Exam-Style Problems

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Nov 2023 p31 q11

2231

In the diagram, OABCDEFG is a cuboid in which OA = 3 units, OC = 2 units and OD = 2 units. Unit vectors i, j and k are parallel to OA, OD and OC respectively. M is the midpoint of EF.

(a) Find the position vector of M.

The position vector of P is i + j + 2k.

(b) Calculate angle PAM.

June 2018 p13 q9

2232

The diagram shows a pyramid OABCD with a horizontal rectangular base OABC. The sides OA and AB have lengths of 8 units and 6 units respectively. The point E on OB is such that OE = 2 units. The point D of the pyramid is 7 units vertically above E. Unit vectors i, j and k are parallel to OA, OC and ED respectively.

(i) Show that \(\overrightarrow{OE} = 1.6\mathbf{i} + 1.2\mathbf{j}\).

(ii) Use a scalar product to find angle BDO.

June 2018 p12 q5

2233

The diagram shows a three-dimensional shape. The base OAB is a horizontal triangle in which angle AOB is 90°. The side OBCD is a rectangle and the side OAD lies in a vertical plane. Unit vectors i and j are parallel to OA and OB respectively and the unit vector k is vertical. The position vectors of A, B and D are given by \(\overrightarrow{OA} = 8\mathbf{i}\), \(\overrightarrow{OB} = 5\mathbf{j}\) and \(\overrightarrow{OD} = 2\mathbf{i} + 4\mathbf{k}\).

(i) Express each of the vectors \(\overrightarrow{DA}\) and \(\overrightarrow{CA}\) in terms of i, j and k.

(ii) Use a scalar product to find angle CAD.

Problem 2234

2234

Fig. 1 shows a rectangle with sides of 7 units and 3 units from which a triangular corner has been removed, leaving a 5-sided polygon OABCD. The sides OA, AB, BC and DO have lengths of 7 units, 3 units, 3 units and 2 units respectively. Fig. 2 shows the polygon OABCD forming the horizontal base of a pyramid in which the point E is 8 units vertically above D. Unit vectors i, j and k are parallel to OA, OD and DE respectively.

(i) Find \(\overrightarrow{CE}\) and the length of \(CE\).

(ii) Use a scalar product to find angle ECA, giving your answer in the form \(\cos^{-1} \left( \frac{m}{\sqrt{n}} \right)\), where m and n are integers.

Nov 2017 p12 q9

2235

The diagram shows a trapezium \(OABC\) in which \(OA\) is parallel to \(CB\). The position vectors of \(A\) and \(B\) relative to the origin \(O\) are given by \(\overrightarrow{OA} = \begin{pmatrix} -2 \\ -2 \\ -1 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 6 \\ 1 \\ 1 \end{pmatrix}\).

(i) Show that angle \(OAB\) is \(90^\circ\).

The magnitude of \(\overrightarrow{CB}\) is three times the magnitude of \(\overrightarrow{OA}\).

(ii) Find the position vector of \(C\).

(iii) Find the exact area of the trapezium \(OABC\), giving your answer in the form \(a\sqrt{b}\), where \(a\) and \(b\) are integers.

Nov 2016 p13 q7

2236

The diagram shows a triangular pyramid ABCD. It is given that \(\overrightarrow{AB} = 3\mathbf{i} + \mathbf{j} + \mathbf{k}\), \(\overrightarrow{AC} = \mathbf{i} - 2\mathbf{j} - \mathbf{k}\), and \(\overrightarrow{AD} = \mathbf{i} + 4\mathbf{j} - 7\mathbf{k}\).

(i) Verify, showing all necessary working, that each of the angles \(DAB\), \(DAC\), and \(CAB\) is \(90^\circ\).

(ii) Find the exact value of the area of the triangle \(ABC\), and hence find the exact value of the volume of the pyramid.

[The volume \(V\) of a pyramid of base area \(A\) and vertical height \(h\) is given by \(V = \frac{1}{3}Ah\).]

Nov 2016 p11 q9

2237

The diagram shows a cuboid OABCDEFG with a horizontal base OABC in which OA = 4 ext{ cm} and AB = 15 ext{ cm}. The height OD of the cuboid is 2 ext{ cm}. The point X on AB is such that AX = 5 ext{ cm} and the point P on DG is such that DP = p ext{ cm}, where p is a constant. Unit vectors i, j and k are parallel to OA, OC and OD respectively.

Find the possible values of p such that angle OPX = 90^0.
For the case where p = 9, find the unit vector in the direction of \(\overrightarrow{XP}\).
A point Q lies on the face CBFG and is such that \(XQ\) is parallel to AG. Find \(\overrightarrow{XQ}\).

Feb/Mar 2016 p12 q7

2238

The diagram shows a pyramid OABC with a horizontal triangular base OAB and vertical height OC. Angles AOB, BOC and AOC are each right angles. Unit vectors i, j and k are parallel to OA, OB and OC respectively, with OA = 4 units, OB = 2.4 units and OC = 3 units. The point P on CA is such that CP = 3 units.

Show that \(\overrightarrow{CP} = 2.4\mathbf{i} - 1.8\mathbf{k}\).
Express \(\overrightarrow{OP}\) and \(\overrightarrow{BP}\) in terms of \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\).
Use a scalar product to find angle BPC.

Nov 2015 p11 q10

2239

The diagram shows a cuboid OABCPQRS with a horizontal base OABC in which AB = 6 cm and OA = a cm, where a is a constant. The height OP of the cuboid is 10 cm. The point T on BR is such that BT = 8 cm, and M is the mid-point of AT. Unit vectors i, j and k are parallel to OA, OC and OP respectively.

(i) For the case where a = 2, find the unit vector in the direction of \(\overrightarrow{PM}\).

(ii) For the case where angle \(ATP = \cos^{-1}\left(\frac{2}{7}\right)\), find the value of a.

Nov 2014 p12 q7

2240

The diagram shows a pyramid \(OABCX\). The horizontal square base \(OABC\) has side 8 units and the centre of the base is \(D\). The top of the pyramid, \(X\), is vertically above \(D\) and \(XD = 10\) units. The mid-point of \(OX\) is \(M\). The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are parallel to \(\overrightarrow{OA}\) and \(\overrightarrow{OC}\) respectively and the unit vector \(\mathbf{k}\) is vertically upwards.

(i) Express the vectors \(\overrightarrow{AM}\) and \(\overrightarrow{AC}\) in terms of \(\mathbf{i}\), \(\mathbf{j}\) and \(\mathbf{k}\).

(ii) Use a scalar product to find angle \(MAC\).

