CP-6 Further vectors — revision question pack

5 specification points · notes, questions, answers and worked methods

Checked against Edexcel 9FM0 section CP-6. Review basis: the qualification registry sourced from the Pearson Edexcel Level 3 Advanced GCE in Further Mathematics (9FM0) specification; registry verification recorded 17 July 2026.

How this checking works

CP-6.1 · Understand and use the vector and Cartesian forms of an equation of a straight line in 3-D.

Explanation

A line through the point with position vector $\mathbf a$ and parallel to non-zero direction $\mathbf d$ has equation $\mathbf r=\mathbf a+\lambda\mathbf d$ . Component equations follow immediately.
When every direction component is non-zero, eliminating $\lambda$ gives Cartesian form $(x-a_1)/d_1=(y-a_2)/d_2=(z-a_3)/d_3$ .
A zero direction component instead gives a fixed coordinate, never a zero denominator.
Through points $A$ and $B$ , a suitable direction is $\overrightarrow{AB}=\mathbf b-\mathbf a$ ; any non-zero scalar multiple describes the same line.
To intersect two lines, equate components and check that one common pair of parameters satisfies all three equations; parallel and skew lines do not meet.

Worked example

Find vector and Cartesian equations of the line through $A(2,1,-1)$ and $B(5,-1,3)$ .

1. $\overrightarrow{AB}=(3,-2,4)$ .
2. $\mathbf r=(2,1,-1)+\lambda(3,-2,4)$ .
3.Eliminate $\lambda$ from the three component equations.

Answer: $\dfrac{x-2}{3}=\dfrac{y-1}{-2}=\dfrac{z+1}{4}$ .

Common mistakes

Don't fall into the trap of using two position vectors as though either one were the line's direction vector.
Don't fall into the trap of writing a zero direction component as a denominator in Cartesian form.
Don't fall into the trap of solving only two component equations when checking whether two three-dimensional lines intersect.

Exam tip

For a line through two points, show the subtraction that produces its direction vector before writing either equation form.

Tier 1 · Easy

1.
Write a vector equation of the line through $(2,-1,4)$ with direction vector $3\mathbf i+2\mathbf j-\mathbf k$ .
(2)
(Total for Question 1 is 2 marks)
2.
The point $P(5,k,1)$ lies on $\mathbf r=(1,-2,3)+\lambda(2,3,-1)$ . Determine $\lambda$ and $k$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
A line has Cartesian equation $\dfrac{x-1}{2}=\dfrac{y+3}{-1}=\dfrac{z-4}{3}$ . Write it in vector form and find the point for which the common parameter is $2$ .
(4)
(Total for Question 1 is 4 marks)
2.
The lines $l_1:\mathbf r=(1,0,2)+s(1,2,-1)$ and $l_2:\mathbf r=(4,p,-1)+t(-1,1,2)$ intersect. Find $p$ and their point of intersection.
(5)
(Total for Question 2 is 5 marks)
3.
The lines $l:\mathbf r=(3,0,-2)+\lambda(1,a,2)$ and $m:\mathbf r=\mu(4,-3,0)$ intersect. Find the value of $a$ and the point of intersection, and write a Cartesian equation of $l$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The line $l$ passes through $A(1,2,-1)$ and $B(4,-2,5)$ . Find vector and Cartesian equations of $l$ , and determine the point on $l$ whose $x$ -coordinate is $7$ .
(6)
(Total for Question 1 is 6 marks)
2.
The lines $l_1$ , $l_2$ and $l_3$ have equations $l_1:\mathbf r=\lambda(1,1,0)$ , $l_2:\mathbf r=(0,1,1)+\mu(1,-1,1)$ and $l_3:\mathbf r=(1,0,0)+t(2,-2,2)$ . Show that $l_1$ and $l_2$ are skew. Show that $l_2$ and $l_3$ are parallel.
(5)
(Total for Question 2 is 5 marks)
3.
A line through the origin meets $l_1:\mathbf r=(1,0,1)+s(1,1,0)$ at $P$ and $l_2:\mathbf r=(0,2,0)+t(0,1,1)$ at $Q$ . Find a vector equation of this line.
(6)
(Total for Question 3 is 6 marks)
4.
The lines $l_1:\mathbf r=(2,-1,4)+\lambda(3,2,-1)$ and $l_2:\mathbf r=(8,a,2)+\mu(b,-4,2)$ are the same line. Find the values of $a$ and $b$ , and write a Cartesian equation of the line.
(6)
(Total for Question 4 is 6 marks)
5.
The line $l$ has equation $\mathbf r=(3,-2,1)+\lambda(2,0,-5)$ . Write a Cartesian equation of $l$ . Determine which of the points $A(7,-2,-9)$ and $B(5,-1,-4)$ lie on $l$ .
(6)
(Total for Question 5 is 6 marks)

CP-6.2 · Understand and use the vector and Cartesian forms of the equation of a plane.

Explanation

A plane through position vector $\mathbf a$ and parallel to independent vectors $\mathbf b,\mathbf c$ has vector equation $\mathbf r=\mathbf a+\lambda\mathbf b+\mu\mathbf c$ . The spanning vectors must not be parallel.
A Cartesian equation has form $px+qy+rz=d$ , where $\mathbf n=(p,q,r)$ is normal to the plane and $d=\mathbf n\cdot\mathbf a$ .
To convert from vector form, find a non-zero vector perpendicular to both spanning directions by solving simultaneous scalar-product equations.
To form a plane through three non-collinear points, subtract one point from the other two to obtain spanning directions.
Substitution of each given point should confirm the final Cartesian equation.

Worked example

Find a Cartesian equation of the plane through $(1,0,2)$ parallel to $(1,1,0)$ and $(0,2,1)$ .

1.Let the normal be $(p,q,r)$ .
2. $p+q=0$ and $2q+r=0$ .
3.Choose $(p,q,r)=(1,-1,2)$ .
4.Use $(1,-1,2)\cdot((x,y,z)-(1,0,2))=0$ .

Answer: $x-y+2z=5$ .

Common mistakes

Don't fall into the trap of using a vector lying in the plane as the plane's normal.
Don't fall into the trap of choosing two parallel spanning vectors, which describe only one direction.
Don't fall into the trap of finding a normal correctly but omitting the given point when determining the constant.

Exam tip

Check the proposed normal has zero scalar product with both spanning vectors before forming the Cartesian equation.

