Pearson Edexcel IAL Mathematics FP3.5 Vectors Question BankPractise vector methods for lines, planes, areas, volumes, intersections, angles and distances in three-dimensional coordinate geometry.SyllabusFirst assessment 2019CourseMathematics YMA01LevelA2
Exam pointsuse vector products and scalar triple products for perpendicular vectors, areas and volumesconvert between line and plane forms to find intersections or shortest distancescalculate angles between lines or planes using direction vectors and normals
FP3.5 - Vectors question 1[Maximum number: 10]The skew lines l1l_{1}l1 and l2l_{2}l2 have equationsl1:r=(i+2j−5k)+λ(5i+j)l_{1}: \mathbf{r}=(\mathbf{i}+2 \mathbf{j}-5 \mathbf{k})+\lambda(5 \mathbf{i}+\mathbf{j})l1:r=(i+2j−5k)+λ(5i+j)andl2:r=(2i−4j+4k)+μ(8i−2j+3k)l_{2}: \mathbf{r}=(2 \mathbf{i}-4 \mathbf{j}+4 \mathbf{k})+\mu(8 \mathbf{i}-2 \mathbf{j}+3 \mathbf{k})l2:r=(2i−4j+4k)+μ(8i−2j+3k)where λ\lambdaλ and μ\muμ are scalar parameters.Question (a)(a)Determine a vector that is perpendicular to both l1l_{1}l1 and l2l_{2}l2[ 2 ]Show Answer(5i+j)×(8i−2j+3k)(5\mathbf i+\mathbf j)\times(8\mathbf i-2\mathbf j+3\mathbf k)(5i+j)×(8i−2j+3k)or solve(ui+vj+wk)⋅(5i+j)=0,(ui+vj+wk)⋅(8i−2j+3k)=0(u\mathbf i+v\mathbf j+w\mathbf k)\cdot(5\mathbf i+\mathbf j)=0, \quad (u\mathbf i+v\mathbf j+w\mathbf k)\cdot(8\mathbf i-2\mathbf j+3\mathbf k)=0(ui+vj+wk)⋅(5i+j)=0,(ui+vj+wk)⋅(8i−2j+3k)=05u+v=0,8u−2v+3w=05u+v=0,\quad 8u-2v+3w=05u+v=0,8u−2v+3w=0n=3i−15j−18k\mathbf n=3\mathbf i-15\mathbf j-18\mathbf kn=3i−15j−18kor any non-zero multiple of i−5j−6k\mathbf i-5\mathbf j-6\mathbf ki−5j−6k.M1 A1Question (b)(b)Determine an equation of the plane parallel to l1l_{1}l1 that contains l2l_{2}l2[ 3 ]Question (i)(i)in the form r=a+s b+t c[ 1 ]Show Answerr=(2i−4j+4k)+s(8i−2j+3k)+t(5i+j)\mathbf r=(2\mathbf i-4\mathbf j+4\mathbf k) +s(8\mathbf i-2\mathbf j+3\mathbf k) +t(5\mathbf i+\mathbf j)r=(2i−4j+4k)+s(8i−2j+3k)+t(5i+j)B1Question (ii)(ii)in the form r.n =p[ 2 ]Show Answer(2i−4j+4k)⋅(3i−15j−18k)=−6(2\mathbf i-4\mathbf j+4\mathbf k)\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6(2i−4j+4k)⋅(3i−15j−18k)=−6M1r⋅(3i−15j−18k)=−6\mathbf r\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6r⋅(3i−15j−18k)=−6orr⋅(−i+5j+6k)=2\mathbf r\cdot(-\mathbf i+5\mathbf j+6\mathbf k)=2r⋅(−i+5j+6k)=2A1Question (c)(c)Determine the shortest distance between l1l_{1}l1 and l2l_{2}l2Give your answer in simplest form.[ 5 ]Show AnswerWay 1:632+152+182=262,2162\frac{6}{\sqrt{3^2+15^2+18^2}}=\frac{2}{\sqrt{62}}, \quad \frac{21}{\sqrt{62}}32+152+1826=622,6221d=262+2162=2362=236262d=\frac{2}{\sqrt{62}}+\frac{21}{\sqrt{62}} =\frac{23}{\sqrt{62}} =\frac{23\sqrt{62}}{62}d=622+6221=6223=622362ORWay 2:AB→=±(−i+6j−9k)\overrightarrow{AB}=\pm(-\mathbf i+6\mathbf j-9\mathbf k)AB=±(−i+6j−9k)d=∣AB→⋅n∣∣n∣=69558=2362d=\frac{|\overrightarrow{AB}\cdot\mathbf n|}{|\mathbf n|} =\frac{69}{\sqrt{558}} =\frac{23}{\sqrt{62}}d=∣n∣∣AB⋅n∣=55869=6223M1 A1 M1 M1 A1Add to Test
