Quick Recall (from W8-L2)
Quick check — pick the correct statement in each question before we meet today's wall.
$\vec{a}\cdot\vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3$ — multiply corresponding components and add. It is a scalar, hence the name.
$\overrightarrow{OA}\cdot\overrightarrow{OB} = (2)(1) + (1)(3) + (-1)(2) = 2 + 3 - 2 = 3$. Watch the sign of the third component: $(-1)(2) = -2$.
$\cos\theta = \dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}$, so $\theta = \cos^{-1}\!\left(\dfrac{\vec{a}\cdot\vec{b}}{|\vec{a}|\,|\vec{b}|}\right)$. Today you will use this on vectors you must first read off a 3D diagram.
The Wall
Last lesson you computed dot products for abstract position vectors like $\overrightarrow{OA} = 2\vec{i} + \vec{j} - \vec{k}$. But real exam problems do not hand you the coordinates — they describe a geometric solid and expect you to figure out the coordinates yourself.
Consider: "OABCDEFG is a cube with side 6. P is on AB with $AP = \dfrac{1}{3}AB$. Q is the midpoint of DF. Find the angle OQP."
You know how to find an angle using the scalar product. But first you must answer: What are the coordinates of every vertex? Where is P? Where is Q? Which vectors form the angle OQP? This translation — from a geometric description to a full coordinate framework — is today's focus.
The Need
We need to extend our scalar product toolkit from abstract vectors to real geometric solids. This means: (1) choosing an origin and aligning axes with edges, (2) labelling every vertex with coordinates, (3) expressing the relevant edges and diagonals as vector sums, and (4) applying the scalar product to find lengths and angles inside the solid.
💡 The Bridge (click to reveal)
For any cuboid OABCDEFG with dimensions $a \times b \times c$, we place O at the origin and align $\vec{i}, \vec{j}, \vec{k}$ with OA, OC, OD. Then every vertex has coordinates that are combinations of $a$, $b$, $c$. Any edge or diagonal is a vector sum like $\overrightarrow{OF} = a\vec{i} + b\vec{j} + c\vec{k}$. The scalar product $\overrightarrow{OA} \cdot \overrightarrow{OB} = |\overrightarrow{OA}||\overrightarrow{OB}|\cos\theta$ then gives angles inside the solid — including the angle of elevation of the spider's journey.
In the last two lessons we learned 3D vector notation and the scalar product. Today we apply them to named geometric solids — cubes, cuboids, prisms, and roof structures. The key skill is translating a geometric description into a coordinate system, then using vector operations to find lengths and angles.
For a cuboid OABCDEFG with dimensions $a \times b \times c$:
• Place O at the origin $(0, 0, 0)$.
• Align $\vec{i}$ with OA, $\vec{j}$ with OC, $\vec{k}$ with OD.
• Then: $A = (a, 0, 0)$, $C = (0, b, 0)$, $D = (0, 0, c)$, $B = (a, b, 0)$, $E = (a, 0, c)$, $F = (a, b, c)$, $G = (0, b, c)$.
For a cube, set $a = b = c = s$ (the side length). For a prism or roof, identify the base, then use symmetry to find the top vertices.
Once vertices have coordinates, any edge or diagonal is found by subtraction:
$$\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP} = (q_1 - p_1)\vec{i} + (q_2 - p_2)\vec{j} + (q_3 - p_3)\vec{k}$$
For a point on an edge: if P divides AB in ratio $t : (1-t)$, then $\overrightarrow{OP} = \overrightarrow{OA} + t\,\overrightarrow{AB}$. This lets us place points anywhere on an edge, not just at vertices.
For two vectors $\vec{a}$ and $\vec{b}$ in 3D:
$$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}$$
In a geometric solid, identify the two vectors forming the angle (e.g. $\overrightarrow{QO}$ and $\overrightarrow{QP}$ for angle OQP), compute their dot product and magnitudes, then use $\theta = \cos^{-1}\!\left(\dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|}\right)$.
- Choose an origin (usually a vertex labelled O) and align $\vec{i}, \vec{j}, \vec{k}$ with three edges.
- Write down the coordinates of every vertex.
- For points on edges, use $\overrightarrow{OP} = \overrightarrow{OA} + t\,\overrightarrow{AB}$.