June 2014 p12 q7

2241

The diagram shows a trapezium ABCD in which BA is parallel to CD. The position vectors of A, B, and C relative to an origin O are given by

\(\overrightarrow{OA} = \begin{pmatrix} 3 \\ 4 \\ 0 \end{pmatrix}, \quad \overrightarrow{OB} = \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}, \quad \text{and} \quad \overrightarrow{OC} = \begin{pmatrix} 4 \\ 5 \\ 6 \end{pmatrix}.\)

Use a scalar product to show that AB is perpendicular to BC.
Given that the length of CD is 12 units, find the position vector of D.

Nov 2022 p31 q11

2242

In the diagram, \(OABCD\) is a solid figure in which \(OA = OB = 4\) units and \(OD = 3\) units. The edge \(OD\) is vertical, \(DC\) is parallel to \(OB\) and \(DC = 1\) unit. The base, \(OAB\), is horizontal and angle \(AOB = 90^\circ\). Unit vectors \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\) are parallel to \(OA, OB\) and \(OD\) respectively. The midpoint of \(AB\) is \(M\) and the point \(N\) on \(BC\) is such that \(CN = 2NB\).

Express vectors \(\overrightarrow{MD}\) and \(\overrightarrow{ON}\) in terms of \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\).
Calculate the angle in degrees between the directions of \(\overrightarrow{MD}\) and \(\overrightarrow{ON}\).
Show that the length of the perpendicular from \(M\) to \(ON\) is \(\sqrt{\frac{22}{5}}\).

Solution

(a) To find \(\overrightarrow{MD}\), we first find \(\overrightarrow{OM}\) which is the midpoint of \(AB\). Since \(OA = 4\mathbf{i}\) and \(OB = 4\mathbf{j}\), \(OM = 2\mathbf{i} + 2\mathbf{j}\). Thus, \(\overrightarrow{MD} = \overrightarrow{OD} - \overrightarrow{OM} = 3\mathbf{k} - (2\mathbf{i} + 2\mathbf{j}) = -2\mathbf{i} - 2\mathbf{j} + 3\mathbf{k}\).

For \(\overrightarrow{ON}\), since \(N\) divides \(BC\) in the ratio 2:1, \(\overrightarrow{ON} = \overrightarrow{OB} + \frac{2}{3}(\overrightarrow{OC} - \overrightarrow{OB}) = 4\mathbf{j} + \frac{2}{3}(\mathbf{i} + \mathbf{k} - 4\mathbf{j}) = 3\mathbf{j} + \mathbf{k}\).

(b) The angle \(\theta\) between \(\overrightarrow{MD}\) and \(\overrightarrow{ON}\) is given by \(\cos \theta = \frac{\overrightarrow{MD} \cdot \overrightarrow{ON}}{|\overrightarrow{MD}| |\overrightarrow{ON}|}\).

\(\overrightarrow{MD} \cdot \overrightarrow{ON} = (-2)(0) + (-2)(3) + (3)(1) = -6 + 3 = -3\).

\(|\overrightarrow{MD}| = \sqrt{(-2)^2 + (-2)^2 + 3^2} = \sqrt{17}\), \(|\overrightarrow{ON}| = \sqrt{3^2 + 1^2} = \sqrt{10}\).

\(\cos \theta = \frac{-3}{\sqrt{17} \cdot \sqrt{10}}\).

\(\theta = \cos^{-1}\left(\frac{-3}{\sqrt{170}}\right) \approx 103.3^\circ\).

(c) To find the perpendicular distance from \(M\) to \(ON\), consider the projection of \(\overrightarrow{OM}\) on \(\overrightarrow{ON}\).

\(\text{Projection of } \overrightarrow{OM} \text{ on } \overrightarrow{ON} = \frac{\overrightarrow{OM} \cdot \overrightarrow{ON}}{|\overrightarrow{ON}|^2} \overrightarrow{ON}\).

\(\overrightarrow{OM} \cdot \overrightarrow{ON} = (2\mathbf{i} + 2\mathbf{j}) \cdot (3\mathbf{j} + \mathbf{k}) = 6\).

\(\text{Projection length} = \frac{6}{10} = \frac{3}{5}\).

\(\text{Perpendicular distance} = \sqrt{|\overrightarrow{OM}|^2 - \left(\frac{3}{5}\right)^2 |\overrightarrow{ON}|^2} = \sqrt{8 - \frac{9}{5}} = \sqrt{\frac{22}{5}}\).

Nov 2013 p13 q4

2243

The diagram shows a pyramid OABC in which the edge OC is vertical. The horizontal base OAB is a triangle, right-angled at O, and D is the mid-point of AB. The edges OA, OB and OC have lengths of 8 units, 6 units and 10 units respectively. The unit vectors i, j and k are parallel to \(\overrightarrow{OA}\), \(\overrightarrow{OB}\) and \(\overrightarrow{OC}\) respectively.

(i) Express each of the vectors \(\overrightarrow{OD}\) and \(\overrightarrow{CD}\) in terms of i, j and k.

(ii) Use a scalar product to find angle ODC.

Nov 2013 p11 q3

2244

The diagram shows a pyramid OABCD in which the vertical edge OD is 3 units in length. The point E is the centre of the horizontal rectangular base OABC. The sides OA and AB have lengths of 6 units and 4 units respectively. The unit vectors i, j and k are parallel to \(\overrightarrow{OA}\), \(\overrightarrow{OC}\) and \(\overrightarrow{OD}\) respectively.

Express each of the vectors \(\overrightarrow{DB}\) and \(\overrightarrow{DE}\) in terms of i, j and k.
Use a scalar product to find angle BDE.

June 2013 p13 q8

2245

The diagram shows a parallelogram \(OABC\) in which

\(\overrightarrow{OA} = \begin{pmatrix} 3 \\ 3 \\ -4 \end{pmatrix}\) and \(\overrightarrow{OB} = \begin{pmatrix} 5 \\ 0 \\ 2 \end{pmatrix}\).

(i) Use a scalar product to find angle \(BOC\).

(ii) Find a vector which has magnitude 35 and is parallel to the vector \(\overrightarrow{OC}\).

June 2011 p13 q5

2246

In the diagram, OABCDEFG is a rectangular block in which OA = OD = 6 cm and AB = 12 cm. The unit vectors i, j and k are parallel to \(\overrightarrow{OA}\), \(\overrightarrow{OC}\) and \(\overrightarrow{OD}\) respectively. The point P is the mid-point of DG, Q is the centre of the square face CBFG and R lies on AB such that AR = 4 cm.

(i) Express each of the vectors \(\overrightarrow{PQ}\) and \(\overrightarrow{RQ}\) in terms of i, j and k.

(ii) Use a scalar product to find angle RQP.

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June 2011 p11 q4

2247

The diagram shows a prism ABCDPQRS with a horizontal square base APSD with sides of length 6 cm. The cross-section ABCD is a trapezium and is such that the vertical edges AB and DC are of lengths 5 cm and 2 cm respectively. Unit vectors i, j and k are parallel to AD, AP and AB respectively.

(i) Express each of the vectors \(\overrightarrow{CP}\) and \(\overrightarrow{CQ}\) in terms of i, j and k.

