Exercise 2 — 5 points Theme: geometry in space Space is equipped with an orthonormal coordinate system $(O; \vec{\imath}, \vec{\jmath}, \vec{k})$. We consider:

• $d_1$ the line passing through point $H(2; 3; 0)$ with direction vector $\vec{u}\left(\begin{array}{c}1\\-1\\1\end{array}\right)$;
• $d_2$ the line with parametric representation:

$$\left\{\begin{aligned} x &= 2k - 3\\ y &= k\\ z &= 5 \end{aligned}\quad\text{where }k\text{ describes }\mathbb{R}.\right.$$ The purpose of this exercise is to determine a parametric representation of a line $\Delta$ that is perpendicular to both lines $d_1$ and $d_2$.

1. a. Determine a direction vector $\vec{v}$ of line $d_2$. b. Prove that lines $d_1$ and $d_2$ are not parallel. c. Prove that lines $d_1$ and $d_2$ are not intersecting. d. What is the relative position of lines $d_1$ and $d_2$?
2. a. Verify that the vector $\vec{w}\left(\begin{array}{c}-1\\2\\3\end{array}\right)$ is orthogonal to both $\vec{u}$ and $\vec{v}$. b. We consider the plane $P$ passing through point $H$ and directed by vectors $\vec{u}$ and $\vec{w}$. We admit that a Cartesian equation of this plane is: $$5x + 4y - z - 22 = 0.$$ Prove that the intersection of plane $P$ and line $d_2$ is the point $M(3; 3; 5)$.
3. Let $\Delta$ be the line with direction vector $\vec{w}$ passing through point $M$.
   A parametric representation of $\Delta$ is therefore given by: $$\left\{\begin{array}{l} x = -r + 3\\ y = 2r + 3\\ z = 3r + 5 \end{array}\text{ where }r\text{ describes }\mathbb{R}.\right.$$ a. Justify that lines $\Delta$ and $d_1$ are perpendicular at a point $L$ whose coordinates you will determine. b. Explain why line $\Delta$ is a solution to the problem posed.

Exercise 4

The objective of this exercise is to determine the distance between two non-coplanar lines. By definition, the distance between two non-coplanar lines in space, $( d _ { 1 } )$ and $( d _ { 2 } )$ is the length of the segment $[\mathrm { EF }]$, where E and F are points belonging respectively to $\left( d _ { 1 } \right)$ and to $( d _ { 2 } )$ such that the line (EF) is orthogonal to $( d _ { 1 } )$ and $( d _ { 2 } )$. The space is equipped with an orthonormal coordinate system $( \mathrm { O } ; \vec { \imath } , \vec { \jmath } , \vec { k } )$. Let $\left( d _ { 1 } \right)$ be the line passing through $\mathrm { A } ( 1 ; 2 ; - 1 )$ with direction vector $\overrightarrow { u _ { 1 } } \left( \begin{array} { l } 1 \\ 2 \\ 0 \end{array} \right)$ and $\left( d _ { 2 } \right)$ the line with parametric representation: $\left\{ \begin{array} { l } x = 0 \\ y = 1 + t \\ z = 2 + t \end{array} , t \in \mathbb { R } \right.$.

1. Give a parametric representation of the line $\left( d _ { 1 } \right)$.
2. Prove that the lines $\left( d _ { 1 } \right)$ and $\left( d _ { 2 } \right)$ are non-coplanar.
3. Let $\mathscr { P }$ be the plane passing through A and directed by the non-collinear vectors $\overrightarrow { u _ { 1 } }$ and $\vec { w } \left( \begin{array} { c } 2 \\ - 1 \\ 1 \end{array} \right)$. Justify that a Cartesian equation of the plane $\mathscr { P }$ is: $- 2 x + y + 5 z + 5 = 0$.
4. a. Without seeking to calculate the coordinates of the intersection point, justify that the line $( d _ { 2 } )$ and the plane $\mathscr { P }$ are secant. b. We denote F the intersection point of the line $( d _ { 2 } )$ and the plane $\mathscr { P }$. Verify that the point F has coordinates $\left( 0 ; - \frac { 5 } { 3 } ; - \frac { 2 } { 3 } \right)$. Let $( \delta )$ be the line passing through F with direction vector $\vec { w }$. It is admitted that the lines $( \delta )$ and $( d _ { 1 } )$ are secant at a point E with coordinates $\left( - \frac { 2 } { 3 } ; - \frac { 4 } { 3 } ; - 1 \right)$.
5. a. Justify that the distance EF is the distance between the lines $\left( d _ { 1 } \right)$ and $\left( d _ { 2 } \right)$. b. Calculate the distance between the lines $\left( d _ { 1 } \right)$ and $\left( d _ { 2 } \right)$.