- Express the required vectors by subtraction: $\overrightarrow{PQ} = \overrightarrow{OQ} - \overrightarrow{OP}$.
- For angles, identify the two vectors at the vertex, compute $\vec{a} \cdot \vec{b}$ and magnitudes, then use $\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$.
Four rounds, alternating teams. Every question uses the same cube with $O$ at the origin, so teams must keep the coordinate system straight. 30 seconds per round; 10 points per correct answer.
Five questions on a side-4 cube, 90 seconds, one attempt each. Immediate feedback and automatic progression. 10 points per correct answer.
In the diagram, OABCDEFG is a cube in which each side has length 6. Unit vectors $\vec{i}$, $\vec{j}$ and $\vec{k}$ are parallel to $\overrightarrow{OA}$, $\overrightarrow{OC}$ and $\overrightarrow{OD}$ respectively. The point P is such that $\overrightarrow{AP} = \dfrac{1}{3}\overrightarrow{AB}$ and the point Q is the midpoint of DF. (i) Express each of the vectors $\overrightarrow{OQ}$ and $\overrightarrow{PQ}$ in terms of $\vec{i}$, $\vec{j}$ and $\vec{k}$. (ii) Find the angle OQP.
👁 Reveal Solution
Side = 6. $O = (0,0,0)$, $A = (6,0,0)$, $B = (6,6,0)$, $C = (0,6,0)$, $D = (0,0,6)$, $E = (6,0,6)$, $F = (6,6,6)$, $G = (0,6,6)$.
$\overrightarrow{AB} = B - A = (0, 6, 0)$, so $\overrightarrow{AP} = \dfrac{1}{3}(0,6,0) = (0,2,0)$.
$P = A + \overrightarrow{AP} = (6, 0, 0) + (0, 2, 0) = (6, 2, 0)$.
Q = midpoint of DF $= \dfrac{(0,0,6) + (6,6,6)}{2} = (3, 3, 6)$.
$\overrightarrow{OQ} = Q - O = (3, 3, 6) = 3\vec{i} + 3\vec{j} + 6\vec{k}$.
$\overrightarrow{PQ} = Q - P = (3-6,\; 3-2,\; 6-0) = (-3, 1, 6) = -3\vec{i} + \vec{j} + 6\vec{k}$.
$\overrightarrow{QO} = O - Q = (-3, -3, -6)$, $\overrightarrow{QP} = P - Q = (3, -1, -6)$.
$\overrightarrow{QO} \cdot \overrightarrow{QP} = (-3)(3) + (-3)(-1) + (-6)(-6) = -9 + 3 + 36 = 30$.
$|\overrightarrow{QO}| = \sqrt{9 + 9 + 36} = \sqrt{54} = 3\sqrt{6}$, $|\overrightarrow{QP}| = \sqrt{9 + 1 + 36} = \sqrt{46}$.
$\cos(\angle OQP) = \dfrac{30}{3\sqrt{6} \cdot \sqrt{46}} = \dfrac{10}{\sqrt{276}} = \dfrac{10}{2\sqrt{69}} = \dfrac{5}{\sqrt{69}}$.
$\overrightarrow{OQ} = 3\vec{i} + 3\vec{j} + 6\vec{k}$, $\overrightarrow{PQ} = -3\vec{i} + \vec{j} + 6\vec{k}$.
$$\angle OQP = \cos^{-1}\!\left(\dfrac{5}{\sqrt{69}}\right) \approx 53.0^{\circ}$$
The diagram shows the roof of a house. The base of the roof, OABC, is rectangular and horizontal with OA = CB = 14 m and OC = AB = 8 m. The top of the roof DE is 5 m above the base and DE = 6 m. The sloping edges OD, CD, AE and BE are all equal in length. Unit vectors $\vec{i}$ and $\vec{j}$ are parallel to OA and OC respectively and the unit vector $\vec{k}$ is vertically upwards. (i) Express the vector $\overrightarrow{OD}$ in terms of $\vec{i}$, $\vec{j}$ and $\vec{k}$, and find its magnitude. (ii) Use a scalar product to find angle DOB.
