Vectors and systems of linear equations – university Year 1 maths worksheet answers to download as PDF

← Back to the worksheet

1. Answer to Question 1

   1. Multiply corresponding components and add: \(\mathbf a\cdot\mathbf b=2(1)+(-1)(4)+3(2)=2-4+6=\) 4. The lengths are \(|\mathbf a|=\sqrt{4+1+9}=\) \(\sqrt{14}\) and \(|\mathbf b|=\sqrt{1+16+4}=\) \(\sqrt{21}\).
   2. From \(\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta\): \(\cos\theta=\dfrac4{\sqrt{14}\sqrt{21}}=\dfrac4{\sqrt{294}}\approx0.2333\), so \(\theta\approx\) 76.5°. The dot product is positive, so the angle is acute, as expected.
   3. \(\mathbf a\cdot\mathbf c=2(1)+(-1)(2)+3(0)=2-2+0=0\). A zero dot product between non-zero vectors means they are perpendicular.
   4. The projection is \(\operatorname{proj}_{\mathbf b}\mathbf a=\dfrac{\mathbf a\cdot\mathbf b}{|\mathbf b|^2}\mathbf b=\dfrac4{21}(1,4,2)=\) \(\left(\tfrac4{21},\tfrac{16}{21},\tfrac8{21}\right)\): the "shadow" of \(\mathbf a\) on the line of \(\mathbf b\).

2. Answer to Question 2

   1. Use \(\mathbf a\times\mathbf b=(a_2b_3-a_3b_2,\ a_3b_1-a_1b_3,\ a_1b_2-a_2b_1)\): \(\mathbf a\times\mathbf b=\big((-1)(2)-(3)(4),\ (3)(1)-(2)(2),\ (2)(4)-(-1)(1)\big)=\) \((-14,-1,9)\).
   2. \((-14,-1,9)\cdot(2,-1,3)=-28+1+27=0\) and \((-14,-1,9)\cdot(1,4,2)=-14-4+18=0\). Both dot products are zero, so the cross product is perpendicular to \(\mathbf a\) and to \(\mathbf b\), as it should be.
   3. The length of the cross product is the area of the parallelogram: \(|\mathbf a\times\mathbf b|=\sqrt{196+1+81}=\sqrt{278}\approx\) 16.67 square units. The triangle is half of the parallelogram: \(\frac12\sqrt{278}\approx\) 8.34 square units.

3. Answer to Question 3

   1. Two direction vectors in the plane are \(\overrightarrow{AB}=(1,1,-2)\) and \(\overrightarrow{AC}=(-1,1,-1)\). Their cross product is perpendicular to both: \(\mathbf n=\overrightarrow{AB}\times\overrightarrow{AC}=(1\cdot(-1)-(-2)(1),\ (-2)(-1)-1\cdot(-1),\ 1\cdot1-1\cdot(-1))=(1,3,2)\). The plane is \(x+3y+2z=d\); substituting \(A\) gives \(d=1+0+4=5\), so \(x+3y+2z=5\). Check with \(B\): \(2+3+0=5\).
   2. Substitute the line into the plane: \((1+t)+3(1-t)+2(1+2t)=6+2t=5\), so \(t=-\frac12\). The point is \((0.5,\ 1.5,\ 0)\).
   3. Use \(d=\dfrac{|ax_0+by_0+cz_0-d|}{\sqrt{a^2+b^2+c^2}}=\dfrac{|3+6+2-5|}{\sqrt{1+9+4}}=\dfrac6{\sqrt{14}}\approx\) 1.60 units.

4. Answer to Question 4

   1. Each row holds the coefficients of one equation, with the right-hand side after the bar: \(\left(\begin{array}{ccc|c}1&2&-1&3\\2&1&1&6\\3&-1&2&7\end{array}\right)\).
   2. Use row 1 to clear the first column: \(R_2\to R_2-2R_1\) gives \((0,-3,3\mid0)\), and \(R_3\to R_3-3R_1\) gives \((0,-7,5\mid-2)\). Then clear the second column below the pivot: \(R_3\to R_3-\frac73R_2\) gives \((0,0,-2\mid-2)\). The matrix is now in echelon form.
   3. Work upwards. The last row says \(-2z=-2\), so \(z=1\). The second row says \(-3y+3z=0\), so \(y=1\). The first row gives \(x+2-1=3\), so \(x=2\). The solution is \((x,y,z)=(2,1,1)\). Check: \(2+2-1=3\), \(4+1+1=6\) and \(6-1+2=7\).

