The resultant of the vectors $\displaystyle \vec{a} = 4\hat{\text{i}} - 3\hat{\text{j}}$ and $\displaystyle \vec{b} = 2p\hat{\text{i}} - p\hat{\text{j}}$ is parallel to the vector $\displaystyle \vec{c} = 2\hat{\text{i}} - 3\hat{\text{j}}$. Find:
(a) the value of $\displaystyle p$
(b) the resultant of vectors $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$.
The vector $\displaystyle \vec{a} = p\hat{\text{i}} + q\hat{\text{j}}$, where $\displaystyle p$ and $\displaystyle q$ are positive constants, is such that $\displaystyle |\vec{a}| = 15$. Given that $\displaystyle \vec{a}$ makes an angle of $\displaystyle 55^\circ$ with $\displaystyle \hat{\text{i}}$, find the values of $\displaystyle p$ and $\displaystyle q$.
The points $\displaystyle A(2, 7, 3)$ and $\displaystyle B(4, 3, 5)$ are joined to form the line segment AB. The point M is the midpoint of AB. Find the distance from M to the point $\displaystyle C(5, 8, 7)$.
Show Solution
Solution
$\displaystyle \begin{aligned}
& \text{Coordinates of } M = \left(\frac{2+4}{2}, \frac{7+3}{2}, \frac{3+5}{2}\right) = (3, 5, 4) \\
& \text{Distance from } M \text{ to } C = \sqrt{(5 - 3)^2 + (8 - 5)^2 + (7 - 4)^2} \\
& \text{Distance} = \sqrt{4 + 9 + 9} \\
& \text{Distance} = \sqrt{22} \\
\end{aligned}$
Problem 7
The coordinates of P and Q are $\displaystyle (2, 3, a)$ and $\displaystyle (a-2, 6, 7)$. Given that the distance from P to Q is $\displaystyle \sqrt{14}$, find the possible values of $\displaystyle a$.
$\displaystyle \overrightarrow{AB}$ is the vector $\displaystyle -3\hat{\text{i}} + t\hat{\text{j}} + 5\hat{\text{k}}$, where $\displaystyle t > 0$. Given that $\displaystyle |\overrightarrow{AB}| = 5\sqrt{2}$, show that $\displaystyle \overrightarrow{AB}$ is parallel to $\displaystyle 6\hat{\text{i}} - 8\hat{\text{j}} - \frac{5}{2}t\hat{\text{k}}$.
The points D, E and F have position vectors $\displaystyle \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix}$, $\displaystyle \begin{pmatrix} 5 \\ 3 \\ 4 \end{pmatrix}$ and $\displaystyle \begin{pmatrix} 2 \\ -1 \\ 8 \end{pmatrix}$ respectively.
(a) Find the vectors $\displaystyle \overrightarrow{DE}$, $\displaystyle \overrightarrow{EF}$ and $\displaystyle \overrightarrow{FD}$.
(b) Find $\displaystyle |\overrightarrow{DE}|$, $\displaystyle |\overrightarrow{EF}|$ and $\displaystyle |\overrightarrow{FD}|$ giving your answers in exact form.
Given that $\displaystyle \overrightarrow{AB} = \begin{pmatrix} 6 \\ -2 \\ 11 \end{pmatrix}$ and $\displaystyle \overrightarrow{AC} = \begin{pmatrix} 15 \\ 8 \\ 5 \end{pmatrix}$, find the area of the parallelogram.
The line $\displaystyle l$ passes through the points A and B with position vectors $\displaystyle \hat{\text{i}} - \hat{\text{j}} + 3\hat{\text{k}}$ and $\displaystyle \hat{\text{i}} + 2\hat{\text{j}} + 2\hat{\text{k}}$ respectively, relative to a fixed origin O.
(a) Find a vector equation of the line $\displaystyle l$.
(b) Find the position vector of the point C which lies on the line segment AB such that $\displaystyle AC = 2CB$.
