Let for a triangle $ABC$ $$\begin{aligned} & \overrightarrow { AB } = - 2 \hat { i } + \hat { j } + 3 \hat { k } \\ & \overrightarrow { CB } = \alpha \hat { i } + \beta \hat { j } + \gamma \hat { k } \\ & \overrightarrow { CA } = 4 \hat { i } + 3 \hat { j } + \delta \hat { k } \end{aligned}$$ If $\delta > 0$ and the area of the triangle $ABC$ is $5 \sqrt { 6 }$ then $\overrightarrow { CB } \cdot \overrightarrow { CA }$ is equal to
(1) 60
(2) 54
(3) 108
(4) 120

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 $\vec { a } , \vec { b }$ 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) $\frac { 1 } { 4 }$

Let $\lambda \in \mathbb{R}$, $\vec{a} = \lambda\hat{i} + 2\hat{j} - 3\hat{k}$, $\vec{b} = \hat{i} - \lambda\hat{j} + 2\hat{k}$. If $((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})) \times (\vec{a} - \vec{b}) = 8\hat{i} - 40\hat{j} - 24\hat{k}$, then $|\lambda(\vec{a} + \vec{b}) \times (\vec{a} - \vec{b})|^{2}$ is equal to
(1) 140
(2) 132
(3) 144
(4) 136

Let $\vec{a}$ and $\vec{b}$ be two vectors. Let $|\vec{a}| = 1$, $|\vec{b}| = 4$ and $\vec{a} \cdot \vec{b} = 2$. If $\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$, then the value of $\vec{b} \cdot \vec{c}$ is
(1) $-24$
(2) $-48$
(3) $-84$
(4) $-60$

Let $\vec { a } = 4 \hat { i } - \hat { j } + \hat { k } , \vec { b } = 11 \hat { i } - \hat { j } + \hat { k }$ and $\vec { c }$ be a vector such that $( \vec { a } + \vec { b } ) \times \vec { c } = \vec { c } \times ( - 2 \vec { a } + 3 \vec { b } )$. If $( 2 \vec { a } + 3 \vec { b } ) \cdot \vec { c } = 1670$, then $| \vec { c } | ^ { 2 }$ is equal to : (1) 1609 (2) 1618 (3) 1600 (4) 1627

Let $\vec { a } = \hat { i } + \hat { j } + \hat { k } , \vec { b } = 2 \hat { i } + 4 \hat { j } - 5 \hat { k }$ and $\vec { c } = x \hat { i } + 2 \hat { j } + 3 \hat { k } , x \in \mathbb { R }$. If $\vec { d }$ is the unit vector in the direction of $\vec { b } + \vec { c }$ such that $\vec { a } \cdot \vec { d } = 1$, then $( \vec { a } \times \vec { b } ) \cdot \vec { c }$ is equal to
(1) 11
(2) 3
(3) 9
(4) 6

Let $\overrightarrow{\mathrm{a}} = 2\hat{i} - \hat{j} + 3\hat{k}$, $\overrightarrow{\mathrm{b}} = 3\hat{i} - 5\hat{j} + \hat{k}$ and $\overrightarrow{\mathrm{c}}$ be a vector such that $\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{c}} = \overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}$ and $(\vec{a} + \vec{c}) \cdot (\vec{b} + \vec{c}) = 168$. Then the maximum value of $|\vec{c}|^2$ is:
(1) 462
(2) 77
(3) 154
(4) 308

Q77. Let $\vec { a } = 4 \hat { i } - \hat { j } + \hat { k } , \vec { b } = 11 \hat { i } - \hat { j } + \hat { k }$ and $\vec { c }$ be a vector such that $( \vec { a } + \vec { b } ) \times \vec { c } = \vec { c } \times ( - 2 \vec { a } + 3 \vec { b } )$. If $( 2 \vec { a } + 3 \vec { b } ) \cdot \vec { c } = 1670$, then $| \vec { c } | ^ { 2 }$ is equal to :
(1) 1609
(2) 1618
(3) 1600
(4) 1627