👁 Reveal Solution
$O = (0,0,0)$, $A = (14,0,0)$, $C = (0,8,0)$, $B = (14,8,0)$.
Check: $OA = 14$, $CB = |B - C| = |(14, 0, 0)| = 14$ ✓, $OC = 8$, $AB = |B - A| = |(0, 8, 0)| = 8$ ✓.
Since $|OD| = |CD|$, D is equidistant from O and C, so D is above the midpoint of OC: $D = (x, 4, 5)$.
Similarly $|AE| = |BE|$ gives E above midpoint of AB: $E = (x', 4, 5)$.
DE = 6 and is parallel to OA (the $\vec{i}$ direction), so $x' - x = 6$.
Since $|OD| = |AE|$: $x^2 + 16 + 25 = (x' - 14)^2 + 16 + 25$, giving $x = 14 - x'$.
From $x' = x + 6$ and $x = 14 - x'$: $x = 14 - x - 6 \Rightarrow 2x = 8 \Rightarrow x = 4$.
So $D = (4, 4, 5)$, $E = (10, 4, 5)$.
$\overrightarrow{OD} = 4\vec{i} + 4\vec{j} + 5\vec{k}$.
$|\overrightarrow{OD}| = \sqrt{16 + 16 + 25} = \sqrt{57}\,\text{m}$.
$\overrightarrow{OD} = (4, 4, 5)$, $\overrightarrow{OB} = (14, 8, 0)$.
$\overrightarrow{OD} \cdot \overrightarrow{OB} = 4(14) + 4(8) + 5(0) = 56 + 32 + 0 = 88$.
$|\overrightarrow{OD}| = \sqrt{57}$, $|\overrightarrow{OB}| = \sqrt{196 + 64} = \sqrt{260} = 2\sqrt{65}$.
$\cos(\angle DOB) = \dfrac{88}{\sqrt{57} \cdot 2\sqrt{65}} = \dfrac{88}{2\sqrt{3705}} = \dfrac{44}{\sqrt{3705}}$.
$\overrightarrow{OD} = 4\vec{i} + 4\vec{j} + 5\vec{k}$, $|\overrightarrow{OD}| = \sqrt{57}\,\text{m} \approx 7.55\,\text{m}$.
$$\angle DOB = \cos^{-1}\!\left(\dfrac{44}{\sqrt{3705}}\right) \approx 43.7^{\circ}$$
Test your understanding of all four lesson objectives. Each question targets one objective — try each before moving on!
In a cuboid $OABCDEFG$, $O$ is the origin with $\overrightarrow{OA}$, $\overrightarrow{OC}$, $\overrightarrow{OD}$ along $\vec{i}$, $\vec{j}$, $\vec{k}$ respectively. If $OA = 4$, $OC = 3$, $OD = 6$, what are the coordinates of $G$ (the vertex diagonally opposite $O$)?
$G$ is the vertex opposite $O$ through the space diagonal, so its coordinates combine all three dimensions: $G = (OA, OC, OD) = (4, 3, 6)$.
$P$ divides $OA$ in the ratio $OP : PA = 1 : 2$. If $\overrightarrow{OA} = 6\vec{i}$, what is $\overrightarrow{OP}$?
$OP = \dfrac{1}{1+2} \times OA = \dfrac{1}{3} \times 6\vec{i} = 2\vec{i}$.
Given $\vec{a} = \vec{i} + 2\vec{j} + 2\vec{k}$ and $\vec{b} = 2\vec{i} + 2\vec{j} + \vec{k}$, what is $\cos\theta$ where $\theta$ is the angle between them?
$\vec{a} \cdot \vec{b} = 1(2) + 2(2) + 2(1) = 8$. $|\vec{a}| = \sqrt{1+4+4} = 3$, $|\vec{b}| = \sqrt{4+4+1} = 3$. So $\cos\theta = \dfrac{8}{3 \times 3} = \dfrac{8}{9}$.
A path goes from $A = (3, 0, 0)$ to $B = (0, 4, 12)$. What is the angle of elevation of this path?
Horizontal distance $= \sqrt{(3-0)^2 + (0-4)^2} = \sqrt{9+16} = 5$. Vertical rise $= 12$. Angle of elevation: $\tan\theta = \dfrac{\text{vertical}}{\text{horizontal}} = \dfrac{12}{5}$, so $\theta = \tan^{-1}\!\left(\dfrac{12}{5}\right)$.
