6 Quick Questions โ Covering All of Week 8
Quick check โ pick the correct option in each question. These cover L1 through L6.
$|\vec{v}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$.
$\overrightarrow{OA} \cdot \overrightarrow{OB} = (1)(3) + (2)(4) + (2)(0) = 3 + 8 + 0 = 11$. Since the dot product is positive, angle AOB is acute.
Space diagonal $= \sqrt{4^2 + 4^2 + 4^2} = \sqrt{48} = 4\sqrt{3}$.
$\vec{r} = \begin{pmatrix}1\\2\\3\end{pmatrix} + \lambda\begin{pmatrix}2\\-1\\4\end{pmatrix}$, i.e. $\vec{r} = \overrightarrow{OA} + \lambda\vec{d}$, where $\vec{a}$ is the position of point $A$ and $\vec{d}$ is the direction vector.
The direction vectors are proportional ($2:4:6 = 1:2:3$), so the lines are parallel. If they were not parallel, you would then check whether they intersect; non-parallel + no intersection = skew.
$$\text{Distance} = \dfrac{|\overrightarrow{AP} \times \vec{d}|}{|\vec{d}|}$$ where $A$ is any point on the line, $\vec{d}$ is the direction vector, and $P$ is the external point. Equivalently, find the foot $N$ on the line using $\lambda = \dfrac{\overrightarrow{AP} \cdot \vec{d}}{|\vec{d}|^2}$, then compute $|\overrightarrow{PN}|$.
A student is asked to find the perpendicular distance from $P(-7, -2, 13)$ to the line $\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 3 \\ -4 \end{pmatrix}$. They write:
$\overrightarrow{AP} = \begin{pmatrix} -8 \\ -4 \\ 8 \end{pmatrix}, \quad \overrightarrow{AP} \times \vec{d} = \begin{pmatrix} -8 \\ -24 \\ -20 \end{pmatrix}$
$\text{Distance} = |\overrightarrow{AP} \times \vec{d}| = \sqrt{64 + 576 + 400} = \sqrt{1040} = 4\sqrt{65}$
What went wrong?
๐ Reveal the Error
The cross product computation is correct: $\overrightarrow{AP} \times \vec{d} = (-8, -24, -20)$ and $|\overrightarrow{AP} \times \vec{d}| = 4\sqrt{65}$.
But the student forgot to divide by $|\vec{d}|$. The perpendicular distance formula is:
$$\text{Distance} = \dfrac{|\overrightarrow{AP} \times \vec{d}|}{|\vec{d}|} = \dfrac{4\sqrt{65}}{\sqrt{26}} = \dfrac{4\sqrt{65}}{\sqrt{26}} = 2\sqrt{10} \approx 6.32$$
The value $4\sqrt{65} \approx 32.25$ is far too large โ it is the area of the parallelogram spanned by $\overrightarrow{AP}$ and $\vec{d}$, not the perpendicular distance. Always check: does your answer have the $|\vec{d}|$ in the denominator?
Secondary trap: Students also confuse skew lines (non-parallel, non-intersecting) with parallel lines. Two lines are parallel only if their direction vectors are proportional. If they are not proportional AND do not intersect, they are skew โ not parallel.
Today's Focus
Consolidate the entire Week 8 vector toolkit โ from components and magnitude through to the perpendicular distance formula. We will apply these skills to exam-style problems involving point-to-line distance, angle conditions, and skew line verification.
A 3D vector is written $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$, or as a column vector $\begin{pmatrix} x \\ y \\ z \end{pmatrix}$. Its magnitude is found by double Pythagoras:
The unit vector in the direction of $\vec{v}$ is $\hat{v} = \dfrac{\vec{v}}{|\vec{v}|}$.
The scalar (dot) product of $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is:
The angle between two vectors: $\cos\theta = \dfrac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$. Two non-zero vectors are perpendicular iff $\vec{a} \cdot \vec{b} = 0$.
In cuboids, cubes, and roofs, express all key vectors (face diagonals, space diagonals, edge vectors) in terms of the three edge vectors $\vec{p}, \vec{q}, \vec{r}$. Use the scalar product to find angles between any two vectors.