Tier 1 · Easy

1.
Find a Cartesian equation of the plane through $(1,2,0)$ with normal vector $2\mathbf i-\mathbf j+3\mathbf k$ .
(3)
(Total for Question 1 is 3 marks)
2.
The plane $\Pi$ has equation $x+ky+2z=d$ . It contains $A(2,-1,3)$ and is parallel to the vector $(1,1,-1)$ . Find $k$ and $d$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
The plane $\Pi$ has vector equation $\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix}+\lambda\begin{pmatrix}1\\2\\0\end{pmatrix}+\mu\begin{pmatrix}0\\1\\1\end{pmatrix}$ . Find a Cartesian equation of $\Pi$ .
(5)
(Total for Question 1 is 5 marks)
2.
The plane $\Pi$ has equation $ax+by+z=4$ , where $a$ and $b$ are constants. The plane contains the line $\mathbf r=(1,1,1)+t(1,-2,1)$ . Find the values of $a$ and $b$ , and hence write down a Cartesian equation of $\Pi$ .
(5)
(Total for Question 2 is 5 marks)
3.
The points $A(1,1,2)$ , $B(-1,2,4)$ and $C(1,-1,1)$ lie in the plane $\Pi$ . The point $D(k,-2,-1)$ also lies in $\Pi$ . Determine a Cartesian equation for $\Pi$ and the value of $k$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
A plane passes through $A(1,1,0)$ , $B(3,0,1)$ and $C(0,2,2)$ . Find both a vector equation and a Cartesian equation of the plane.
(6)
(Total for Question 1 is 6 marks)
2.
The plane $\Pi$ contains the line of intersection of $x+y+z=1$ and $2x-y+z=3$ . It is perpendicular to the plane $x+2y-z=0$ . Find a Cartesian equation of $\Pi$ .
(5)
(Total for Question 2 is 5 marks)
3.
The plane $\Sigma$ contains the line $\mathbf r=(2,0,1)+\lambda(1,3,-1)$ and the point $A(0,2,3)$ . Find a Cartesian equation of $\Sigma$ and the perpendicular distance from $P(4,-1,0)$ to $\Sigma$ .
(6)
(Total for Question 3 is 6 marks)
4.
The plane $\Pi$ contains the line of intersection of $2x-y=2$ and $x+z=2$ , and it passes through $C(0,0,1)$ . Find a Cartesian equation of $\Pi$ .
(6)
(Total for Question 4 is 6 marks)
5.
The lines $l_1:\mathbf r=(1,0,2)+s(1,2,3)$ and $l_2:\mathbf r=(0,k,6)+t(2,1,-1)$ lie in the same plane. Find $k$ and a Cartesian equation of this plane.
(6)
(Total for Question 5 is 6 marks)

CP-6.3 · Calculate the scalar product and use it to express the equation of a plane, and to calculate the angle between two lines, two planes and between a line and a plane.

Explanation

The scalar product satisfies $\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta$ .
It gives the angle between lines from their direction vectors and the angle between planes from their normal vectors; use an absolute value when the acute angle is required.
A plane through $\mathbf a$ with normal $\mathbf n$ has equation $\mathbf n\cdot(\mathbf r-\mathbf a)=0$ , equivalently $\mathbf r\cdot\mathbf n=k$ .
For a line direction $\mathbf d$ and plane normal $\mathbf n$ , the acute line-plane angle $\alpha$ satisfies $\sin\alpha=|\mathbf d\cdot\mathbf n|/(|\mathbf d||\mathbf n|)$ because it complements the direction-normal angle.
The vector roles must therefore be identified before choosing sine or cosine.

Worked example

Find the acute angle between lines with directions $(1,2,2)$ and $(2,1,-2)$ .

1.The scalar product is $2+2-4=0$ .
2.Both direction vectors are non-zero.
3.A zero scalar product gives $\cos\theta=0$ .

Answer: $\theta=\pi/2$ .

Common mistakes

Don't fall into the trap of using plane direction vectors rather than normals for the angle between two planes.
Don't fall into the trap of reporting the direction-normal angle instead of its complement for a line and a plane.
Don't fall into the trap of omitting the vector magnitudes from the scalar-product angle formula.

Exam tip

Name each vector as a direction or a normal before substituting into an angle formula.

Tier 1 · Easy

1.
The plane $\Pi$ passes through $(1,-1,2)$ and has normal vector $(2,-1,2)$ . Use a scalar-product equation to find a Cartesian equation of $\Pi$ .
(3)
(Total for Question 1 is 3 marks)
2.
The lines $l_1$ and $l_2$ have direction vectors $(k,1,0)$ and $(1,0,0)$ , where $k>0$ . The acute angle between $l_1$ and $l_2$ is $45^\circ$ . Find $k$ .
(3)
(Total for Question 2 is 3 marks)

Tier 2 · Standard

1.
Planes $\Pi_1$ and $\Pi_2$ have normals $(1,2,2)$ and $(2,-1,2)$ respectively. Find the exact cosine of the acute angle between the planes.
(4)
(Total for Question 1 is 4 marks)
2.
Planes with normals $(1,k,2)$ and $(1,1,0)$ meet at an acute angle of $60^\circ$ . Given that $k>0$ , find the exact value of $k$ .
(4)
(Total for Question 2 is 4 marks)
3.
The plane $\Pi_1$ has equation $\mathbf r=(1,0,2)+\lambda(1,1,0)+\mu(0,1,2)$ , and $\Pi_2$ has equation $x+2y-z=5$ . Find the acute angle between the planes, giving your answer in degrees to one decimal place.
(4)
(Total for Question 3 is 4 marks)

Tier 3 · Hard

1.
A line has direction vector $(2,-1,2)$ and the plane $3x+5y-z=8$ has normal $(3,5,-1)$ . Determine the exact sine of the acute angle $\alpha$ between the line and the plane.
(5)
(Total for Question 1 is 5 marks)
2.
The line $l$ has equation $\mathbf r=(1,-1,2)+\lambda(1,2,-2)$ and the plane $\Pi$ has equation $2x+3y+6z=5$ . Find the acute angle between $l$ and $\Pi$ , giving your answer in degrees to one decimal place.
(4)
(Total for Question 2 is 4 marks)
3.
The planes $\Pi_1:x+2y+z=3$ and $\Pi_2:3x-y+2z=1$ meet in the line $l$ . The plane $\Pi_3$ has equation $x+2y+2z=0$ . Find a direction vector for $l$ and calculate in degrees the acute angle made by $l$ with $\Pi_3$ . Give your answer to one decimal place.
(6)
(Total for Question 3 is 6 marks)
4.
The plane $\Pi$ passes through $C(1,0,-1)$ and has normal vector $(2,-2,1)$ . Write its equation in the form $\mathbf r\cdot\mathbf n=k$ . Find the acute angle between $\Pi$ and the line through $A(0,1,1)$ and $B(3,3,0)$ , giving your answer in degrees to one decimal place.
(6)
(Total for Question 4 is 6 marks)
5.
The planes $\Pi_1:x+2y-2z=1$ and $\Pi_2:2x-y+2z=3$ meet in the line $l$ . The plane $\Pi_k$ has equation $kx+y+2z=0$ , where $k>0$ . Given that the sine of the acute angle between $l$ and $\Pi_k$ is $14/\sqrt{390}$ , find $k$ .
(6)
(Total for Question 5 is 6 marks)

CP-6.4 · Check whether vectors are perpendicular by using the scalar product.