The diagram shows a room with rectangular walls, floor and ceiling. A string has been stretched in a straight line between the corners A and G. The corner O is taken as the origin. A is (5, 0, 0), C is (0, 4, 0) and D is (0, 0, 3), where the lengths are in metres. A spider walks up the string, starting at A.
(i) Write down the coordinates of G.
(ii) Find the vector $\overrightarrow{AG}$ and the distance the spider walks along the string from A to G.
(iii) Find the angle of elevation of the spider's journey along the string.
💡 Hint
The room is a cuboid with $OA = 5$, $OC = 4$, $OD = 3$. G is the vertex opposite A through the space diagonal: $G = (0, 4, 3)$.
For the angle of elevation: the horizontal component of $\overrightarrow{AG}$ is $\sqrt{(-5)^2 + 4^2} = \sqrt{41}$, and the vertical component is 3. Use $\tan\theta = \dfrac{\text{vertical}}{\text{horizontal}}$.
👁 Reveal Solution
The cuboid has dimensions $5 \times 4 \times 3$. G is diagonally opposite A: $G = (0, 4, 3)$.
$\overrightarrow{AG} = G - A = (0-5,\; 4-0,\; 3-0) = (-5, 4, 3) = -5\vec{i} + 4\vec{j} + 3\vec{k}$.
$|\overrightarrow{AG}| = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}\,\text{m}$.
The horizontal component: $\sqrt{(-5)^2 + 4^2} = \sqrt{41}$.
The vertical component: $3$.
$\tan\theta = \dfrac{3}{\sqrt{41}}$, so $\theta = \tan^{-1}\!\left(\dfrac{3}{\sqrt{41}}\right) \approx 25.1^{\circ}$.
$G = (0, 4, 3)$; $\overrightarrow{AG} = -5\vec{i} + 4\vec{j} + 3\vec{k}$; distance $= 5\sqrt{2}\,\text{m} \approx 7.07\,\text{m}$; angle of elevation $\approx 25.1^{\circ}$.
Relative to an origin O, the position vectors of the points A and B are given by $\overrightarrow{OA} = 2\vec{i} - 8\vec{j} + 4\vec{k}$ and $\overrightarrow{OB} = 7\vec{i} + 2\vec{j} - \vec{k}$.
(i) Find the value of $\overrightarrow{OA} \cdot \overrightarrow{OB}$ and hence state whether angle AOB is acute, obtuse or a right angle.
(ii) The point X is such that $\overrightarrow{AX} = \dfrac{2}{5}\overrightarrow{AB}$. Find the unit vector in the direction of OX.
💡 Hint
(i) Compute the dot product: $(2)(7) + (-8)(2) + (4)(-1)$. If positive, the angle is acute; if negative, obtuse; if zero, a right angle.
(ii) First find $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$. Then $\overrightarrow{OX} = \overrightarrow{OA} + \dfrac{2}{5}\overrightarrow{AB}$. The unit vector is $\dfrac{\overrightarrow{OX}}{|\overrightarrow{OX}|}$.
👁 Reveal Solution
$\overrightarrow{OA} \cdot \overrightarrow{OB} = (2)(7) + (-8)(2) + (4)(-1) = 14 - 16 - 4 = -6$.
Since $-6 < 0$, the angle AOB is obtuse.
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (7-2,\; 2-(-8),\; -1-4) = (5, 10, -5)$.
$\overrightarrow{AX} = \dfrac{2}{5}\overrightarrow{AB} = \dfrac{2}{5}(5, 10, -5) = (2, 4, -2)$.
$\overrightarrow{OX} = \overrightarrow{OA} + \overrightarrow{AX} = (2, -8, 4) + (2, 4, -2) = (4, -4, 2)$.
$|\overrightarrow{OX}| = \sqrt{16 + 16 + 4} = \sqrt{36} = 6$.
Unit vector $= \dfrac{\overrightarrow{OX}}{|\overrightarrow{OX}|} = \dfrac{1}{6}(4\vec{i} - 4\vec{j} + 2\vec{k}) = \dfrac{2}{3}\vec{i} - \dfrac{2}{3}\vec{j} + \dfrac{1}{3}\vec{k}$.