A line through point $A$ with position vector $\vec{a}$, in the direction $\vec{d}$, has equation:
The parameter $\lambda$ determines the position along the line. A point $P$ lies on the line iff $\overrightarrow{OP} = \vec{a} + \lambda \vec{d}$ for some $\lambda$.
Two lines $\vec{r} = \vec{a}_1 + \lambda \vec{d}_1$ and $\vec{r} = \vec{a}_2 + \mu \vec{d}_2$:
Parallel iff $\vec{d}_1 = k\vec{d}_2$ (direction vectors proportional).
Intersecting iff the system $\vec{a}_1 + \lambda \vec{d}_1 = \vec{a}_2 + \mu \vec{d}_2$ has a consistent solution for $\lambda, \mu$.
Skew iff not parallel AND no intersection (the system is inconsistent).
The perpendicular distance from external point $P$ to the line $\vec{r} = \vec{a} + \lambda \vec{d}$ is:
where $A$ is any point on the line (use $\vec{a}$), $\vec{d}$ is the direction vector, and $\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA}$.
Alternative method: Find the foot $N$ on the line using $\lambda_N = \dfrac{\overrightarrow{AP} \cdot \vec{d}}{|\vec{d}|^2}$, then $N = \vec{a} + \lambda_N \vec{d}$, and the distance is $|\overrightarrow{PN}|$.
Watch out: Never forget the $|\vec{d}|$ in the denominator! $|\overrightarrow{AP} \times \vec{d}|$ alone gives the area of a parallelogram, not a distance.
Four rounds, alternating teams. Each round asks for a perpendicular distance or the formula behind it; teams have 30 seconds and score 10 points for a correct answer.
Five distance questions, 90 seconds, one attempt each. Immediate colour feedback and automatic progression. 10 points per correct answer.
Find the perpendicular distance of the point $P(-7, -2, 13)$ to the line $\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 3 \\ -4 \end{pmatrix}$.
๐ Reveal Solution
$A = (1, 2, 5)$, โ $\vec{d} = \begin{pmatrix} 1 \\ 3 \\ -4 \end{pmatrix}$, โ $P = (-7, -2, 13)$.
$\overrightarrow{AP} = \overrightarrow{OP} - \overrightarrow{OA} = \begin{pmatrix} -7-1 \\ -2-2 \\ 13-5 \end{pmatrix} = \begin{pmatrix} -8 \\ -4 \\ 8 \end{pmatrix}$.
$\overrightarrow{AP} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -8 & -4 & 8 \\ 1 & 3 & -4 \end{vmatrix}$
$= \hat{i}[(-4)(-4) - (8)(3)] - \hat{j}[(-8)(-4) - (8)(1)] + \hat{k}[(-8)(3) - (-4)(1)]$
$= \hat{i}(16 - 24) - \hat{j}(32 - 8) + \hat{k}(-24 + 4)$
$= \begin{pmatrix} -8 \\ -24 \\ -20 \end{pmatrix}$.
$|\overrightarrow{AP} \times \vec{d}| = \sqrt{(-8)^2 + (-24)^2 + (-20)^2} = \sqrt{64 + 576 + 400} = \sqrt{1040} = \sqrt{16 \times 65} = 4\sqrt{65}$.
$|\vec{d}| = \sqrt{1 + 9 + 16} = \sqrt{26}$.
$$\text{Distance} = \dfrac{4\sqrt{65}}{\sqrt{26}} = \dfrac{4\sqrt{65 \div 26}}{1} = 4\sqrt{\dfrac{5}{2}} = \dfrac{4\sqrt{10}}{2} = 2\sqrt{10}$$
$$\boxed{\text{Perpendicular distance} = 2\sqrt{10} \approx 6.32}$$
Find the foot $N$ using $\lambda_N = \dfrac{\overrightarrow{AP} \cdot \vec{d}}{|\vec{d}|^2}$:
$\overrightarrow{AP} \cdot \vec{d} = (-8)(1) + (-4)(3) + (8)(-4) = -8 - 12 - 32 = -52$.
$|\vec{d}|^2 = 26$, โ so $\lambda_N = \dfrac{-52}{26} = -2$.