Explanation

Two non-zero vectors are perpendicular exactly when their scalar product is zero. In three dimensions, multiply corresponding components and add all three products.
The vectors may originate at different points; only their directions matter, so displacement vectors must first be formed in a consistent order when points are given.
An unknown component or parameter can be found by setting the scalar product equal to zero and solving the resulting equation.
A diagram cannot prove perpendicularity, and one pair of components is insufficient.
The zero vector has zero scalar product with every vector but has no direction, so it must not be described as perpendicular in the ordinary geometric sense.

Worked example

Find $k$ so that $(2,k,-1)$ is perpendicular to $(3,-2,4)$ .

1.Set the scalar product equal to zero.
2. $2(3)+k(-2)+(-1)(4)=0$ .
3. $2-2k=0$ .

Answer: $k=1$ .

Common mistakes

Don't fall into the trap of adding vector components instead of multiplying corresponding components.
Don't fall into the trap of omitting the third component product in a three-dimensional scalar product.
Don't fall into the trap of calling the zero vector perpendicular merely because its scalar product is zero.

Exam tip

For 'show perpendicular', display the full scalar product and its value zero.

Tier 1 · Easy

1.
Show that the vectors $(1,2,-1)$ and $(3,-1,1)$ are perpendicular.
(2)
(Total for Question 1 is 2 marks)
2.
Determine whether the vectors $3\mathbf i-\mathbf j+2\mathbf k$ and $2\mathbf i+4\mathbf j+\mathbf k$ are perpendicular.
(2)
(Total for Question 2 is 2 marks)

Tier 2 · Standard

1.
Find the value of $k$ for which $(k,2,-1)$ is perpendicular to $(3,1,5)$ .
(3)
(Total for Question 1 is 3 marks)
2.
The vectors $(2,k,-1)$ and $(k,3,4)$ are perpendicular. Find $k$ .
(3)
(Total for Question 2 is 3 marks)
3.
Show that the planes $3x+y-2z=7$ and $2x+4y+5z=1$ are perpendicular.
(3)
(Total for Question 3 is 3 marks)

Tier 3 · Hard

1.
Points $A(1,0,2)$ , $B(3,-1,5)$ , $C(-2,4,1)$ and $D(-1,9,2)$ define the vectors $\overrightarrow{AB}$ and $\overrightarrow{CD}$ . Prove that these vectors are perpendicular.
(4)
(Total for Question 1 is 4 marks)
2.
Show that the vectors $(1,2,2)$ , $(2,-2,1)$ and $(2,1,-2)$ are mutually perpendicular.
(4)
(Total for Question 2 is 4 marks)
3.
The planes $\Pi_1:kx+(k+1)y+2z=1$ and $\Pi_2:2x+ky-3z=4$ are perpendicular. Find the set of possible values of $k$ .
(4)
(Total for Question 3 is 4 marks)
4.
The point $P$ lies on the line $\mathbf r=t(1,-1,1)$ . Given $A(1,-3,2)$ and $B(4,-1,1)$ , find the possible coordinates of $P$ for which $\overrightarrow{AP}$ is perpendicular to $\overrightarrow{BP}$ .
(4)
(Total for Question 4 is 4 marks)

CP-6.5 · Find the intersection of a line and a plane. Calculate the perpendicular distance between two lines, from a point to a line and from a point to a plane.

Explanation

To intersect a line and plane, substitute the line's parametric coordinates into the plane equation, solve for the parameter and recover all three coordinates. If the parameter equation is inconsistent the line is parallel to the plane; if it is an identity the line lies in the plane.
The point-to-plane distance is the absolute value obtained by substituting the point into the plane expression, divided by the normal's magnitude.
For point-to-line distance, choose a general point on the line and make its connector to the fixed point perpendicular to the line direction.
For two lines, the shortest connector is perpendicular to both directions; this also handles skew lines.
Every distance is non-negative.

A line-plane intersection and a perpendicular point-plane distance.

Worked example

Find the distance from $(1,2,-1)$ to the plane $2x-y+2z-4=0$ .

1.Substitution gives $2(1)-2+2(-1)-4=-6$ .
2.Take the absolute value: $|-6|=6$ .
3.The normal magnitude is $\sqrt{2^2+(-1)^2+2^2}=3$ .

Answer: The perpendicular distance is $6/3=2$ .

Common mistakes

Don't fall into the trap of using the signed plane expression as a negative distance instead of taking its absolute value.
Don't fall into the trap of dividing by the squared normal magnitude rather than by the magnitude.
Don't fall into the trap of measuring between arbitrary points on two lines instead of constructing their perpendicular connector.

Exam tip

For every distance question, identify the perpendicular direction and state why the chosen connector is shortest.

Tier 1 · Easy

1.
The line $\mathbf r=\begin{pmatrix}1\\0\\2\end{pmatrix}+t\begin{pmatrix}2\\1\\-1\end{pmatrix}$ meets the plane $x+y+z=6$ . Find the point of intersection.
(4)
(Total for Question 1 is 4 marks)
2.
Find the point on $l:\mathbf r=(1,2,0)+t(2,0,0)$ closest to $P(4,-1,3)$ , and find the distance from $P$ to $l$ .
(4)
(Total for Question 2 is 4 marks)

Tier 2 · Standard

1.
Find the perpendicular distance from $P(2,-1,4)$ to the plane $2x-y+2z-9=0$ .
(4)
(Total for Question 1 is 4 marks)
2.
Show that the line $l:\mathbf r=(1,2,3)+t(2,1,-1)$ does not meet the plane $\Pi:x+y+3z=4$ , and find the distance between $l$ and $\Pi$ .
(5)
(Total for Question 2 is 5 marks)
3.
Find the perpendicular distance from $P(8,3,12)$ to the line $\mathbf r=(1,2,0)+t(3,4,12)$ .
(5)
(Total for Question 3 is 5 marks)

Tier 3 · Hard

1.
The skew lines $l_1$ and $l_2$ have equations $\mathbf r=t(1,0,1)$ and $\mathbf r=(1,2,0)+s(0,1,1)$ . Calculate the perpendicular distance between them.
(6)
(Total for Question 1 is 6 marks)
2.
The skew lines are $l_1:\mathbf r=(1,0,0)+s(1,2,0)$ and $l_2:\mathbf r=(0,2,1)+t(0,1,2)$ . Find their common perpendicular segment, its length, and a vector equation of the common perpendicular line.
(7)
(Total for Question 2 is 7 marks)
3.
The plane $\Pi_c$ has equation $2x-y+2z=c$ . The perpendicular distances from $A(1,4,-2)$ and $B(7,-2,4)$ to $\Pi_c$ are in the ratio $1:2$ . Find the possible values of $c$ and the two distances for each value.
(7)
(Total for Question 3 is 7 marks)
4.
The lines are $l_1:\mathbf r=(1,0,-1)+s(2,-1,2)$ and $l_2:\mathbf r=(4,2,3)+t(2,-1,2)$ . Find the point on $l_1$ closest to the fixed point $(4,2,3)$ on $l_2$ , and hence find the perpendicular distance between the lines.
(7)
(Total for Question 4 is 7 marks)
5.
The line $l:\mathbf r=(1,-2,3)+t(2,1,-1)$ meets the plane $\Pi_1:x+2y-z=4$ at $P$ . Find the coordinates of $P$ and the perpendicular distance from $P$ to the plane $\Pi_2:x+2y+3z=20$ .
(7)
(Total for Question 5 is 7 marks)

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

CP-6.1 · Understand and use the vector and Cartesian forms of an equation of a straight line in 3-D.