(i) $\overrightarrow{OA} \cdot \overrightarrow{OB} = -6$; angle AOB is obtuse.
(ii) Unit vector $= \dfrac{2}{3}\vec{i} - \dfrac{2}{3}\vec{j} + \dfrac{1}{3}\vec{k}$.
The diagram shows a cube OABCDEFG in which the length of each side is 4 units. The unit vectors $\vec{i}$, $\vec{j}$ and $\vec{k}$ are parallel to $\overrightarrow{OA}$, $\overrightarrow{OC}$ and $\overrightarrow{OD}$ respectively. The midpoints of OA and DG are P and Q respectively and R is the centre of the square face ABFE.
(i) Express each of the vectors $\overrightarrow{PR}$ and $\overrightarrow{PQ}$ in terms of $\vec{i}$, $\vec{j}$ and $\vec{k}$.
(ii) Use a scalar product to find angle QPR.
(iii) Find the perimeter of triangle PQR, giving your answer correct to 1 decimal place.
💡 Hint 1
Set up coordinates (side 4): $O=(0,0,0)$, $A=(4,0,0)$, $B=(4,4,0)$, $C=(0,4,0)$, $D=(0,0,4)$, $E=(4,0,4)$, $F=(4,4,4)$, $G=(0,4,4)$.
P = midpoint of OA $= (2, 0, 0)$. Q = midpoint of DG $= (0, 2, 4)$. R = centre of ABFE: the face has vertices $A=(4,0,0)$, $B=(4,4,0)$, $F=(4,4,4)$, $E=(4,0,4)$, so $R = (4, 2, 2)$.
💡 Hint 2
For angle QPR (vertex at P), use $\overrightarrow{PQ}$ and $\overrightarrow{PR}$.
For the perimeter, you also need $|\overrightarrow{QR}|$: $R - Q = (4, 0, -2)$.
👁 Reveal Solution
P = midpoint of OA $= (2, 0, 0)$.
Q = midpoint of DG $= \dfrac{(0,0,4) + (0,4,4)}{2} = (0, 2, 4)$.
R = centre of ABFE $= \dfrac{(4,0,0) + (4,4,0) + (4,4,4) + (4,0,4)}{4} = (4, 2, 2)$.
$\overrightarrow{PR} = R - P = (4-2,\; 2-0,\; 2-0) = (2, 2, 2) = 2\vec{i} + 2\vec{j} + 2\vec{k}$.
$\overrightarrow{PQ} = Q - P = (0-2,\; 2-0,\; 4-0) = (-2, 2, 4) = -2\vec{i} + 2\vec{j} + 4\vec{k}$.
$\overrightarrow{PQ} \cdot \overrightarrow{PR} = (-2)(2) + (2)(2) + (4)(2) = -4 + 4 + 8 = 8$.
$|\overrightarrow{PQ}| = \sqrt{4+4+16} = \sqrt{24} = 2\sqrt{6}$, $|\overrightarrow{PR}| = \sqrt{4+4+4} = \sqrt{12} = 2\sqrt{3}$.
$\cos(\angle QPR) = \dfrac{8}{2\sqrt{6} \cdot 2\sqrt{3}} = \dfrac{8}{4\sqrt{18}} = \dfrac{8}{12\sqrt{2}} = \dfrac{2}{3\sqrt{2}} = \dfrac{\sqrt{2}}{3}$.
$\angle QPR = \cos^{-1}\!\left(\dfrac{\sqrt{2}}{3}\right) \approx 61.9^{\circ}$.
$|\overrightarrow{PQ}| = 2\sqrt{6} \approx 4.899$, $|\overrightarrow{PR}| = 2\sqrt{3} \approx 3.464$.
$\overrightarrow{QR} = R - Q = (4-0,\; 2-2,\; 2-4) = (4, 0, -2)$, $|\overrightarrow{QR}| = \sqrt{16+0+4} = \sqrt{20} = 2\sqrt{5} \approx 4.472$.