$N = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix} + (-2)\begin{pmatrix} 1 \\ 3 \\ -4 \end{pmatrix} = \begin{pmatrix} -1 \\ -4 \\ 13 \end{pmatrix}$.
$\overrightarrow{PN} = N - P = \begin{pmatrix} -1-(-7) \\ -4-(-2) \\ 13-13 \end{pmatrix} = \begin{pmatrix} 6 \\ -2 \\ 0 \end{pmatrix}$.
$|\overrightarrow{PN}| = \sqrt{36 + 4 + 0} = \sqrt{40} = 2\sqrt{10}$. โ $\checkmark$ Both methods agree.
The points A and B have position vectors, relative to the origin O, given by $\overrightarrow{OA} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\overrightarrow{OB} = 2\hat{i} + \hat{j} + 3\hat{k}$. The line $l$ has vector equation $\vec{r} = (1 - 2t)\hat{i} + (5 + t)\hat{j} + (2 - t)\hat{k}$.
(i) Show that $l$ does not intersect the line passing through A and B.
(ii) The point P lies on $l$ and is such that angle PAB is equal to $60^\circ$. Given that the position vector of P is $(1 - 2t)\hat{i} + (5 + t)\hat{j} + (2 - t)\hat{k}$, show that $3t^2 + 7t + 2 = 0$. Hence find the only possible position vector of P.
๐ Reveal Solution
Line through A and B: $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (2-1)\hat{i} + (1-2)\hat{j} + (3-3)\hat{k} = \hat{i} - \hat{j}$.
Equation: $\vec{r} = \begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix} + s\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$.
Line $l$: $\vec{r} = \begin{pmatrix} 1 \\ 5 \\ 2 \end{pmatrix} + t\begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix}$.
Direction vectors $\begin{pmatrix} 1 \\ -1 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} -2 \\ 1 \\ -1 \end{pmatrix}$ are not proportional, so the lines are not parallel.
For intersection, set equal:
$1 + s = 1 - 2t \implies s = -2t$ โ (1)
$2 - s = 5 + t \implies s = -3 - t$ โ (2)
$3 = 2 - t \implies t = -1$ โ (3)
From (3): $t = -1$. From (1): $s = -2(-1) = 2$. From (2): $s = -3-(-1) = -2$.
But $s = 2$ and $s = -2$ is a contradiction. The system is inconsistent, so the lines do not intersect.
Since the lines are not parallel and do not intersect, they are skew. $\blacksquare$
$\overrightarrow{AP} = P - A = (1-2t-1)\hat{i} + (5+t-2)\hat{j} + (2-t-3)\hat{k} = -2t\,\hat{i} + (3+t)\,\hat{j} + (-1-t)\,\hat{k}$.
$\overrightarrow{AB} = \hat{i} - \hat{j}$, โ $|\overrightarrow{AB}| = \sqrt{2}$.
$\overrightarrow{AP} \cdot \overrightarrow{AB} = (-2t)(1) + (3+t)(-1) + (-1-t)(0) = -2t - 3 - t = -3(t+1)$.
$|\overrightarrow{AP}|^2 = 4t^2 + (3+t)^2 + (1+t)^2 = 4t^2 + 9 + 6t + t^2 + 1 + 2t + t^2 = 6t^2 + 8t + 10$.
Using $\cos 60^\circ = \dfrac{1}{2}$:
$$\dfrac{-3(t+1)}{\sqrt{6t^2+8t+10}\,\sqrt{2}} = \dfrac{1}{2}$$
$$-6(t+1) = \sqrt{2}\,\sqrt{6t^2+8t+10}$$
Squaring both sides:
$$36(t+1)^2 = 2(6t^2+8t+10)$$
$$36t^2 + 72t + 36 = 12t^2 + 16t + 20$$
$$24t^2 + 56t + 16 = 0$$
$$\boxed{3t^2 + 7t + 2 = 0} \quad \checkmark$$
$3t^2 + 7t + 2 = 0 \implies (3t+1)(t+2) = 0 \implies t = -\dfrac{1}{3}$ or $t = -2$.
