**Question**

The planes m and n have equations 3x + y − 2z = 10 and x − 2y + 2z = 5 respectively. The line l has

equation r = 4i + 2j + k + \lambda (i + j + 2k).

**(i)** Show that l is parallel to m.

**(ii)** Calculate the acute angle between the planes m and n.

**(iii)** A point P lies on the line l. The perpendicular distance of P from the plane n is equal to 2. Find

the position vectors of the two possible positions of P.

**Answer/Explanation**

**(i)** EITHER: Expand scalar product of a normal to m and a direction vector of l

Verify scalar product is zero

Verify that one point of l does not lie in the plane

OR: Substitute coordinates of a general point of l in the equation of the plane m

Obtain correct equation in

λ in any form Verify that the equation is not satisfied for any value of λ 3

(ii) Use correct method to evaluate a scalar product of normal vectors to m and n

Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result

Obtain answer 74.5° or 1.30 radians 3

**(iii)** EITHER: Using the components of a general point

P of l form an equation in

λ by

equating the perpendicular distance from n to 2

OR: Take a point Q on l, e.g. (5, 3, 3) and form an equation in λ by equating the

length of the projection of QP onto a normal to plane n to 2

Obtain a correct modular or non-modular equation in any form Solve for λ and obtain a position vector for P, e.g. 7i + 5j + 7j from λ = 3

Obtain position vector of the second point, e.g. 3i + j – k from λ = – 1

**Question**

Two lines l and m have equations \(r=ai+2j+3k+\lambda (i-2j+3k) and r=2i+j+2k+\mu (2i-j+k)\) respectively, where a is a constant. It is given that the lines intersect.**(i)** Find the value of a.

**(ii) **When a has this value, find the equation of the plane containing l and m.

**Answer/Explanation**

**(i)** Express general point of l or m in component form

e.g. (a +λ, 2 – 2λ, 3 + 3λ) or(2 + 2μ, 1 –μ, 2 +μ)

Equate at least two pairs of corresponding components and solve for λ or for μ

Obtain either λ = – 2 or μ = – 5

or\( \lambda = \frac{1}{3}a or \mu =\frac{2}{3}a-1\)

or\( \lambda =\frac{1}{5}(a-4) or \mu =\frac{1}{5}(3a-7)\) Obtain a = – 6

**(ii)** Use scalar product to obtain a relevant equation in a, b and c, e.g. a – 2b + 3c = 0

Obtain a second equation, e.g. 2a – b + c = 0 and solve for one ratio

Obtain a : b : c = 1 : 5 : 3

Substitute a relevant point and values of a, b, c in general equation and find d

Obtain correct answer x + 5y + 3z = 13

**Alternative method for question (ii)**

Attempt to calculate vector product of relevant vectors,

Obtain two correct components

Obtain correct answer, e.g. i + 5j + 3k

Substitute a relevant point and find d

Obtain correct answer x + 5y + 3z = 13

**(ii)** Alternative method for question **(ii)**

Using a relevant point and relevant vectors, form a 2-parameter equation for the plane

State a correct equation, e.g. r = 2i + j + 2k +λ(i – 2j + 3k) +μ(2i – j + k)

State three correct equations in x, y, z, λ and μ

Eliminate λ and μ

Obtain correct answer x + 5y + 3z = 13

**Question**

The diagram shows a set of rectangular axes Ox, Oy and Oz, and four points A, B, C and D with position vectors \(\vec{OA}=3i,\vec{OB}=3i+4j,\vec{OC}=i+3j and \vec{OD}=2i+3j+5k.\)

**(i)** Find the equation of the plane BCD, giving your answer in the form ax + by + cz = d.

**(ii)** Calculate the acute angle between the planes BCD and OABC.

**Answer/Explanation**

**(i) **Obtain a vector parallel to the plane, e.g.\(\overrightarrow{CB}=2i+j\)

Use scalar product to obtain an equation in a,b,c

Obtain two correct equation in a,b,c

Solve to obtain a:b:c,

Obtain a:b:c:-10:-1,

Obtain equation 5x-10y-z=-25

**Alternative method 1 **

Obtain a vector p[arallel to the plane ,eg..\(\overrightarrow{CB}=2i+j\)

Obtain a second such vector and calculate thier vector product ,e.g. \(\left ( 2i+J \right )\times \left ( i+5k \right )\)