Q78. Let $\vec { a } = \hat { i } + \hat { j } + \hat { k } , \vec { b } = 2 \hat { i } + 4 \hat { j } - 5 \hat { k }$ and $\vec { c } = x \hat { i } + 2 \hat { j } + 3 \hat { k } , x \in \mathbb { R }$. If $\vec { d }$ is the unit vector in the direction of $\vec { b } + \vec { c }$ such that $\vec { a } \cdot \vec { d } = 1$, then $( \vec { a } \times \vec { b } ) \cdot \vec { c }$ is equal to
(1) 11
(2) 3
(3) 9
(4) 6

Q89. Let ABC be a triangle of area $15 \sqrt { 2 }$ and the vectors $\overrightarrow { \mathrm { AB } } = \hat { i } + 2 \hat { j } - 7 \hat { k } , \overrightarrow { \mathrm { BC } } = \mathrm { a } \hat { i } + \mathrm { b } \hat { j } + \mathrm { ck }$ and $\overrightarrow { \mathrm { AC } } = 6 \hat { i } + \mathrm { d } \hat { j } - 2 \hat { k } , \mathrm {~d} > 0$. Then the square of the length of the largest side of the triangle ABC is $\_\_\_\_$

Q90. Let $\overrightarrow { \mathrm { a } } = \hat { i } - 3 \hat { j } + 7 \hat { k } , \overrightarrow { \mathrm {~b} } = 2 \hat { i } - \hat { j } + \hat { k }$ and $\overrightarrow { \mathrm { c } }$ be a vector such that $\overrightarrow { \mathrm { a } } + 2 \overrightarrow { \mathrm {~b} } ) \times \overrightarrow { \mathrm { c } } = 3 ( \overrightarrow { \mathrm { c } } \times \overrightarrow { \mathrm { a } } )$. If $\vec { a } \cdot \vec { c } = 130$, then $\vec { b } \cdot \vec { c }$ is equal to $\_\_\_\_$

ANSWER KEYS

1. (2)	2. (3)
9. (2)	10. (3)
17. (2)	18. (4)
25. (4)	26. (5)
33. (4)	34. (3)
41. (1)	42. (2)
49. (4)	50. (2)
57. (5)	58. (2)
65. (2)	66. (3)
73. (1)	74. (4)
81. (125)	82. (910)
89. (72)	90. (30)

1. (4)
2. (2)
3. (2)
4. (5)
5. (3)
6. (3)
7. (80)
8. (14)
9. (1)
10. (4)
11. (54)
12. (2)
13. (2)
14. (3)
15. (50)
16. (3)
17. (2)
18. (22)
19. (6)
20. (4)
21. (1)
22. (108)
23. (4)
24. (2)
25. (19)
26. (10)
27. (2)
28. (3)
29. (8)
30. (1)
31. (3)
32. (3)
33. (24)
34. (4)
35. (1)
36. (240)
37. (727)
38. (4)
39. (2)
40. (150)
41. (2)
42. (2)
43. (4)
44. (56)
45. (4)
46. (4)
47. (600)
48. (3)
49. (2)
50. (3)
51. (25)
52. (3)
53. (1)
54. (4)
55. (3)
56. (1)
57. (2)
58. (86)
59. (2)
60. (2)
61. (1)
62. (3)
63. (2)
64. (4)
65. (2)
66. (18)

In coordinate space, consider two vectors $\vec{u}$ and $\vec{v}$ satisfying the dot product $\vec{u} \cdot \vec{v} = \sqrt{15}$ and the cross product $\vec{u} \times \vec{v} = (-1, 0, 3)$. Select the correct options.
(1) The angle $\theta$ between $\vec{u}$ and $\vec{v}$ (where $0 \leq \theta \leq \pi$, $\pi$ is the circumference ratio) is greater than $\frac{\pi}{4}$
(2) $\vec{u}$ could be $(1, 0, -1)$
(3) $|\vec{u}| + |\vec{v}| \geq 2\sqrt{5}$
(4) If $\vec{v}$ is known, then $\vec{u}$ can be uniquely determined
(5) If $|\vec{u}| + |\vec{v}|$ is known, then $|\vec{v}|$ can be uniquely determined

