3D coordinates: distance and midpoint Problem set B One question at a time

Points A and B are \(A(-5, 3, 9)\) and \(B(7, -11, -1)\). M is the midpoint of AB.

1. (a) Find the coordinates of M.

   • Add each pair of coordinates and halve
   • \(M = (1, -4, 4)\)

2. (b) Find the distance from the origin to M. Give your answer to 3 significant figures.

   • \(OM = \sqrt{1+16+16} = 5.74\)

3. (c) P and Q are \(P(8, -4)\) and \(Q(-2, 5)\) in two dimensions. Find the midpoint of PQ.

   • \(\left(\frac{8+(-2)}{2}, \frac{-4+(5)}{2}\right) = (3, 0.5)\)

\(P(-3, 4, 1)\) and \(Q(2, -2, 8)\) are two points.

1. (a) Find the distance PQ to 3 significant figures.

   • \(PQ = \sqrt{(2-(-3))^2+(-2-(4))^2+(8-(1))^2}\)
   • \(= \sqrt{110} = 10.5\)

2. (b) \(R(2, k, 3)\) is a distance 8 from \(S(-1, 5, -2)\). Find the possible values of k, to 3 significant figures.

   • \(RS^2 = 64\): \((k-(5))^2 + 34 = 64\)
   • \((k-(5))^2 = 30\)
   • \(k = −0.477\) or \(k = 10.5\) (two values: the point can be on either side of S)

The midpoint of \(A(a, 3 b, -2)\) and \(B(2 b, a, 8)\) is \(M(\frac{11}{2}, 7, 3)\). Find the values of a and b.

• Each coordinate of M is the average: \(\frac{a+2 b}{2} = 11/2\) and \(\frac{3 b+a}{2} = 7\)
• \(a + 2 b = 11\) and \(a + 3 b = 14\)
• Solve together: \(a = 5\), \(b = 3\)
• Check with the third coordinate: \(\frac{-2+8}{2} = 3\)

Find the equation of the line through \(P(-2, -5)\) and \(Q(4, 3)\). Give it in the form \(y = mx + c\) and in the form \(ax + by + d = 0\) with integer coefficients.

• Gradient \(= \frac{3-(-5)}{4-(-2)} = \frac{4}{3}\)
• \(y - (-5) = \frac{4}{3}(x - (-2))\)
• \(y = \frac{4}{3}x - \frac{7}{3}\)
• Multiply by 3 and collect terms: \(4x - 3y - 7 = 0\)

A line L has equation \(2x + 3y - 12 = 0\). The point P is \((-1, 4)\).

1. (a) Write down the gradient of L.

   • Rearrange: \(y = - \frac{2}{3}x + 4\)
   • Gradient \(- \frac{2}{3}\)

2. (b) Find the equation of the line through P perpendicular to L.

   • Perpendicular gradient \(= -1 \div - \frac{2}{3} = \frac{3}{2}\)
   • \(y = \frac{3}{2}x + \frac{11}{2}\)

3. (c) Find the coordinates of the point where this line meets L.

   • Solve the two equations together on the calculator
   • Meeting point \((- \frac{9}{13}, \frac{58}{13})\)

4. (d) Find the shortest distance from P to L.

   • \(\sqrt{(-1-- \frac{9}{13})^2+(4-\frac{58}{13})^2} = 0.555\)

A submarine travels in a straight line from \(A(-200, 350, -40)\) to \(B(160, -100, -180)\), where all coordinates are in metres and the z-coordinate is the depth relative to sea level (negative below it). The journey takes 2 minutes.

1. (a) Find the distance travelled.

   • \(\sqrt{(160-(-200))^2+(-100-(350))^2+(-180-(-40))^2}\)
   • \(= \sqrt{351700} = 593\) m

2. (b) Find the average speed in metres per second.

   • Speed = distance ÷ time = \(593 \div 120\)
   • \(4.94\) m/s

A greenhouse has a rectangular floor OABC and a horizontal ridge. The diagram shows the frame. O is the origin. The x-axis is along OA, the y-axis is along OC and the z-axis is vertical. All lengths are in metres. OA = 4, OC = 2.5. D and E are the ends of the ridge, 2.2 m above the ground, directly above the points 2 m along OA from O and C.

1. (a) Write down the coordinates of B and E.

   • B is above nothing: on the ground, \((4, 2.5, 0)\)
   • E is at the same height as D and the same x: \((2, 2.5, 2.2)\)

2. (b) M is the midpoint of the ridge DE. Find the coordinates of M.

   • \(D = (2, 0, 2.2)\), \(E = (2, 2.5, 2.2)\)
   • \(M = (2, 1.25, 2.2)\)

3. (c) A brace runs from M to a peg at \((2, 5.5, 0)\). Find the length of the brace.

   • \(\sqrt{(2-2)^2+(5.5-1.25)^2+(0-2.2)^2} = 4.79\) m

A packing crate is a cuboid 1.6 m long, 0.9 m wide and 0.7 m high. A straight rod rests with one end at corner A on the bottom and the other at the opposite corner G on the top.

1. (a) Find the length of the diagonal AC of the base.

   • \(AC^2 = 1.6^2 + 0.9^2 = 3.370\)
   • \(AC = 1.84\) m

2. (b) Find the length of the longest rod, the straight line AG.

   • \(AG^2 = AC^2 + 0.7^2 = 3.860\)
   • \(AG = 1.96\) m

A cleaner wipes along the surface of a closed storage box 1.5 m long, 1 m wide and 0.6 m high, from corner A to the opposite corner G, in the shortest way. The cleaner stays on the surface. Find that distance.

• Unfold two faces into a rectangle. Three different rectangles are possible.
• \(\sqrt{(1+0.6)^2+1.5^2} = 2.19\); \(\sqrt{(1.5+0.6)^2+1^2} = 2.33\); \(\sqrt{(1.5+1)^2+0.6^2} = 2.57\)
• The shortest is 2.19 m

ABCD is a square. A is \((3, -2)\) and C is \((-3, 4)\), and AC is a diagonal. B is the vertex with the larger x-coordinate.

1. (a) Find the midpoint of AC.

   • \(M = \left(\frac{3+(-3)}{2}, \frac{-2+(4)}{2}\right) = (0, 1)\)

2. (b) Find the coordinates of B and D.

   • M to C: \((-3, 3)\). Turn through 90°: \((3, 3)\)
   • \(B = M + (3, 3) = (3, 4)\)
   • \(D = M - (3, 3) = (-3, -2)\)

3. (c) Find the area of the square.

   • \(AB^2 = (3-(3))^2+(4-(-2))^2 = 36.0\)
   • Area \(= AB^2 = 36.0\)

ABCD is a rhombus with \(A(-3, 2)\) and \(C(1, 6)\). The x-coordinate of B is 3.

1. (a) Find the equation of the perpendicular bisector of AC.

   • Midpoint of AC: \(M = (-1, 4)\)
   • Gradient of AC \(= 1\), so the perpendicular gradient is \(-1\)
   • \(y = -x + 3\)

2. (b) The diagonal BD lies on this line. Find the coordinates of B and D.

   • At \(x = 3\): \(y = 0\), so \(B = (3, 0)\)
   • The diagonals bisect each other, so \(D = 2M - B = (-5, 8)\)

3. (c) Find the side length of the rhombus.

   • \(AB = \sqrt{(3-(-3))^2+(0-(2))^2} = 6.32\)