Vectors

6(a) 0 t  −1 B1 Finds vector from any point on l 1 to any point on l 2 .       1 − 1 =t 0             t 0  1  i j k  −1 M1 A1 Finds common perpendicular.   −2 −1 0 = 2   0 −2 1   4    −1  −1 M1 A1 Uses formula for shortest distance. t     0  2 = 21t = 21=4.2 21     5      1   4  5 6(b) r= 21i+j+λ( 2i+j )+μ(−2j+k ) B1 FT Using their value of t. 5 1 6(c)  0   5  M1 A1 Uses dot product of −2j+k and normal.     19 −2  −6 = 5 110cosαleading to cosα=         5 110  1   7  Acute angle between l and Π is 90−α=54.1 ° A1 2 2 3 © UCLES 2021 Page 11 of 14 6(d)  −1  5  M1 Dot product using their normal to Π . 1  2    −6  = 21 110cosαleading to cosα= 11 A1 FT         21 110  4   7  76.8 ° A1 3 Question Answer Marks Guidance

6(a) 0 t  −1 B1 Finds vector from any point on l 1 to any point on l 2 .       1 − 1 =t 0             t 0  1  i j k  −1 M1 A1 Finds common perpendicular.   −2 −1 0 = 2   0 −2 1   4    −1  −1 M1 A1 Uses formula for shortest distance. t     0  2 = 21t = 21=4.2 21     5      1   4  5 6(b) r= 21i+j+λ( 2i+j )+μ(−2j+k ) B1 FT Using their value of t. 5 1 6(c)  0   5  M1 A1 Uses dot product of −2j+k and normal.     19 −2  −6 = 5 110cosαleading to cosα=         5 110  1   7  Acute angle between l and Π is 90−α=54.1 ° A1 2 2 3 © UCLES 2021 Page 11 of 14 6(d)  −1  5  M1 Dot product using their normal to Π . 1  2    −6  = 21 110cosαleading to cosα= 11 A1 FT         21 110  4   7  76.8 ° A1 3 Question Answer Marks Guidance

6(a) −(−2i−2j+4k )+(−i−2j+k )=i−3k B1 OE. Finds direction vector from P to a point of l. 1 r=−2i−2j+4k+λ( 2i−3j )+μ( i−3k ) B1 FT OE. FT their i – 3k r =−i−2j+k+λ( 2i−3j )+μ( i−3k ) 2 © UCLES 2021 Page 12 of 16 6(b) i j k  −9 M1 A1 OE. Finds vector perpendicular to Π . 2   2 −3 0 = −6   3 −1 3  7   −9(3)−6(0)+7(−2)=−41 M1 Uses point on Π . 2 9x+6y−7z=41 A1 4 6(c) i j k 9 3 M1 A1 Finds vector perpendicular to Π . 1     2 −3 0 = 6 ~ 2     1 0 −3  3    1   3  9  DM1 Dot product using their normal vectors.  2    6  = 14 166cosαleading to cosα= 32 A1 FT      1     −7   14 166 48.4° A1 0.845 rad 5 © UCLES 2021 Page 13 of 16 6(d)  −2 9   10+9t  M1 A1 FT Scales OP by a factor of ±5 and uses multiple of    OF =OQ+QF =−5  −2  +t  6  =  10+6t  their normal to Π .       2   4     −7     −20−7t   9 ( 10+9t )+6 ( 10+6t )−7 (−20−7t )=41leading to290+166t=41 DM1 Substitutes into the equation of Π . 2 −7  A1  2   t=−3 leading toOF = 1   2 −19   2 4 Question Answer Marks Guidance

5(a) i j k  −3 M1 A1 Finds vector perpendicular to Π.   1 0 1 = 2     2 3 0 3   −3(−2)+2(3)+3(3)=21 M1 Substitutes point on Π. −3x+2y+3z=21 A1 4 5(b)  2  B1 Forms general point on line (given as a single vector).   −3    5+t   −3(2)+2(−3)+3(5+t)=21 leading to t=6 M1 Substitutes into the equation forΠ.  2  A1   −3     11 3 5(c) 0  −3 M1 Uses dot product of k and their normal.  0    2  = 1 22cosαleading to cosα= 3 A1 FT         22 1 3     Acute angle between l and Π is 90−α=39.8 ° A1 3 © UCLES 2021 Page 10 of 14 5(d) 0  2   −2 B1 Finds direction vector from P to plane.       0 − −3 = 3             7  5   2   −2  −3 M1 A1 Uses dot product of their direction and normal vectors. 1     18 3  2 = =3.84     22 22     2 3     3