Obtain two correct components

Obtain correct answer e.g. 5i-10j-k

Substitute to find d

Obtain equation 5x-10y-z=-25

**Alternative method ****2**

Obtain a vector parallel to the palne e.g,\(\overrightarrow{DB}=i+j-5k\)

Obtain a second vector aqnd form correctly a 2-parameter equation for the plane

State a correct equation in x,y,z,\(\Lambda \) and \(\mu \)

Eliminate \(\Lambda \) and \(\mu \)

Obtain equation 5x-10y-z=-25

**Alternative Method 3**

Substitute for B and C and obtain 3a+4b=d and a+3b=d

Substitute for D to obtain a third equation and eliminatye one unknown 9a,b<ord) entirely

Obtain two correct equation in two unknowns,e.g.a:b:c

Solve to obtain their ratio ,e.g.a:b:c

Obtain a:b:c=5:-10:-1

a:c:d=5:-1:-25 or b:c:d=10:1:25

Obtain equation 5x-10y-z=-25

**Alternate Method 4**

Substitute for B and C and obtain 3a+4b=d and a+3b+d

Solve to obtain a:b:d

Obtain a:b:d=1:-2:-5

Substitute for C to obtain c

Obtain equation 5x-10y-z=-25.

**(ii)** State or imply a normal vector for the plane OABC is k

Carry out correct process for evaluating a scalar product of relevant the inverse consine of the result

Obtain answer\( 84.9^{\circ}\)( or 1.48 radian

**Question**

Two planes have equations 2x+ 3y − z = 1 and x − 2y + z = 3.**(i)** Find the acute angle between the planes.

**(ii)** Find a vector equation for the line of intersection of the planes.

**Answer/Explanation**

**(i)** State or imply a correct normal vector to either plane, e.g. 2i + 3j – k, or i –2j +k

Carry out correct process for evaluating the scalar product of two normal vectors

Using the correct process for the moduli, divide the scalar product of the two normal vectors by the product of their moduli and evaluate the inverse cosine of the result

Obtain answer 56.9° or 0.994 radians

**(ii)** **EITHER:** Carry out a complete strategy for finding a point on the line (call the line l)

Obtain such a point, e.g. (1, 1, 4)

**EITHER:** State a correct equation for a direction vector ai +bj +ck for l,e.g. 2a+ 3b – c = 0

State a second equation, e.g. a – 2b + c = 0, and solve for one ratio, e.g. a: b

Obtain a : b : c = 1 : – 3: – 7, or equivalent

State a correct answer, e.g. r= i + j + 4k +λ(i – 3j –7k)

**OR1: ** Attempt to calculate the vector product of the two normal vectors

Obtain two correct components

Obtain i – 3j – 7k, or equivalent

State a correct answer, e.g. r= i + j + 4k +λ(i – 3j – 7k), or equivalent

**OR2:** Obtain a second point on l e.g. (0, 4, 11)

Subtract position vectors and obtain a direction vector for l

Obtain i – 3j – 7k, or equivalent

State a correct answer, e.g. r = 4j + 11k +μ(i – 3j – 7k), or equivalent

**OR3:** Express one variable in terms of a second

Obtain a correct simplified expression, e.g. y x = 4 −3x

Express the third variable in terms of the second

Obtain a correct simplified expression, e.g. z x = 11 − 7x

Form a vector equation for the line

State a correct answer, e.g. r= 4j + 11k +λ(i – 3j – 7k), or equivalent

**OR4: ** Express one variable in terms of a second

Obtain a correct simplified expression, e.g. \(x=\frac{4}{3}-\frac{y}{3}\)

Express the same variable in terms of the third

Obtain a correct simplified expression, e.g. \(x=\frac{11}{7}-\frac{z}{7}\)

Form a vector equation for the line

Obtain a correct answer, e.g.r = 4j + 11k+μ(i – 3j –7k), or equivalent

*Question*

The equations of two linesl and m are \(r=3i-j-2k+\lambda (-i+j+4k)\) and \(r=4i+4j-3k+\mu (2i+j-2k)\) respectively.

(i) Show that the lines do not intersect.

(ii) Calculate the acute angle between the directions of the lines.

(iii) Find the equation of the plane which passes through the point (3, −2, −1) and which is parallel to both l and m. Give your answer in the form ax + by + cz = d

**Answer/Explanation**

.