In coordinate space, there are two mutually perpendicular planes $\alpha$ and $\beta$. For two points $\mathrm { A }$ and $\mathrm { B }$ on plane $\alpha$, $\overline { \mathrm { AB } } = 3 \sqrt { 5 }$, and line AB is parallel to plane $\beta$. The distance between point A and plane $\beta$ is 2, and the distance between a point P on plane $\beta$ and plane $\alpha$ is 4. Find the area of triangle PAB. [4 points]

In coordinate space, there are three points $\mathrm { A } , \mathrm { B } , \mathrm { C }$ not on the same line. For a plane $\alpha$ satisfying the following conditions, let $d ( \alpha )$ be the minimum distance among the distances from each point $\mathrm { A } , \mathrm { B } , \mathrm { C }$ to plane $\alpha$.
(a) Plane $\alpha$ intersects segment AC and also intersects segment BC.
(b) Plane $\alpha$ does not intersect segment AB.
Among planes $\alpha$ satisfying the above conditions, let $\beta$ be the plane where $d ( \alpha )$ is maximized. Which of the following statements in are correct? [4 points]
$\text{ㄱ}$. Plane $\beta$ is perpendicular to the plane passing through the three points $\mathrm { A } , \mathrm { B } , \mathrm { C }$. $\text{ㄴ}$. Plane $\beta$ passes through the midpoint of segment AC or the midpoint of segment BC. $\text{ㄷ}$. When the three points are $\mathrm { A } ( 2,3,0 ) , \mathrm { B } ( 0,1,0 ) , \mathrm { C } ( 2 , - 1,0 )$, $d ( \beta )$ equals the distance between point B and plane $\beta$.
(1) ㄱ
(2) ㄷ
(3) ㄱ, ㄴ
(4) ㄴ, ㄷ
(5) ㄱ, ㄴ, ㄷ

In the coordinate plane, for a parallelogram OACB with $\overline { \mathrm { OA } } = \sqrt { 2 } , \overline { \mathrm { OB } } = 2 \sqrt { 2 }$ and $\cos ( \angle \mathrm { AOB } ) = \frac { 1 } { 4 }$, point P satisfies the following conditions. (가) $\overrightarrow { \mathrm { OP } } = s \overrightarrow { \mathrm { OA } } + t \overrightarrow { \mathrm { OB } } ( 0 \leq s \leq 1, 0 \leq t \leq 1 )$ (나) $\overrightarrow { \mathrm { OP } } \cdot \overrightarrow { \mathrm { OB } } + \overrightarrow { \mathrm { BP } } \cdot \overrightarrow { \mathrm { BC } } = 2$
For a point X moving on a circle centered at O and passing through point A, let $M$ and $m$ be the maximum and minimum values of $| 3 \overrightarrow { \mathrm { OP } } - \overrightarrow { \mathrm { OX } } |$ respectively. When $M \times m = a \sqrt { 6 } + b$, find the value of $a ^ { 2 } + b ^ { 2 }$. (Given that $a$ and $b$ are rational numbers.) [4 points]

As shown in the figure, in the triangular pyramid $P - A B C$, $A B = B C = 2 \sqrt { 2 }$, $P A = P B = P C = A C = 4$, and $O$ is the midpoint of $A C$.
(1) Prove: $P O \perp$ plane $A B C$;
(2) If point $M$ is on edge $B C$ such that $M C = 2 M B$, find the distance from point $C$ to plane $P O M$.