7(a) i j k  2   1  M1 A1 Finds common perpendicular.     n= 2 2 0 = −2 ~ −1     0 −1 1  −2   −1   1  A1   r. −1 =0     −1   3 7(b) 1 1 B1 Divides by magnitude of the normal to Π = 0.267 12 +32 +22 14 1 7(c)  1   1  M1 Takes dot product of normal vectors.  −1    −3  = 3 14cosαleading tocosα= 6 A1 FT     3 14   −1     −2   22.2° A1 Accept 0.388 radians. Mark final answer. 3 © UCLES 2021 Page 15 of 16 7(d) i j k  −20  5  M1 A1 Finds direction of common perpendicular.     2 1 −7 = 12 ~ −3     −2 −3 1   −4     1     2  2−2μ  2−2μ−2λ  M1 A1 Finds PQ.          OP=λ 1 ,OQ= 2−3μ leading toPQ= 2−3μ−λ         −7     μ     μ+7λ    2−2μ−2λ   −2 2−2μ−2λ   5  M1 Uses that dot product of PQ with line direction is zero,          2−3μ−λ  −3 =0 or 2−3μ−λ =k −3         or, alternatively, PQ is a multiple of the common   μ+7λ     1     μ+7λ     1   perpendicular (parameter k not 1). 14μ+14λ=10 A1 Deduces one equation. 2−2μ−2λ   2  A1 Deduces second equation.     2−3μ−λ  1 =014μ+54λ=6       μ+7λ     −7    2  M1 A1 Solves for λ and substitutes into OP. λ=− 1 leading to O  P  =− 1  1    10 10   −7    −0.2  5  B1 FT FT using their common perpendicular.     r= −0.1 +k −3          0.7   1  10 © UCLES 2021 Page 16 of 16

5(a) i j k  −3 M1 A1 Finds vector perpendicular to Π.   1 0 1 = 2     2 3 0 3   −3(−2)+2(3)+3(3)=21 M1 Substitutes point on Π. −3x+2y+3z=21 A1 4 5(b)  2  B1 Forms general point on line (given as a single vector).   −3    5+t   −3(2)+2(−3)+3(5+t)=21 leading to t=6 M1 Substitutes into the equation forΠ.  2  A1   −3     11 3 5(c) 0  −3 M1 Uses dot product of k and their normal.  0    2  = 1 22cosαleading to cosα= 3 A1 FT         22 1 3     Acute angle between l and Π is 90−α=39.8 ° A1 3 © UCLES 2021 Page 10 of 14 5(d) 0  2   −2 B1 Finds direction vector from P to plane.       0 − −3 = 3             7  5   2   −2  −3 M1 A1 Uses dot product of their direction and normal vectors. 1     18 3  2 = =3.84     22 22     2 3     3

2(a) AB 8i7j5k AC3j3k B1 Finds direction vectors of two lines in the plane. i j k  6  1 M1 A1 Finds normal to the plane ABC.     8 7 5  24 ~ 4         0 3 3 24 4 (4)4(4)4(1) 8x4y4z8 M1 A1 Substitutes point. Alternative method for question 2(a) Setting up 3 equations using the points given M1 x4y4z 8 A1 A1 A1 A1 5 2(b) 8 M1 A1 Divides their constant by magnitude of 1.39 their normal vector. 12 42 42 8 CAO 33 2 2(c)  2  B1 Equation of line OD.   rt 3     3 2t12t12t 8 M1 Substitutes into equation of plane. 8,12,12 A1 3 © UCLES 2022 Page 7 of 15

2(a) AB 8i7j5k AC3j3k B1 Finds direction vectors of two lines in the plane. i j k  6  1 M1 A1 Finds normal to the plane ABC.     8 7 5  24 ~ 4         0 3 3 24 4 (4)4(4)4(1) 8x4y4z8 M1 A1 Substitutes point. Alternative method for question 2(a) Setting up 3 equations using the points given M1 x4y4z 8 A1 A1 A1 A1 5 2(b) 8 M1 A1 Divides their constant by magnitude of 1.39 their normal vector. 12 42 42 8 CAO 33 2 2(c)  2  B1 Equation of line OD.   rt 3     3 2t12t12t 8 M1 Substitutes into equation of plane. 8,12,12 A1 3 © UCLES 2022 Page 7 of 15

7(a) AB4ijk CDj(3)k B1 Finds direction vectors of the two lines. i j k  2  M1 A1 Finds common perpendicular.   This may also be done by setting up simultaneous 4 1 1  124   equations and solving them.   0 1 3  4  5  2  M1 A1 Uses formula for perpendicular distance. 1     2  124     (2)2 (412)2 16     4   4  M1 Sets equal to 3 and forms in quadratic in . 303 39(10)2 9  172 100164  172 100164 162 80640 leading to 2 540 A1 AG 7 7(b)(i) r7i4jks4ijkt5i3j2k M1 A1 OE M1 for using a correct point and attempting to find relevant direction vectors. 2 7(b)(ii) i j k  8  M1 A1 Finds normal to the plane  . 2   4 1 1  25     5 3 5  7  8(7)25(4)7(1) leading to 8x25y7z163 M1 A1 Substitutes point. 4 © UCLES 2022 Page 16 of 17 7(c) i j k  5  M1 A1 Finds normal to the plane  . 1   4 1 1  13     5 3 2  7   5   8  M1 A1 Uses dot product of normal vectors.     414 cos0.9776176... 13  25  243 738cos leading to cos     243 738      7   7  12.1 A1 Mark final answer. Accept 0.212c 5 © UCLES 2022 Page 17 of 17

