Using the result from task $a$, give the width of the gate to the nearest meter. Justify using the statement from task $d$ that the straight line of travel of the skier is inclined at less than $30 ^ { \circ }$ to the horizontal.

Justify by calculation that the skier actually passes through the gate.

We consider four distinct points $A, B, C$ and $D$ in the canonical Euclidean space $\mathbb{R}^3$ such that $AB = BC = CD = DA = 1$, $AC = a > 0$ and $BD = b > 0$. We assume that the four distinct points $A, B, C$ and $D$ are not coplanar. We denote $I$ the midpoint of $[AC]$ and $J$ the midpoint of $[BD]$.
a) Show that $(IJ)$ is the common perpendicular to the lines $(AC)$ and $(BD)$.
b) By projecting the points $B$ and $D$ onto the plane containing $(AC)$ and perpendicular to $(IJ)$, show that $a^2 + b^2 < 4$.

We consider four distinct points $A, B, C$ and $D$ in the canonical Euclidean space $\mathbb{R}^3$ such that $AB = BC = CD = DA = 1$, $AC = a > 0$ and $BD = b > 0$.
Show that if strictly positive reals $a$ and $b$ satisfy the relation $a^2 + b^2 \leqslant 4$, then there indeed exist four distinct points $A, B, C$ and $D$ in the canonical Euclidean space $\mathbb{R}^3$ satisfying $AB = BC = CD = DA = 1$, $AC = a$ and $BD = b$.

133- Three points $A(2,1,\circ)$, $B(3,-1,2)$, $C(-1,1,3)$ are vertices of a triangle. What is $\cos A$?
(1) $\dfrac{\sqrt{2}}{6}$ (2) $\dfrac{\sqrt{2}}{4}$ (3) $\dfrac{\sqrt{2}}{6}$ (4) $\dfrac{\sqrt{3}}{4}$

133. Points $O(0,0,0)$, $B(-1,2,4)$, $A(5,-4,1)$ are given, and $\overrightarrow{AM} = \dfrac{2}{3}\overrightarrow{AB}$ and $\overrightarrow{AB}$ are known. The value of $|\overrightarrow{OM}|$ is:

(2) $\sqrt{11}$	(1) $\sqrt{10}$
(4) $\sqrt{14}$	(3) $\sqrt{13}$

134. The distance between the two lines $\dfrac{x-1}{2} = \dfrac{y+2}{1} = \dfrac{z}{-1}$ and $\left(x = 2y+1,\ z = -y+2\right)$ is:

(2) $2\sqrt{2}$	(1) $\sqrt{6}$
(4) $3\sqrt{2}$	(3) $2\sqrt{3}$

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134- What is the initial distance from the line passing through point $(1, 2, -3)$ parallel to the vector with components $(4, -3, -5)$?
(1) $\dfrac{\sqrt{5}}{2}$ (2) $\sqrt{3}$ (3) $\sqrt{5}$ (4) $2\sqrt{3}$

135- What is the length of the common perpendicular of the line $\dfrac{x-1}{4} = \dfrac{y+2}{3} = \dfrac{z-3}{1}$ and the $z$-axis?
(1) $2/2$ (2) $2/4$ (3) $2/5$ (4) $2/6$

134. The distance from point $(1,3,2)$ to the line of intersection of the plane $2x - y - z = 4$ with the plane $xOy$ is:
(1) $2$ (2) $\sqrt{6}$ (3) $3$ (4) $\sqrt{10}$

133- Vector $\mathbf{a}$ makes an angle of $60°$ with each of the axes $Ox$ and $Oy$, and makes a right angle with the axis $z$. This vector is perpendicular to which plane? Which equation does the plane satisfy?
(1) $x - \sqrt{2}y + z = 0$ (2) $2x + 2y + \sqrt{2}z = 0$
(4) $x + y - \sqrt{2}z = 0$ (3) $x + y + \sqrt{2}z = 0$

136- For which value of $m$, the three vectors $\vec{a} = (-1,2,3)$, $\vec{b} = (2,\circ,1)$, $\vec{c} = (-4,m,5)$ are coplanar?
(1) $-2$ (2) $2$ (3) $3$ (4) $4$

155 -- Suppose point $M$ is the intersection of the line passing through points $A(2,2,1)$ and $B(2,-1,5)$ with the plane $x + y + z = 1$. What is the distance from point $M$ to point $B$?

• [(1)] $25$
• [(2)] $27$
• [(3)] $10\sqrt{5}$
• [(4)] $5\sqrt{13}$

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The points $( 4,7 , - 1 ) , ( 1,2 , - 1 ) , ( - 1 , - 2 , - 1 )$ and $( 2,3 , - 1 )$ in $\mathbb { R } ^ { 3 }$ are the vertices of a
(A) rectangle which is not a square.
(B) rhombus.
(C) parallelogram which is not a rectangle.
(D) trapezium which is not a parallelogram.

9. Given in three-dimensional space the points $A(3, 1, 0)$, $B(3, -1, 2)$, $C(1, 1, 2)$. After verifying that $ABC$ is an equilateral triangle and that it is contained in the plane $\alpha$ with equation $x + y + z - 4 = 0$, establish which are the points $P$ such that $ABCP$ is a regular tetrahedron.

Consider the line $r$ passing through the two points $A(1, -2, 0)$ and $B(2, 3, -1)$, determine the Cartesian equation of the spherical surface with center $C(1, -6, 7)$ and tangent to $r$.

3. In space with orthogonal Cartesian coordinate system $O x y z$, the plane $\pi : 3 x - 2 y + 5 = 0$ is given. －Determine the coordinates of point $H$, the orthogonal projection of $P ( 4,2,1 )$ onto the plane $\pi$; －Determine the intersection of the line $s$ : $\left\{ \begin{array} { l } x - y + 1 = 0 \\ z - 2 = 0 \end{array} \right.$ with the plane $\pi$.