Find a vector equation of the straight line which passes through the point A with position vector $\displaystyle 2\hat{\text{i}} + 3\hat{\text{j}} - 4\hat{\text{k}}$, and is parallel to the vector $\displaystyle 2\hat{\text{j}} + 3\hat{\text{k}}$.
Show Solution
Solution
$\displaystyle \begin{aligned}
& \text{Let the position vector of point A be } \vec{a}. \\
& \vec{a} = 2\hat{\text{i}} + 3\hat{\text{j}} - 4\hat{\text{k}} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} \\
& \text{Let the direction vector be } \vec{b}. \\
& \vec{b} = 2\hat{\text{j}} + 3\hat{\text{k}} = \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \\
& \text{The vector equation of a line is given by } \vec{r} = \vec{a} + \lambda\vec{b} \\
& \vec{r} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix} + \lambda \begin{pmatrix} 0 \\ 2 \\ 3 \end{pmatrix} \\
& \text{or equivalently, } \\
& \vec{r} = 2\hat{\text{i}} + 3\hat{\text{j}} - 4\hat{\text{k}} + \lambda(2\hat{\text{j}} + 3\hat{\text{k}}) \\
\end{aligned}$
Problem 16
A straight line $\displaystyle l$ has vector equation $\displaystyle \vec{r} = (\hat{\text{i}} + 2\hat{\text{j}} - \hat{\text{k}}) + \lambda(3\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}})$.
Show that another vector equation of $\displaystyle l$ is $\displaystyle \vec{r} = (7\hat{\text{i}} + 4\hat{\text{j}} - 5\hat{\text{k}}) + \mu(9\hat{\text{i}} + 3\hat{\text{j}} - 6\hat{\text{k}})$.
Show Solution
Solution
$\displaystyle \begin{aligned}
& \text{When } \lambda = 2: \\
& \vec{r} = \hat{\text{i}} + 2\hat{\text{j}} - \hat{\text{k}} + 2(3\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}}) \\
& \vec{r} = 7\hat{\text{i}} + 4\hat{\text{j}} - 5\hat{\text{k}} \\
& \text{This shows the point } (7, 4, -5) \text{ lies on the original line.} \\
& \text{To show the original point } (1, 2, -1) \text{ lies on the new line, substitute it into the new equation:} \\
& \hat{\text{i}} + 2\hat{\text{j}} - \hat{\text{k}} = (7\hat{\text{i}} + 4\hat{\text{j}} - 5\hat{\text{k}}) + \mu(9\hat{\text{i}} + 3\hat{\text{j}} - 6\hat{\text{k}}) \\
& \text{Equating components:} \\
& 1 = 7 + 9\mu \implies 9\mu = -6 \implies \mu = \displaystyle -\frac{2}{3} \\
& \text{Check y-component: } 4 + 3\left(\displaystyle -\frac{2}{3}\right) = 4 - 2 = 2 \\
& \text{Check z-component: } -5 - 6\left(\displaystyle -\frac{2}{3}\right) = -5 + 4 = -1 \\
& \text{Since } \mu = \displaystyle -\frac{2}{3} \text{ satisfies all components, the point } (1, 2, -1) \text{ lies on the new line.} \\
& \text{Also, the direction vectors are \parallel since:} \\
& 9\hat{\text{i}} + 3\hat{\text{j}} - 6\hat{\text{k}} = 3(3\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}}) \\
& \text{Therefore, the two equations represent the same line.} \\
\end{aligned}$
Problem 17
With respect to an origin O, the position vectors of the points L, M and N are $\displaystyle \begin{pmatrix} 4 \\ 7 \\ 7 \end{pmatrix}$, $\displaystyle \begin{pmatrix} 1 \\ 3 \\ 2 \end{pmatrix}$ and $\displaystyle \begin{pmatrix} 2 \\ 4 \\ 6 \end{pmatrix}$ respectively.
(a) Find the vectors $\displaystyle \overrightarrow{ML}$ and $\displaystyle \overrightarrow{MN}$.
(b) Prove that $\displaystyle \cos \angle LMN = \frac{9}{10}$.