Consider the lines
$$\begin{aligned} & L _ { 1 } : \frac { x + 1 } { 3 } = \frac { y + 2 } { 1 } = \frac { z + 1 } { 2 } \\ & L _ { 2 } : \frac { x - 2 } { 1 } = \frac { y + 2 } { 2 } = \frac { z - 3 } { 3 } \end{aligned}$$
The distance of the point $( 1,1,1 )$ from the plane passing through the point $( - 1 , - 2 , - 1 )$ and whose normal is perpendicular to both the lines $L _ { 1 }$ and $L _ { 2 }$ is
(A) $\frac { 2 } { \sqrt { 75 } }$
(B) $\frac { 7 } { \sqrt { 75 } }$
(C) $\frac { 13 } { \sqrt { 75 } }$
(D) $\frac { 23 } { \sqrt { 75 } }$

If the distance between the plane $\mathrm { Ax } - 2 \mathrm { y } + \mathrm { z } = \mathrm { d }$ and the plane containing the lines $\frac { x - 1 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 3 } { 4 }$ and $\frac { x - 2 } { 3 } = \frac { y - 3 } { 4 } = \frac { z - 4 } { 5 }$ is $\sqrt { 6 }$, then $| d |$ is

Match the statements in Column-I with the values in Column-II.
Column I
A) A line from the origin meets the lines $\frac { x - 2 } { 1 } = \frac { y - 1 } { - 2 } = \frac { z + 1 } { 1 }$ and $\frac { x - \frac { 8 } { 3 } } { 2 } = \frac { y + 3 } { - 1 } = \frac { z - 1 } { 1 }$ at $P$ and $Q$ respectively. If length $\mathrm { PQ } = d$, then $d ^ { 2 }$ is
B) The values of $x$ satisfying $\tan ^ { - 1 } ( x + 3 ) - \tan ^ { - 1 } ( x - 3 ) = \sin ^ { - 1 } \left( \frac { 3 } { 5 } \right)$ are
C) Non-zero vectors $\vec { a } , \vec { b }$ and $\vec { c }$ satisfy $\vec { a } \cdot \vec { b } = 0$, $( \overrightarrow { \mathrm { b } } - \overrightarrow { \mathrm { a } } ) \cdot ( \overrightarrow { \mathrm { b } } + \overrightarrow { \mathrm { c } } ) = 0$ and $2 | \overrightarrow { \mathrm {~b} } + \overrightarrow { \mathrm { c } } | = | \overrightarrow { \mathrm { b } } - \overrightarrow { \mathrm { a } } |$. If $\vec { a } = \mu \vec { b } + 4 \vec { c }$, then the possible values of $\mu$ are
D) Let f be the function on $[ - \pi , \pi ]$ given by $f ( 0 ) = 9$ and $f ( x ) = \sin \left( \frac { 9 x } { 2 } \right) / \sin \left( \frac { x } { 2 } \right)$ for $x \neq 0$. The value of $\frac { 2 } { \pi } \int _ { - \pi } ^ { \pi } f ( x ) d x$ is
Column II p) $-4$ q) $0$ r) $4$ s) $-1$ (or as given in paper) t) $6$

A line $l$ passing through the origin is perpendicular to the lines $l _ { 1 } : ( 3 + t ) \hat { i } + ( - 1 + 2 t ) \hat { j } + ( 4 + 2 t ) \hat { k } , - \infty < t < \infty$ $l _ { 2 } : ( 3 + 2 s ) \hat { i } + ( 3 + 2 s ) \hat { j } + ( 2 + s ) \hat { k } , - \infty < s < \infty$ Then, the coordinate(s) of the point(s) on $l _ { 2 }$ at a distance of $\sqrt { 17 }$ from the point of intersection of $l$ and $l _ { 1 }$ is (are)
(A) $\left( \frac { 7 } { 3 } , \frac { 7 } { 3 } , \frac { 5 } { 3 } \right)$
(B) $( - 1 , - 1,0 )$
(C) $( 1,1,1 )$
(D) $\left( \frac { 7 } { 9 } , \frac { 7 } { 9 } , \frac { 8 } { 9 } \right)$