Tier 1 · Easy

Mark scheme for CP-6.1 Tier 1 · Easy
Question	Scheme	Marks
1	$\mathbf r=\begin{pmatrix}2\\-1\\4\end{pmatrix}+\lambda\begin{pmatrix}3\\2\\-1\end{pmatrix}$	2
(2 marks)		2
Notes
Use $\mathbf r=\mathbf a+\lambda\mathbf d$ with $\mathbf a=(2,-1,4)$ and $\mathbf d=(3,2,-1)$ .
2	$1+2\lambda=5$ $\lambda=2$ $k=4$	3
(3 marks)		3
Notes
The $x$ -coordinate gives $1+2\lambda=5$ , so $\lambda=2$ . The $y$ -coordinate is then $k=-2+3(2)=4$ . The $z$ -coordinate checks this value because $3-2=1$ .

Tier 2 · Standard

Mark scheme for CP-6.1 Tier 2 · Standard
Question	Scheme	Marks
1	$\mathbf r=\begin{pmatrix}1\\-3\\4\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}$ Point: $(5,-5,10)$	4
(4 marks)		4
Notes
Set each Cartesian ratio equal to $\lambda$ . Then $x=1+2\lambda$ , $y=-3-\lambda$ and $z=4+3\lambda$ , giving the vector equation. At $\lambda=2$ the point is $(1+4,-3-2,4+6)=(5,-5,10)$ .
2	$s+t=3$ $s+2t=3$ $s=3$ and $t=0$ $p=6$ Intersection: $(4,6,-1)$	5
(5 marks)		5
Notes
Equating $x$ gives $s+t=3$ , while equating $z$ gives $s+2t=3$ . Hence $t=0$ and $s=3$ . The $y$ -coordinates then give $2s=p+t$ , so $p=6$ , and both lines give the point $(4,6,-1)$ .
3	Equating $z$ gives $-2+2\lambda=0$ , so $\lambda=1$ . Equating $x$ then gives $4=4\mu$ , so $\mu=1$ . Equating $y$ gives $a=-3$ . The lines meet at $(4,-3,0)$ . $\dfrac{x-3}{1}=\dfrac{y}{-3}=\dfrac{z+2}{2}$	5
(5 marks)		5
Notes
At the intersection, the $z$ -coordinates give $-2+2\lambda=0$ , so $\lambda=1$ . The $x$ -coordinates then give $3+1=4\mu$ , so $\mu=1$ . Equating the $y$ -coordinates gives $a=-3$ . Both lines therefore give $(4,-3,0)$ . Using the point $(3,0,-2)$ and direction $(1,-3,2)$ gives the stated Cartesian equation.

Tier 3 · Hard

Mark scheme for CP-6.1 Tier 3 · Hard
Question	Scheme	Marks
1	$\mathbf r=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}3\\-4\\6\end{pmatrix}$ $\dfrac{x-1}{3}=\dfrac{y-2}{-4}=\dfrac{z+1}{6}$ Point: $(7,-6,11)$	6
(6 marks)		6
Notes
$\overrightarrow{AB}=(4-1,-2-2,5-(-1))=(3,-4,6)$ , which gives both forms. On the line, $x=1+3\lambda$ . Setting this equal to $7$ gives $\lambda=2$ , so $y=2-8=-6$ and $z=-1+12=11$ .
2	$(1,1,0)$ and $(1,-1,1)$ are not parallel. Equating $x$ and $y$ gives $\lambda=\mu=\dfrac12$ . Equating $z$ would require $\mu=-1$ , so the lines do not intersect. Therefore $l_1$ and $l_2$ are skew. $l_2\parallel l_3$ because $(2,-2,2)=2(1,-1,1)$ .	5
(5 marks)		5
Notes
The direction vectors of $l_1$ and $l_2$ are not scalar multiples. If the lines met, their $x$ - and $y$ -coordinates would give $\lambda=\mu$ and $\lambda=1-\mu$ , so $\lambda=\mu=1/2$ . Their $z$ -coordinates instead require $0=1+\mu$ , a contradiction. The lines are therefore non-parallel and non-intersecting, so they are skew. The direction of $l_3$ is twice the direction of $l_2$ , so $l_2\parallel l_3$ .
3	$P=(1+s,s,1)$ and $Q=(0,2+t,t)$ . Since $O,P,Q$ are collinear, $Q=kP$ for some non-zero $k$ . The $x$ -components give $s=-1$ . The $y$ - and $z$ -components then give $2+t=-k$ and $t=k$ . $t=k=-1$ , so $P=(0,-1,1)$ and $Q=(0,1,-1)$ . $\mathbf r=\alpha(0,-1,1)$ , where $\alpha$ is a scalar parameter.	6
(6 marks)		6
Notes
Write $P=(1+s,s,1)$ and $Q=(0,2+t,t)$ . A line through the origin contains both points exactly when $Q=kP$ for some non-zero scalar $k$ . The $x$ -component gives $s=-1$ . The remaining components give $2+t=-k$ and $t=k$ , so $t=k=-1$ . Thus $P=(0,-1,1)$ and $Q=-P$ , and the required line has direction $(0,-1,1)$ .
4	The direction $(b,-4,2)$ is a scalar multiple of $(3,2,-1)$ . The multiplier is $-2$ because $2=-2(-1)$ . $b=-6$ $2+3\lambda=8$ , so the point $(8,a,2)$ corresponds to $\lambda=2$ on $l_1$ . $a=-1+2(2)=3$ , with $4-2=2$ checking the third coordinate. $\dfrac{x-2}{3}=\dfrac{y+1}{2}=\dfrac{z-4}{-1}$	6
(6 marks)		6
Notes
Coincident lines have parallel direction vectors, so $(b,-4,2)=-2(3,2,-1)$ and $b=-6$ . The fixed point of $l_2$ must lie on $l_1$ . Its $x$ -coordinate gives $\lambda=2$ , hence its $y$ -coordinate is $-1+2(2)=3$ and its $z$ -coordinate is $4-2=2$ . Eliminating $\lambda$ from $l_1$ gives the stated Cartesian equation.
5	$x=3+2\lambda$ , $y=-2$ and $z=1-5\lambda$ . $\lambda=\dfrac{x-3}{2}$ $\lambda=\dfrac{z-1}{-5}$ $\dfrac{x-3}{2}=\dfrac{z-1}{-5},\quad y=-2$ For $A$ , $\lambda=2$ from the $x$ -coordinate, and this also gives $y=-2$ and $z=-9$ , so $A$ lies on $l$ . $B$ does not lie on $l$ because its $y$ -coordinate is not $-2$ .	6
(6 marks)		6
Notes
The zero $y$ -component of the direction fixes $y=-2$ ; it must not appear as a zero denominator. Eliminating $\lambda$ from the $x$ - and $z$ -components gives $(x-3)/2=(z-1)/(-5)$ . Substitution shows that $A$ corresponds to $\lambda=2$ , whereas $B$ fails the fixed-coordinate condition.