Referred to a fixed origin O, the points A, B and C have position vectors $\displaystyle 9\hat{\text{i}} - 2\hat{\text{j}} + \hat{\text{k}}$, $\displaystyle 6\hat{\text{i}} + 2\hat{\text{j}} + 6\hat{\text{k}}$ and $\displaystyle 3\hat{\text{i}} + p\hat{\text{j}} + q\hat{\text{k}}$ respectively, where p and q are constants.
(a) Find, in vector form, an equation of the line $\displaystyle l$ which passes through A and B.
Given that C lies on $\displaystyle l$,
(b) find the value of p and the value of q.
(c) calculate, in degrees, the acute angle between OC and AB.
The point D lies on AB and is such that OD is perpendicular to AB.
Referred to a fixed origin O, the points A and B have position vectors $\displaystyle \begin{pmatrix} 1 \\ 2 \\ -3 \end{pmatrix}$ and $\displaystyle \begin{pmatrix} 5 \\ 0 \\ -3 \end{pmatrix}$ respectively.
(a) Find, in vector form, an equation of the line $\displaystyle l_1$ which passes through A and B.
The line $\displaystyle l_2$ has equation $\displaystyle \vec{r} = \begin{pmatrix} 4 \\ -4 \\ 3 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -2 \\ 2 \end{pmatrix}$, where $\displaystyle \mu$ is a scalar parameter.
(b) Show that A lies on $\displaystyle l_2$.
(c) Find, in degrees, the acute angle between the lines $\displaystyle l_1$ and $\displaystyle l_2$.
The point C with position vector $\displaystyle \begin{pmatrix} 0 \\ 4 \\ -5 \end{pmatrix}$ lies on $\displaystyle l_2$.
(d) Find the shortest distance from C to the line $\displaystyle l_1$.
Two submarines are travelling in straight lines through the ocean. Relative to a fixed origin, the vector equations of the two lines, $\displaystyle l_1$ and $\displaystyle l_2$, along which they travel are
The line $\displaystyle l_1$ has vector equation $\displaystyle \vec{r} = 6\hat{\text{i}} + 8\hat{\text{j}} + 5\hat{\text{k}} + \lambda(\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}})$ where $\displaystyle \lambda$ is a scalar parameter.
The point A has coordinates (3, a, 2), where a is a constant. The point B has coordinates (8, 6, b), where b is a constant. Points A and B lie on the line $\displaystyle l_1$.
(a) Find the values of a and b.
Given that the point O is the origin, and that the point P lies on $\displaystyle l_1$ such that OP is perpendicular to $\displaystyle l_1$,
(b) find the coordinates of P.
(c) Hence find the distance OP, giving your answer in surd form.
Show Solution
Solution
$\displaystyle \begin{aligned}
& \text{(a) If A lies on } l_1: \\
& 6 + \lambda = 3 \implies \lambda = -3 \\
& 8 - \lambda = a \implies 8 - (-3) = a \implies a = 11 \\
& \text{If B lies on } l_1: \\
& 6 + \lambda = 8 \implies \lambda = 2 \\
& 5 + \lambda = b \implies b = 5 + 2 \implies b = 7 \\
& \text{(b) Let P be the point } (x, y, z). \\
& \overrightarrow{OP} \text{ is perpendicular to } l_1: \\
& \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \cdot \begin{pmatrix} x \\ y \\ z \end{pmatrix} = 0 \implies x - y + z = 0 \\
& \text{Since P lies on } l_1: \\
& x = 6 + \lambda \\
& y = 8 - \lambda \\
& z = 5 + \lambda \\
& \text{Substitute into equation:} \\
& (6 + lambda) - (8 - lambda) + (5 + lambda) = 0 \\
& 3\lambda + 3 = 0 \implies \lambda = -1 \\
& x = 6 + (-1) = 5 \\
& y = 8 - (-1) = 9 \\
& z = 5 + (-1) = 4 \\
& \text{So P has coordinates } (5, 9, 4). \\
& \text{(c)} \\
& |\overrightarrow{OP}| = \sqrt{5^2 + 9^2 + 4^2} = \sqrt{122} \\
\end{aligned}$
Problem 23
Relative to a fixed origin O, the point A has position vector $\displaystyle 6\hat{\text{i}} + 3\hat{\text{j}} + 4\hat{\text{k}}$ and the point B has position vector $\displaystyle 5\hat{\text{i}} + 2\hat{\text{j}} + 6\hat{\text{k}}$. The line $\displaystyle l$ passes through the points A and B.