4(a) i j k −4  2  M1 A1 Finds vector perpendicular to Π.     −1 2 1 = 2 ~ −1         3 2 −1 −8  4  2(3)−(−2)+4(1)=12 M1 Substitutes point on Π. 2x−y+4z=12 A1 4 4(b) 0  2  M1 A1 FT Uses dot product of j+k and their normal.     3 1 −1 = 2 21cosleading to cos= May be implied by angle 62.4°     42     1  4  Acute angle between l and Π is 90−=27.6 A1 0.48 radians 3 4(c)  2  2+2t M1 A1 Forms OFusing parameter.     OF =OP+t −1 = 3−t          4  1+4t 21t+5=12 M1 Substitutes into equation of plane. 8 A1 1  OF = 8   3   7 4 © UCLES 2022 Page 9 of 15

6(a) 0 2 −2 B1 Finds a vector joining any point of l 1 to       any point of l . 2 − 0 = 2 2             6 1  5  . i j k  2  M1 A1 Finds common perpendicular.   1 2 −2 = −4     1 −1 2 −3 −2  2  M1 A1 Uses formula for shortest distance. 1     27 2 −4 = =5.01     29     29  5  −3 5 6(b)(i) r=2i+k+s(i−j+2k)+t(2i−4j−3k) B1 FT their common perpendicular from part (a). 1 6(b)(ii) i j k −14 M1 A1FT Finds normal to the plane  . FT their 2   1 2 −2 = −1 common perpendicular from part (a).     2 −4 −3  −8  14(0)+(2)+8(6)=5014x+ y+8z=50 M1 A1 Substitutes point. 4 © UCLES 2022 Page 12 of 15 6(c) i j k 11 M1 A1 FT Finds normal to the plane  . FT their 1   1 −1 2 = 7 part (b)(i).     2 −4 −3 −2 11 14 M1 A1 Uses dot product of normal vectors.     145 7 1 = 174 261cos leading to cos=     174 261     −2  8  47.1 A1 0.822 radians 5 Question Answer Marks Guidance

4(a) i j k −4  2  M1 A1 Finds vector perpendicular to Π.     −1 2 1 = 2 ~ −1         3 2 −1 −8  4  2(3)−(−2)+4(1)=12 M1 Substitutes point on Π. 2x−y+4z=12 A1 4 4(b) 0  2  M1 A1 FT Uses dot product of j+k and their normal.     3 1 −1 = 2 21cosleading to cos= May be implied by angle 62.4°     42     1  4  Acute angle between l and Π is 90−=27.6 A1 0.48 radians 3 4(c)  2  2+2t M1 A1 Forms OFusing parameter.     OF =OP+t −1 = 3−t          4  1+4t 21t+5=12 M1 Substitutes into equation of plane. 8 A1 1  OF = 8   3   7 4 © UCLES 2022 Page 9 of 15

7(a) i j k 0 0 M1 A1 Finds common perpendicular.     1 1 1  2 ~ 1         1 1 1 2 1 (4)(3)7 yz7 M1 A1 Substitutes point. 4 7(b) 7 B1 FT   States point common to both planes e.g. 2 .     5 i j k  2  M1 A1FT Finds direction of line.   0 1 1  5     5 3 5  5  7  2  A1 OE.     r 2  5         5  5  4 7(c) aa77 M1 Uses correct formula for distance from A to .  2 1 2 a1 A1 2 © UCLES 2023 Page 15 of 16 7(d) M1 A1 Uses correct formula. bb  1 2 2 (1b)2 2b2 2 M1 Solves for b. 2b 1b2 2b2 b2 2b10 b1 2 A1 CAO 4 © UCLES 2023 Page 16 of 16

7(a) i j k 0 0 M1 A1 Finds common perpendicular.     1 1 1  2 ~ 1         1 1 1 2 1 (4)(3)7 yz7 M1 A1 Substitutes point. 4 7(b) 7 B1 FT   States point common to both planes e.g. 2 .     5 i j k  2  M1 A1FT Finds direction of line.   0 1 1  5     5 3 5  5  7  2  A1 OE.     r 2  5         5  5  4 7(c) aa77 M1 Uses correct formula for distance from A to .  2 1 2 a1 A1 2 © UCLES 2023 Page 15 of 16 7(d) M1 A1 Uses correct formula. bb  1 2 2 (1b)2 2b2 2 M1 Solves for b. 2b 1b2 2b2 b2 2b10 b1 2 A1 CAO 4 © UCLES 2023 Page 16 of 16

6(a) AB2ij4k AC3i3k BCijk B1 Finds direction vectors of two lines in the plane. i j k  3   1  M1 A1FT Finds normal to the plane ABC.     2 1 4   ~ 2         3 0 3     1  1(1)2(1)1(0)1x2yz1 M1 A1 Substitutes point. CAO 5 6(b) 1 1 M1 A1FT Divides by magnitude of normal vector.  0.408 FT their (a). 12 22 12 6 2 © UCLES 2023 Page 12 of 16 6(c) 2 12 122 M1 A1 Finds PQ, where P is a point on OC and Q is a point on          OP  ,OQ 1 PQ 1 AB.              3  4   43   122 2 M1* Uses that dot product of PQ with line direction is zero.     1  1 0          43   3  17141 dM1 Deduces one equation. 122 2 dM1 Deduces second equation.     1  1 021171          43   4   2 dM1 A1 Solves for or  and substitutes into OP. 4  4   OP 1   5 5    3  2 1 A1 OE 4    r 1 k 2     5      3  0 8 © UCLES 2023 Page 13 of 16