CP-6.2 · Understand and use the vector and Cartesian forms of the equation of a plane.

Tier 1 · Easy

Mark scheme for CP-6.2 Tier 1 · Easy
Question	Scheme	Marks
1	$2x-y+3z=0$	3
(3 marks)		3
Notes
Use $2(x-1)-(y-2)+3(z-0)=0$ . Expanding gives $2x-y+3z=0$ .
2	$1+k-2=0$ $k=1$ $d=7$	3
(3 marks)		3
Notes
The plane has normal $(1,k,2)$ . Since $(1,1,-1)$ is parallel to the plane, $(1,k,2)\cdot(1,1,-1)=1+k-2=0$ , so $k=1$ . Substituting $A(2,-1,3)$ gives $d=2-1+6=7$ .

Tier 2 · Standard

Mark scheme for CP-6.2 Tier 2 · Standard
Question	Scheme	Marks
1	$2x-y+z=4$	5
(5 marks)		5
Notes
Let a normal be $(p,q,r)$ . Perpendicularity to $(1,2,0)$ and $(0,1,1)$ gives $p+2q=0$ and $q+r=0$ . Taking $q=-1$ gives the normal $(2,-1,1)$ . Using the point $(1,0,2)$ gives $2(x-1)-y+(z-2)=0$ , hence $2x-y+z=4$ .
2	$a+b=3$ $a-2b+1=0$ $a=\dfrac53$ $b=\dfrac43$ $5x+4y+3z=12$	5
(5 marks)		5
Notes
The point $(1,1,1)$ gives $a+b+1=4$ . The line direction lies in the plane, so $(a,b,1)\cdot(1,-2,1)=0$ . Solving $a+b=3$ and $a-2b=-1$ gives $a=5/3$ and $b=4/3$ . Multiplying the plane equation by $3$ gives $5x+4y+3z=12$ .
3	$\overrightarrow{AB}=(-2,1,2)$ and $\overrightarrow{AC}=(0,-2,-1)$ . Solving $-2p+q+2r=0$ and $-2q-r=0$ gives a normal $(3,-2,4)$ . $\Pi:3x-2y+4z=9$ Substituting $D$ gives $3k+4-4=9$ . $k=3$	5
(5 marks)		5
Notes
The independent directions $AB=(-2,1,2)$ and $AC=(0,-2,-1)$ lie in the plane. A normal $(p,q,r)$ must satisfy $-2p+q+2r=0$ and $-2q-r=0$ ; solving these simultaneous perpendicularity equations gives $(3,-2,4)$ . Using $A$ gives $3x-2y+4z=9$ . Since $D$ lies in the plane, $3k-2(-2)+4(-1)=9$ , so $k=3$ .

Tier 3 · Hard

Mark scheme for CP-6.2 Tier 3 · Hard
Question	Scheme	Marks
1	$\mathbf r=\begin{pmatrix}1\\1\\0\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\1\end{pmatrix}+\mu\begin{pmatrix}-1\\1\\2\end{pmatrix}$ $3x+5y-z=8$	6
(6 marks)		6
Notes
$\overrightarrow{AB}=(2,-1,1)$ and $\overrightarrow{AC}=(-1,1,2)$ give the vector form. For a normal $(p,q,r)$ , the equations $2p-q+r=0$ and $-p+q+2r=0$ have the solution $(p,q,r)=(3,5,-1)$ . Through $A$ , the Cartesian equation is $3(x-1)+5(y-1)-z=0$ , which simplifies to $3x+5y-z=8$ .
2	Setting $z=0$ gives $A=\left(\dfrac43,-\dfrac13,0\right)$ . Setting $x=0$ gives $B=(0,-1,2)$ . $\overrightarrow{BA}=\dfrac23(2,1,-3)$ , so a direction is $(2,1,-3)$ . $\mathbf n_\Pi=(5,-1,3)$ $\Pi:5x-y+3z=7$	5
(5 marks)		5
Notes
With $z=0$ , the two original planes give $A=(4/3,-1/3,0)$ . With $x=0$ , they give $B=(0,-1,2)$ . Hence the intersection line has direction $(2,1,-3)$ . For a normal $(p,q,r)$ , $2p+q-3r=0$ and $p+2q-r=0$ give $(p,q,r)=(5,-1,3)$ . Using $A$ gives $5x-y+3z=7$ .
3	The vectors $(1,3,-1)$ and $(-2,2,2)$ lie in $\Sigma$ . A normal $(p,q,r)$ satisfies $p+3q-r=0$ and $-2p+2q+2r=0$ . Solving gives $\mathbf n=(1,0,1)$ , and $(1,3,-1)\cdot\mathbf n=0$ . $\Sigma:x+z=3$ ; both $(2,0,1)$ and $A(0,2,3)$ satisfy this equation. $d(P,\Sigma)=\dfrac{\|4+0-3\|}{\sqrt{1^2+1^2}}$ $d(P,\Sigma)=\dfrac{1}{\sqrt2}=\dfrac{\sqrt2}{2}$	6
(6 marks)		6
Notes
Let $B=(2,0,1)$ be the fixed point on the line. Then $\overrightarrow{BA}=(-2,2,2)$ also lies in $\Sigma$ . A normal $(p,q,r)$ is perpendicular to both in-plane directions, so $p+3q-r=0$ and $-2p+2q+2r=0$ . Solving gives $(p,q,r)=(1,0,1)$ up to scale. The checks $(1,3,-1)\cdot(1,0,1)=0$ , $2+1=3$ and $0+3=3$ verify the direction and both stated points. Hence $\Sigma:x+z=3$ . Applying the point-to-plane distance formula to $P(4,-1,0)$ gives $\|4+0-3\|/\sqrt2=\sqrt2/2$ .
4	$P=(1,0,1)$ lies on the line of intersection. $Q=(2,2,0)$ also lies on the line of intersection. $\overrightarrow{PQ}=(1,2,-1)$ $\overrightarrow{PC}=(-1,0,0)$ A normal $(p,q,r)$ satisfies $p+2q-r=0$ and $-p=0$ , so take $(p,q,r)=(0,1,2)$ . $\Pi:y+2z=2$	6
(6 marks)		6
Notes
Find two points on the intersection line directly: $P=(1,0,1)$ and $Q=(2,2,0)$ . The vectors $PQ=(1,2,-1)$ and $PC=(-1,0,0)$ lie in $\Pi$ . A normal is perpendicular to both, so simultaneous perpendicularity gives $p+2q-r=0$ and $-p=0$ . Choosing $(p,q,r)=(0,1,2)$ and using $P$ gives $y+2z=2$ .
5	At the intersection, $1+s=2t$ and $2+3s=6-t$ . Solving gives $s=1$ and $t=1$ . The $y$ -coordinates give $2=k+1$ , so $k=1$ and the intersection is $(2,2,5)$ . A normal $(p,q,r)$ satisfies $p+2q+3r=0$ and $2p+q-r=0$ . Solving gives $(p,q,r)=(5,-7,3)$ up to scale. Using $(2,2,5)$ gives $5x-7y+3z=11$ .	6
(6 marks)		6
Notes
The direction vectors are not parallel, so coplanarity requires the lines to intersect. Equating the $x$ - and $z$ -coordinates gives $s=t=1$ ; the $y$ -coordinates then give $k=1$ . A normal $(p,q,r)$ is perpendicular to both $(1,2,3)$ and $(2,1,-1)$ , so simultaneous perpendicularity gives a normal $(5,-7,3)$ . Through $(2,2,5)$ , the plane is $5x-7y+3z=11$ .

