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A) Prove (1+cosx)/sinx + sinx/(1+cosx) = 2cosecx. b) Solve the complex equation (x−2yj) − (y−xj) = 2 + j. c) X is a point with coordinates (1,-2,4) and Y is a point with coordinates (2,-1,-2). Determine using column vectors i. XY ii. the midpoint of XY.

Mathematics

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Answers

2 parts
c(i)

[11−6]\begin{bmatrix} 1 \\ 1 \\ -6 \end{bmatrix}

c(ii)

(32,−32,1)\left( \frac{3}{2}, -\frac{3}{2}, 1 \right)

Step-by-step solution

3 steps

You've provided an image with several questions. I will solve Question 6, part c) i) and ii) which asks to determine XY⃗\vec{XY} and the midpoint of XY using column vectors.

Part c· 1 step

X is a point with coordinates (1,−2,4)(1, -2, 4) and Y is a point with coordinates (2,−1,−2)(2, -1, -2).

  1. Write the coordinates of points X and Y as column vectors.

    OX⃗=[1−24]\vec{OX} = \begin{bmatrix} 1 \\ -2 \\ 4 \end{bmatrix} OY⃗=[2−1−2]\vec{OY} = \begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix}

Part c(i)

[11−6]\begin{bmatrix} 1 \\ 1 \\ -6 \end{bmatrix}

Determine XY⃗\vec{XY}.

  1. To find the vector XY⃗\vec{XY}, subtract the position vector of X from the position vector of Y.

    XY⃗=OY⃗−OX⃗\vec{XY} = \vec{OY} - \vec{OX} XY⃗=[2−1−2]−[1−24]\vec{XY} = \begin{bmatrix} 2 \\ -1 \\ -2 \end{bmatrix} - \begin{bmatrix} 1 \\ -2 \\ 4 \end{bmatrix} XY⃗=[2−1−1−(−2)−2−4]\vec{XY} = \begin{bmatrix} 2 - 1 \\ -1 - (-2) \\ -2 - 4 \end{bmatrix} XY⃗=[11−6]\vec{XY} = \begin{bmatrix} 1 \\ 1 \\ -6 \end{bmatrix}

    The column vector XY⃗\vec{XY} is:

    [11−6]\boxed{\begin{bmatrix} 1 \\ 1 \\ -6 \end{bmatrix}}

Part c(ii)

(32,−32,1)\left( \frac{3}{2}, -\frac{3}{2}, 1 \right)

Determine the midpoint of XY.

  1. To find the midpoint of the line segment XY, average the coordinates of X and Y.

    Let the midpoint be M.

    M=(x1+x22,y1+y22,z1+z22)M = \left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2} \right) M=(1+22,−2+(−1)2,4+(−2)2)M = \left( \frac{1+2}{2}, \frac{-2+(-1)}{2}, \frac{4+(-2)}{2} \right) M=(32,−32,22)M = \left( \frac{3}{2}, \frac{-3}{2}, \frac{2}{2} \right) M=(32,−32,1)M = \left( \frac{3}{2}, -\frac{3}{2}, 1 \right)

    The midpoint of XY is:

    (32,−32,1)\boxed{\left( \frac{3}{2}, -\frac{3}{2}, 1 \right)}
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