4(a)  2  −2 4 B1 Finds direction of one line to another.       −2 − −3 = 1              3  −5 8 i j k 18 9 M1 A1 Find common perpendicular.     −4 3 5 = 14 ~ 7         2 −3 1  6  3 4 9 M1 A1 Uses formula for shortest distance. 1     67 1 7 = (=5.68)     139     139 8 3 5 4(b) i j k  3   1  M1 A1 Finds vector perpendicular to the plane.     1 0 1 = 9 ~ 3         −4 3 5 −3 −1 1(−1)+3(−3)−1(−4)=−6 x+3y−z=−6 M1 A1 Uses point in the plane. 4 © UCLES 2023 Page 9 of 15

5(a) i j k  2   1  M1 A1 Finds perpendicular to . 1     1 −2 −3 = −8 ~ −4         3 0 −1  6   3  1(1)−4(−1)+3(−2)=−1 M1 Uses point on . 1 x−4y+3z=−1 A1 4 5(b)  1 1 M1 A1 Shows dot product with direction of    line is 0. −4 1 =1−4+3=0        3 1 2 5(c) −4  1  −3  1   M1 A1 Uses correct formula for distance from 1     1      point on l to . 1 −4 or 0 −4 +1 1          12 +42 +32   3     3   12 +42 +32     1     3     1 (−3.1+ 0.−4+1.3+1) 12 +42 +32 1 A1 (=0.196) 26 3 © UCLES 2023 Page 10 of 15 5(d)  1  B1    15   5   0  States point common to both planes e.g.  4 .  7    15   5     0  0  or or alternative.   17   −4     1    7    17 i j k −17 M1 A1 Finds direction of line.   1 −4 3 = 7     3 3 2  15   5  −17 A1 OE. 7     r= 0 +  7       − 7 4   15  4 © UCLES 2023 Page 11 of 15

4(a)  2  −2 4 B1 Finds direction of one line to another.       −2 − −3 = 1              3  −5 8 i j k 18 9 M1 A1 Find common perpendicular.     −4 3 5 = 14 ~ 7         2 −3 1  6  3 4 9 M1 A1 Uses formula for shortest distance. 1     67 1 7 = (=5.68)     139     139 8 3 5 4(b) i j k  3   1  M1 A1 Finds vector perpendicular to the plane.     1 0 1 = 9 ~ 3         −4 3 5 −3 −1 1(−1)+3(−3)−1(−4)=−6 x+3y−z=−6 M1 A1 Uses point in the plane. 4 © UCLES 2023 Page 9 of 15

5(a) AB2j5k AC 5i5j BC 5i7j5k B1 Finds direction vectors of two lines in the plane. i j k 25 5 M1 A1 Finds normal to the plane ABC.     0 2 5  25 ~ 5         5 5 0  10   2  5(2)5(2)2(4)85x5y2z8 M1 A1 Substitutes point. AEF for final answer. 5 5(b) 82ij3k5i5j2k 7 M1 A1 Correct formula for distance from point to plane.  0.953 52 52 22 54 2 5(c) 2 3  5  B1        CD 1  3  4             3  4  1 i j k  18  M1 A1 Find common perpendicular.   0 2 5  25     5 4 1 10 5  18  M1 A1 Uses formula for shortest distance. 1     35 5  25  1.08     1049 1049     0 10     5 © Cambridge University Press & Assessment 2024 Page 10 of 16

5(a) AB2j5k AC5i5j BC5i7j5k B1 Finds direction vectors of two lines in the plane. i j k 25 5 M1 A1 Finds normal to the plane ABC.     0 2 5  25 ~ 5         5 5 0  10   2  5(2)5(2)2(4)85x5y2z8 M1 A1 Substitutes point. AEF for final answer. 5 5(b) 82ij3k5i5j2k 7 M1 A1 Correct formula for distance from point to plane.  0.953 52 52 22 54 2 5(c) 2 3  5  B1        CD 1  3  4             3  4  1 i j k  18  M1 A1 Find common perpendicular.   0 2 5  25     5 4 1 10 5  18  M1 A1 Uses formula for shortest distance. 1     35 5  25  1.08     1049 1049      0  10 5 © Cambridge University Press & Assessment 2024 Page 10 of 16

5(a) 3  1  4 B1 OE.       Finds direction of one line to another. 4  4  0             0 1 1       i j k  5  M1 A1 Find common perpendicular.   0 1 2  2     1 2 1 1 4  5  M1 A1 Uses formula for shortest distance. 1     21 0  2  (3.83)     30 30      1  1 5 5(b) 5(1)2(4)1(1)2 5x2yz2 M1 Uses point in the plane. A1 FT Follow through their normal in part 5(a) 2 © Cambridge University Press & Assessment 2024 Page 11 of 16 5(c) ijk B1 Finding direction between the two given points. i j k 1 M1 Find common perpendicular. Allow use of   their normal from part (b). 1 1 1  2     1 4 3 3 x2y3z6 A1 OE 3 Question Answer Marks Guidance