3. Verify that the points $O ( 0,0,0 ) , A ( 1,4,8 ) , B ( - 6,0,12 )$ and $C ( - 7 , - 4,4 )$ are coplanar. Calculate the area and perimeter of the quadrilateral $O A B C$ and classify it.

25. For three vectors $\vec { u } , \vec { v } , \vec { w }$ which of the following expressions is not equal to any of the remaining three?
(A) $\vec { u } \cdot ( \vec { v } \times \vec { w } )$
(B) $( \vec { v } \times \vec { w } ) \cdot \vec { u }$
(C) $\vec { v } \cdot ( \vec { v } \times \vec { w } )$
(D) $( \vec { u } \times \vec { v } ) \cdot \vec { w }$

21. Let $\vec { a } = \overrightarrow { 2 } \hat { \jmath } + \hat { k } , \vec { b } = \hat { \imath } + \overrightarrow { 2 } \hat { \jmath } - \hat { k }$ and a unit vector $\vec { c }$ be coplanar. If $\vec { c }$ is perpendicular to $\vec { a }$, then $\vec { c } =$
(A) $\frac { 1 } { \sqrt { 2 } } ( - \hat { \jmath } + \hat { k } )$
(B) $\frac { 1 } { \sqrt { 3 } } ( - \hat { \imath } - \hat { \jmath } - \hat { k } )$
(C) $\frac { 1 } { \sqrt { 5 } } ( \hat { \imath } - 2 \hat { \jmath } )$
(D) $\frac { 1 } { \sqrt { 3 } } ( \hat { \imath } - \hat { \jmath } - \hat { k } )$

29. Let $\vec { a } = \vec { \imath } - \vec { k } , \vec { b } = x \vec { \imath } + \vec { \jmath } + ( 1 - x ) \vec { k }$ and $\vec { c } = y \vec { \imath } + x \vec { \jmath } + ( 1 + x - y ) \vec { k }$. Then $[ \vec { a } , \vec { b } , \vec { c } ]$ depends on :
(A) Only x
(B) Only $y$
(C) Neither $x$ nor $y$
(D) Both $x$ and $y$

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7. The value of $k$ such that $( x - 4 ) / 1 = ( y - 2 ) / 1 = ( z - k ) / 2$ lies in the plane $2 x - 4 y + z = 7$, is:
(a) 7
(b) - 7
(c) no real value
(d) 4

(i) Find the equation of the plane passing through the points $( 2,1,0 ) , ( 5,0,1 )$ and $( 4,1,1 )$ (ii) If P is the point $( 2,1,6 )$ then the point Q such that PQ is perpendicular to the plane in (i) and the mid point of PQ lies on it.

If \overrightarrow { \mathrm { u } } , \mathrm { v } ^ { \rightarrow } , \mathrm { w } ^ { \rightarrow } are three noncoplanar unit vectors and $\alpha , \beta , \gamma$ are the angles between $\mathrm { u } \rightarrow$ are and $\mathrm { v } ^ { \rightarrow } , \mathrm { v } ^ { \rightarrow } are and $\overrightarrow { \mathrm { w } } , \mathrm { w } ^ { \rightarrow } and $\overrightarrow { \mathrm { u } }$ are respectively and $\overrightarrow { \mathrm { x } } , \mathrm { y } ^ { \rightarrow } , \mathrm { z }$ are unit vectors along the bisectors of the angles $\mathrm { a } , \mathrm { b } , \mathrm { g }$ respectively. Prove that $$\left[ \begin{array} { l l l } \vec { x } \times \vec { y } & \vec { y } \times \vec { z } & \vec { z } \times \vec { x } \end{array} \right] = \frac { 1 } { 16 } \left[ \begin{array} { l l l } \vec { u } & \vec { v } & \vec { w } \end{array} \right] ^ { 2 } \sec ^ { 2 } \frac { \alpha } { 2 } \sec ^ { 2 } \frac { \beta } { 2 } \sec ^ { 2 } \frac { \gamma } { 2 }$$

2. $\vec { a } , \vec { b } , \vec { c } , \vec { d }$ are four distinct vectors satisfying the conditions $\vec { a } \times \vec { b } = \vec { c } \times \vec { d }$ and $\vec { a } \times \vec { c } = \vec { b } \times \vec { d }$, then prove that $\vec { a } \cdot \vec { b } + \vec { c } \cdot \vec { d } \neq \vec { a } \cdot \vec { c } + \vec { b } \cdot \vec { d }$.
Sol. Given that $\vec { a } \times \vec { b } = \vec { c } \times \vec { d }$ and $\vec { a } \times \vec { c } = \vec { b } \times \vec { d }$ $\Rightarrow \overrightarrow { \mathrm { a } } \times ( \overrightarrow { \mathrm { b } } - \overrightarrow { \mathrm { c } } ) = ( \overrightarrow { \mathrm { c } } - \overrightarrow { \mathrm { b } } ) \times \overrightarrow { \mathrm { d } } = \overrightarrow { \mathrm { d } } \times ( \overrightarrow { \mathrm { b } } - \overrightarrow { \mathrm { c } } ) \Rightarrow \overrightarrow { \mathrm { a } } - \overrightarrow { \mathrm { d } } \| \overrightarrow { \mathrm { b } } - \overrightarrow { \mathrm { c } }$ $\Rightarrow ( \vec { a } - \vec { d } ) \cdot ( \vec { b } - \vec { c } ) \neq 0 \Rightarrow \vec { a } \cdot \vec { b } + \vec { d } \cdot \vec { c } \neq \vec { d } \cdot \vec { b } + \vec { a } \cdot \vec { c }$.