Question (a)(a)Determine a vector that is perpendicular to both l1l_{1}l1 and l2l_{2}l2[ 2 ]Show Answer(5i+j)×(8i−2j+3k)(5\mathbf i+\mathbf j)\times(8\mathbf i-2\mathbf j+3\mathbf k)(5i+j)×(8i−2j+3k)or solve(ui+vj+wk)⋅(5i+j)=0,(ui+vj+wk)⋅(8i−2j+3k)=0(u\mathbf i+v\mathbf j+w\mathbf k)\cdot(5\mathbf i+\mathbf j)=0, \quad (u\mathbf i+v\mathbf j+w\mathbf k)\cdot(8\mathbf i-2\mathbf j+3\mathbf k)=0(ui+vj+wk)⋅(5i+j)=0,(ui+vj+wk)⋅(8i−2j+3k)=05u+v=0,8u−2v+3w=05u+v=0,\quad 8u-2v+3w=05u+v=0,8u−2v+3w=0n=3i−15j−18k\mathbf n=3\mathbf i-15\mathbf j-18\mathbf kn=3i−15j−18kor any non-zero multiple of i−5j−6k\mathbf i-5\mathbf j-6\mathbf ki−5j−6k.M1 A1
Question (b)(b)Determine an equation of the plane parallel to l1l_{1}l1 that contains l2l_{2}l2[ 3 ]Question (i)(i)in the form r=a+s b+t c[ 1 ]Show Answerr=(2i−4j+4k)+s(8i−2j+3k)+t(5i+j)\mathbf r=(2\mathbf i-4\mathbf j+4\mathbf k) +s(8\mathbf i-2\mathbf j+3\mathbf k) +t(5\mathbf i+\mathbf j)r=(2i−4j+4k)+s(8i−2j+3k)+t(5i+j)B1Question (ii)(ii)in the form r.n =p[ 2 ]Show Answer(2i−4j+4k)⋅(3i−15j−18k)=−6(2\mathbf i-4\mathbf j+4\mathbf k)\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6(2i−4j+4k)⋅(3i−15j−18k)=−6M1r⋅(3i−15j−18k)=−6\mathbf r\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6r⋅(3i−15j−18k)=−6orr⋅(−i+5j+6k)=2\mathbf r\cdot(-\mathbf i+5\mathbf j+6\mathbf k)=2r⋅(−i+5j+6k)=2A1
Question (i)(i)in the form r=a+s b+t c[ 1 ]Show Answerr=(2i−4j+4k)+s(8i−2j+3k)+t(5i+j)\mathbf r=(2\mathbf i-4\mathbf j+4\mathbf k) +s(8\mathbf i-2\mathbf j+3\mathbf k) +t(5\mathbf i+\mathbf j)r=(2i−4j+4k)+s(8i−2j+3k)+t(5i+j)B1
Question (ii)(ii)in the form r.n =p[ 2 ]Show Answer(2i−4j+4k)⋅(3i−15j−18k)=−6(2\mathbf i-4\mathbf j+4\mathbf k)\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6(2i−4j+4k)⋅(3i−15j−18k)=−6M1r⋅(3i−15j−18k)=−6\mathbf r\cdot(3\mathbf i-15\mathbf j-18\mathbf k)=-6r⋅(3i−15j−18k)=−6orr⋅(−i+5j+6k)=2\mathbf r\cdot(-\mathbf i+5\mathbf j+6\mathbf k)=2r⋅(−i+5j+6k)=2A1
Question (c)(c)Determine the shortest distance between l1l_{1}l1 and l2l_{2}l2Give your answer in simplest form.[ 5 ]Show AnswerWay 1:632+152+182=262,2162\frac{6}{\sqrt{3^2+15^2+18^2}}=\frac{2}{\sqrt{62}}, \quad \frac{21}{\sqrt{62}}32+152+1826=622,6221d=262+2162=2362=236262d=\frac{2}{\sqrt{62}}+\frac{21}{\sqrt{62}} =\frac{23}{\sqrt{62}} =\frac{23\sqrt{62}}{62}d=622+6221=6223=622362ORWay 2:AB→=±(−i+6j−9k)\overrightarrow{AB}=\pm(-\mathbf i+6\mathbf j-9\mathbf k)AB=±(−i+6j−9k)d=∣AB→⋅n∣∣n∣=69558=2362d=\frac{|\overrightarrow{AB}\cdot\mathbf n|}{|\mathbf n|} =\frac{69}{\sqrt{558}} =\frac{23}{\sqrt{62}}d=∣n∣∣AB⋅n∣=55869=6223M1 A1 M1 M1 A1