In $\mathbb { R } ^ { 3 }$, consider the planes $P _ { 1 } : y = 0$ and $P _ { 2 } : x + z = 1$. Let $P _ { 3 }$ be a plane, different from $P _ { 1 }$ and $P _ { 2 }$, which passes through the intersection of $P _ { 1 }$ and $P _ { 2 }$. If the distance of the point $( 0,1,0 )$ from $P _ { 3 }$ is 1 and the distance of a point $( \alpha , \beta , \gamma )$ from $P _ { 3 }$ is 2, then which of the following relations is (are) true?
(A) $2 \alpha + \beta + 2 \gamma + 2 = 0$
(B) $2 \alpha - \beta + 2 \gamma + 4 = 0$
(C) $2 \alpha + \beta - 2 \gamma - 10 = 0$
(D) $2 \alpha - \beta + 2 \gamma - 8 = 0$

Let $\ell _ { 1 }$ and $\ell _ { 2 }$ be the lines $\vec { r } _ { 1 } = \lambda ( \hat { i } + \hat { j } + \hat { k } )$ and $\vec { r } _ { 2 } = ( \hat { j } - \hat { k } ) + \mu ( \hat { i } + \hat { k } )$, respectively. Let $X$ be the set of all the planes $H$ that contain the line $\ell _ { 1 }$. For a plane $H$, let $d ( H )$ denote the smallest possible distance between the points of $\ell _ { 2 }$ and $H$. Let $H _ { 0 }$ be a plane in $X$ for which $d \left( H _ { 0 } \right)$ is the maximum value of $d ( H )$ as $H$ varies over all planes in $X$.
Match each entry in List-I to the correct entries in List-II.
List-I
(P) The value of $d \left( H _ { 0 } \right)$ is
(Q) The distance of the point $( 0,1,2 )$ from $H _ { 0 }$ is
(R) The distance of origin from $H _ { 0 }$ is
(S) The distance of origin from the point of intersection of planes $y = z , x = 1$ and $H _ { 0 }$ is
List-II
(1) $\sqrt { 3 }$
(2) $\frac { 1 } { \sqrt { 3 } }$
(3) 0
(4) $\sqrt { 2 }$
(5) $\frac { 1 } { \sqrt { 2 } }$
The correct option is:
(A) $( P ) \rightarrow ( 2 )$ $( Q ) \rightarrow ( 4 )$ $( R ) \rightarrow ( 5 )$ $( S ) \rightarrow ( 1 )$
(B) $( P ) \rightarrow ( 5 )$ $( Q ) \rightarrow ( 4 )$ $( R ) \rightarrow ( 3 )$ $( S ) \rightarrow ( 1 )$
(C) $( P ) \rightarrow ( 2 )$ $( Q ) \rightarrow ( 1 )$ $( R ) \rightarrow ( 3 )$ $( S ) \rightarrow ( 2 )$
(D) $( P ) \rightarrow ( 5 )$ $( Q ) \rightarrow ( 1 )$ $( R ) \rightarrow ( 4 )$ $( S ) \rightarrow ( 2 )$

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is:
(1) $3\sqrt{10}$
(2) $10\sqrt{3}$
(3) $\frac{10}{\sqrt{3}}$
(4) $\frac{20}{3}$

The distance of the point $(1, -5, 9)$ from the plane $x - y + z = 5$ measured along the line $x = y = z$ is: (1) $3\sqrt{10}$ (2) $10\sqrt{3}$ (3) $\frac{10}{\sqrt{3}}$ (4) $\frac{20}{3}$

The distance of the point $(1, 3, -7)$ from the plane passing through the point $(1, -1, -1)$, having normal perpendicular to both the lines $$\frac { x - 1 } { 1 } = \frac { y + 2 } { - 2 } = \frac { z - 4 } { 3 } \quad \text{and} \quad \frac { x - 2 } { 2 } = \frac { y + 1 } { - 1 } = \frac { z + 7 } { - 1 }$$ is:
(1) $\frac { 10 } { \sqrt { 74 } }$
(2) $\frac { 20 } { \sqrt { 74 } }$
(3) $\frac { 5 } { \sqrt { 83 } }$
(4) $\frac { 10 } { \sqrt { 83 } }$