Check $t = -\frac{1}{3}$: $\overrightarrow{AP} \cdot \overrightarrow{AB} = -3(-\frac{1}{3}+1) = -3(\frac{2}{3}) = -2 < 0$, so $\cos\theta < 0$, giving $\theta = 120^\circ \neq 60^\circ$. Rejected.
Check $t = -2$: $\overrightarrow{AP} \cdot \overrightarrow{AB} = -3(-2+1) = -3(-1) = 3 > 0$, so $\cos\theta > 0$, giving $\theta = 60^\circ$. Valid.
$P = (1-2(-2))\hat{i} + (5+(-2))\hat{j} + (2-(-2))\hat{k} = 5\hat{i} + 3\hat{j} + 4\hat{k}$.
$$\boxed{\overrightarrow{OP} = 5\hat{i} + 3\hat{j} + 4\hat{k}}$$
Attempt each question on paper first, then click Reveal Solution to check your work. Questions are ordered from easiest to hardest.
For each point P and line l find (a) the coordinates of the point N on the line such that PN is perpendicular to the line (b) the distance PN.
(i) $P(-2, 11, 5)$ and $\vec{r} = \begin{pmatrix} 0 \\ 2 \\ -3 \end{pmatrix} + t\begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix}$
๐ก Hint
Use $\lambda_N = \dfrac{\overrightarrow{AP} \cdot \vec{d}}{|\vec{d}|^2}$ to find the parameter of the foot $N$, then compute $N = A + \lambda_N \vec{d}$ and $|\overrightarrow{PN}|$.
๐ Reveal Solution
$A = (0, 2, -3)$, โ $\vec{d} = \begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix}$, โ $P = (-2, 11, 5)$.
$\overrightarrow{AP} = \begin{pmatrix} -2-0 \\ 11-2 \\ 5-(-3) \end{pmatrix} = \begin{pmatrix} -2 \\ 9 \\ 8 \end{pmatrix}$.
$\overrightarrow{AP} \cdot \vec{d} = (-2)(-1) + (9)(2) + (8)(5) = 2 + 18 + 40 = 60$.
$|\vec{d}|^2 = 1 + 4 + 25 = 30$.
$\lambda_N = \dfrac{60}{30} = 2$.
$N = \begin{pmatrix} 0 \\ 2 \\ -3 \end{pmatrix} + 2\begin{pmatrix} -1 \\ 2 \\ 5 \end{pmatrix} = \begin{pmatrix} 0-2 \\ 2+4 \\ -3+10 \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ 7 \end{pmatrix}$.
$$\boxed{N = (-2, 6, 7)}$$
$\overrightarrow{PN} = N - P = \begin{pmatrix} -2-(-2) \\ 6-11 \\ 7-5 \end{pmatrix} = \begin{pmatrix} 0 \\ -5 \\ 2 \end{pmatrix}$.
$|\overrightarrow{PN}| = \sqrt{0 + 25 + 4} = \sqrt{29}$.
$$\boxed{PN = \sqrt{29} \approx 5.39}$$
Find the distance of the point $C(0, 6, 0)$ to the line joining the points $A(-4, 2, -3)$ and $B(-2, 0, 1)$.
๐ก Hint
The line through A and B has direction $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$. Use the perpendicular distance formula $\dfrac{|\overrightarrow{AC} \times \overrightarrow{AB}|}{|\overrightarrow{AB}|}$.
๐ Reveal Solution
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \begin{pmatrix} -2-(-4) \\ 0-2 \\ 1-(-3) \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 4 \end{pmatrix}$.
$\overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \begin{pmatrix} 0-(-4) \\ 6-2 \\ 0-(-3) \end{pmatrix} = \begin{pmatrix} 4 \\ 4 \\ 3 \end{pmatrix}$.
$\overrightarrow{AC} \times \overrightarrow{AB} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 4 & 4 & 3 \\ 2 & -2 & 4 \end{vmatrix}$
$= \hat{i}(16+6) - \hat{j}(16-6) + \hat{k}(-8-8) = \begin{pmatrix} 22 \\ -10 \\ -16 \end{pmatrix}$.