(ii) Use a scalar product to calculate angle PCQ.

Solution

(i) To find \(\overrightarrow{CP}\), note that \(C\) is at \((6, 0, 2)\) and \(P\) is at \((0, 6, 0)\). Thus, \(\overrightarrow{CP} = (0 - 6)\mathbf{i} + (6 - 0)\mathbf{j} + (0 - 2)\mathbf{k} = -6\mathbf{i} + 6\mathbf{j} - 2\mathbf{k}\).

For \(\overrightarrow{CQ}\), \(Q\) is at \((0, 6, 5)\). Thus, \(\overrightarrow{CQ} = (0 - 6)\mathbf{i} + (6 - 0)\mathbf{j} + (5 - 2)\mathbf{k} = -6\mathbf{i} + 6\mathbf{j} + 3\mathbf{k}\).

(ii) The scalar product \(\overrightarrow{CP} \cdot \overrightarrow{CQ} = (-6)(-6) + (6)(6) + (-2)(3) = 36 + 36 - 6 = 66\).

The magnitudes are \(|\overrightarrow{CP}| = \sqrt{(-6)^2 + 6^2 + (-2)^2} = \sqrt{76}\) and \(|\overrightarrow{CQ}| = \sqrt{(-6)^2 + 6^2 + 3^2} = \sqrt{81}\).

Using the formula for the angle, \(66 = \sqrt{76} \times \sqrt{81} \times \cos \theta\), we find \(\cos \theta = \frac{66}{\sqrt{76} \times \sqrt{81}}\).

Thus, \(\theta = \cos^{-1}\left(\frac{66}{\sqrt{76} \times \sqrt{81}}\right) \approx 32.7^{\circ}\) or \(0.571 \text{ rad}\).

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Nov 2010 p13 q10

2248

The diagram shows triangle OAB, in which the position vectors of A and B with respect to O are given by \(\overrightarrow{OA} = 2\mathbf{i} + \mathbf{j} - 3\mathbf{k}\) and \(\overrightarrow{OB} = -3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}\).

C is a point on OA such that \(\overrightarrow{OC} = p \overrightarrow{OA}\), where p is a constant.

Find angle AOB. [4]

Find \(\overrightarrow{BC}\) in terms of p and vectors \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\). [1]

Find the value of p given that BC is perpendicular to OA. [4]

Solution

(i) To find angle AOB, use the dot product formula:

\(\overrightarrow{OA} \cdot \overrightarrow{OB} = (-6) + 2 + 12 = 8\)

The magnitudes are \(|\overrightarrow{OA}| = \sqrt{14}\) and \(|\overrightarrow{OB}| = \sqrt{29}\).

\(\cos AOB = \frac{8}{\sqrt{14} \sqrt{29}}\)

\(AOB = \cos^{-1}\left(\frac{8}{\sqrt{14} \sqrt{29}}\right) \approx 66.6^\circ\)

(ii) \(\overrightarrow{BC} = \overrightarrow{OB} - \overrightarrow{OC}\)

\(\overrightarrow{OC} = p(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\)

\(\overrightarrow{BC} = (-3\mathbf{i} + 2\mathbf{j} - 4\mathbf{k}) - p(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\)

\(\overrightarrow{BC} = 3\mathbf{i} - 2\mathbf{j} + 4\mathbf{k} + p(2\mathbf{i} + \mathbf{j} - 3\mathbf{k})\)

(iii) For \(BC\) to be perpendicular to \(OA\), \(\overrightarrow{BC} \cdot \overrightarrow{OA} = 0\).

\((3 + 2p) \cdot 2 + (p - 2) \cdot 1 + (4 - 3p) \cdot (-3) = 0\)

\(2(3 + 2p) + (p - 2) - 3(4 - 3p) = 0\)

\(6 + 4p + p - 2 - 12 + 9p = 0\)

\(14p - 8 = 0\)

\(p = \frac{4}{7}\) or 0.571

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Nov 2010 p12 q9

2249

The diagram shows a pyramid OABCP in which the horizontal base OABC is a square of side 10 cm and the vertex P is 10 cm vertically above O. The points D, E, F, G lie on OP, AP, BP, CP respectively and DEFG is a horizontal square of side 6 cm. The height of DEFG above the base is a cm. Unit vectors i, j and k are parallel to OA, OC and OD respectively.

Show that a = 4.

Express the vector \(\overrightarrow{BG}\) in terms of i, j and k.

Use a scalar product to find angle GBA.

Solution

(i) To find \(a\), we use the similarity of triangles. The triangle \(OPD\) is similar to \(OPA\). The ratio of the sides gives:

\(\frac{10 - a}{10} = \frac{6}{10}\)

Solving for \(a\):

\(10 - a = 6\)

\(a = 4\)

(ii) To express \(\overrightarrow{BG}\), we find the coordinates of \(B\) and \(G\). \(B\) is at \((10, 10, 0)\) and \(G\) is at \((0, 6, 4)\). Thus,

\(\overrightarrow{BG} = (0 - 10)\mathbf{i} + (6 - 10)\mathbf{j} + (4 - 0)\mathbf{k}\)

\(\overrightarrow{BG} = -10\mathbf{i} - 4\mathbf{j} + 4\mathbf{k}\)

(iii) To find angle \(GBA\), we use the scalar product:

\(\overrightarrow{BG} \cdot \overrightarrow{BA} = 40\)

\(\cos GBA = \frac{40}{\sqrt{132} \times \sqrt{100}}\)

\(GBA = \cos^{-1}\left(\frac{40}{\sqrt{13200}}\right)\)

\(GBA = 69.6^\circ\)

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Nov 2010 p11 q5

2250

The diagram shows a pyramid OABC with a horizontal base OAB where OA = 6 cm, OB = 8 cm and angle AOB = 90°. The point C is vertically above O and OC = 10 cm. Unit vectors i, j and k are parallel to OA, OB and OC as shown. Use a scalar product to find angle ACB.