*Question*

The points P and Q have position vectors, relative to the origin O, given by

\(\underset{OP}{\rightarrow}=7i+7j-5k\) and \(\underset{OQ}{\rightarrow}=5i+j+k.

The mid-point of PQ is the point A. The plane \(\mathbb{I}\) is perpendicular to the line PQ and passes through A.

**(i)** Find the equation of \(\mathbb{I}\), giving your answer in the form ax + by + cz = d. [4]

**(ii) The straight line through P parallel to the x-axis meets \(\mathbb{I}\) at the point B. Find the distance AB, correct to 3 significant figures. [5]**

**Answer/Explanation**

Ans:

** (i)** State or imply A is (1, 4, –2)

State or imply \(\underset{QP}{\rightarrow}=12i+6j-6k\) or equivalent

Use QP as normal and A as mid-point to find equation of plane

Obtain 12x + 6y – 6z = 48 or equivalent

** (ii)** Either State equation of PB is **r** = 7**i** + 7**j** – 5**k** + λ**i**

Set up and solve a relevant equation for λ .

Obtain λ = −9 and hence B is (–2, 7, –5)

Use correct method to find distance between A and B.

Obtain 5.20

Or Obtain 12 for result of scalar product of QP and **i** or equivalent

Use correct method involving moduli, scalar product and cosine to find angle APB

Obtain 35.26° or equivalent

Use relevant trigonometry to find AB

Obtain 5.20

### Question

The plane with equation 2x + 2y − z = 5 is denoted by m. Relative to the origin O, the points A and B have coordinates (3, 4, 0) and (−1, 0, 2) respectively.

(i) Show that the plane m bisects AB at right angles.

A second plane p is parallel to m and nearer to O. The perpendicular distance between the planes is 1.

(ii) Find the equation of p, giving your answer in the form ax + by + cz = d.

**Answer/Explanation**

6(i) State or obtain coordinates (1, 2, 1) for the mid-point of AB

Verify that the midpoint lies on m

State or imply a correct normal vector to the plane, e.g. 2 i+2 j-k

State or imply a direction vector for the segment AB, e.g. −4i-4j+2k

Confirm that m is perpendicular to AB

6(ii) State or imply that the perpendicular distance of m from the origin is \(\frac{5}{3}\) , or

unsimplified equivalent

State or imply that n has an equation of the form 2x+2y-z=k

Obtain answer 2x+2y-z=2

**Question**

The line l has equation r = i + 2j − 3k + ,2i − j + k. The plane p has equation 3x + y − 5z = 20.**(i)** Show that the line l lies in the plane p.

**(ii)** A second plane is parallel to l, perpendicular to p and contains the point with position vector

3i − j + 2k. Find the equation of this plane, giving your answer in the form ax + by + cz = d.

**Answer/Explanation**

**(i)** Verify that the point with position vector i jk + 2 3 − lies in the plane

EITHER:

Find a second point on l and substitute its coordinates in the equation of p

Verify that the second point, e.g. (3, 1, – 2), lies in the plane

OR:

Expand scalar product of a normal to p and the direction vector of l

Verify scalar product is zero

Obtain answer 4 13 5 9 x yz + += , or equivalent

### Question

Two lines have equations \(r=i+2j+k+\lambda (ai+2j-k)\ and \ r=21+j-k+\mu (2i-j+k),\) , where a is a

constant.

(a) Given that the two lines intersect, find the value of a and the position vector of the point of

intersection. [5]

(b) Given instead that the acute angle between the directions of the two lines is \(\cos ^{-1}(\frac{1}{6})\). find the

two possible values of a. [6]

**Answer/Explanation**

(b) Use correct process for finding the scalar product of direction

Using the correct process for the moduli, divide the scalar

product by the product of the moduli and equate the result to

\(\pm \frac{1}{6}\)

State a correct equation in a in any form, \(e.g.\frac{2a-2-1}{\sqrt{6}\sqrt{(a^{2}+5)}}=\pm \frac{1}{6}\)

Solve for a

Obtain a = 1

Obtain \(a=\frac{49}{23}\)

**(b) Alternative method for question(b)**

\(\cos (\theta )=[|a^{2}+2^{2}+(-1)^{2}|^{2}+|2^{2}+(-1)^{2}+1^{2}|^{2}-|(a-2)^{2}+3^{2}+(-2)^{2}|^{2}]/[2|a^{2}+2^{2}+(-1)^{2}|.|2^{2}+(-1)^{2}+1^{2}|]\)