CP-6.3 · Calculate the scalar product and use it to express the equation of a plane, and to calculate the angle between two lines, two planes and between a line and a plane.

Tier 1 · Easy

Mark scheme for CP-6.3 Tier 1 · Easy
Question	Scheme	Marks
1	$2x-y+2z=7$	3
(3 marks)		3
Notes
The scalar-product form is $(2,-1,2)\cdot((x,y,z)-(1,-1,2))=0$ . Thus $2(x-1)-(y+1)+2(z-2)=0$ , so $2x-y+2z=7$ .
2	$\dfrac{k}{\sqrt{k^2+1}}=\dfrac1{\sqrt2}$ $2k^2=k^2+1$ $k=1$	3
(3 marks)		3
Notes
Since $k>0$ , the scalar-product equation is $k/\sqrt{k^2+1}=1/\sqrt2$ . Squaring gives $2k^2=k^2+1$ , so $k^2=1$ and the condition $k>0$ gives $k=1$ .

Tier 2 · Standard

Mark scheme for CP-6.3 Tier 2 · Standard
Question	Scheme	Marks
1	$\cos\theta=\dfrac49$ $\theta=\arccos\left(\dfrac49\right)$	4
(4 marks)		4
Notes
The scalar product of the normals is $1(2)+2(-1)+2(2)=4$ . Each normal has magnitude $3$ , so $\cos\theta=\|4\|/(3\times3)=4/9$ .
2	$\dfrac{\|1+k\|}{\sqrt{2}\sqrt{k^2+5}}=\dfrac12$ $k^2+4k-3=0$ $k=-2\pm\sqrt7$ $k=\sqrt7-2$	4
(4 marks)		4
Notes
Using the two normals gives $\|1+k\|/(\sqrt2\sqrt{k^2+5})=1/2$ . Squaring and simplifying gives $k^2+4k-3=0$ , so $k=-2\pm\sqrt7$ . Since $k>0$ , the required value is $k=\sqrt7-2$ .
3	A normal to $\Pi_1$ satisfies $p+q=0$ and $q+2r=0$ . A normal to $\Pi_1$ is $(2,-2,1)$ ; a normal to $\Pi_2$ is $(1,2,-1)$ . The scalar product is $-3$ , the normal magnitudes are $3$ and $\sqrt6$ , and hence $\cos\theta=\dfrac{1}{\sqrt6}$ . $\theta=65.9^\circ$ to one decimal place.	4
(4 marks)		4
Notes
For a normal $(p,q,r)$ to $\Pi_1$ , perpendicularity to $(1,1,0)$ and $(0,1,2)$ gives $p+q=0$ and $q+2r=0$ , so take $(2,-2,1)$ . The other normal is $(1,2,-1)$ . Their scalar product is $-3$ and their magnitudes are $3$ and $\sqrt6$ , hence $\cos\theta=3/(3\sqrt6)=1/\sqrt6$ . Therefore $\theta=65.9^\circ$ .

Tier 3 · Hard

Mark scheme for CP-6.3 Tier 3 · Hard
Question	Scheme	Marks
1	$\sin\alpha=\dfrac{1}{3\sqrt{35}}$ $\alpha=\arcsin\left(\dfrac{1}{3\sqrt{35}}\right)$	5
(5 marks)		5
Notes
The scalar product is $(2,-1,2)\cdot(3,5,-1)=6-5-2=-1$ . The magnitudes are $3$ and $\sqrt{35}$ . Therefore $\sin\alpha=\|-1\|/(3\sqrt{35})=1/(3\sqrt{35})$ .
2	$\mathbf d=(1,2,-2),\ \mathbf n=(2,3,6)$ $\mathbf d\cdot\mathbf n=-4,\ \|\mathbf d\|=3,\ \|\mathbf n\|=7$ $\sin\alpha=\dfrac4{21}$ $\alpha=11.0^\circ$	4
(4 marks)		4
Notes
Use the line direction $\mathbf d=(1,2,-2)$ and the plane normal $\mathbf n=(2,3,6)$ . Their scalar product is $-4$ , while their magnitudes are $3$ and $7$ . Therefore $\sin\alpha=\|-4\|/(3\times7)=4/21$ , so $\alpha=\arcsin(4/21)=11.0^\circ$ to one decimal place.
3	A direction $(a,b,c)$ of $l$ satisfies $a+2b+c=0$ and $3a-b+2c=0$ . Solving gives a direction $\mathbf d=(5,1,-7)$ . A normal to $\Pi_3$ is $\mathbf n=(1,2,2)$ . $\mathbf d\cdot\mathbf n=-7$ , $\|\mathbf d\|=5\sqrt3$ and $\|\mathbf n\|=3$ . $\sin\alpha=\dfrac{7}{15\sqrt3}$ $\alpha=15.6^\circ$ to one decimal place.	6
(6 marks)		6
Notes
A direction of the intersection line is perpendicular to both plane normals. Solving $a+2b+c=0$ and $3a-b+2c=0$ gives $(a,b,c)=(5,1,-7)$ up to scale. With the normal $(1,2,2)$ to $\Pi_3$ , the line-plane formula gives $\sin\alpha=\|-7\|/(5\sqrt3\cdot3)=7/(15\sqrt3)$ . Hence $\alpha=15.630\ldots^\circ=15.6^\circ$ to one decimal place.
4	$\mathbf n=(2,-2,1)$ $k=(1,0,-1)\cdot(2,-2,1)=1$ $\Pi:\mathbf r\cdot(2,-2,1)=1$ $\overrightarrow{AB}=(3,2,-1)$ $\sin\alpha=\dfrac{\|(3,2,-1)\cdot(2,-2,1)\|}{\sqrt{14}\sqrt9}=\dfrac{1}{3\sqrt{14}}$ $\alpha=5.1^\circ$ to one decimal place.	6
(6 marks)		6
Notes
Using $C$ in $\mathbf r\cdot\mathbf n=k$ gives $k=1$ . The line direction is $AB=(3,2,-1)$ . Its scalar product with the plane normal is $1$ , while the magnitudes are $\sqrt{14}$ and $3$ . Hence the sine of the acute line-plane angle is $1/(3\sqrt{14})$ , giving $5.1^\circ$ to one decimal place.
5	A direction $(a,b,c)$ of $l$ satisfies $a+2b-2c=0$ and $2a-b+2c=0$ . A direction vector for $l$ is $(2,-6,-5)$ . $\dfrac{\|2k-16\|}{\sqrt{65}\sqrt{k^2+5}}=\dfrac{14}{\sqrt{390}}$ Squaring and simplifying gives $43k^2+96k-139=0$ . The roots are $k=1$ and $k=-\dfrac{139}{43}$ . $k=1$ because $k>0$ .	6
(6 marks)		6
Notes
A direction of the intersection line is perpendicular to both plane normals. Solving the two scalar-product equations gives $(2,-6,-5)$ , of magnitude $\sqrt{65}$ . A normal to $\Pi_k$ is $(k,1,2)$ , so the line-plane angle equation is the one shown. After squaring it reduces to $43k^2+96k-139=0=(k-1)(43k+139)$ , and positivity selects $k=1$ .