Solution

First, determine the vectors \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\):
\(\overrightarrow{AC} = -6\mathbf{i} + 10\mathbf{k}\)
\(\overrightarrow{BC} = -8\mathbf{j} + 10\mathbf{k}\)
Calculate the dot product \(\overrightarrow{AC} \cdot \overrightarrow{BC}\):
\(\overrightarrow{AC} \cdot \overrightarrow{BC} = (-6)(0) + (0)(-8) + (10)(10) = 100\)
Find the magnitudes of \(\overrightarrow{AC}\) and \(\overrightarrow{BC}\):
\(|\overrightarrow{AC}| = \sqrt{(-6)^2 + 0^2 + 10^2} = \sqrt{136}\)
\(|\overrightarrow{BC}| = \sqrt{0^2 + (-8)^2 + 10^2} = \sqrt{164}\)
Use the scalar product formula to find \(\cos \angle ACB\):
\(\overrightarrow{AC} \cdot \overrightarrow{BC} = |\overrightarrow{AC}| |\overrightarrow{BC}| \cos \angle ACB\)
\(100 = \sqrt{136} \sqrt{164} \cos \angle ACB\)
\(\cos \angle ACB = \frac{100}{\sqrt{136} \sqrt{164}}\)
Calculate \(\angle ACB\):
\(\angle ACB = \cos^{-1}\left(\frac{100}{\sqrt{136} \sqrt{164}}\right) \approx 48.0^{\circ}\)

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June 2010 p11 q10

2251

The diagram shows the parallelogram OABC. Given that \(\overrightarrow{OA} = \mathbf{i} + 3\mathbf{j} + 3\mathbf{k}\) and \(\overrightarrow{OC} = 3\mathbf{i} - \mathbf{j} + \mathbf{k}\), find

the unit vector in the direction of \(\overrightarrow{OB}\),
the acute angle between the diagonals of the parallelogram,
the perimeter of the parallelogram, correct to 1 decimal place.

Solution

(i) To find \(\overrightarrow{OB}\), use \(\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{OC} = (\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) + (3\mathbf{i} - \mathbf{j} + \mathbf{k}) = 4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\).
The magnitude of \(\overrightarrow{OB}\) is \(\sqrt{4^2 + 2^2 + 4^2} = 6\).
The unit vector in the direction of \(\overrightarrow{OB}\) is \(\frac{1}{6}(4\mathbf{i} + 2\mathbf{j} + 4\mathbf{k})\).
(ii) To find \(\overrightarrow{AC}\), use \(\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = (3\mathbf{i} - \mathbf{j} + \mathbf{k}) - (\mathbf{i} + 3\mathbf{j} + 3\mathbf{k}) = 2\mathbf{i} - 4\mathbf{j} - 2\mathbf{k}\).
The magnitudes are \(|\overrightarrow{OB}| = 6\) and \(|\overrightarrow{AC}| = \sqrt{24}\).
The dot product \(\overrightarrow{AC} \cdot \overrightarrow{OB} = 8 - 8 - 8 = -8\).
Using \(\overrightarrow{AC} \cdot \overrightarrow{OB} = |\overrightarrow{AC}| |\overrightarrow{OB}| \cos \theta\), we have \(-8 = 6 \times \sqrt{24} \times \cos \theta\).
Solving for \(\theta\), \(\theta = 105.8^\circ\), so the acute angle is \(74.2^\circ\).
(iii) The magnitudes are \(|\overrightarrow{OA}| = \sqrt{19}\) and \(|\overrightarrow{OC}| = \sqrt{11}\).
The perimeter is \(2(\sqrt{19} + \sqrt{11}) \approx 15.4\).

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Nov 2009 p12 q6

2252

In the diagram, \(OABCDEFG\) is a cube in which each side has length 6. Unit vectors \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\) are parallel to \(\overrightarrow{OA}, \overrightarrow{OC}\) and \(\overrightarrow{OD}\) respectively. The point \(P\) is such that \(\overrightarrow{AP} = \frac{1}{3} \overrightarrow{AB}\) and the point \(Q\) is the mid-point of \(DF\).

(i) Express each of the vectors \(\overrightarrow{OQ}\) and \(\overrightarrow{PQ}\) in terms of \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\).

(ii) Find the angle \(OQP\).

Solution

(i) To find \(\overrightarrow{OQ}\), note that \(Q\) is the midpoint of \(DF\). Since \(D = (0, 6, 6)\) and \(F = (6, 6, 6)\), the midpoint \(Q\) is \((3, 6, 6)\). Therefore, \(\overrightarrow{OQ} = 3\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\).

For \(\overrightarrow{PQ}\), since \(P\) is such that \(\overrightarrow{AP} = \frac{1}{3} \overrightarrow{AB}\), and \(A = (6, 0, 0)\), \(B = (6, 6, 0)\), we have \(P = (6, 2, 0)\). Thus, \(\overrightarrow{PQ} = (3 - 6)\mathbf{i} + (6 - 2)\mathbf{j} + (6 - 0)\mathbf{k} = -3\mathbf{i} + \mathbf{j} + 6\mathbf{k}\).

(ii) To find the angle \(OQP\), use the dot product formula: \(\overrightarrow{OQ} \cdot \overrightarrow{PQ} = |\overrightarrow{OQ}| |\overrightarrow{PQ}| \cos \theta\).

Calculate \(\overrightarrow{OQ} \cdot \overrightarrow{PQ} = (3\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) \cdot (-3\mathbf{i} + \mathbf{j} + 6\mathbf{k}) = -9 + 3 + 36 = 30\).

Find magnitudes: \(|\overrightarrow{OQ}| = \sqrt{3^2 + 3^2 + 6^2} = \sqrt{54}\), \(|\overrightarrow{PQ}| = \sqrt{(-3)^2 + 1^2 + 6^2} = \sqrt{46}\).

Thus, \(30 = \sqrt{54} \sqrt{46} \cos \theta\).

\(\cos \theta = \frac{30}{\sqrt{54} \sqrt{46}}\).

\(\theta = \cos^{-1}\left(\frac{30}{\sqrt{54} \sqrt{46}}\right) \approx 53.0^\circ\).

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Nov 2019 p12 q7

2253

The diagram shows a three-dimensional shape OABCDEFG. The base OABC and the upper surface DEFG are identical horizontal rectangles. The parallelograms OAED and CBFG both lie in vertical planes. Points P and Q are the mid-points of OD and GF respectively. Unit vectors i and j are parallel to \(\overrightarrow{OA}\) and \(\overrightarrow{OC}\) respectively and the unit vector k is vertically upwards. The position vectors of A, C and D are given by \(\overrightarrow{OA} = 6\mathbf{i}\), \(\overrightarrow{OC} = 8\mathbf{j}\) and \(\overrightarrow{OD} = 2\mathbf{i} + 10\mathbf{k}\).

(i) Express each of the vectors \(\overrightarrow{PB}\) and \(\overrightarrow{PQ}\) in terms of i, j and k.

(ii) Determine whether P is nearer to Q or to B.

(iii) Use a scalar product to find angle BPQ.