5. Answer to Question 5

   1. \(R_2\to R_2-R_1\) gives \((0,1,2\mid1)\) and \(R_3\to R_3-R_1\) gives \((0,2,k-1\mid2)\). Then \(R_3\to R_3-2R_2\) gives \((0,0,k-5\mid0)\), as required.
   2. If \(k\ne5\), the last row gives \((k-5)z=0\), so \(z=0\). Then \(y+2z=1\) gives \(y=1\), and \(x+y+z=1\) gives \(x=0\). The unique solution is \((0,1,0)\).
   3. If \(k=5\), the last row is \(0=0\), so \(z=t\) is a free parameter. Then \(y=1-2t\) and \(x=1-y-z=t\). There are infinitely many solutions \((t,\ 1-2t,\ t)\): the three planes meet in a line.
   4. With right-hand side 4, the last row becomes \((0,0,0\mid1)\), which says \(0=1\). The system is inconsistent: no solution.

6. Answer to Question 6

   1. \(R_2\to R_2-2R_1\) gives a zero row, because row 2 is twice row 1. \(R_3\to R_3-R_1\) gives \((0,-2,-2)\). Two non-zero rows remain in echelon form, so rank 2.
   2. The reduced system is \(x+2y+3z=0\) and \(-2y-2z=0\). From the second, \(y=-z\); then \(x=-2y-3z=2z-3z=-z\). Letting \(z=-t\), the solutions are \(t(1,1,-1)\), \(t\in\mathbb R\). Check: \(A(1,1,-1)^T=(0,0,0)^T\).
   3. Each non-zero row fixes one unknown, so the number of free parameters is the number of unknowns minus the rank: \(3-2=1\). One parameter means the solutions form a line through the origin.

7. Answer to Question 7

   1. The trolley moves from \((0,0,0)\) to \((5,2,0)\), so \(\mathbf d=(5,2,0)\) and \(|\mathbf d|=\sqrt{25+4}=\sqrt{29}\approx\) 5.39 m.
   2. \(W=\mathbf F\cdot\mathbf d=3(5)+4(2)+0=15+8=\) 23 J. Only the part of the force along the motion does work.
   3. \(|\mathbf F|=\sqrt{9+16}=5\), so \(\cos\theta=\dfrac{\mathbf F\cdot\mathbf d}{|\mathbf F||\mathbf d|}=\dfrac{23}{5\sqrt{29}}\approx0.854\), and \(\theta\approx\) 31.3°.
   4. The component of \(\mathbf F\) along \(\mathbf d\) is \(\dfrac{\mathbf F\cdot\mathbf d}{|\mathbf d|}=\dfrac{23}{\sqrt{29}}\approx\) 4.27 N, out of a total force of 5 N. Equivalently \(5\cos31.3^\circ\approx4.27\).

8. Answer to Question 8

   1. \(x+y+z=10\), \(1.2x+0.8y+2z=12\), \(90x+100y+140z=1040\).
   2. Multiply the second by 5: \(6x+4y+10z=60\); subtract \(4\times\) the first: \(2x+6z=20\), so \(x=10-3z\). Divide the third by 10 and subtract \(10\times\) the first: \(-x+4z=4\). Then \(-(10-3z)+4z=4\) gives \(7z=14\), so \(z=2\), \(x=4\), \(y=4\). 4 l orange, 4 l apple, 2 l mango.
   3. \(4+4+2=10\), \(4.8+3.2+4=12\), \(360+400+280=1040\).
   4. Now \(-x+4z=0\); with \(x=10-3z\): \(7z=10\), \(z\approx1.43\), \(x\approx5.71\), \(y\approx2.86\) litres.

9. Answer to Question 9

   1. Each element must balance, so left minus right is zero: carbon \(3a-c=0\), hydrogen \(8a-2d=0\), oxygen \(2b-2c-d=0\).
   2. There are 3 equations in 4 unknowns, so the rank is at most 3 and there is at least one free parameter. This makes sense chemically: if \((a,b,c,d)\) balances the equation, so does any multiple.
   3. Take \(a=1\). Then \(c=3\) and \(d=4\), and the oxygen equation gives \(2b=6+4\), so \(b=5\). The balanced equation is \(\mathrm{C_3H_8}+5\mathrm{O_2}\to3\mathrm{CO_2}+4\mathrm{H_2O}\). Check oxygen: \(10=6+4\).
   4. Multiply the solution by 10: \(a=10\) gives \(b=50\), so 50 molecules of oxygen are needed.

10. Answer to Question 10

    1. \(A\): \(x_4+400=x_1+250\); \(B\): \(x_1+200=x_2+300\); \(C\): \(x_2+350=x_3+400\); \(D\): \(x_3+300=x_4+300\).
    2. Adding the first three gives the fourth, so the rank is 3 and there is one free parameter: \(x_1=150+t\), \(x_2=50+t\), \(x_3=t\), \(x_4=t\).
    3. \(t=0\): \(x_1=150\), \(x_2=50\), \(x_3=0\).
    4. Flows cannot be negative, so \(t\ge0\) and \(x_1\ge\) 150 cars per hour.