If $L _ { 1 }$ is the line of intersection of the planes $2 x - 2 y + 3 z - 2 = 0 , x - y + z + 1 = 0$ and $L _ { 2 }$ is the line of intersection of the planes $x + 2 y - z - 3 = 0,3 x - y + 2 z - 1 = 0$, then the distance of the origin from the plane, containing the lines $L _ { 1 }$ and $L _ { 2 }$ is
(1) $\frac { 1 } { \sqrt { 2 } }$
(2) $\frac { 1 } { 4 \sqrt { 2 } }$
(3) $\frac { 1 } { 3 \sqrt { 2 } }$
(4) $\frac { 1 } { 2 \sqrt { 2 } }$

The length of the perpendicular from the point $(2,-1,4)$ on the straight line $\frac{x+3}{10} = \frac{y-2}{-7} = \frac{z}{1}$ is
(1) greater than 3 but less than 4
(2) greater than 4
(3) less than 2
(4) greater than 2 but less than 3

Let $P$ be the plane, which contains the line of intersection of the planes, $x + y + z - 6 = 0$ and $2 x + 3 y + z + 5 = 0$ and it is perpendicular to the $x y$-plane. Then the distance of the point $( 0,0,256 )$ from $P$ is equal to:
(1) $205 \sqrt { 5 }$ units
(2) $\frac { 17 } { \sqrt { 5 } }$ units
(3) $\frac { 11 } { \sqrt { 5 } }$ units
(4) $63 \sqrt { 5 }$ units

The distance of the point $( 1 , - 2,3 )$ from the plane $x - y + z = 5$ measured parallel to the line $\frac { x } { 2 } = \frac { y } { 3 } = \frac { z } { - 6 }$ is:
(1) $\frac { 7 } { 5 }$
(2) 1
(3) $\frac { 1 } { 7 }$
(4) 7

Let the foot of the perpendicular from the point $( 1,2,4 )$ on the line $\frac { x + 2 } { 4 } = \frac { y - 1 } { 2 } = \frac { z + 1 } { 3 }$ be $P$. Then the distance of $P$ from the plane $3 x + 4 y + 12 z + 23 = 0$ is
(1) $\frac { 50 } { 13 }$
(2) $\frac { 63 } { 13 }$
(3) $\frac { 65 } { 13 }$
(4) 4

The shortest distance between the lines $\frac { x - 3 } { 2 } = \frac { y - 2 } { 3 } = \frac { z - 1 } { - 1 }$ and $\frac { x + 3 } { 2 } = \frac { y - 6 } { 1 } = \frac { z - 5 } { 3 }$ is
(1) $\frac { 18 } { \sqrt { 5 } }$
(2) $\frac { 22 } { 3 \sqrt { 5 } }$
(3) $\frac { 46 } { 3 \sqrt { 5 } }$
(4) $6 \sqrt { 3 }$

Let $P$ be the plane containing the straight line $\frac { x - 3 } { 9 } = \frac { y + 4 } { - 1 } = \frac { z - 7 } { - 5 }$ and perpendicular to the plane containing the straight lines $\frac { x } { 2 } = \frac { y } { 3 } = \frac { z } { 5 }$ and $\frac { x } { 3 } = \frac { y } { 7 } = \frac { z } { 8 }$. If $d$ is the distance of $P$ from the point $(2,-5,11)$, then $d ^ { 2 }$ is equal to
(1) $\frac { 147 } { 2 }$
(2) 96
(3) $\frac { 32 } { 3 }$
(4) 54

Let $N$ be the foot of perpendicular from the point $P ( 1 , - 2 , 3 )$ on the line passing through the points $( 4 , 5 , 8 )$ and $( 1 , - 7 , 5 )$. Then the distance of $N$ from the plane $2x - 2y + z + 5 = 0$ is $\_\_\_\_$.

Let $S$ be the set of all values of $\lambda$, for which the shortest distance between the lines $\frac { x - \lambda } { 0 } = \frac { y - 3 } { 4 } = \frac { z + 6 } { 1 }$ and $\frac { x + \lambda } { 3 } = \frac { y } { - 4 } = \frac { z - 6 } { 0 }$ is 13. Then $8 \left| \sum _ { \lambda \in S } \lambda \right|$ is equal to
(1) 306
(2) 304
(3) 308
(4) 302

The shortest distance between the lines $x + 1 = 2 y = - 12 z$ and $x = y + 2 = 6 z - 6$ is
(1) 2
(2) 3
(3) $\frac { 5 } { 2 }$
(4) $\frac { 3 } { 2 }$