7(a) i j k  6  −2 M1 A1 Finds common perpendicular. Allow one error.     2 1 1 = −3 ~ 1         1 −4 2 −9  3  −2(1)+(3)+3(−2)=−5 M1 Substitutes point on l . 1 2x−y−3z=5 A1 CAO. 4 © Cambridge University Press & Assessment 2024 Page 13 of 15 7(b) i j k 6 M1 A1 Finds the normal to  . 2   1 −4 2 = 4     1 1 −2 5  2  6 M1 Uses dot product of normal vectors.     −7 −1 4 = 14 77coscos=         14 77 −3 5 77.7 A1 No ISW. Accept 1.36 rad. 4 © Cambridge University Press & Assessment 2024 Page 14 of 15 7(c) 1+2  1+   −2  M1 A1 Finds PQ.       OP= 3+ ,OQ= −2−4 PQ= −5−4−             −2+  9+2 11+2−  −2  2  −2   6  M1 Uses that dot product of PQ with line direction         −5−4− 1 =0 or −5−4− =k −3         is zero, or, alternatively, PQ is a multiple of the         11+2− 1 11+2− −9 common perpendicular. −6+6=0 A1 Deduces one equation.  −2   1  A1 Deduces second equation.     −5−4− −4 =021+42=0         11+2−  2   3  −1 M1 Solves for  or  and substitutes into OP or     =1OP= 4 or =−2OQ= 6     OQ     −1  5  −1  6  A1 FT OE. FT using their common perpendicular. Must     have"r=". r= 6 +t −3          5  −9 7 © Cambridge University Press & Assessment 2024 Page 15 of 15

2(a) i j k 24 M1 A1 Finds vector perpendicular to the plane.   1 −1 −4 = −4     2 5 −4 7   24(1)−4(3)+7(−1)=d 24x−4y+7z=5 M1 A1 Uses point in the plane. 4 2(b) 24  5  M1A1FT Uses dot product of 5i−5j−2k and     126 their normal. −4 −5 = 641 54coscos=     641 54     7 −2     Acute angle between l and Π is 90−=42.6 A1 0.744 radians 2 3 © Cambridge University Press & Assessment 2024 Page 6 of 16

7(a) i j k  6  −2 M1 A1 Finds common perpendicular. Allow one error.     2 1 1 = −3 ~ 1         1 −4 2 −9  3  −2(1)+(3)+3(−2)=−5 M1 Substitutes point on l . 1 2x−y−3z=5 A1 CAO. 4 © Cambridge University Press & Assessment 2024 Page 13 of 15 7(b) i j k 6 M1 A1 Finds the normal to  . 2   1 −4 2 = 4     1 1 −2 5  2  6 M1 Uses dot product of normal vectors.     −7 −1 4 = 14 77coscos=         14 77 −3 5 77.7 A1 No ISW. Accept 1.36 rad. 4 © Cambridge University Press & Assessment 2024 Page 14 of 15 7(c) 1+2  1+   −2  M1 A1 Finds PQ.       OP= 3+ ,OQ= −2−4 PQ= −5−4−             −2+  9+2 11+2−  −2  2  −2   6  M1 Uses that dot product of PQ with line direction         −5−4− 1 =0 or −5−4− =k −3         is zero, or, alternatively, PQ is a multiple of the         11+2− 1 11+2− −9 common perpendicular. −6+6=0 A1 Deduces one equation.  −2   1  A1 Deduces second equation.     −5−4− −4 =021+42=0         11+2−  2   3  −1 M1 Solves for  or  and substitutes into OP or     =1OP= 4 or =−2OQ= 6     OQ     −1  5  −1  6  A1 FT OE. FT using their common perpendicular. Must     have"r=". r= 6 +t −3          5  −9 7 © Cambridge University Press & Assessment 2024 Page 15 of 15

6(a) B1 Finds direction vectors of two lines in the plane. AB=2j+4k AC=i−j+k i j k  6   3  M1 A1 Finds normal to the plane ABC.     0 2 4 = 4 ~ 2         1 −1 1 −2 −1 3(1)+2(0)−(−2)=53x+2y−z=5 M1 A1 Substitutes point. 5 © Cambridge University Press & Assessment 2025 Page 14 of 18 6(b) i j k 2t+4 1 M1 A1 Finds normal to the plane ABD.     0 2 4 = 0 ~ 0         0 0 t+2  0  0  3  1 M1 Uses dot product of normal vectors in correct     formula. 2 0 = 14cos         −1 0 (t −2) =36.7 A1 0.641 rad. 4 6(c) i j k 2t−2 t−1 M1 A1 Finds common perpendicular.     0 2 4 = −4 ~ −2         −1 1 t+1  2   1   1  t−1 M1 A1 Uses formula for perpendicular distance. 1     t+2 −1 −2 =     t2 −2t+6    t2 −2t+6  1   1  t+2 = 2(t+2)2 =2 ( t2 −2t+6 ) M1 Sets equal to 2 and solves for t. t2 −2t+6 t2 −8t+8=0t=2 ( 2+ 2 ) ,t=2 ( 2− 2 ) A1 A1 t =6.83,t =1.17. 7 © Cambridge University Press & Assessment 2025 Page 15 of 18