11. Answer to Question 11

    1. Substitute each measurement into \(h=a+bt+ct^2\): \(t=0\) gives \(a=1\); \(t=1\) gives \(a+b+c=6\); \(t=2\) gives \(a+2b+4c=9\).
    2. Using \(a=1\): \(b+c=5\) and \(2b+4c=8\), that is \(b+2c=4\). Subtracting, \(c=-1\), and then \(b=6\). The model is \(h=1+6t-t^2\).
    3. Complete the square: \(h=1-(t^2-6t)=10-(t-3)^2\). The square is never negative, so the maximum height is 10 m at \(t=3\) s.
    4. The ball lands when \(h=0\): \(t^2-6t-1=0\), so \(t=3\pm\sqrt{10}\). Only the positive root makes sense: \(t=3+\sqrt{10}\approx\) 6.16 s.

12. Answer to Question 12

    1. \(\overrightarrow{AP}=P-A=(5,1,2)\). Then \(\overrightarrow{AP}\times\mathbf d=(1\cdot0-2\cdot1,\ 2\cdot1-5\cdot0,\ 5\cdot1-1\cdot1)=(-2,2,4)\).
    2. The cross product has length \(|\overrightarrow{AP}||\mathbf d|\sin\theta\), so dividing by \(|\mathbf d|\) gives the perpendicular distance: \(\dfrac{\sqrt{4+4+16}}{\sqrt2}=\sqrt{12}\approx\) 3.46 m.
    3. The closest point is \(A+t\mathbf d\) with \(t=\dfrac{\overrightarrow{AP}\cdot\mathbf d}{|\mathbf d|^2}=\dfrac{5+1}2=3\), which is \((3,3,10)\). Check: the distance from \(P\) is \(\sqrt{4+4+4}=\sqrt{12}\), as in part (b).
    4. Since \(3.46>3\), the drone is not breaking the rule, but the margin is less than half a metre.

13. Answer to Question 13

    1. Each performance gives one equation: \(100a+50c+30s=2200\), \(80a+70c+20s=1960\), \(120a+30c+50s=2540\). Dividing each by 10 gives smaller numbers: \(10a+5c+3s=220\), \(8a+7c+2s=196\), \(12a+3c+5s=254\).
    2. Eliminating \(a\) from the second and third equations using the first, and then eliminating \(c\), leaves one equation in \(s\), giving \(s=10\). Back substitution gives \(c=8\) and \(a=15\). The prices are \(a=£15\), \(c=£8\), \(s=£10\). Check Friday: \(1500+400+300=2200\).
    3. Sunday: \(120(15)+30(8)+50(10)=1800+240+500=2540\), which matches the takings.
    4. 100 adults and 40 seniors bring in \(1500+400=1900\), leaving \(2500-1900=600\). At £8 each, the theatre must sell \(600/8=\) 75 children's tickets.

14. Answer to Question 14

    1. The coefficients of \(x,y,z\) give the normals: \(\mathbf n_1=(1,0,1)\) and \(\mathbf n_2=(0,1,1)\).
    2. The angle between the planes is the angle between their normals: \(\cos\theta=\dfrac{\mathbf n_1\cdot\mathbf n_2}{|\mathbf n_1||\mathbf n_2|}=\dfrac{1}{\sqrt2\sqrt2}=\dfrac12\), so the angle is 60°.
    3. The ridge lies in both planes, so it is perpendicular to both normals: \(\mathbf n_1\times\mathbf n_2=(0\cdot1-1\cdot1,\ 1\cdot0-1\cdot1,\ 1\cdot1-0\cdot0)=\) \((-1,-1,1)\).
    4. The point \((6,6,0)\) satisfies both equations, so the ridge is \(\mathbf r=(6,6,0)+t(-1,-1,1)\). Above \((2,2,0)\) we need \(6-t=2\), so \(t=4\), and the height of the ridge there is 4 m. Check: \(2+4=6\) in both planes.

15. Answer to Question 15

    1. Adding the equations: \(2x+0y+3z=11\).
    2. Unique point: \(a\ne3\). Line: \(a=3\), \(b=11\). None: \(a=3\), \(b\ne11\).
    3. \(2x+3z=11\) and \(2x+z=8\): \(z=1.5\), \(x=3.25\), \(y=6-4.75=1.25\): \((3.25,\ 1.25,\ 1.5)\).
    4. Let \(z=t\): \(x=\dfrac{11-3t}2\), \(y=6-x-t=\dfrac{1+t}2\): \(\mathbf r=\left(\tfrac{11}2,\tfrac12,0\right)+t\left(-\tfrac32,\tfrac12,1\right)\).
    5. Each pair of planes meets in a line, and the three lines are parallel but do not coincide.

