SECTION B (20 MARKS) 2. a. A line passes through points A(2,-1,5) and B(3,6,-4) i. Write a vector equation of the line. (5marks) ii. Write parametric equations for the line. (5marks) iii Determine if the point C (0,-15,9) lies on the line. (5marks) b. Find the foci,vertices and asymptotes of a hyperbola with equation ((x+2)^2)/(9)=1+((y-1)^2)/(4) (5marks) 3. a. Find the axis, vertex ,focus and directrix of the parabola 2y^2+16x-12y+2=0 (6 mark) b. Let z_(1)=2+2sqrt (3i) and z_(2)=-1-sqrt (3i) Evaluate 3(z_(1)z_(2)) (4 marks) c. Given that overrightarrow (A)=i-3j+2k and overrightarrow (B)=2i+6j find i. overrightarrow (A)X3overrightarrow (B) (5 marks) ii. overrightarrow (B)Xoverrightarrow (A) (5 marks) 4. a. Check if the two line 3x-5=2y and 4x+5y=1 are parallel (4 marks) b. Find the foci vertices and asymptotes of a hyperbola with equation ((x+2)^2)/(9)-1=-((y-1)^2)/(4) (6 marks) c. Find a unit vector that is perpendicular to both overrightarrow (A)=2i-2j-k and overrightarrow (B)=i+j+k What is the area of the parallelogram with overrightarrow (A) and overrightarrow (B) as its sides? (5 marks) d. Find the axis.vertex,focus and directrix of the parabola y^2+8x-6y+1=0 (5 marks) 5. a. Write the equation of the circle 6x^2+6y^2-24x+36y+30=0 in standard form (4 marks) b. Derive the equation of an ellipse and use it to find the foci and vertices of the ellipse (x^2)/(16)=1-(y^2)/(25) (10 marks) c. Given overrightarrow (A)=3i-2j-5k and overrightarrow (B)=2i+j+k Find; Vert -7AVert (2 marks) ii Vert 2overrightarrow (A)-overrightarrow (3BVert ) (2 marks) ii,overrightarrow (A),overrightarrow (B) (2 marks)

2.<br />a. i. To write a vector equation of the line passing through points A(2,-1,5) and B(3,6,-4), we first find the direction vector by subtracting the coordinates of A from B: $\overrightarrow{AB} = (3-2, 6-(-1), -4-5) = (1, 7, -9)$. The vector equation of the line is then given by $\mathbf{r} = \mathbf{A} + t\overrightarrow{AB}$, where $\mathbf{A}$ is the position vector of point A and $t$ is a scalar parameter.<br /><br />ii. The parametric equations for the line can be obtained by expressing the coordinates of any point on the line in terms of the parameter $t$. Using the position vector of point A and the direction vector, we have:<br />$x = 2 + t$<br />$y = -1 + 7t$<br />$z = 5 - 9t$<br /><br />iii. To determine if point C(0,-15,9) lies on the line, we substitute the coordinates of C into the parametric equations and see if we get a consistent value for $t$. Substituting $x = 0$, $y = -15$, and $z = 9$ into the equations, we get:<br />$0 = 2 + t$<br />$-15 = -1 + 7t$<br />$9 = 5 - 9t$<br /><br />Solving these equations simultaneously, we find that $t = -3$. Substituting $t = -3$ into the equation for $z$, we get $9 = 5 - 9(-3)$, which is true. Therefore, point C lies on the line.<br /><br />b. To find the foci, vertices, and asymptotes of the hyperbola with equation $\frac{(x+2)^2}{9} = 1 + \frac{(y-1)^2}{4}$, we first rewrite it in the standard form of a hyperbola equation: $\frac{(x+2)^2}{9} - \frac{(y-1)^2}{4} = 1$. From this equation, we can identify the center as $(-2, 1)$, $a = 3$, and $b = 2$. The foci are located at a distance of $c = \sqrt{a^2 + b^2}$ from the center, so the foci are at $(-2 \pm \sqrt{13}, 1)$. The vertices are located at a distance of $a$ from the center, so the vertices are at $(-2 \pm 3, 1)$. The asymptotes are given by the equations $y - 1 = \pm \frac{b}{a}(x + 2)$, which simplify to $y = 1 \pm \frac{2}{3}(x + 2)$.<br /><br />3.<br />a. To find the axis, vertex, focus, and directrix of the parabola $2y^2 + 16x - 12y + 2 = 0$, we first rewrite it in the standard form of a parabola equation: $y^2 = -8x + 6$. The axis is the line $y = 0$, the vertex is at $(0, 0)$, the focus is at $(-1, 0)$, and the directrix is the line $x = 1$.<br /><br />b. To evaluate $3(z_1z_2)$, where $z_1 = 2 + 2\sqrt{3i}$ and $z_2 = -1 - \sqrt{3i}$, we first multiply $z_1$ and $z_2$ together:<br />$z_1z_2 = (2 + 2\sqrt{3i})(-1 - \sqrt{3i}) = -2 - 2\sqrt{3i} - 2\sqrt{3i} - 6i^2 = -2 - 4\sqrt{3i} + 6 = 4 - 4\sqrt{3i}$<br />Then, we multiply the result by 3:<br />$3(z_1z_2) = 3(4 - 4\sqrt{3i}) = 12 - 12\sqrt{3i}$<br /><br />c. i. To find $\overrightarrow{A} \times 3\overrightarrow{B}$, where $\overrightarrow{A} = i - 3j + 2k$ and $\overrightarrow{B} = 2i + 6j$, we first multiply $\overrightarrow{B}$ by 3:<br />$3\overrightarrow{B} = 6i + 18j$<br />Then, we find the cross product of $\