6(a) B1 Finds direction vectors of two lines in the plane. AB=2j+4k AC=i−j+k i j k  6   3  M1 A1 Finds normal to the plane ABC.     0 2 4 = 4 ~ 2         1 −1 1 −2 −1 3(1)+2(0)−(−2)=53x+2y−z=5 M1 A1 Substitutes point. 5 © Cambridge University Press & Assessment 2025 Page 14 of 18 6(b) i j k 2t+4 1 M1 A1 Finds normal to the plane ABD.     0 2 4 = 0 ~ 0         0 0 t+2  0  0  3  1 M1 Uses dot product of normal vectors in correct     formula. 2 0 = 14cos         −1 0 (t −2) =36.7 A1 0.641 rad. 4 6(c) i j k 2t−2 t−1 M1 A1 Finds common perpendicular.     0 2 4 = −4 ~ −2         −1 1 t+1  2   1   1  t−1 M1 A1 Uses formula for perpendicular distance. 1     t+2 −1 −2 =     t2 −2t+6    t2 −2t+6  1   1  t+2 = 2(t+2)2 =2 ( t2 −2t+6 ) M1 Sets equal to 2 and solves for t. t2 −2t+6 t2 −8t+8=0t=2 ( 2+ 2 ) ,t=2 ( 2− 2 ) A1 A1 t =6.83,t =1.17. 7 © Cambridge University Press & Assessment 2025 Page 15 of 18

5(a) i j k  6  M1 A1 Finds vector perpendicular to Π.   1 −2 −1 = −1     3 2 −2  8  6(2)−(3)+8(−2)=−7 M1 Substitutes point on Π. 6x−y+8z=−7 A1 4 © Cambridge University Press & Assessment 2025 Page 13 of 19 5(b) 4  6  4+6t M1 A1FT Uses OP and multiple of their normal toΠ.       OF =OP+PF = 2 +t −1 = 2−t             9  8  9+8t 6(4+6t)−(2−t)+8(9+8t)=−7101t+94=−7 M1 Substitutes into the equation forΠ. −2 A1   t=−1OF = 3      1  4 5(c)  3   6  M1 A1FT Uses dot product of 3i+5j−k and their normal     5 5 −1 = 35 101coscos= in a correct formula.         35 101 −1  8  Acute angle between l and Π is 90−=4.8 A1 Mark final answer. No ISW 3 Question Answer Marks Guidance

3(a) AB=−5i+j−2k AC=i+2k B1 Finds direction vectors of two lines in the plane. i j k  2  M1 A1 Finds normal to the plane ABC.   −5 1 −2 = 8     1 0 2 −1   2(0)+8(2)−1(3)=132x+8y−z=13 M1 A1 Substitutes point. 5 3(b) 13 M1 Divides their d by magnitude of their normal vector. =1.57 22 +82 +12 A1 FT 13 Or oe 69 2 3(c) 2 0 M1 Dot product of normal and line direction.     13 8 2 = 69 13coscos=     69 13     -1 3     A1 FT 90−=25.7ο A1 0.449 radians CAO 3 © Cambridge University Press & Assessment 2025 Page 11 of 18

5(a) i j k −5  5  M1 A1 Finds perpendicular to . 1     0 1 2 = 2 ~ −2         1 3 1 −1  1  5(−3)−2(−1)+(−1)=−14 M1 Uses point on . 1 5x−2y+z=−14 A1 4 © Cambridge University Press & Assessment 2025 Page 12 of 17 5(b) −2 −5 M1 A1 Uses correct formula for perpendicular 1     distance from point to .  −1    2  1 52 +22 +12      1  −1 7 A1 Accept 1.28. 30 3 5(c) 0 B1  0      States point common to both planes e.g. 7 . Or 0 .         0 −14 i j k  0  0 M1 A1 Finds direction of line.     5 −2 1 = 8  1         3 2 −1 16 2 0 0 A1 OE     r= 7 + 1         0 2 4 © Cambridge University Press & Assessment 2025 Page 13 of 17

6(a) 1 1 M1 A1 Allow 0.267 or better = 12 +32 +22 14 2 © Cambridge University Press & Assessment 2025 Page 14 of 19 6(b) i j k 1 M1 A1 Finds normal to OAB. Allow 1 error for   the M mark to be awarded. 0 −1 2 = 4     2 0 −1 2   1 1 M1 Takes dot product of normal vectors.     17 4  3 = 21 14coscos=     21 14     2 2     7.5° A1 Accept 0.131 radians. Mark final answer of 7.5°or better. 4 © Cambridge University Press & Assessment 2025 Page 15 of 19 6(c)  2   2   2−2  M1 A1 Finds PQ or direction of common        OP= −1 ,OQ= −1+ PQ= −1++ perpendicular.             −1 2−3 2−3+         2−2   2  M1 Uses that dot product of PQ with line       −1++    −1  =0 direction is zero, or, alternatively, PQ     2−3+ −1 is a multiple of the common     1   perpendicular 1 .     1 6−6=1 A1 Deduces one equation.  2−2   2  A1 Deduces second equation.     −1++  1 =014−6=7         2−3+ −3       3   6  M1 A1 Solves for  and substitutes into OQ 3   2     = 4 OQ=−1 4 = 1 4 −1  . and PQ or   −1 4    −1   7    2   = OP= 7 −1 . 12 12    −1    6  1 B1 FT OE. FT using their common     perpendicular. r= 1 −1 +k 1 4    Must have r =      −1  1 8 © Cambridge University Press & Assessment 2025 Page 16 of 19

