(Course 2) Q1 For $\mathbf { A } , \mathbf { B } , \mathbf { D } , \mathbf { E }$ and $\mathbf { G }$ in the following sentences, choose the correct answer from among choices (0) $\sim$ (9) below, and for the other $\square$, enter the correct number.
Given a sphere of radius 2 with the center at point O, we have a tetrahedron ABCD whose four vertices are on the sphere. Let $\mathrm { AB } = \mathrm { BC } = \mathrm { CA } = 2$ and side BD be a diameter of the sphere.
Set $\overrightarrow { \mathrm { OA } } = \vec { a } , \overrightarrow { \mathrm { OB } } = \vec { b }$ and $\overrightarrow { \mathrm { OC } } = \vec { c }$.
(1) Let M and N denote the midpoints of segments DA and BC, respectively. Then we have
$$\overrightarrow { \mathrm { DA } } = \mathbf { A } , \quad \overrightarrow { \mathrm { MN } } = \frac { \mathbf { B } } { \mathbf { C } } + \mathbf { D }$$
(2) When the midpoint of segment MN is denoted by P and the center of gravity of triangle BCD is denoted by G, we see that
$$\overrightarrow { \mathrm { OP } } = \frac { \mathbf { E } } { \mathbf { F } } , \quad \overrightarrow { \mathrm { OG } } = \frac { \mathbf { G } } { \mathbf { F } } , \quad | \overrightarrow { \mathrm { PG } } | = \frac { \sqrt { \mathbf { H } } } { \mathbf { I } }$$
Also, since $\overrightarrow { \mathrm { AG } } = \frac { \mathbf { K } } { \mathbf { L } } \overrightarrow { \mathrm { AP } }$, we see that the three points A, P and G are on a straight line. (0) $\vec { a }$
(1) $\vec { b }$
(2) $\vec { c }$
(3) $\vec { a } - \vec { b }$
(4) $\vec { b } - \vec { c }$
(5) $\vec { c } - \vec { a }$ (6) $\vec { a } + \vec { b }$ (7) $\vec { b } + \vec { c }$ (8) $\vec { c } + \vec { a }$ (9) $\vec { a } + \vec { b } + \vec { c }$

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

Given the lines $r _ { 1 } \equiv \left\{ \begin{array} { l } 6 x - y - z = 1 , \\ 2 x - y + z = 1 \end{array} \quad \right.$ and $r _ { 2 } \equiv \left\{ \begin{array} { l } 3 x - 5 y - 2 z = 3 , \\ 3 x + y + 4 z = 3 \end{array} \right.$ it is requested:\ a) (1 point) Study the relative position of $r _ { 1 }$ and $r _ { 2 }$.\ b) (1 point) Calculate the distance between the two lines.\ c) (1 point) Find the equation of the plane that contains $r _ { 1 }$ and the point $\mathrm { P } ( 1,2,3 )$.

Given the points $P ( 1 , - 2,1 ) , Q ( - 4,0,1 ) , R ( - 3,1,2 ) , S ( 0 , - 3,0 )$, it is requested:
a) (1 point) Find the equation of the plane containing $\mathrm { P } , \mathrm { Q }$ and R.
b) (1 point) Study the relative position of line $r$, which passes through points P and Q, and line $s$, which passes through R and S.
c) (1 point) Find the area of the triangle formed by points $\mathrm { P } , \mathrm { Q }$ and R.

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 $\overrightarrow { O A }$.

Given the planes $\pi _ { 1 } \equiv 4 x + 6 y - 12 z + 1 = 0 , \pi _ { 2 } \equiv - 2 x - 3 y + 6 z - 5 = 0$, it is requested:
a) (1 point) Calculate the volume of a cube that has two of its faces in these planes.
b) (1.5 points) For the square with consecutive vertices ABCD , with $\mathrm { A } ( 2,1,3 )$ and $\mathrm { B } ( 1,2,3 )$, calculate the vertices C and D , knowing that C belongs to the planes $\pi _ { 2 } \mathrm { and } \pi _ { 3 } \equiv x - y + z = 2$.

Given the points $\mathrm { A } ( 1,1,1 ) , \mathrm { B } ( 1,3 , - 3 )$ and $\mathrm { C } ( - 3 , - 1,1 )$, it is requested:
a) (1 point) Determine the equation of the plane containing the three points.
b) ( 0.5 points) Obtain a point D (different from $\mathrm { A } , \mathrm { B }$ and C ) such that the vectors $\overrightarrow { A B } , \overrightarrow { A C } , \overrightarrow { A D }$ are linearly dependent.
c) (1 point) Find a point P on the OX axis, such that the volume of the tetrahedron with vertices A, B, C and P equals 1.