16. Answer to Question 16

    1. Rewrite the junction law as \(I_1-I_2-I_3=0\). Then \(A=\) \(\begin{pmatrix}1&-1&-1\\2&4&0\\0&4&-6\end{pmatrix}\) and \(\mathbf b=(0,10,0)\).
    2. Loop 2 gives \(I_3=\frac23I_2\). The junction law then gives \(I_1=I_2+\frac23I_2=\frac53I_2\). Substituting into loop 1: \(\frac{10}3I_2+4I_2=10\), so \(\frac{22}3I_2=10\) and \(I_2=\dfrac{15}{11}\). Hence \(I_1=\dfrac{25}{11}\approx2.27\), \(I_2\approx1.36\), \(I_3=\dfrac{10}{11}\approx0.91\) A.
    3. \(I_2+I_3=\dfrac{15}{11}+\dfrac{10}{11}=\dfrac{25}{11}=I_1\), so the junction law holds.
    4. \(P=10I_1=\dfrac{250}{11}\approx\) 22.7 W supplied by the battery.

17. Answer to Question 17

    1. Rows \((1,x_i)\): \(A=\) \(\begin{pmatrix}1&0\\1&1\\1&2\\1&3\end{pmatrix}\), \(\mathbf y=(1,3,4,4)\).
    2. \(A^TA=\) \(\begin{pmatrix}4&6\\6&14\end{pmatrix}\), \(A^T\mathbf y=(12,23)\).
    3. Determinant \(56-36=20\): \(a=\dfrac{14\times12-6\times23}{20}=1.5\), \(b=\dfrac{4\times23-6\times12}{20}=1\): \(y=1.5+x\).
    4. Residuals \(-0.5,\ 0.5,\ 0.5,\ -0.5\); sum of squares 1.
    5. The normal equations say \(A^T(\mathbf y-A\mathbf u)=\mathbf 0\): the error is orthogonal to the column space, so \(A\mathbf u\) is the projection of \(\mathbf y\).

18. Answer to Question 18

    1. \(\mathbf v\times\mathbf w=(12,-4,-2)\); \(\mathbf u\cdot(\mathbf v\times\mathbf w)=\) 24.
    2. \(\det=2(12-0)-0+0=24\).
    3. Cell: 24 Å\(^3\); tetrahedron: \(\frac16\times24=\) 4 Å\(^3\).
    4. All three have \(z=0\), so the triple product is 0. Likewise \(\mathbf u\), \(\mathbf v\), \((3,3,0)\) lie in the plane \(z=0\): the volume is 0, so no.

19. Answer to Question 19

    1. The directions are not multiples. Equating: \(s=0\), \(0=1+t\), \(0=2+t\): impossible, so they do not meet.
    2. \(\mathbf n=(1,0,0)\times(0,1,1)=\) \((0,-1,1)\).
    3. With \(A=(0,0,0)\), \(B=(0,1,2)\): \(d=\dfrac{|\overrightarrow{AB}\cdot\mathbf n|}{|\mathbf n|}=\dfrac{|-1+2|}{\sqrt2}=\) \(\dfrac1{\sqrt2}\approx0.707\) km.
    4. \(P_2-P_1=(-s,1+t,2+t)\) must be perpendicular to both directions: \(s=0\) and \(2t+3=0\), \(t=-1.5\). Points \((0,0,0)\) and \((0,-0.5,0.5)\).
    5. \(0.707<1\): the separation would be broken if both aircraft were at those points at the same time; but the times also matter.

20. Answer to Question 20

    1. \(q(t)=|\mathbf b|^2t^2+2(\mathbf a\cdot\mathbf b)t+|\mathbf a|^2\).
    2. A quadratic that is never negative has discriminant \(\le0\): \(4(\mathbf a\cdot\mathbf b)^2-4|\mathbf a|^2|\mathbf b|^2\le0\).
    3. When the discriminant is 0: \(q\) has a root, so \(\mathbf a=-t\mathbf b\), i.e. the vectors are parallel.
    4. \(|\mathbf a+\mathbf b|^2=|\mathbf a|^2+2\mathbf a\cdot\mathbf b+|\mathbf b|^2\le|\mathbf a|^2+2|\mathbf a||\mathbf b|+|\mathbf b|^2=(|\mathbf a|+|\mathbf b|)^2\).
    5. \(\mathbf a\cdot\mathbf b=11\le3\times5=15\); \(|\mathbf a+\mathbf b|=|(4,2,6)|=\sqrt{56}\approx7.48\le8\).