5(a) i j k −5  5  M1 A1 Finds perpendicular to . 1     0 1 2 = 2 ~ −2         1 3 1 −1  1  5(−3)−2(−1)+(−1)=−14 M1 Uses point on . 1 5x−2y+z=−14 A1 4 © Cambridge University Press & Assessment 2025 Page 12 of 17 5(b) −2 −5 M1 A1 Uses correct formula for perpendicular 1     distance from point to .  −1    2  1 52 +22 +12      1  −1 7 A1 Accept 1.28. 30 3 5(c) 0 B1  0      States point common to both planes e.g. 7 . Or 0 .         0 −14 i j k  0  0 M1 A1 Finds direction of line.     5 −2 1 = 8  1         3 2 −1 16 2 0 0 A1 OE     r= 7 + 1         0 2 4 © Cambridge University Press & Assessment 2025 Page 13 of 17

4(a) B1 Finds direction vectors of two lines in the plane. AB=−i−3j−3k AC=−i−4j−6k i j k  6  M1 A1 Finds normal to the plane ABC.   −1 −3 −3 = −3     −1 −4 −6  1  6(3)−3(5)+(5)=86x−3y+z=8 M1 A1 Substitutes point. CAO 5 4(b)  1   2  −1 B1        CD= −1 − 1 = −2              3  −1  4  i j k −18 M1 A1 Find common perpendicular.   −1 −3 −3 = 7     −1 −2 4  −1  −1 −18 M1 A1 Uses formula for shortest distance. 1     4 −4  7 = =0.207     374 374     −6  −1  5 © Cambridge University Press & Assessment 2025 Page 12 of 20

5(a) EITHER © Cambridge University Press & Assessment 2026 Page 15 of 23  − 3 − 2 4  −  − 2 4 4  =  − − 0 1 6  (B1 Finds direction of one line to another. i 3 2 − j 4 4 k 2 3 =  − 2 0 − 5 2 0  ~  − 4 − 1 4  M1 Find common perpendicular by taking cross product of the two correct direction vectors. A1 Correct. −1 −4 M1 Uses formula for shortest distance using 1     10 their common perpendicular and direction −6  −1 =     vector. Scalar product and magnitudes 33 33      0   4  must be evaluated. A1) Correct answer. Accept 1.74 OR Alternative method for 5(a) Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(a) 1 1 0 6 1 5  = 1 1 3 6 9 5  = A1 Solves equations from using perpendicular to both lines to find a n d   −40 M1 Finds PQ 33   −10  33  40    33 M1 Finds length of PQ PQ= ( −40)2 + ( −10)2 + (40)2 33 33 33 = 1 0 3 3 A1) 5 Trace ID: PC-1563EA0F9E Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a1 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/11 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(b) © Cambridge University Press & Assessment 2026 Page 17 of 23 i − 3 3 j 5 − 4 k 3 2 =  2 1 − 2 5 3  *M1 Finds the normal to 1  by taking cross product of two correct direction vectors. A1 Correct. i j k −22 *M1 Finds the normal to  by taking cross 2   −3 −2 4 = 17 product of two correct direction vectors.     2 4 3 −8   A1 Correct 2 1 2 5 3 1 2 2 7 8 7 1 8 8 3 7 c o s c o s 7 1 2 8 0 5 8 3 7    −    − −  =  = − DM1 Uses dot product of their normal vectors. Scalar product and magnitudes must be evaluated. 1 8 0 7 4 .7  − =  A1 Or 1.30 rad. Mark final answer. AWRT 74.7 (1dp) or 1.3 (1dp) 6 Question Answer Marks Guidance

5(a) EITHER © Cambridge University Press & Assessment 2026 Page 15 of 23  − 3 − 2 4  −  − 2 4 4  =  − − 0 1 6  (B1 Finds direction of one line to another. i 3 2 − j 4 4 k 2 3 =  − 2 0 − 5 2 0  ~  − 4 − 1 4  M1 Find common perpendicular by taking cross product of the two correct direction vectors. A1 Correct. −1 −4 M1 Uses formula for shortest distance using 1     10 their common perpendicular and direction −6  −1 =     vector. Scalar product and magnitudes 33 33      0   4  must be evaluated. A1) Correct answer. Accept 1.74 OR Alternative method for 5(a) Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(a) 1 1 0 6 1 5  = 1 1 3 6 9 5  = A1 Solves equations from using perpendicular to both lines to find a n d   −40 M1 Finds PQ 33   −10  33  40    33 M1 Finds length of PQ PQ= ( −40)2 + ( −10)2 + (40)2 33 33 33 = 1 0 3 3 A1) 5 Trace ID: PC-F469E54369 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a2 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/12 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(b) © Cambridge University Press & Assessment 2026 Page 17 of 23 i − 3 3 j 5 − 4 k 3 2 =  2 1 − 2 5 3  *M1 Finds the normal to 1  by taking cross product of two correct direction vectors. A1 Correct. i j k −22 *M1 Finds the normal to  by taking cross 2   −3 −2 4 = 17 product of two correct direction vectors.     2 4 3 −8   A1 Correct 2 1 2 5 3 1 2 2 7 8 7 1 8 8 3 7 c o s c o s 7 1 2 8 0 5 8 3 7    −    − −  =  = − DM1 Uses dot product of their normal vectors. Scalar product and magnitudes must be evaluated. 1 8 0 7 4 .7  − =  A1 Or 1.30 rad. Mark final answer. AWRT 74.7 (1dp) or 1.3 (1dp) 6 Question Answer Marks Guidance