CP-6.4 · Check whether vectors are perpendicular by using the scalar product.

Tier 1 · Easy

Mark scheme for CP-6.4 Tier 1 · Easy
Question	Scheme	Marks
1	$(1,2,-1)\cdot(3,-1,1)=0$ , so the vectors are perpendicular.	2
(2 marks)		2
Notes
Their scalar product is $1(3)+2(-1)+(-1)(1)=3-2-1=0$ . Since both vectors are non-zero, they are perpendicular.
2	$(3,-1,2)\cdot(2,4,1)=6-4+2=4$ No; the scalar product is not zero, so the vectors are not perpendicular.	2
(2 marks)		2
Notes
Their scalar product is $3(2)+(-1)(4)+2(1)=6-4+2=4$ . Since it is not zero, the vectors are not perpendicular.

Tier 2 · Standard

Mark scheme for CP-6.4 Tier 2 · Standard
Question	Scheme	Marks
1	$k=1$	3
(3 marks)		3
Notes
Perpendicularity requires $3k+2(1)+(-1)(5)=0$ . Thus $3k-3=0$ , giving $k=1$ .
2	$2k+3k-4=0$ $5k=4$ $k=\dfrac45$	3
(3 marks)		3
Notes
Perpendicularity requires $(2,k,-1)\cdot(k,3,4)=0$ . Thus $2k+3k-4=0$ , so $k=4/5$ .
3	The normals are $(3,1,-2)$ and $(2,4,5)$ . $(3,1,-2)\cdot(2,4,5)=6+4-10=0$ Since both normals are non-zero, the planes are perpendicular.	3
(3 marks)		3
Notes
The coefficient vectors $(3,1,-2)$ and $(2,4,5)$ are normals to the planes. Their scalar product is $6+4-10=0$ . Since both normals are non-zero, the planes are perpendicular.

Tier 3 · Hard

Mark scheme for CP-6.4 Tier 3 · Hard
Question	Scheme	Marks
1	$\overrightarrow{AB}=(2,-1,3)$ and $\overrightarrow{CD}=(1,5,1)$ $\overrightarrow{AB}\perp\overrightarrow{CD}$	4
(4 marks)		4
Notes
$\overrightarrow{AB}=B-A=(2,-1,3)$ and $\overrightarrow{CD}=D-C=(1,5,1)$ . Their scalar product is $2(1)+(-1)(5)+3(1)=2-5+3=0$ , so the two vectors are perpendicular.
2	$(1,2,2)\cdot(2,-2,1)=2-4+2=0$ $(1,2,2)\cdot(2,1,-2)=2+2-4=0$ $(2,-2,1)\cdot(2,1,-2)=4-2-2=0$ Hence the three vectors are mutually perpendicular.	4
(4 marks)		4
Notes
The three pairwise scalar products are $2-4+2=0$ , $2+2-4=0$ and $4-2-2=0$ . All three vectors are non-zero, so each pair is perpendicular and the vectors are mutually perpendicular.
3	The normals are $(k,k+1,2)$ and $(2,k,-3)$ . $2k+k(k+1)-6=0$ $k^2+3k-6=0$ $k=\dfrac{-3\pm\sqrt{33}}{2}$	4
(4 marks)		4
Notes
Perpendicular planes have perpendicular normals. Setting the scalar product of $(k,k+1,2)$ and $(2,k,-3)$ equal to zero gives $2k+k(k+1)-6=0$ , or $k^2+3k-6=0$ . The quadratic formula gives $k=(-3\pm\sqrt{33})/2$ . Neither value makes either normal zero.
4	$\overrightarrow{AP}=(t-1,3-t,t-2)$ and $\overrightarrow{BP}=(t-4,1-t,t-1)$ . $\overrightarrow{AP}\cdot\overrightarrow{BP}=3t^2-12t+9=0$ $3(t-1)(t-3)=0$ , so $t=1$ or $t=3$ . $P=(1,-1,1)$ or $P=(3,-3,3)$ .	4
(4 marks)		4
Notes
A general point is $P=(t,-t,t)$ . Forming the two displacement vectors and setting their scalar product equal to zero gives $3t^2-12t+9=3(t-1)(t-3)=0$ . Thus $t=1$ or $3$ , producing the two stated points. Neither displacement vector is zero at either value.

CP-6.5 · Find the intersection of a line and a plane. Calculate the perpendicular distance between two lines, from a point to a line and from a point to a plane.

Tier 1 · Easy

Mark scheme for CP-6.5 Tier 1 · Easy
Question	Scheme	Marks
1	Point of intersection: $\left(4,\dfrac32,\dfrac12\right)$	4
(4 marks)		4
Notes
On the line, $x=1+2t$ , $y=t$ and $z=2-t$ . Substitution into the plane gives $1+2t+t+2-t=6$ , so $2t=3$ and $t=3/2$ . The coordinates are $(4,3/2,1/2)$ .
2	$1+2t=4$ $t=\dfrac32$ Closest point: $(4,2,0)$ Distance: $3\sqrt2$	4
(4 marks)		4
Notes
A shortest connector must be perpendicular to the line direction, so its $x$ -component is zero. Hence $1+2t=4$ and $t=3/2$ , giving $Q=(4,2,0)$ . Then $\overrightarrow{PQ}=(0,3,-3)$ has length $3\sqrt2$ .