Problem 4
Answer the following questions on shapes in the three-dimensional orthogonal coordinate system $xyz$.
I. Consider the surface $S _ { 1 }$ represented by the equation $x ^ { 2 } + 2 y ^ { 2 } - z ^ { 2 } = 0$. Find the equations expressed in $x , y$, and $z$ of the normal line and the tangent plane $T$ to the surface $S _ { 1 }$ at the point $\mathrm { A } ( 2, 0, 2 )$.
II. Consider the surface $S _ { 2 }$ represented by the following set of equations with the parameters $u$ and $v$: $$\left\{ \begin{array} { l } x = \frac { 1 } { \sqrt { 2 } } \cosh u \cos v \\ y = \frac { 1 } { 2 } \cosh u \sin v - \frac { 1 } { \sqrt { 2 } } \sinh u \\ z = \frac { 1 } { 2 } \cosh u \sin v + \frac { 1 } { \sqrt { 2 } } \sinh u \end{array} \right.$$ where $u$ and $v$ are real numbers, and $0 \leq v < 2 \pi$.
Let $S _ { 3 }$ be the surface obtained by rotating the surface $S _ { 2 }$ around the $x$-axis by $- \pi / 4$. Here, the positive direction of rotation is the direction of the semi-circular arrow on the $yz$-plane shown in Figure 4.1.
Answer the following questions.

1. Find the matrix $\boldsymbol { R }$ that represents the linear transformation rotating a shape around the $x$-axis by $- \pi / 4$.
2. Find an equation expressed in $x , y$, and $z$ for the surface $S _ { 3 }$.
3. Find an equation expressed in $x , y$, and $z$ for the surface $S _ { 2 }$.

III. Consider the solid $V$ that is enclosed by the surface $S _ { 3 }$ obtained in Question II.2 and by the two planes $z = 1$ and $z = - 1$. Answer the following questions.

1. Calculate the area of the cross section obtained by cutting the solid $V$ with the $xz$-plane.
2. Calculate the area of the cross section obtained by cutting the solid $V$ with the plane $T$ obtained in Question I.

Problem 4

In the three-dimensional orthogonal coordinate system $x y z$, consider the surface $S$ defined by the following equation:
$$\left( \begin{array} { c } x ( \theta , \phi ) \\ y ( \theta , \phi ) \\ z ( \theta , \phi ) \end{array} \right) = \left( \begin{array} { c c c } \cos \theta & - \sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1 \end{array} \right) \left( \begin{array} { c } \cos \phi + 2 \\ 0 \\ \sin \phi \end{array} \right)$$
where $\theta$ and $\phi$ are parameters of the surface $S$, and $0 \leq \theta < 2 \pi , 0 \leq \phi < 2 \pi$. Let $V$ be the region surrounded by the surface $S$, and let $W$ be the region satisfying the inequality $x ^ { 2 } + y ^ { 2 } \leq 4$. Answer the following questions for the surface $S$.
I. Find the unit normal vector oriented inward the region $V$ at the point $P \left( \begin{array} { c } \frac { 1 } { \sqrt { 2 } } \\ \frac { 1 } { \sqrt { 2 } } \\ 0 \end{array} \right)$ on the surface $S$.
II. Find the area of the surface $S$ included in the region $W$.
III. Find the overlapping volume created by the two regions $V$ and $W$.
IV. Consider the three-dimensional curve $C$ on the surface $S$, which is defined by setting $\theta = \phi$. Find the curvature of the curve $C$ at the point $Q \left( \begin{array} { l } 0 \\ 2 \\ 1 \end{array} \right)$ on the curve $C$. Note that, in general, given a three-dimensional curve defined by $c ( t ) = \left( \begin{array} { c } x ( t ) \\ y ( t ) \\ z ( t ) \end{array} \right)$ represented by a parameter $t$, the curvature $\kappa ( t )$ of the curve at the point $c ( t )$ on the curve is given by the following equation:
$$\kappa ( t ) = \frac { \left| \frac { \mathrm { d } c ( t ) } { \mathrm { d } t } \times \frac { \mathrm { d } ^ { 2 } c ( t ) } { \mathrm { d } t ^ { 2 } } \right| } { \left| \frac { \mathrm { d } c ( t ) } { \mathrm { d } t } \right| ^ { 3 } }$$