Solution

(i) To find \(\overrightarrow{PB}\), we first find \(\overrightarrow{P}\) as the midpoint of \(\overrightarrow{OD}\):

\(\overrightarrow{P} = \frac{1}{2}(\overrightarrow{O} + \overrightarrow{D}) = \frac{1}{2}(2\mathbf{i} + 10\mathbf{k}) = \mathbf{i} + 5\mathbf{k}\)

\(\overrightarrow{B} = \overrightarrow{OA} + \overrightarrow{OC} = 6\mathbf{i} + 8\mathbf{j}\)

\(\overrightarrow{PB} = \overrightarrow{B} - \overrightarrow{P} = (6\mathbf{i} + 8\mathbf{j}) - (\mathbf{i} + 5\mathbf{k}) = 5\mathbf{i} + 8\mathbf{j} - 5\mathbf{k}\)

To find \(\overrightarrow{PQ}\), we first find \(\overrightarrow{Q}\) as the midpoint of \(\overrightarrow{GF}\):

\(\overrightarrow{Q} = \frac{1}{2}(\overrightarrow{G} + \overrightarrow{F}) = \frac{1}{2}(6\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}) = 3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\)

\(\overrightarrow{PQ} = \overrightarrow{Q} - \overrightarrow{P} = (3\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) - (\mathbf{i} + 5\mathbf{k}) = 4\mathbf{i} + 8\mathbf{j} + 5\mathbf{k}\)

(ii) Calculate the lengths:

\(\text{Length of } PB = \sqrt{5^2 + 8^2 + (-5)^2} = \sqrt{114} \approx 10.7\)

\(\text{Length of } PQ = \sqrt{4^2 + 8^2 + 5^2} = \sqrt{105} \approx 10.2\)

Since \(10.2 < 10.7\), P is nearer to Q.

(iii) Use the scalar product to find angle \(BPQ\):

\(\overrightarrow{PB} \cdot \overrightarrow{PQ} = 5 \times 4 + 8 \times 8 + (-5) \times 5 = 20 + 64 - 25 = 59\)

\(\text{Magnitude of } \overrightarrow{PB} = \sqrt{114}\), \(\text{Magnitude of } \overrightarrow{PQ} = \sqrt{105}\)

\(\cos BPQ = \frac{59}{\sqrt{114} \times \sqrt{105}}\)

\(BPQ = \cos^{-1}\left(\frac{59}{\sqrt{114} \times \sqrt{105}}\right) \approx 57.4^\circ \text{ or } 1.00 \text{ rad}\)

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Nov 2008 p1 q4

2254

The diagram shows a semicircular prism with a horizontal rectangular base \(ABCD\). The vertical ends \(AED\) and \(BFC\) are semicircles of radius 6 cm. The length of the prism is 20 cm. The mid-point of \(AD\) is the origin \(O\), the mid-point of \(BC\) is \(M\) and the mid-point of \(DC\) is \(N\). The points \(E\) and \(F\) are the highest points of the semicircular ends of the prism. The point \(P\) lies on \(EF\) such that \(EP = 8\) cm.

Unit vectors \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\) are parallel to \(OD, OM\) and \(OE\) respectively.

(i) Express each of the vectors \(\overrightarrow{PA}\) and \(\overrightarrow{PN}\) in terms of \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\).

(ii) Use a scalar product to calculate angle \(APN\).

Solution

(i) To find \(\overrightarrow{PA}\), note that \(P\) is 8 cm along \(EF\) from \(E\), so \(P\) has coordinates \((8, 0, 6)\). \(A\) is at \((0, 0, 0)\). Thus, \(\overrightarrow{PA} = (0 - 8)\mathbf{i} + (0 - 0)\mathbf{j} + (0 - 6)\mathbf{k} = -8\mathbf{i} - 6\mathbf{k}\).

For \(\overrightarrow{PN}\), \(N\) is at \((20, 0, 0)\). Thus, \(\overrightarrow{PN} = (20 - 8)\mathbf{i} + (0 - 0)\mathbf{j} + (0 - 6)\mathbf{k} = 12\mathbf{i} - 6\mathbf{k}\).

(ii) The dot product \(\overrightarrow{PA} \cdot \overrightarrow{PN} = (-8)(12) + (0)(0) + (-6)(-6) = -96 + 36 = -60\).

The magnitudes are \(|\overrightarrow{PA}| = \sqrt{(-8)^2 + 0^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\).

\(|\overrightarrow{PN}| = \sqrt{(12)^2 + 0^2 + (-6)^2} = \sqrt{144 + 36} = \sqrt{180} = 6\sqrt{5}\).

\(\cos \angle APN = \frac{-60}{10 \times 6\sqrt{5}} = \frac{-60}{60\sqrt{5}} = -\frac{1}{\sqrt{5}}\).

\(\angle APN = \cos^{-1}\left(-\frac{1}{\sqrt{5}}\right) \approx 99^\circ\).

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Nov 2007 p1 q10

2255

The diagram shows a cube OABCDEFG in which the length of each side is 4 units. The unit vectors i, j, and k are parallel to \(\overrightarrow{OA}\), \(\overrightarrow{OC}\), and \(\overrightarrow{OD}\) respectively. The mid-points of OA and DG are P and Q respectively and R is the centre of the square face ABFE.

Express each of the vectors \(\overrightarrow{PR}\) and \(\overrightarrow{PQ}\) in terms of i, j, and k.

Use a scalar product to find angle QPR.

Find the perimeter of triangle PQR, giving your answer correct to 1 decimal place.

Solution

(i) The coordinates of P are \((2, 0, 0)\) since it is the midpoint of OA. The coordinates of Q are \((0, 2, 4)\) since it is the midpoint of DG. The coordinates of R are \((2, 2, 2)\) since it is the center of the face ABFE.

\(\overrightarrow{PR} = (2 - 2)\mathbf{i} + (2 - 0)\mathbf{j} + (2 - 0)\mathbf{k} = 2\mathbf{j} + 2\mathbf{k}\)

\(\overrightarrow{PQ} = (0 - 2)\mathbf{i} + (2 - 0)\mathbf{j} + (4 - 0)\mathbf{k} = -2\mathbf{i} + 2\mathbf{j} + 4\mathbf{k}\)

(ii) The dot product \(\overrightarrow{PQ} \cdot \overrightarrow{PR} = (-2)(0) + (2)(2) + (4)(2) = 8\)

The magnitudes are \(|\overrightarrow{PQ}| = \sqrt{(-2)^2 + 2^2 + 4^2} = \sqrt{24}\) and \(|\overrightarrow{PR}| = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{8}\).

Using the dot product formula, \(\cos \theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PR}}{|\overrightarrow{PQ}| |\overrightarrow{PR}|} = \frac{8}{\sqrt{24} \cdot \sqrt{8}}\)

\(\theta = \cos^{-1}\left(\frac{8}{\sqrt{24} \cdot \sqrt{8}}\right) \approx 61.9^\circ\)

(iii) \(\overrightarrow{QR} = (2 - 0)\mathbf{i} + (2 - 2)\mathbf{j} + (2 - 4)\mathbf{k} = 2\mathbf{i} - 2\mathbf{k}\)

\(|\overrightarrow{QR}| = \sqrt{2^2 + 0^2 + (-2)^2} = \sqrt{8}\)

Perimeter of \(\triangle PQR = |\overrightarrow{PR}| + |\overrightarrow{PQ}| + |\overrightarrow{QR}| = \sqrt{8} + \sqrt{24} + \sqrt{8} \approx 12.8 \text{ cm}\)

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June 2006 p1 q8

2256

The diagram shows the roof of a house. The base of the roof, \(OABC\), is rectangular and horizontal with \(OA = CB = 14 \, \text{m}\) and \(OC = AB = 8 \, \text{m}\). The top of the roof \(DE\) is 5 m above the base and \(DE = 6 \, \text{m}\). The sloping edges \(OD, CD, AE\) and \(BE\) are all equal in length.

Unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are parallel to \(OA\) and \(OC\) respectively and the unit vector \(\mathbf{k}\) is vertically upwards.

Express the vector \(\overrightarrow{OD}\) in terms of \(\mathbf{i}, \mathbf{j}\) and \(\mathbf{k}\), and find its magnitude. [4]

Use a scalar product to find angle \(DOB\). [4]

Solution

(i) To express \(\overrightarrow{OD}\), note that \(OD\) is a diagonal of the rectangular base \(OABC\) and is elevated by 5 m. The horizontal components are half the lengths of \(OA\) and \(OC\), so \(\overrightarrow{OD} = 4\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\).

The magnitude of \(\overrightarrow{OD}\) is calculated using the Pythagorean theorem: \(\sqrt{4^2 + 4^2 + 5^2} = \sqrt{57} \approx 7.55 \, \text{m}\).

(ii) Vector \(\overrightarrow{OB} = 14\mathbf{i} + 8\mathbf{j}\).

The dot product \(\overrightarrow{OD} \cdot \overrightarrow{OB} = 4 \times 14 + 4 \times 8 = 88\).

The magnitudes are \(\sqrt{57}\) for \(\overrightarrow{OD}\) and \(\sqrt{260}\) for \(\overrightarrow{OB}\).

Using the dot product formula: \(88 = \sqrt{57} \times \sqrt{260} \times \cos \theta\).

Solving for \(\theta\), \(\cos \theta = \frac{88}{\sqrt{57} \times \sqrt{260}}\), giving \(\theta \approx 43.7^\circ\).

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Nov 2003 p1 q7

2257

The diagram shows a triangular prism with a horizontal rectangular base ADFC, where CF = 12 units and DF = 6 units. The vertical ends ABC and DEF are isosceles triangles with AB = BC = 5 units. The mid-points of BE and DF are M and N respectively. The origin O is at the mid-point of AC.

Unit vectors i, j and k are parallel to OC, ON and OB respectively.

Find the length of OB.

Express each of the vectors \(\overrightarrow{MC}\) and \(\overrightarrow{MN}\) in terms of i, j and k.

Evaluate \(\overrightarrow{MC} \cdot \overrightarrow{MN}\) and hence find angle CMN, giving your answer correct to the nearest degree.

Solution

(i) To find the length of OB, use the Pythagorean theorem in triangle OBC. Since O is the midpoint of AC, the height OB is perpendicular to AC. Given AB = 5 units and BC = 5 units, the height OB can be calculated as:

\(OB = \sqrt{AB^2 - \left(\frac{AC}{2}\right)^2} = \sqrt{5^2 - 3^2} = \sqrt{16} = 4\)

(ii) To express \(\overrightarrow{MC}\) and \(\overrightarrow{MN}\) in terms of i, j, and k:

Since M is the midpoint of BE, and N is the midpoint of DF, we have:

\(\overrightarrow{MC} = 3\mathbf{i} - 6\mathbf{j} - 4\mathbf{k}\)

\(\overrightarrow{MN} = 6\mathbf{j} - 4\mathbf{k}\)

(iii) To evaluate \(\overrightarrow{MC} \cdot \overrightarrow{MN}\) and find angle CMN:

Calculate the dot product:

\(\overrightarrow{MC} \cdot \overrightarrow{MN} = (3)(0) + (-6)(6) + (-4)(-4) = -36 + 16 = -20\)

Find the magnitudes:

\(|\overrightarrow{MC}| = \sqrt{3^2 + (-6)^2 + (-4)^2} = \sqrt{61}\)

\(|\overrightarrow{MN}| = \sqrt{0^2 + 6^2 + (-4)^2} = \sqrt{52}\)

Use the dot product formula to find the angle:

\(\overrightarrow{MC} \cdot \overrightarrow{MN} = |\overrightarrow{MC}| |\overrightarrow{MN}| \cos \theta\)

\(-20 = \sqrt{61} \sqrt{52} \cos \theta\)

\(\cos \theta = \frac{-20}{\sqrt{61} \sqrt{52}}\)

\(\theta = \cos^{-1}\left(\frac{-20}{\sqrt{61} \sqrt{52}}\right) \approx 111^\circ\)

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June 2002 p1 q5

2258

The diagram shows a solid cylinder standing on a horizontal circular base, centre O and radius 4 units. The line BA is a diameter and the radius OC is at 90° to OA. Points O', A', B' and C' lie on the upper surface of the cylinder such that OO', AA', BB' and CC' are all vertical and of length 12 units. The mid-point of BB' is M.

Unit vectors i, j and k are parallel to OA, OC and OO' respectively.

(i) Express each of the vectors \(\overrightarrow{MO}\) and \(\overrightarrow{MC}\) in terms of i, j and k.

(ii) Hence find the angle OMC.

Solution

(i) The vector \(\overrightarrow{MO}\) is expressed as \(4\mathbf{i} - 6\mathbf{k}\).

The vector \(\overrightarrow{MC}\) is expressed as \(4\mathbf{i} + 4\mathbf{j} + 6\mathbf{k}\).

(ii) To find the angle \(\angle OMC\), we use the dot product formula:

\(\overrightarrow{MO} \cdot \overrightarrow{MC} = 4 \times 4 + 0 \times 4 + (-6) \times 6 = 16 - 36 = -20\).

The magnitudes are \(|\overrightarrow{MO}| = \sqrt{4^2 + 6^2} = \sqrt{52}\) and \(|\overrightarrow{MC}| = \sqrt{4^2 + 4^2 + 6^2} = \sqrt{68}\).

Using the cosine formula: \(\overrightarrow{MO} \cdot \overrightarrow{MC} = |\overrightarrow{MO}| |\overrightarrow{MC}| \cos \theta\).

\(-20 = \sqrt{52} \times \sqrt{68} \times \cos \theta\).

\(\cos \theta = \frac{-20}{\sqrt{52} \times \sqrt{68}}\).

\(\theta = \cos^{-1}\left(\frac{-20}{\sqrt{52} \times \sqrt{68}}\right) \approx 109.7^\circ\).

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Nov 2019 p11 q10

2259

Relative to an origin O, the position vectors of the points A, B, C and D, shown in the diagram, are given by

\(\overrightarrow{OA} = \begin{pmatrix} -1 \\ 3 \\ -4 \end{pmatrix}, \overrightarrow{OB} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix}, \overrightarrow{OC} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} \text{ and } \overrightarrow{OD} = \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix}.\)

Show that AB is perpendicular to BC.