Tier 2 · Standard

Mark scheme for CP-6.5 Tier 2 · Standard
Question	Scheme	Marks
1	$2(2)-(-1)+2(4)-9=4$ $\text{Distance}=\dfrac{\|2x-y+2z-9\|}{\sqrt{2^2+(-1)^2+2^2}}$ The normal has magnitude $\sqrt{2^2+(-1)^2+2^2}=3$ . Distance $=\dfrac43$	4
(4 marks)		4
Notes
Substitute $P$ into the left side and take the absolute value: $\|2(2)-(-1)+2(4)-9\|=\|4\|=4$ . The normal has magnitude $\sqrt{2^2+(-1)^2+2^2}=3$ , so the distance is $4/3$ .
2	Substitution gives $12=4$ , so there is no solution. $\mathbf d\cdot\mathbf n=2+1-3=0$ , so $l\parallel\Pi$ . Take the point $(1,2,3)$ on $l$ . $\dfrac{\|12-4\|}{\sqrt{1^2+1^2+3^2}}=\dfrac8{\sqrt{11}}$ Distance $=\dfrac{8\sqrt{11}}{11}$	5
(5 marks)		5
Notes
A general point on $l$ is $(1+2t,2+t,3-t)$ . Substitution into the plane gives $(1+2t)+(2+t)+3(3-t)=12$ , which would require $12=4$ , so the line does not meet the plane. Its direction $(2,1,-1)$ has zero scalar product with the plane normal $(1,1,3)$ , confirming that $l\parallel\Pi$ . Using $(1,2,3)$ on $l$ , the distance is $\|1+2+9-4\|/\sqrt{11}=8/\sqrt{11}=8\sqrt{11}/11$ .
3	A general point on the line is $Q=(1+3t,2+4t,12t)$ . $\overrightarrow{PQ}\cdot(3,4,12)=169t-169=0$ $t=1$ , so the foot of the perpendicular is $Q=(4,6,12)$ . $\overrightarrow{PQ}=(-4,3,0)$ $\|PQ\|=\sqrt{(-4)^2+3^2}=5$	5
(5 marks)		5
Notes
Write a general point on the line as $Q=(1+3t,2+4t,12t)$ . At the foot of the perpendicular, $\overrightarrow{PQ}=(3t-7,4t-1,12t-12)$ is perpendicular to the line direction $(3,4,12)$ . Their scalar product is $169t-169$ , so $t=1$ and $Q=(4,6,12)$ . Hence $\overrightarrow{PQ}=(-4,3,0)$ and the distance is $5$ .

Tier 3 · Hard

Mark scheme for CP-6.5 Tier 3 · Hard
Question	Scheme	Marks
1	Distance $=\sqrt3$	6
(6 marks)		6
Notes
Take $P=(t,0,t)$ on $l_1$ and $Q=(1,2+s,s)$ on $l_2$ . Then $\overrightarrow{PQ}=(1-t,2+s,s-t)$ . Perpendicularity to $(1,0,1)$ and $(0,1,1)$ gives $1+s-2t=0$ and $2+2s-t=0$ . Solving gives $t=0$ and $s=-1$ , so the shortest connector is $(1,1,-1)$ and its length is $\sqrt{1^2+1^2+(-1)^2}=\sqrt3$ .
2	$3+2t-5s=0$ $4+5t-2s=0$ $s=\dfrac13$ and $t=-\dfrac23$ $P=\left(\dfrac43,\dfrac23,0\right)$ $Q=\left(0,\dfrac43,-\dfrac13\right)$ $\|PQ\|=\dfrac{\sqrt{21}}3$ $\mathbf r=\left(\dfrac43,\dfrac23,0\right)+\lambda(4,-2,1)$	7
(7 marks)		7
Notes
Take $P=(1+s,2s,0)$ and $Q=(0,2+t,1+2t)$ . Requiring $Q-P$ to be perpendicular to $(1,2,0)$ and $(0,1,2)$ gives $3+2t-5s=0$ and $4+5t-2s=0$ . Thus $s=1/3$ and $t=-2/3$ , so $P=(4/3,2/3,0)$ and $Q=(0,4/3,-1/3)$ . Their connector is $(-1/3)(4,-2,1)$ , giving length $\sqrt{21}/3$ and the stated common perpendicular line.
3	$d(A,\Pi_c)=\dfrac{\|c+6\|}{3}$ $d(B,\Pi_c)=\dfrac{\|24-c\|}{3}$ $2\|c+6\|=\|24-c\|$ $4(c+6)^2=(24-c)^2$ , which gives $(c-4)(c+36)=0$ . $c=4$ or $c=-36$ For $c=4$ , the distances are $\dfrac{10}{3}$ and $\dfrac{20}{3}$ . For $c=-36$ , the distances are $10$ and $20$ .	7
(7 marks)		7
Notes
The normal magnitude is $3$ . The plane expressions at $A$ and $B$ are $-6-c$ and $24-c$ , giving distances $\|c+6\|/3$ and $\|24-c\|/3$ . The ratio condition gives $2\|c+6\|=\|24-c\|$ . Squaring and simplifying gives $3(c-4)(c+36)=0$ , so $c=4$ or $-36$ . Direct substitution gives the stated pairs of distances and verifies the ratio in both cases.
4	Both lines have direction $(2,-1,2)$ , so they are parallel. A general point on $l_1$ is $P=(1+2s,-s,-1+2s)$ . From $B=(4,2,3)$ , $\overrightarrow{BP}=(-3+2s,-2-s,-4+2s)$ . $\overrightarrow{BP}\cdot(2,-1,2)=9s-12=0$ $s=\dfrac43$ $P=\left(\dfrac{11}{3},-\dfrac43,\dfrac53\right)$ $\|BP\|=\dfrac13\sqrt{1+100+16}=\sqrt{13}$ .	7
(7 marks)		7
Notes
For parallel lines, the shortest connector is perpendicular to their common direction. Writing a general point $P$ on $l_1$ and setting $BP\cdot(2,-1,2)=0$ gives $9s-12=0$ , so $s=4/3$ . The connector is $(-1/3,-10/3,-4/3)$ , whose length is $\sqrt{117}/3=\sqrt{13}$ .
5	A general point on $l$ is $(1+2t,-2+t,3-t)$ . Substitution in $\Pi_1$ gives $-6+5t=4$ . $t=2$ $P=(5,0,1)$ The absolute plane expression for $P$ in $\Pi_2$ is $\|5+0+3-20\|=12$ . The normal to $\Pi_2$ has magnitude $\sqrt{1^2+2^2+3^2}=\sqrt{14}$ . The distance is $\dfrac{12}{\sqrt{14}}=\dfrac{6\sqrt{14}}7$ .	7
(7 marks)		7
Notes
Substituting the line coordinates into $\Pi_1$ gives $-6+5t=4$ , so $t=2$ and $P=(5,0,1)$ . Substitution of $P$ into the expression for $\Pi_2$ gives an absolute value of $12$ . Dividing by the normal magnitude $\sqrt{14}$ gives $12/\sqrt{14}=6\sqrt{14}/7$ .