(a) Find the vector $\displaystyle \overrightarrow{AB}$.
(b) Find a vector equation for the line $\displaystyle l$.
The point C has position vector $\displaystyle 4\hat{\text{i}} + 10\hat{\text{j}} + 2\hat{\text{k}}$. The point P lies on $\displaystyle l$. Given that the vector CP is perpendicular to $\displaystyle l$,
Two aeroplanes are modelled as travelling in straight lines. Aeroplane A travels from a point with position vector $\displaystyle \begin{pmatrix} 120 \\ -80 \\ 13 \end{pmatrix} \text{ km}$ to a point with position vector $\displaystyle \begin{pmatrix} 200 \\ 20 \\ 5 \end{pmatrix} \text{ km}$, relative to a fixed origin O. Aeroplane B starts at a point with position vector $\displaystyle \begin{pmatrix} -20 \\ 35 \\ 5 \end{pmatrix} \text{ km}$ relative to O, and flies in the direction of $\displaystyle \begin{pmatrix} 10 \\ -2 \\ 0.1 \end{pmatrix}$.
(a) Show that the flight paths of the two aeroplanes will intersect, and determine the position vector of the point of intersection.
An air traffic controller states that this means that the planes will collide.
(b) Explain why this conclusion is not necessarily correct.
Obtain the shortest distance between the lines with equations
$\displaystyle \vec{r} = (3s - 3)\hat{\text{i}} - s\hat{\text{j}} + (s + 1)\hat{\text{k}}$
and $\displaystyle \vec{r} = (3 + t)\hat{\text{i}} + (2t - 2)\hat{\text{j}} + \hat{\text{k}}$
where $\displaystyle s, t$ are parameters.
The position vectors of the points A, B, C and D relative to a fixed origin O, are $\displaystyle (-\hat{\text{j}} + 2\hat{\text{k}})$, $\displaystyle (\hat{\text{i}} - 3\hat{\text{j}} + 5\hat{\text{k}})$, $\displaystyle (2\hat{\text{i}} - 2\hat{\text{j}} + 7\hat{\text{k}})$ and $\displaystyle (\hat{\text{j}} + 2\hat{\text{k}})$ respectively.
Relative to a fixed origin O, the point M has position vector $\displaystyle -4\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}}$.
The straight line $\displaystyle l$ has equation $\displaystyle \vec{r} \times \overrightarrow{OM} = 5\hat{\text{i}} - 10\hat{\text{k}}$.
(a) Express the equation of the line $\displaystyle l$ in the form $\displaystyle \vec{r} = \vec{a} + t\vec{b}$, where $\displaystyle \vec{a}$ and $\displaystyle \vec{b}$ are constant vectors and $\displaystyle t$ is a parameter.
(b) Verify that the point N with coordinates $\displaystyle (2, -3, 1)$ lies on $\displaystyle l$ and find the area of triangle OMN.