Show that ABCD is a trapezium.

Find the area of ABCD, giving your answer correct to 2 decimal places.

Solution

(i) To show that \(AB\) is perpendicular to \(BC\), calculate the vectors \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\):

\(\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} - \begin{pmatrix} -1 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} 3 \\ -6 \\ 9 \end{pmatrix}\)

\(\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ -3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 1 \\ 0 \end{pmatrix}\)

Calculate the dot product:

\(\overrightarrow{AB} \cdot \overrightarrow{BC} = 3 \times 2 + (-6) \times 1 + 9 \times 0 = 6 - 6 = 0\)

Since the dot product is zero, \(AB\) is perpendicular to \(BC\).

(ii) To show that \(ABCD\) is a trapezium, calculate \(\overrightarrow{DC}\):

\(\overrightarrow{DC} = \overrightarrow{OC} - \overrightarrow{OD} = \begin{pmatrix} 4 \\ -2 \\ 5 \end{pmatrix} - \begin{pmatrix} 2 \\ 2 \\ -1 \end{pmatrix} = \begin{pmatrix} 2 \\ -4 \\ 6 \end{pmatrix}\)

Check if \(\overrightarrow{AB}\) is parallel to \(\overrightarrow{DC}\):

\(\overrightarrow{AB} = k \overrightarrow{DC}\) where \(k = \frac{3}{2}\)

Since \(AB \parallel DC\), \(ABCD\) is a trapezium.

(iii) To find the area of \(ABCD\), calculate the magnitudes:

\(|\overrightarrow{AB}| = \sqrt{9 + 36 + 81} = \sqrt{126} = 11.22\)

\(|\overrightarrow{DC}| = \sqrt{4 + 16 + 36} = \sqrt{56} = 7.483\)

\(|\overrightarrow{BC}| = \sqrt{4 + 1 + 0} = \sqrt{5} = 2.236\)

Area = \(\frac{1}{2} (|\overrightarrow{AB}| + |\overrightarrow{DC}|) \times |\overrightarrow{BC}| = \frac{1}{2} (11.22 + 7.483) \times 2.236 = 20.92\)

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June 2019 p13 q6

2260

The diagram shows a solid figure ABCDEF in which the horizontal base ABC is a triangle right-angled at A. The lengths of AB and AC are 8 units and 4 units respectively and M is the mid-point of AB. The point D is 7 units vertically above A. Triangle DEF lies in a horizontal plane with DE, DF and FE parallel to AB, AC and CB respectively and N is the mid-point of FE. The lengths of DE and DF are 4 units and 2 units respectively. Unit vectors i, j and k are parallel to \overrightarrow{AB}, \overrightarrow{AC} and \overrightarrow{AD} respectively.

Find \overrightarrow{MF} in terms of i, j and k.

Find \overrightarrow{FN} in terms of i and j.

Find \overrightarrow{MN} in terms of i, j and k.

Use a scalar product to find angle FMN.

Solution

(i) To find \(\overrightarrow{MF}\), note that M is the midpoint of AB, so M is at \((4, 0, 0)\). F is directly above D at \((0, 4, 7)\). Thus, \(\overrightarrow{MF} = (0 - 4)\mathbf{i} + (4 - 0)\mathbf{j} + (7 - 0)\mathbf{k} = -4\mathbf{i} + 2\mathbf{j} + 7\mathbf{k}\).

(ii) N is the midpoint of FE. Since F is at \((0, 4, 7)\) and E is at \((8, 0, 0)\), N is at \((4, 2, 3.5)\). Thus, \(\overrightarrow{FN} = (4 - 0)\mathbf{i} + (2 - 4)\mathbf{j} = 2\mathbf{i} - \mathbf{j}\).

(iii) \(\overrightarrow{MN} = \overrightarrow{MF} + \overrightarrow{FN} = (-4\mathbf{i} + 2\mathbf{j} + 7\mathbf{k}) + (2\mathbf{i} - \mathbf{j}) = -2\mathbf{i} + \mathbf{j} + 7\mathbf{k}\).

(iv) To find angle FMN, use the scalar product: \(\overrightarrow{MF} \cdot \overrightarrow{MN} = -4 \times -2 + 2 \times 1 + 7 \times 7 = 59\).

The magnitudes are \(|\overrightarrow{MF}| = \sqrt{(-4)^2 + 2^2 + 7^2} = \sqrt{69}\) and \(|\overrightarrow{MN}| = \sqrt{(-2)^2 + 1^2 + 7^2} = \sqrt{54}\).

Thus, \(\cos FMN = \frac{59}{\sqrt{69} \times \sqrt{54}} = \frac{59}{\sqrt{3726}}\).

Therefore, \(FMN = \cos^{-1}\left(\frac{59}{\sqrt{3726}}\right) \approx 14.9^\circ \text{ or } 0.259 \text{ radians}\).

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June 2019 p11 q7

2261

The diagram shows a three-dimensional shape in which the base OABC and the upper surface DEFG are identical horizontal squares. The parallelograms OAED and CBFG both lie in vertical planes. The point M is the mid-point of AF.

Unit vectors i and j are parallel to OA and OC respectively and the unit vector k is vertically upwards. The position vectors of A and D are given by \(\overrightarrow{OA} = 8\mathbf{i}\) and \(\overrightarrow{OD} = 3\mathbf{i} + 10\mathbf{k}\).

(i) Express each of the vectors \(\overrightarrow{AM}\) and \(\overrightarrow{GM}\) in terms of i, j and k. [3]

(ii) Use a scalar product to find angle GMA correct to the nearest degree. [4]

Solution

(i) To find \(\overrightarrow{AM}\), note that \(M\) is the midpoint of \(AF\). Since \(A = 8\mathbf{i}\) and \(F = 8\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}\), the midpoint \(M\) is:

\(M = \frac{1}{2}(A + F) = \frac{1}{2}(8\mathbf{i} + (8\mathbf{i} + 8\mathbf{j} + 10\mathbf{k})) = \frac{1}{2}(16\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}) = 8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}\)

Thus, \(\overrightarrow{AM} = M - A = (8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) - 8\mathbf{i} = 4\mathbf{j} + 5\mathbf{k}\).