$|\overrightarrow{AC} \times \overrightarrow{AB}| = \sqrt{484 + 100 + 256} = \sqrt{840} = 2\sqrt{210}$.
$|\overrightarrow{AB}| = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$.
$$\text{Distance} = \dfrac{2\sqrt{210}}{2\sqrt{6}} = \sqrt{\dfrac{210}{6}} = \sqrt{35}$$
$$\boxed{\text{Distance} = \sqrt{35} \approx 5.92}$$
With respect to the origin O, the position vectors of two points A and B are given by $\overrightarrow{OA} = \hat{i} + 2\hat{j} + 2\hat{k}$ and $\overrightarrow{OB} = 3\hat{i} + 4\hat{j}$. The point P lies on the line through A and B, and $\overrightarrow{AP} = \lambda \overrightarrow{AB}$.
(i) Show that $\overrightarrow{OP} = (1 + 2\lambda)\hat{i} + (2 + 2\lambda)\hat{j} + (2 - 2\lambda)\hat{k}$.
(ii) By equating expressions for $\cos \text{AOP}$ and $\cos \text{BOP}$ in terms of $\lambda$, find the value of $\lambda$ for which OP bisects the angle AOB.
(iii) When $\lambda$ has this value, verify that AP : PB = OA : OB.
๐ก Hint 1
(i) Use $\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \overrightarrow{OA} + \lambda\overrightarrow{AB}$. First compute $\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA}$.
(ii) If OP bisects $\angle AOB$, then $\angle AOP = \angle BOP$, so $\cos(\angle AOP) = \cos(\angle BOP)$. Use the dot product formula for each and set them equal.
๐ก Hint 2
(iii) When $\lambda = \frac{3}{8}$: $AP = \lambda|AB|$ and $PB = (1-\lambda)|AB|$, so $AP:PB = \lambda:(1-\lambda)$. Compare with $OA:OB = |\overrightarrow{OA}|:|\overrightarrow{OB}|$.
๐ Reveal Solution
$\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = (3-1)\hat{i} + (4-2)\hat{j} + (0-2)\hat{k} = 2\hat{i} + 2\hat{j} - 2\hat{k}$.
$\overrightarrow{OP} = \overrightarrow{OA} + \overrightarrow{AP} = \overrightarrow{OA} + \lambda\overrightarrow{AB}$
$= (\hat{i} + 2\hat{j} + 2\hat{k}) + \lambda(2\hat{i} + 2\hat{j} - 2\hat{k})$
$= (1+2\lambda)\hat{i} + (2+2\lambda)\hat{j} + (2-2\lambda)\hat{k}$. โ $\checkmark$
$\overrightarrow{OA} \cdot \overrightarrow{OP} = 1(1+2\lambda) + 2(2+2\lambda) + 2(2-2\lambda) = 1+2\lambda+4+4\lambda+4-4\lambda = 9+2\lambda$.
$\overrightarrow{OB} \cdot \overrightarrow{OP} = 3(1+2\lambda) + 4(2+2\lambda) + 0(2-2\lambda) = 3+6\lambda+8+8\lambda = 11+14\lambda$.
$|\overrightarrow{OA}| = \sqrt{1+4+4} = 3$, โ $|\overrightarrow{OB}| = \sqrt{9+16} = 5$.
$\cos(\angle AOP) = \dfrac{9+2\lambda}{3|\overrightarrow{OP}|}$, โ $\cos(\angle BOP) = \dfrac{11+14\lambda}{5|\overrightarrow{OP}|}$.
Setting equal (OP bisects the angle):
$$\dfrac{9+2\lambda}{3} = \dfrac{11+14\lambda}{5}$$
$$5(9+2\lambda) = 3(11+14\lambda)$$
$$45 + 10\lambda = 33 + 42\lambda$$
$$12 = 32\lambda$$
$$\boxed{\lambda = \dfrac{3}{8}}$$
When $\lambda = \frac{3}{8}$:
$AP = \lambda|\overrightarrow{AB}| = \frac{3}{8}|\overrightarrow{AB}|$, โ $PB = (1-\lambda)|\overrightarrow{AB}| = \frac{5}{8}|\overrightarrow{AB}|$.