Show Solution
Solution
$\displaystyle \begin{aligned}
& \text{(a) Let } \vec{r} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \text{ and } \overrightarrow{OM} = \begin{pmatrix} -4 \\ 1 \\ -2 \end{pmatrix}. \\
& \vec{r} \times \overrightarrow{OM} = \begin{pmatrix} x \\ y \\ z \end{pmatrix} \times \begin{pmatrix} -4 \\ 1 \\ -2 \end{pmatrix} = \begin{pmatrix} -2y - z \\ -4z + 2x \\ x + 4y \end{pmatrix} \\
& \text{We are given } \vec{r} \times \overrightarrow{OM} = \begin{pmatrix} 5 \\ 0 \\ -10 \end{pmatrix}. \text{ Equating components:} \\
& -2y - z = 5 \quad \text{--- (1)} \\
& 2x - 4z = 0 \quad \text{--- (2)} \\
& x + 4y = -10 \quad \text{--- (3)} \\
& \text{Let } z = 1. \\
& \text{From (2): } 2x - 4(1) = 0 \implies x = 2. \\
& \text{From (1): } -2y - 1 = 5 \implies y = -3. \\
& \text{So a point on the line is } \vec{a} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}. \\
& \text{Since } \vec{r} \times \overrightarrow{OM} \text{ is a constant, the direction vector } \vec{b} \text{ is \parallel to } \overrightarrow{OM}. \\
& \vec{b} = \begin{pmatrix} -4 \\ 1 \\ -2 \end{pmatrix} \\
& \text{The equation of the line is:} \\
& \vec{r} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix} + t \begin{pmatrix} -4 \\ 1 \\ -2 \end{pmatrix} \\
& \text{(b) When } t = 0, \vec{r} = \begin{pmatrix} 2 \\ -3 \\ 1 \end{pmatrix}, \text{ which matches the position vector of N(2, -3, 1). Thus, N lies on } l. \\
& \text{Area of } triangle OMN = \frac{1}{2} |\overrightarrow{ON} \times \overrightarrow{OM}| \\
& \text{Since N lies on the line } l, \text{ we already know } \overrightarrow{ON} \times \overrightarrow{OM} = 5\hat{\text{i}} - 10\hat{\text{k}}. \\
& \text{Area} = \frac{1}{2} left| \begin{pmatrix} 5 \\ 0 \\ -10 \end{pmatrix} right| = \frac{1}{2} \sqrt{25 + 0 + 100} = \frac{\sqrt{125}}{2} = \frac{5\sqrt{5}}{2} \\
\end{aligned}$
Problem 30
The line $\displaystyle l_1$ has equation $\displaystyle \vec{r} = \hat{\text{i}} - \hat{\text{j}} + \lambda(\hat{\text{i}} + 2\hat{\text{j}} + 3\hat{\text{k}})$ and the line $\displaystyle l_2$ has equation $\displaystyle \vec{r} = 2\hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}} + \mu(2\hat{\text{i}} - \hat{\text{j}} + \hat{\text{k}})$.
(a) Find a vector which is perpendicular to both $\displaystyle l_1$ and $\displaystyle l_2$.
The point A lies on $\displaystyle l_1$ and the point B lies on $\displaystyle l_2$.
Given that AB is also perpendicular to $\displaystyle l_1$ and $\displaystyle l_2$,
A plane passes through the three points A, B, C, whose position vectors, referred to an origin O, are $\displaystyle (\hat{\text{i}} + 3\hat{\text{j}} + 3\hat{\text{k}})$, $\displaystyle (3\hat{\text{i}} + \hat{\text{j}} + 4\hat{\text{k}})$, $\displaystyle (2\hat{\text{i}} + 4\hat{\text{j}} + \hat{\text{k}})$ respectively.
(a) Find, in the form $\displaystyle (l\hat{\text{i}} + m\hat{\text{j}} + n\hat{\text{k}})$, a unit vector normal to this plane.
(b) Find also a Cartesian equation of the plane.
(c) Find the perpendicular distance from the origin to this plane.
(a) Show that the vector $\displaystyle \hat{\text{i}} + \hat{\text{k}}$ is perpendicular to the plane with vector equation $\displaystyle \vec{r} = \hat{\text{i}} + s\hat{\text{j}} + t(\hat{\text{i}} - \hat{\text{k}})$.
(b) Find the perpendicular distance from the origin to this plane.
(c) Hence or otherwise obtain a Cartesian equation of the plane.