For \(\overrightarrow{GM}\), since \(G = 3\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}\), we have:

\(\overrightarrow{GM} = M - G = (8\mathbf{i} + 4\mathbf{j} + 5\mathbf{k}) - (3\mathbf{i} + 8\mathbf{j} + 10\mathbf{k}) = 5\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}\)

(ii) The scalar product \(\overrightarrow{AM} \cdot \overrightarrow{GM}\) is:

\(\overrightarrow{AM} \cdot \overrightarrow{GM} = (4\mathbf{j} + 5\mathbf{k}) \cdot (5\mathbf{i} - 4\mathbf{j} - 5\mathbf{k}) = 0 \cdot 5 + 4 \cdot (-4) + 5 \cdot (-5) = -16 - 25 = -41\)

The magnitudes are:

\(|\overrightarrow{AM}| = \sqrt{0^2 + 4^2 + 5^2} = \sqrt{41}\)

\(|\overrightarrow{GM}| = \sqrt{5^2 + (-4)^2 + (-5)^2} = \sqrt{66}\)

Using the cosine formula:

\(\overrightarrow{AM} \cdot \overrightarrow{GM} = |\overrightarrow{AM}| \cdot |\overrightarrow{GM}| \cdot \cos \theta\)

\(-41 = \sqrt{41} \cdot \sqrt{66} \cdot \cos \theta\)

\(\cos \theta = \frac{-41}{\sqrt{41} \cdot \sqrt{66}}\)

\(\theta = \cos^{-1}\left(\frac{-41}{\sqrt{41} \cdot \sqrt{66}}\right) \approx 121^{\circ}\)

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Nov 2018 p13 q6

2262

The diagram shows a solid figure OABCDEFG with a horizontal rectangular base OABC in which OA = 8 units and AB = 6 units. The rectangle DEFG lies in a horizontal plane and is such that D is 7 units vertically above O and DE is parallel to OA. The sides DE and DG have lengths 4 units and 2 units respectively. Unit vectors i, j and k are parallel to OA, OC and OD respectively. Use a scalar product to find angle OBF, giving your answer in the form cos^-1(\frac{a}{b}), where a and b are integers.

Solution

First, find the vectors BO and BF:
\(\mathbf{BO} = -8\mathbf{i} - 6\mathbf{j}\)
\(\mathbf{BF} = -6\mathbf{j} - 8\mathbf{i} + 7\mathbf{k} + 4\mathbf{i} + 2\mathbf{j} = -4\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\)
Calculate the dot product \(\mathbf{BF} \cdot \mathbf{BO}\):
\(\mathbf{BF} \cdot \mathbf{BO} = (-4)(-8) + (-4)(-6) = 32 + 24 = 56\)
Calculate the magnitudes:
\(|\mathbf{BF}| = \sqrt{(-4)^2 + (-4)^2 + 7^2} = \sqrt{16 + 16 + 49} = \sqrt{81} = 9\)
\(|\mathbf{BO}| = \sqrt{(-8)^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10\)
Use the cosine formula:
\(\cos \theta = \frac{\mathbf{BF} \cdot \mathbf{BO}}{|\mathbf{BF}| \times |\mathbf{BO}|} = \frac{56}{9 \times 10} = \frac{56}{90} = \frac{28}{45}\)
Thus, the angle \(\theta = \cos^{-1}\left( \frac{28}{45} \right)\).

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Nov 2018 p12 q7

2263

The diagram shows a solid cylinder standing on a horizontal circular base with centre O and radius 4 units. Points A, B and C lie on the circumference of the base such that AB is a diameter and angle BOC = 90^ ext{o}. Points P, Q and R lie on the upper surface of the cylinder vertically above A, B and C respectively. The height of the cylinder is 12 units. The mid-point of CR is M and N lies on BQ with BN = 4 units.

Unit vectors i and j are parallel to OB and OC respectively and the unit vector k is vertically upwards.

Evaluate \(\overrightarrow{PN} \cdot \overrightarrow{PM}\) and hence find angle MPN.

Solution

First, calculate the vectors \(\overrightarrow{PN}\) and \(\overrightarrow{PM}\).

\(\overrightarrow{PN} = 8\mathbf{i} - 8\mathbf{k}\)

\(\overrightarrow{PM} = 4\mathbf{i} + 4\mathbf{j} - 6\mathbf{k}\)

Calculate the dot product \(\overrightarrow{PN} \cdot \overrightarrow{PM}\):

\(\overrightarrow{PN} \cdot \overrightarrow{PM} = (8)(4) + (0)(4) + (-8)(-6) = 32 + 0 + 48 = 80\)

Calculate the magnitudes:

\(|\overrightarrow{PN}| = \sqrt{8^2 + 0^2 + (-8)^2} = \sqrt{128}\)

\(|\overrightarrow{PM}| = \sqrt{4^2 + 4^2 + (-6)^2} = \sqrt{68}\)

Use the cosine formula:

\(\sqrt{128} \times \sqrt{68} \times \cos M \hat{P} N = 80\)

\(\cos M \hat{P} N = \frac{80}{\sqrt{128} \times \sqrt{68}}\)

Calculate \(M \hat{P} N\):

\(M \hat{P} N = \cos^{-1}\left(\frac{80}{\sqrt{128} \times \sqrt{68}}\right) \approx 31.0^{\circ}\) awrt

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Nov 2018 p11 q8

2264

The diagram shows a solid figure OABCDEF having a horizontal rectangular base OABC with OA = 6 units and AB = 3 units. The vertical edges OF, AD and BE have lengths 6 units, 4 units and 4 units respectively. Unit vectors i, j and k are parallel to OA, OC and OF respectively.

Find \(\overrightarrow{DF}\).

Find the unit vector in the direction of \(\overrightarrow{EF}\).

Use a scalar product to find angle EFD.

Solution

(i) To find \(\overrightarrow{DF}\), note that \(D\) is at \((6, 0, 4)\) and \(F\) is at \((0, 0, 6)\). Thus, \(\overrightarrow{DF} = (0 - 6)\mathbf{i} + (0 - 0)\mathbf{j} + (6 - 4)\mathbf{k} = -6\mathbf{i} + 2\mathbf{k}\).

(ii) To find \(\overrightarrow{EF}\), note that \(E\) is at \((6, 3, 4)\) and \(F\) is at \((0, 0, 6)\). Thus, \(\overrightarrow{EF} = (0 - 6)\mathbf{i} + (0 - 3)\mathbf{j} + (6 - 4)\mathbf{k} = -6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}\).

The magnitude of \(\overrightarrow{EF}\) is \(\sqrt{(-6)^2 + (-3)^2 + 2^2} = \sqrt{49} = 7\).

The unit vector in the direction of \(\overrightarrow{EF}\) is \(\frac{1}{7}(-6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k})\).

(iii) To find angle \(EFD\), use the scalar product: \(\overrightarrow{DF} \cdot \overrightarrow{EF} = (-6\mathbf{i} + 2\mathbf{k}) \cdot (-6\mathbf{i} - 3\mathbf{j} + 2\mathbf{k}) = 36 + 4 = 40\).

The magnitudes are \(|\overrightarrow{DF}| = \sqrt{40}\) and \(|\overrightarrow{EF}| = 7\).

Thus, \(\cos \angle EFD = \frac{40}{7\sqrt{40}}\).

Therefore, \(\angle EFD = 25.4^\circ\).

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