$AP:PB = \frac{3}{8}:\frac{5}{8} = 3:5$.
$OA:OB = 3:5$. โ $\checkmark$
$$\boxed{AP:PB = OA:OB = 3:5}$$
The straight line $l_1$ passes through the points $(0, 1, 5)$ and $(2, -2, 1)$. The straight line $l_2$ has equation $\vec{r} = 7\hat{i} + \hat{j} + \hat{k} + \mu(\hat{i} + 2\hat{j} + 5\hat{k})$.
(i) Show that the lines $l_1$ and $l_2$ are skew.
(ii) Find the acute angle between the direction of the line $l_2$ and the direction of the $x$-axis.
๐ก Hint 1
(i) First find the direction of $l_1$: $\vec{d}_1 = (2-0, -2-1, 1-5) = (2, -3, -4)$. The direction of $l_2$ is $\vec{d}_2 = (1, 2, 5)$. Check they are not proportional. Then set up the intersection system with two parameters $\lambda, \mu$ and show it is inconsistent.
๐ก Hint 2
(ii) The $x$-axis has direction $\hat{i} = (1, 0, 0)$. Use $\cos\theta = \dfrac{\vec{d}_2 \cdot \hat{i}}{|\vec{d}_2||\hat{i}|} = \dfrac{1}{\sqrt{30}}$.
๐ Reveal Solution
Direction of $l_1$: $\vec{d}_1 = (2-0, -2-1, 1-5) = \begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}$.
Direction of $l_2$: $\vec{d}_2 = \begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}$.
Check proportionality: $\dfrac{2}{1} = 2$, $\dfrac{-3}{2} = -1.5$. Since $2 \neq -1.5$, the direction vectors are not proportional, so the lines are not parallel.
Equation of $l_1$: $\vec{r} = \begin{pmatrix} 0 \\ 1 \\ 5 \end{pmatrix} + \lambda\begin{pmatrix} 2 \\ -3 \\ -4 \end{pmatrix}$.
Equation of $l_2$: $\vec{r} = \begin{pmatrix} 7 \\ 1 \\ 1 \end{pmatrix} + \mu\begin{pmatrix} 1 \\ 2 \\ 5 \end{pmatrix}$.
For intersection, set components equal:
$2\lambda = 7 + \mu$ โ (1)
$1 - 3\lambda = 1 + 2\mu \implies -3\lambda = 2\mu \implies \mu = -\frac{3}{2}\lambda$ โ (2)
$5 - 4\lambda = 1 + 5\mu$ โ (3)
Substitute (2) into (1): $2\lambda = 7 - \frac{3}{2}\lambda \implies \frac{7}{2}\lambda = 7 \implies \lambda = 2$.
Then $\mu = -\frac{3}{2}(2) = -3$.
Check in (3): LHS $= 5 - 4(2) = -3$. RHS $= 1 + 5(-3) = -14$.
$-3 \neq -14$ โ inconsistent! The system has no solution, so the lines do not intersect.
Not parallel + no intersection $\implies$ $l_1$ and $l_2$ are skew. $\blacksquare$
The $x$-axis has direction $\hat{i} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$.
$\vec{d}_2 \cdot \hat{i} = (1)(1) + (2)(0) + (5)(0) = 1$.
$|\vec{d}_2| = \sqrt{1+4+25} = \sqrt{30}$, โ $|\hat{i}| = 1$.
$\cos\theta = \dfrac{\vec{d}_2 \cdot \hat{i}}{|\vec{d}_2||\hat{i}|} = \dfrac{1}{\sqrt{30}}$.
$$\theta = \cos^{-1}\!\left(\dfrac{1}{\sqrt{30}}\right) \approx 79.5^\circ$$
Since $79.5^\circ < 90^\circ$, this is already the acute angle.
$$\boxed{\theta = \cos^{-1}\!\left(\dfrac{1}{\sqrt{30}}\right) \approx 79.5^\circ}$$
Exit Ticket
Before you leave, tick the objectives you are confident you can achieve. Be honest โ this guides future review.