QUESTION ONE (30 MARKS) a. Find the equation of a line through the point (5,-3) which is inclined at (pi )/(3) to the positive direction of the x-axis. (3marks) b. Find the rectangular co-ordinates of a point B(4,(pi )/(2)) (3marks) C. Change x^2+y^2-4y=0 to polar form. (3marks) d. Find a vector equation and the cartesian equation of the perpendicular bisectors of PQ, where P and Q are the points with position vectors: i) -3i-j and 7i+j (4marks) ii)ai+bj and 2ai+3bj (3marks) e. If z_(1)=2(cos320^circ +jsin320^circ ) and z_(2)=2(cos120^circ +jsin120^circ ) Determine modulus and argument of z_(2)z_(1) (6marks) f.Given A = A=[} 1&1 3&4&5 3&6&5 ] Find i) vert B-Avert (4marks) ii) A X B (4marks) QUESTION TWO (20 MARKS) a. Find an equation of the line parallel to the line with equation 6x-2y=8 and which passes through the point (2,-3) (3marks) b. i) Find a vector equation of the line passing through the points A(2,-2,-1) and B(4,-3,1) (4marks) ii) Hence find the Cartesian equation of the line AB. (3marks)

QUESTION ONE (30 MARKS)<br /><br />a. To find the equation of a line through the point $(5,-3)$ which is inclined at $\frac{\pi}{3}$ to the positive direction of the x-axis, we can use the point-slope form of a line equation:<br /><br />$y - y_1 = m(x - x_1)$<br /><br />where $(x_1, y_1)$ is the given point and $m$ is the slope of the line. The slope of the line can be found using the tangent function:<br /><br />$m = \tan(\theta) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$<br /><br />Substituting the values, we get:<br /><br />$y + 3 = \sqrt{3}(x - 5)$<br /><br />Simplifying, we have:<br /><br />$y = \sqrt{3}x - 5\sqrt{3} - 3$<br /><br />b. To find the rectangular coordinates of a point $B(4, \frac{\pi}{2})$, we can use the polar to Cartesian coordinate conversion formulas:<br /><br />$x = r \cos(\theta)$<br />$y = r \sin(\theta)$<br /><br />where $r$ is the radius and $\theta$ is the angle in radians. Substituting the given values, we get:<br /><br />$x = 4 \cos\left(\frac{\pi}{2}\right) = 0$<br />$y = 4 \sin\left(\frac{\pi}{2}\right) = 4$<br /><br />Therefore, the rectangular coordinates of point B are $(0, 4)$.<br /><br />c. To change the equation $x^2 + y^2 - 4y = 0$ to polar form, we can use the relationships between Cartesian and polar coordinates:<br /><br />$x = r \cos(\theta)$<br />$y = r \sin(\theta)$<br /><br />Substituting these into the given equation, we get:<br /><br />$r^2 \cos^2(\theta) + r^2 \sin^2(\theta) - 4r \sin(\theta) = 0$<br /><br />Simplifying, we have:<br /><br />$r^2 = 4r \sin(\theta)$<br /><br />Dividing both sides by $r$, we get:<br /><br />$r = 4 \sin(\theta)$<br /><br />Therefore, the polar form of the given equation is $r = 4 \sin(\theta)$.<br /><br />d. i) To find the vector equation and the Cartesian equation of the perpendicular bisector of PQ, where P and Q have position vectors $-3i - j$ and $7i + j$, we can use the formula for the midpoint of a line segment:<br /><br />Midpoint = $\frac{1}{2}(P + Q)$<br /><br />Substituting the given vectors, we get:<br /><br />Midpoint = $\frac{1}{2}((-3i - j) + (7i + j)) = 2i$<br /><br />The vector equation of the perpendicular bisector is:<br /><br />$\vec{r} = 2i + t\vec{u}$<br /><br />where $\vec{u}$ is a unit vector perpendicular to PQ. The Cartesian equation can be found by substituting the vector equation into the equation of a plane.<br /><br />ii) For the points $ai + bj$ and $2ai + 3bj$, the same process can be followed to find the vector equation and the Cartesian equation of the perpendicular bisector.<br /><br />e. To determine the modulus and argument of $z_2z_1$, we can first find the product of $z_1$ and $z_2$:<br /><br />$z_2z_1 = 2(\cos320^\circ + j\sin320^\circ) \cdot 2(\cos120^\circ + j\sin120^\circ)$<br /><br />Using the angle addition formula for cosine and sine, we can simplify the expression:<br /><br />$z_2z_1 = 4(\cos(320^\circ + 120^\circ) + j\sin(320^\circ + 120^\circ))$<br /><br />$z_2z_1 = 4(\cos440^\circ + j\sin440^\circ)$<br /><br />Since $440^\circ$ is equivalent to $80^\circ$, we can rewrite the expression as:<br /><br />$z_2z_1 = 4(\cos80^\circ + j\sin80^\circ)$<br /><br />The modulus of $z_2z_1$ is given by:<br /><br />$\vert z_2z_1 \vert = \sqrt{4^2 + 4^2} = 4\sqrt{2}$<br /><br />The argument of $z_2z_1$ is given by:<br /><br />$\arg(z_2z_1) = \tan^{-1}\left(\frac{4\sin80^\circ}{4\cos80^\circ}\right) = 45^\circ$<br /><br />Therefore, the modulus of $z_2z_1$ is $4\sqrt{