Given the point $P(3,3,0)$ and the line $r \equiv \frac{x-2}{-1} = \frac{y}{1} = \frac{z+1}{0}$, find:\ a) (0.75 points) Write the equation of the plane that contains point $P$ and line $r$.\ b) (1 point) Calculate the point symmetric to $P$ with respect to $r$.\ c) (0.75 points) Find two points $A$ and $B$ on $r$ such that triangle $ABP$ is right-angled, has area $\frac{3}{\sqrt{2}}$ and the right angle is at $A$.

For the parallelogram $ABCD$, the consecutive vertices $A(1,0,-1)$, $B(2,1,0)$ and $C(4,3,-2)$ are known. Find:\ a) (1 point) Calculate an equation of the line that passes through the midpoint of segment $AC$ and is perpendicular to segments $AC$ and $BC$.\ b) (1 point) Find the coordinates of vertex $D$ and the area of the resulting parallelogram.\ c) (0.5 points) Calculate the cosine of the angle formed by vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$.

Given the lines $r \equiv \left\{ \begin{array} { l } x - y = 2 \\ 3 x - z = - 1 \end{array} \right. , s \equiv \left\{ \begin{array} { l } x = - 1 + 2 \lambda \\ y = - 4 - \lambda \\ z = \lambda \end{array} \right.$ it is requested:\ a) (1 point) Calculate the relative position of lines $r$ and $s$.\ b) (0.5 points) Find the equation of the plane perpendicular to line $r$ and passing through point $P ( 2 , - 1,5 )$.\ c) (1 point) Find the equation of the plane parallel to line $r$ that contains line s.

With a laser device located at point $\mathrm { P } ( 1,1,1 )$ it has been possible to follow the trajectory of a particle that moves along the line with equations $r \equiv \left\{ \begin{array} { l } 2 x - y = 10 \\ x - z = - 90 \end{array} \right.$. a) ( 0.5 points) Calculate a direction vector of $r$ and the position of the particle when its trajectory intersects the plane $z = 0$. b) (1.25 points) Calculate the closest position of the particle to the laser device. c) ( 0.75 points) Determine the angle between the plane with equation $x + y = 2$ and the line $r$.

Let the plane $\pi \equiv x + y + z = 1$, the line $r _ { 1 } \equiv \left\{ \begin{array} { l } x = 1 + \lambda \\ y = 1 - \lambda \\ z = - 1 \end{array} , \lambda \in \mathbb { R } \right.$ and the point $P ( 0,1,0 )$. a) ( 0.5 points) Verify that the line $r _ { 1 }$ is contained in the plane $\pi$ and that the point P belongs to the same plane. b) ( 0.75 points) Find an equation of the line contained in the plane $\pi$ that passes through P and is perpendicular to $r _ { 1 }$. c) (1.25 points) Calculate an equation of the line, $r _ { 2 }$, that passes through P and is parallel to $r _ { 1 }$. Find the area of a square that has two of its sides on the lines $r _ { 1 }$ and $r _ { 2 }$.

Let the plane $\pi \equiv z = x$ and the points $\mathrm { A } ( 0 , - 1,0 )$ and $\mathrm { B } ( 0,1,0 )$ belonging to the plane $\pi$. a) ( 1.25 points) If points A and B are adjacent vertices of a square with vertices $\{ \mathrm { A } , \mathrm { B } , \mathrm { C } , \mathrm { D } \}$ that lies in the plane $\pi$, find the possible points C and D. b) ( 1.25 points) If points A and B are opposite vertices of a square that lies in the plane $\pi$, determine the other two vertices of it.

Let the lines $r \equiv \left\{ \begin{array} { l } x + y + 2 = 0 \\ y - 2 z + 1 = 0 \end{array} \right.$ and $s \equiv \left\{ \begin{array} { l } x = 2 - 2 t \\ y = 5 + 2 t \\ z = t \end{array} , t \in \mathbb { R } \right.$. a) (1.5 points) Study the relative position of the given lines and calculate the distance between them. b) ( 0.5 points) Determine an equation of the plane $\pi$ that contains the lines $r$ and $s$. c) (0.5 points) Let P and Q be the points on the lines $r$ and $s$, respectively, that are contained in the plane with equation $z = 0$. Calculate an equation of the line passing through points $P$ and $Q$.

Given the points $\mathrm { A } ( 1 , - 2,3 ) , \mathrm { B } ( 0,2 , - 1 )$ and $\mathrm { C } ( 2,1,0 )$. Find:\ a) (1.25 points) Verify that they form a triangle $T$ and find an equation of the plane containing them.\ b) ( 0.75 points) Calculate the intersection of the line passing through points A and B with the plane $z = 1$.\ c) ( 0.5 points) Determine the perimeter of triangle T.