Equate the result to \(\pm \frac{1}{6}\)

Solve for *a*

Obtain *a* = 1

Obtain \(a=\frac{49}{23}\)

**Question**

**Question**

In the diagram, OABCDEFG is a cuboid in which OA=2 units, OC = 3 units and OD = 2 units. Unit vectors i, j and k are parellel to OA, OC and OD respectively. The point M on AB is such that MD = 2AM. The midpoint of FG is N.**(a) Express the vectors \(\overrightarrow{OM}\) and \(\overrightarrow{MN}\) in terms of i, j and k****(b) Find a vector equation for the line through M and N.**

(c) Find the position vector of P, the foot of the perpendicular from D to the line through M and N.

**Answer/Explanation**

**Ans:**

(a) Obtain \(\overrightarrow{OM}=2i+j\)

Use a correct method to find \(\overrightarrow{MN}\)

Obtain \(\overrightarrow{MN}=-i+2j+2k\)

(b) Use a correct method to form an equation for MN

Obtain \(r=2i+j+ \lamdba(-i+2j+2k)\), or equivalent

(c) Find \(\overrightarrow{DP}\) for a point P on MN with parameter \(\lambda, e.g.(2- \lambda,1+2\lambda,-2+2\lambda)\)

Equate scalar product of \(\overrightarrow{DP}\) and a direction vector for MN to zero and solve for \(\lambda\).

Obtain\(\lambda=\frac{4}{9}\)

State that the position vector of P is \(\frac{14}{9}i+\frac{17}{9}j+\frac{8}{9}k\)

Two lines l and m have equations r = 3i + 2j + 5k + s(4i − j + 3k) and r = i − j − 2k + t(−i + 2j + 2k) respectively.

### (a) *Question*

**Show that l and m are perpendicular. **

**(b) ***Question*

*Question*

**Show that l and m intersect and state the position vector of the point of intersection.**

### (c) *Question*

Show that the length of the perpendicular from the origin to the line m is \(\frac{1}{3}\sqrt{5}\) .

**Answer/Explanation**

**Ans:(a)**

Use correct method to evaluate the scalar product of relevant vectors

Obtain answer zero and deduce the given statement

**Ans:(b)**

Express general point of l or m in component form, e.g. (3 + 4s, 2 – s, 5 + 3s) or (1 – t, – 1 +2t, – 2 + 2t)

Equate at least two pairs of components and solve for s or for t

Obtain correct answer s = – 1 and t = 2

Verify that all three equations are satisfied

State position vector of the intersection – i + 3j +2k, or equivalent

**Ans:(c)**

Taking a general point P on m, form an equation in t by either equating a relevant scalar product to zero, or equating the derivative of \(\left | \overrightarrow{OP} \right |\) to zero, or taking a specific point Q on m, e.g. (1, – 1, – 2), using Pythagoras in triangle OPQ

Obtain t = \(\frac{7}{9}\)

Carry out correct method to find OP

Obtain \(\frac{\sqrt{5}}{3}\)

### Question

With respect to the origin O, the points A and B have position vectors given by \(\overrightarrow{OA}=\begin{pmatrix}1\\2\\ 1\end{pmatrix}\ and \ \overrightarrow{OB}=\begin{pmatrix}3\\1\\ -2\end{pmatrix}\). The line l has equation \(r=\begin{pmatrix}2\\3\\1\end{pmatrix}+\lambda \begin{pmatrix}1\\-2\\ 1\end{pmatrix}\).

** (a) Find the acute angle between the directions of AB and l. ** [4]

(b) Find the position vector of the point P on l such that AP = BP. [5]

**Answer/Explanation**

Ans

(a) State or imply \(\overrightarrow{AB}=\begin{pmatrix}2\\-1 \\ -3\end{pmatrix}\)

Use the correct process to calculate the scalar product of a pair of relevant vectors, e.g. their

\(\overrightarrow{AB}\) and a direction vector for *l*

Using the correct process for the moduli, divide the scalar product by the

product of the moduli of the two vectors and evaluate the inverse cosine of

the result

Obtain answer 83.7° or 1.46 radians

(b) State or imply \(\pm \overrightarrow{AP}\ and \ \pm \overrightarrow{BP}\) in component form, i.e.