Perimeter $= 2\sqrt{6} + 2\sqrt{3} + 2\sqrt{5} = 2(\sqrt{6} + \sqrt{3} + \sqrt{5}) \approx 2(2.449 + 1.732 + 2.236) \approx 12.8$.
(i) $\overrightarrow{PR} = 2\vec{i} + 2\vec{j} + 2\vec{k}$, $\overrightarrow{PQ} = -2\vec{i} + 2\vec{j} + 4\vec{k}$.
(ii) $\angle QPR \approx 61.9^{\circ}$.
(iii) Perimeter $\approx 12.8$ units.
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 midpoint of AD is the origin O, the midpoint of BC is M and the midpoint 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 $\vec{i}$, $\vec{j}$ and $\vec{k}$ are parallel to OD, OM and OE respectively.
(i) Express each of the vectors $\overrightarrow{PA}$ and $\overrightarrow{PN}$ in terms of $\vec{i}$, $\vec{j}$ and $\vec{k}$.
(ii) Use a scalar product to calculate angle APN.
Cambridge International AS & A Level Mathematics 9709 Paper 1 Q4 November 2008
💡 Hint 1
Radius 6 means the diameter AD = 12. O is the midpoint of AD, so $A = (-6, 0, 0)$, $D = (6, 0, 0)$.
The prism length is 20 (along $\vec{j}$): $B = (-6, 20, 0)$, $C = (6, 20, 0)$, $M = (0, 20, 0)$.
E is the highest point above O: $E = (0, 0, 6)$. F above M: $F = (0, 20, 6)$.
N = midpoint of DC: $D = (6,0,0)$, $C = (6,20,0)$, so $N = (6, 10, 0)$.
💡 Hint 2
P lies on EF with EP = 8. Since $|EF| = 20$, P is $\dfrac{8}{20} = \dfrac{2}{5}$ of the way from E to F.
$P = E + \dfrac{2}{5}(F - E) = (0, 0, 6) + \dfrac{2}{5}(0, 20, 0) = (0, 8, 6)$.
For angle APN (vertex at P), use $\overrightarrow{PA}$ and $\overrightarrow{PN}$.
👁 Reveal Solution
Radius 6, so AD = BC = 12 (diameter). Prism length = 20.
$O = (0,0,0)$, $A = (-6, 0, 0)$, $D = (6, 0, 0)$.
$B = (-6, 20, 0)$, $C = (6, 20, 0)$, $M = (0, 20, 0)$.
$E = (0, 0, 6)$, $F = (0, 20, 6)$.
$N = \dfrac{D + C}{2} = (6, 10, 0)$.
EP = 8, $|EF| = 20$. $P = E + \dfrac{8}{20}(F - E) = (0, 0, 6) + \dfrac{2}{5}(0, 20, 0) = (0, 8, 6)$.
$\overrightarrow{PA} = A - P = (-6-0,\; 0-8,\; 0-6) = (-6, -8, -6) = -6\vec{i} - 8\vec{j} - 6\vec{k}$.
$\overrightarrow{PN} = N - P = (6-0,\; 10-8,\; 0-6) = (6, 2, -6) = 6\vec{i} + 2\vec{j} - 6\vec{k}$.
$\overrightarrow{PA} \cdot \overrightarrow{PN} = (-6)(6) + (-8)(2) + (-6)(-6) = -36 - 16 + 36 = -16$.
$|\overrightarrow{PA}| = \sqrt{36+64+36} = \sqrt{136} = 2\sqrt{34}$.
$|\overrightarrow{PN}| = \sqrt{36+4+36} = \sqrt{76} = 2\sqrt{19}$.
$\cos(\angle APN) = \dfrac{-16}{2\sqrt{34} \cdot 2\sqrt{19}} = \dfrac{-16}{4\sqrt{646}} = \dfrac{-4}{\sqrt{646}}$.
$\angle APN = \cos^{-1}\!\left(\dfrac{-4}{\sqrt{646}}\right) \approx 99.1^{\circ}$.
(i) $\overrightarrow{PA} = -6\vec{i} - 8\vec{j} - 6\vec{k}$, $\overrightarrow{PN} = 6\vec{i} + 2\vec{j} - 6\vec{k}$.
(ii) $\angle APN \approx 99.1^{\circ}$.
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest — this guides next lesson's review.