Show Solution
Solution
$\displaystyle \begin{aligned}
& \text{(a) The plane is defined by direction vectors } \vec{d}_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \text{ and } \vec{d}_2 = \begin{pmatrix} 1 \\ 0 \\ -1 \end{pmatrix}. \\
& \text{The normal vector } \vec{n} \text{ is:} \\
& \vec{n} = \vec{d}_1 \times \vec{d}_2 = \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} \\
& \text{The given vector is } \hat{\text{i}} + \hat{\text{k}} = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}. \\
& \text{Since } \begin{pmatrix} -1 \\ 0 \\ -1 \end{pmatrix} = -1 \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix}, \text{ the vector } \hat{\text{i}} + \hat{\text{k}} \text{ is a scalar \multiple of the normal vector.} \\
& \text{Therefore, it is perpendicular to the plane.} \\
& \text{(b) Using } \vec{n}_0 = \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} \text{ as the normal vector and point } \vec{a} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \text{ on the plane.} \\
& \text{The perpendicular distance is } d = \frac{|\vec{a} \cdot \vec{n}_0|}{|\vec{n}_0|}. \\
& |\vec{a} \cdot \vec{n}_0| = left| \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} right| = 1 \\
& |\vec{n}_0| = \sqrt{1^2 + 0^2 + 1^2} = \sqrt{2} \implies d = \frac{1}{\sqrt{2}} = \frac{\sqrt{2}}{2} \\
& \text{(c) The equation of the plane is } \vec{r} \cdot \vec{n}_0 = \vec{a} \cdot \vec{n}_0. \\
& \begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 1 \\ 0 \\ 1 \end{pmatrix} = 1 \implies x + z = 1 \\
& \text{Cartesian equation: } x + z = 1. \\
\end{aligned}$
Problem 33
The points A, B and C have position vectors $\displaystyle \hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}$, $\displaystyle 5\hat{\text{i}} - 2\hat{\text{j}} + \hat{\text{k}}$ and $\displaystyle 3\hat{\text{i}} + 2\hat{\text{j}} + 6\hat{\text{k}}$ respectively, referred to an origin O.
(a) Find a vector perpendicular to the plane containing the points A, B and C.
(b) Hence, or otherwise, find an equation for the plane which contains the points A, B and C, in the form $\displaystyle ax + by + cz + d = 0$.
Given that P and Q are the points with position vectors $\displaystyle \vec{p}$ and $\displaystyle \vec{q}$ respectively, relative to an origin O, and that
$\displaystyle \vec{p} = 3\hat{\text{i}} - \hat{\text{j}} + 2\hat{\text{k}}$
$\displaystyle \vec{q} = 2\hat{\text{i}} + \hat{\text{j}} - \hat{\text{k}}$
(a) find $\displaystyle \vec{p} \times \vec{q}$.
(b) Hence, or otherwise, find an equation of the plane containing O, P and Q in the form $\displaystyle ax + by + cz = d$.
The line with equation $\displaystyle (\vec{r} - \vec{p}) \times \vec{q} = \vec{0}$ meets the plane with equation $\displaystyle \vec{r} \cdot (\hat{\text{i}} + \hat{\text{j}} + \hat{\text{k}}) = 2$ at the point T.
The planes $\displaystyle \Pi_1$ and $\displaystyle \Pi_2$ are defined by the equations $\displaystyle 2x + 2y - z = 9$ and $\displaystyle x - 2y = 7$ respectively.
(a) Find the acute angle between $\displaystyle \Pi_1$ and $\displaystyle \Pi_2$ giving your answer to the nearest degree.
(b) Find in the form $\displaystyle \vec{r} \times \vec{u} = \vec{v}$ an equation of the line of intersection of $\displaystyle \Pi_1$ and $\displaystyle \Pi_2$.
A pyramid has a square base OPQR and vertex S. Referred to O, the points P, Q, R and S have position vectors $\displaystyle \overrightarrow{OP} = 2\hat{\text{i}}$, $\displaystyle \overrightarrow{OQ} = 2\hat{\text{i}} + 2\hat{\text{j}}$, $\displaystyle \overrightarrow{OR} = 2\hat{\text{j}}$, $\displaystyle \overrightarrow{OS} = \hat{\text{i}} + \hat{\text{j}} + 4\hat{\text{k}}$.
(a) Express $\displaystyle \overrightarrow{PS}$ in terms of $\displaystyle \hat{\text{i}}, \hat{\text{j}}$ and $\displaystyle \hat{\text{k}}$.