(1 +λ, 1 – 2λ, λ) and (– 1 +λ, 2 – 2λ, 3 +λ), or equivalent

Form an equation in λ by equating moduli or by using cos BAP = cos ABP

Obtain a correct equation in any form

(1+ λ)^{2}+ (1- 2λ)^{2 }+ λ^{2}=(λ – 1)^{2} + (2 – 2λ)^{2} + (λ + 3)^{2}

Solve for λ and obtain position vector

Obtain correct position vector for P in any form, e.g. (8, – 9, 7) or 8i –9j + 7k

**Question**

**Question**

Two lines have equations

(a) Show that the lines are skew.

(b) Find the acute angle between the directions of the two lines.

**Answer/Explanation**

**Ans:**

- Express general point of a line in component for, e.g.

(1+2s,3-s,2+3x) or (2+t,1-t,4+4t)

Equate at least two pairs of components and solve for s of for t

Obtain correct answer for s or for t (possible answers are -1, 6, \(\frac{2}{5}\) for s and -3, 4, -\(\frac{1}{5}\) for t)

Verify that all three component equations are not satisfied

Show that the lines are not parallel and are thus skew - Carry out correct process for evaluating the scalar product of the direction vectors

Using the correct process for the moduli, divide the scalar product by the product of the moduli and evaluate the inverse cosine of the result

Obtain answer \(19.1^o\) or 0.333 radians

**Question**

**Question**

Two planes have equations 3x + y − z = 2 and x − y + 2z = 3.**(i)** Show that the planes are perpendicular. **(ii)** Find a vector equation for the line of intersection of the two planes.

**Answer/Explanation**

**(i)**State or imply a correct normal vector to either plane, e.g. 3i jk + − or ij k − + 2

Use correct method to calculate their scalar product

Show value is zero and planes are perpendicular

**(ii) **

EITHER: Carry out a complete strategy for finding a point on l the line of intersection

Obtain such a point, e.g. (0, 7, 5) , (1, 0, 1), (5/4, –7/4, 0)

EITHER: State two equations for a direction vector ai+bj+ck for l,

e.g. 3a+b-c=0 and a-b+2c=0

Solve for one ratio, e.g. a : b

Obtain a : b : c = 1 : −7 : −4, or equivalent

State a correct answer, e.g. r=7j+5k+Λ(i-7j+4k)

OR1: Obtain a second point on l, e.g. (1, 0, 1)

Subtract vectors and obtain a direction vector for l

Obtain −i+7j+4k, or equivalent

State a correct answer, e.g. r=i+k+Λ(-i+7j+4k)

OR2: Attempt to find the vector product of the two normal vectors

Obtain two correct components of the product

Obtain i-7j-4k , or equivalent

State a correct answer, e.g. r=i+k+Λ(-i+7j+4k)

OR1: Express one variable in terms of a second variable

Obtain a correct simplified expression, e.g. y = 7 – 7x

Express the third variable in terms of the second

Obtain a correct simplified expression, e.g. z = 5 – 4x

Form a vector equation for the line

Obtain a correct equation, e.g. r=7j+5k+Λ(i-7j-4k)

OR2: Express one variable in terms of a second variable

Obtain a correct simplified expression, e.g. z=5-4x

Express the same variable in terms of the third

Obtain a correct simplified expression e.g. z(7+4y)/7

Form a vector equation for the line

Obtain a correct equation, e.g. \(r=\frac{5}{4}i-\frac{7}{4}j+\Lambda (-\frac{1}{4}i+\frac{7}{4}j+k)\)

*Question*

The points A and B have position vectors 2i + j + k and i − 2j + 2k respectively. The line l has vector equation r = i + 2j − 3k + μ(i – 3j – 2k).

**(a)**Find a vector equation for the line through A and B.

**(b)** Find the acute angle between the directions of AB and l, giving your answer in degrees.

**(c) Show that the line through A and B does not intersect the line l.**

**Answer/Explanation**

Ans:

(a)Obtain direction vector −i− 3j+k

Use a correct method to form a vector equation

Obtain answer r = 2i + j + k + λ (-i-3j+k)

or r = i 2j + 2k + λ (-i-3j+k)

(b) Carry out the correct process for evaluating the scalar product of the direction vectors.