7(a) i j k  1  M1 A1 Finds normal to the plane . For the method   mark allow one arithmetic error in calculating 1 −1 0 = 1   the cross product but must be using the correct   −3 1 −1  −2  direction vectors. © Cambridge University Press & Assessment 2026 Page 18 of 20 ( 0 ) + ( 1 ) − 2 ( 2 ) = − 3  x + y − 2 z = − 3 M1 Substitutes point using their cross product but must be from using two correct direction vectors. A1 Answer must be in the form a x + b y + c z = d . 4 7(b) 1 6 3 1 1 2 6 1 2 6 5 3 6 6     − −  −    + + +  = − *M1 A1 Forms perpendicular distance of P from plane  in terms of . For only the method mark allow follow through of their d from part (a). 3 6 6  −  DM1 Forms and solves linear inequality in . 04 A1 4 Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) © Cambridge University Press & Assessment 2026 Page 19 of 20 O Q 0 1 2 , O R 1 2 3 Q R 2 1 3 2             =         =  − −   =  − − − −  M1 Allow one arithmetic error in calculating one element.  A1 FindsQR or direction of common perpendicular.   1−3  0 M1 Uses that dot product of QR with line direction       −    1  =0 is zero, or, alternatively, QRis a multiple of the     2−−2 2     common perpendicular  1 2 − 1  , the multiple is not 1. 5 4   + = A1 Deduces one equation. 2 1 3 2 1 3 1 0 1 1 5         − − − −    − −  =  + = M1 Deduces second equation for their Q R   −1 −3 M1 A1 Solves for and substitutes into 7  6    = OR= 7 = 1 7 18  18  18    29     29   18  1  −  6     1 QR= −  3   1      6  O R   and Q R   or  1 −   6 0   13      1 = OQ= 13 1 and QR= − 18 18   3   2     1      6  Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 7(c) −3  1  B1 FT OE. FT using their common perpendicular.     Answer must be in equation form. r = 1 7 +k 2 18        29 −1     8 © Cambridge University Press & Assessment 2026 Page 20 of 20 Trace ID: PC-39A0636DD7 Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a3 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/13 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/

5(a) A B  =  − 2   i − j − 2 k A C = i − j − k B C = 3 i + k B1 Finds direction vectors of two lines in the plane. i j k −1 M1 Finds normal to the plane ABC by   taking cross product of two of their −2 −1 −2 = −4   vectors from above. Allow one   1 −1 −1  3  numerical or sign error in cross product for M1. A1FT Allow FT their vectors from above. − ( 4 ) − 4 ( 3 ) + 3 ( 3 ) = − 7  − x − 4 y + 3 z = − 7 M1 Substitutes a correct point. A1 CAO 5 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(b) © Cambridge University Press & Assessment 2026 Page 17 of 25 2 1 3 1 4 3 1 4 2 6 c o s c o s 1 4 3 2 6       − −  =  = M1 Uses scalar product of O D   and their normal. For M1 the scalar product and lengths must be evaluated but need not be simplified. A1FT Lengths and scalar product simplified correctly. Acute angle between O D and Π is 90−=9.05 A1 Mark final answer. Accept 9 .0  9 .1  or 0.16 rad. But not 9 3 Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(c) EITHER: (B1 2 5 −3        CD= 1 − 2 = −1             3 2  1  i j k −3 *M1 Find common perpendicular of their   CD and AB. Again, at most one error in −2 −1 −2 = 8   cross product.   −3 −1 1 −1 A1 Correct answer © Cambridge University Press & Assessment 2026 Page 18 of 25 1 7 4  − 1 1 1    − 8 − 3 1  = 1 0 7 4 = 1 .1 6 DM1 Uses formula for shortest distance. May use any vector joining a point of AB to a point of CD. For M1 the scalar product and length must be evaluated but need not be simplified. A1) CAO Surd or decimal. OR: Alternative method for 5(c) 4 −2     r= 3 + −1         3 −2 r 5 2 2 1 3 1  =   +  − −  (B1 Write down a general point on each of AB and CD. −1−2+3 M1 Finds vector joining them.   1−+      1−2−  Trace ID: PC-4F45C0CD5B Re-uploading, mirroring or re-hosting this file on any other website or domain is unauthorised and traceable. Licensed for hosting on papacambridge.com only. P92a31p/1a4 CMaarkm Scbhermide gJuene 2·0 2p6 a| Spoaurccea: mpapbarciadmgbreid.gceo.comm 9231/14 Mark Scheme June 2026 Downloaded from PapaCambridge - https://papacambridge.com/ 5(c) © Cambridge University Press & Assessment 2026 Page 19 of 25 1 9 5 0   − + − = 3 5 1 1 0   + − = M1 Scalar product of each line direction with this vector is zero. Both equations needed but can be unsimplified. 1 3 3 7  = 1 3 6 7  = A1 Solves the equations.  − − 43 1 5 3 7 0753 7  has magnitude 1 0 7 4 = 1 .1 6 A1) CAO Surd or decimal. 5 Question Answer Marks Guidance