(b) Show that the vector $\displaystyle -4\hat{\text{j}} + \hat{\text{k}}$ is perpendicular to OS and PS.
(c) Find to the nearest degree the acute angle between the line SQ and the plane OSP.
The plane $\displaystyle \Pi$ has equation $\displaystyle 2x + y + 3z = 21$ and the origin is O.
The line $\displaystyle l$ passes through the point $\displaystyle P(1, 2, 1)$ and is perpendicular to $\displaystyle \Pi$.
(a) Find a vector equation of $\displaystyle l$.
The line $\displaystyle l$ meets the plane $\displaystyle \Pi$ at the point M.
With respect to a fixed origin O, the straight lines $\displaystyle l_1$ and $\displaystyle l_2$ are given by
$\displaystyle l_1 : \vec{r} = \hat{\text{i}} - \hat{\text{j}} + \lambda(2\hat{\text{i}} + \hat{\text{j}} - 2\hat{\text{k}})$
$\displaystyle l_2 : \vec{r} = \hat{\text{i}} + 2\hat{\text{j}} + 2\hat{\text{k}} + \mu(-3\hat{\text{i}} + 4\hat{\text{k}})$
where $\displaystyle \lambda$ and $\displaystyle \mu$ are scalar parameters.
(a) Show that the lines intersect.
(b) Find the position vector of their point of intersection.
(c) Find the cosine of the acute angle contained between the lines.
(d) Find a vector equation of the plane containing the lines.
Relative to an origin O, the points A and B have position vectors $\displaystyle \vec{a}$ metres and $\displaystyle \vec{b}$ metres respectively, where
$\displaystyle \vec{a} = 5\hat{\text{i}} + 2\hat{\text{j}}, \quad \vec{b} = 2\hat{\text{i}} - \hat{\text{j}} - 3\hat{\text{k}}$
The point C moves such that the volume of the tetrahedron OABC is always $\displaystyle 5 \text{ m}^3$.
Determine Cartesian equations of the locus of the point C.
(a) Verify that the point P with position vector $\displaystyle 3\hat{\text{i}} - \hat{\text{j}} + 2\hat{\text{k}}$ lies on both $\displaystyle L_1$ and $\displaystyle L_2$.
(c) Find a Cartesian equation of the plane containing $\displaystyle L_1$ and $\displaystyle L_2$.
The points with position vectors $\displaystyle \vec{a}_1$ and $\displaystyle \vec{a}_2$ are $\displaystyle A_1$ and $\displaystyle A_2$ respectively.
(d) By expressing $\displaystyle \overrightarrow{A_1P}$ and $\displaystyle \overrightarrow{A_2P}$ as multiples of $\displaystyle \vec{b}_1$ and $\displaystyle \vec{b}_2$ respectively, or otherwise, find the area of the triangle $\displaystyle PA_1A_2$.
With respect to the origin O the points A, B, C have position vectors
$\displaystyle a(5\hat{\text{i}} - \hat{\text{j}} - 3\hat{\text{k}}), \quad a(-4\hat{\text{i}} + 4\hat{\text{j}} - \hat{\text{k}}), \quad a(5\hat{\text{i}} - 2\hat{\text{j}} + 11\hat{\text{k}})$
respectively, where $\displaystyle a$ is a non-zero constant. Find
(a) a vector equation for the line BC.
(b) a vector equation for the plane OAB.
(c) the cosine of the acute angle between the lines OA and OB.
(d) Obtain, in the form $\displaystyle \vec{r} \cdot \vec{n} = p$ a vector equation for $\displaystyle \Pi$, the plane which passes through A and is perpendicular to BC.
In a tetrahedron ABCD the coordinates of the vertices B, C, D are respectively (1, 2, 3), (2, 3, 3), (3, 2, 4). Find
(a) the equation of the plane BCD,
(b) the sine of the angle between BC and the plane $\displaystyle x + 2y + 3z = 4$.
(c) If AC and AD are perpendicular to BD and BC respectively and if $\displaystyle AB = \sqrt{26}$, find the coordinates of the two possible positions of A.