Using the correct process for the moduli, divide the scalar product by the product of the moduli and find the inverse cosine of the result for any 2 vectors

Obtain answer 61.1 °

(c)

Express general point of AB or l in component form, e.g. (2 –λ, 1 – 3λ, 1 + λ) or (1 + µ, 2 – 3 µ, – 3 – 2µ)

Equate at least two pairs of components and solve for λ or for µ

Obtain a correct answer for λ or µ, e.g. λ = \(= 6, \frac{1}{6}, or -\frac{14}{9}; \mu = -5, \frac{2}{3} or – \frac{11}{9}\)

Verify that all three equations are not satisfied, and the lines do not intersect

Express general point of AB or l in component form,

e.g. (1 –λ*,− 2 – 3λ*, 2 +λ*) or (1 + µ*, 2 – 3µ*, – 3 – 2µ*)

*Question*

*Question*

The line *l* has equation** r** = \(\begin{pmatrix}1\\2\\-1\\\end{pmatrix}+\lambda \begin{pmatrix}2\\1\\3\\\end{pmatrix}\) . The plane *p* has equation **r**. \(\begin{pmatrix}2\\-1\\-1\\\end{pmatrix}\) = 6.

** (i) Show that l is parallel to p. [3]**

** (ii)** A line *m* lies in the plane *p* and is perpendicular to *l*. The line *m* passes through the point with coordinates (5, 3, 1). Find a vector equation for *m*. [6]

**Answer/Explanation**

Ans:

** (i)** *EITHER*: Substitute for **r** in the given equation of *p* and expand scalar product

Obtain equation in λ in any correct form

Verify this is not satisfied for any value of λ

*OR1:* Substitute coordinates of a general point of *l* in the Cartesian equation of plane *p*

Obtain equation in λ in any correct form

Verify this is not satisfied for any value of λ

* OR2:* Expand scalar product of the normal to *p* and the direction vector of *l *

Verify scalar product is zero

Verify that one point of *l* does not lie in the plane

* OR3: *Use correct method to find the perpendicular distance of a general point of *l* from *p*

Obtain a correct unsimplified expression in terms of λ

Show that the perpendicular distance is 5/ √6 , or equivalent, for all λ

* OR4:* Use correct method to find the perpendicular distance of a particular point of *l* from *p*

Show that the perpendicular distance is 5/ √6 , or equivalent

Show that the perpendicular distance of a second point is also 5/ √6 , or equivalent

**(ii)** *EITHER:* Calling the unknown direction vector *a***i** + *b***j **+ *c***k** state equation 2*a* + *b* + 3*c* = 0

State equation 2*a* – *b* – *c* = 0

Solve for one ratio, e.g.* a : b*

Obtain ratio* a : b : c* = 1 : 4 : − 2, or equivalent

* *OR: Attempt to calculate the vector product of the direction vector of *l* and the normal vector of the plane *p*, e.g. (2**i** + **j **+ 3**k** ) x (2**i** – **j **– **k)**

Obtain two correct components of the product

Obtain answer 2**i** + 8**j –** 4**k** , or equivalent

Form line equation with relevant vectors

Obtain answer **r** = 5**i** + 3**j +** **k + ***µ *(**i** + 4**j** – 2**k**), or equivalent

**Question**

**Question**

With respect to the origin O, the points A, B and C have position vectors given by

The plane m is parallel to \(\vec{OC}\) and contains A and B.

**(i) Find the equation of m, giving your answer in the form ax + by + cß = d. ****(ii)** Find the length of the perpendicular from C to the line through A and B.

**Answer/Explanation**

**(i)** EITHER Use scalar product of relevant vectors, or subtract point equations to form two

equations in a,b,c, e.g. a – 5b – 3c = 0 and a – b – 3c = 0

State two correct equations in a,b,c

Solve simultaneous equations and find one ratio, e.g. a : c, or b = 0

Obtain a : b : c = 3 : 0 : 1, or equivalent

Substitute a relevant point in 3x + z = d and evaluate d

Obtain equation 3x + z = 13, or equivalent

OR 1 Attempt to calculate vector product of relevant vectors,

e.g. (i – 5j – 3k) × (i – j – 3k)

Obtain 2 correct components of the product

Obtain correct product, e.g. 12i + 4k

Substitute a relevant point in 12x + 4z = d and evaluate d

Obtain 3x + z =13, or equivalent

OR 2 Attempt to form 2–parameter equation for the plane with relevant vectors

State a correct equation e.g. r = 3i – 2j + 4k + λ(i – 5j – 3k) + μ(i – j –3k)

State 3 equations in x, y, z, λ and μ

Eliminate λ and μ

Obtain equation 3x + z =13, or equivalent