Further vectors – further maths worksheet answers to download as PDF

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1. Answer to Question 1

   Method: the scalar product is \(\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta\). Two non-zero vectors are perpendicular exactly when their scalar product is 0.

   1. Multiply corresponding components and add; use Pythagoras for the magnitudes. \(\mathbf a\cdot\mathbf b=2-4-6=\) \(-8\); \(|\mathbf a|=\sqrt{4+1+9}=\) \(\sqrt{14}\); \(|\mathbf b|=\sqrt{1+16+4}=\) \(\sqrt{21}\)
   2. Rearrange the scalar product formula for \(\cos\theta\). \(\cos\theta=\dfrac{-8}{\sqrt{14}\sqrt{21}}=\dfrac{-8}{\sqrt{294}}=-0.4666\), so \(\theta=\) 117.8\(^\circ\) The angle is obtuse because the scalar product is negative.
   3. Compute the scalar product. \(3-1-2=0\), so the vectors are perpendicular. So the two vectors are at \(90^\circ\).
   4. Set the scalar product equal to zero and solve for \(k\). \(3k+2k-4=0\), so \(k=\dfrac45\) Check: \(\tfrac{12}5+\tfrac85-4=0\).

2. Answer to Question 2

   1. \(\mathbf a\times\mathbf b=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\2&-1&3\\1&4&-2\end{vmatrix}=\mathbf i(2-12)-\mathbf j(-4-3)+\mathbf k(8+1)=\) \(-10\mathbf i+7\mathbf j+9\mathbf k\)
   2. \(\mathbf a\cdot(\mathbf a\times\mathbf b)=-20-7+27=0\) and \(\mathbf b\cdot(\mathbf a\times\mathbf b)=-10+28-18=0\). ✓
   3. Parallelogram: \(|\mathbf a\times\mathbf b|=\sqrt{100+49+81}=\sqrt{230}\approx\) 15.2. Triangle: half of this, 7.58.

3. Answer to Question 3

   1. \(\mathbf r=\begin{pmatrix}1\\2\\-1\end{pmatrix}+\lambda\begin{pmatrix}2\\-1\\3\end{pmatrix}\)
   2. \(\dfrac{x-1}2=\dfrac{y-2}{-1}=\dfrac{z+1}3\)
   3. Direction \(\overrightarrow{PQ}=\begin{pmatrix}2\\-2\\3\end{pmatrix}\), so \(\mathbf r=\begin{pmatrix}3\\0\\1\end{pmatrix}+\mu\begin{pmatrix}2\\-2\\3\end{pmatrix}\)
   4. For \((7,-4,7)\): \(3+2\mu=7\) gives \(\mu=2\); then \(y=-4\) ✓ and \(z=1+6=7\) ✓. For \((1,2,-1)\): \(3+2\mu=1\) gives \(\mu=-1\), but then \(y=2\) ✓ and \(z=1-3=-2\ne-1\), so it does not lie on the line.

4. Answer to Question 4

   1. \(\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n=2+6=8\), so \(2x-y+3z=8\). For \((3,1,1)\): \(6-1+3=8\) ✓.
   2. \(\overrightarrow{AB}=(-1,2,0)\) and \(\overrightarrow{AC}=(-1,0,3)\). Normal \(=\overrightarrow{AB}\times\overrightarrow{AC}=(6,3,2)\). Then \(6x+3y+2z=6\times1=6\): \(6x+3y+2z=6\). (Check \(B\): \(6\) ✓, \(C\): \(6\) ✓.)
   3. \(\mathbf r\cdot\begin{pmatrix}2\\-1\\3\end{pmatrix}=8\)

5. Answer to Question 5

   Method: the angle between a line with direction \(\mathbf d\) and a plane with normal \(\mathbf n\) satisfies \(\sin\varphi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}\) (the complement of the angle with the normal). The angle between two planes is the angle between their normals.

   1. Here \(\mathbf d=(1,2,2)\), \(\mathbf n=(2,-1,2)\) and both have magnitude 3. \(\sin\varphi=\dfrac{|\mathbf d\cdot\mathbf n|}{|\mathbf d||\mathbf n|}=\dfrac{|2-2+4|}{3\times3}=\dfrac49\), so \(\varphi=\) 26.4\(^\circ\) So the line meets the plane at about \(26.4^\circ\).
   2. The normals are \((1,2,2)\) with magnitude 3 and \((3,0,4)\) with magnitude 5. The angle between the planes equals the angle between their normals: \(\cos\theta=\dfrac{|(1,2,2)\cdot(3,0,4)|}{3\times5}=\dfrac{11}{15}\), so \(\theta=\) 42.8\(^\circ\) So the planes meet at about \(42.8^\circ\).

6. Answer to Question 6

   Method: the distance from the point \((x_1,y_1,z_1)\) to the plane \(ax+by+cz=d\) is \(\dfrac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}\). Alternatively, follow the normal line from the point until it meets the plane.

   1. Substitute the coordinates of \(P\) into the formula. \(\dfrac{|2(2)-2(3)+(-1)-5|}{\sqrt{4+4+1}}=\dfrac{|{-8}|}3=\) \(\dfrac83\) So \(P\) is \(\tfrac83\approx2.67\) units from the plane.
   2. A line perpendicular to the plane has the normal vector \((2,-2,1)\) as its direction. The normal is the direction: \(\mathbf r=\begin{pmatrix}2\\3\\-1\end{pmatrix}+t\begin{pmatrix}2\\-2\\1\end{pmatrix}\)
   3. Substitute the general point of the line into the equation of the plane. Substituting: \(2(2+2t)-2(3-2t)+(-1+t)=5\) gives \(4+4t-6+4t-1+t=5\), \(9t=8\), \(t=\tfrac89\). The foot is \(\left(\tfrac{34}9,\tfrac{11}9,-\tfrac19\right)\). Check: \(\tfrac{68}9-\tfrac{22}9-\tfrac19=\tfrac{45}9=5\) ✓. Check: the distance from \(P\) to the foot is \(\tfrac89\times|(2,-2,1)|=\tfrac89\times3=\tfrac83\), as in part a).

7. Answer to Question 7

   1. Speed \(=|\mathbf d|=\sqrt{4+16+1}=\) \(\sqrt{21}\approx4.58\) m s\(^{-1}\).
   2. The closest point has \(\mathbf r\cdot\mathbf d=0\): \((10+2t)2+(-20+4t)4+(30-t)(-1)=0\), \(20+4t-80+16t-30+t=0\), \(21t=90\), \(t=\) \(\tfrac{30}7\approx4.29\) s.
   3. \(\mathbf a\times\mathbf d=(-100,70,80)\), magnitude \(\sqrt{21300}=145.9\). Distance \(=\dfrac{145.9}{\sqrt{21}}=\) 31.8 m.
   4. \(31.8>30\), so the drone does stay at least 30 m away.

8. Answer to Question 8

   1. \(\overrightarrow{PQ}=(6,0,0)\), \(\overrightarrow{PR}=(0,4,2)\), product \(=(0\cdot2-0\cdot4,\ 0\cdot0-6\cdot2,\ 6\cdot4-0)=\) \((0,-12,24)\)
   2. The normal is \((0,-1,2)\). \(\mathbf r\cdot\mathbf n=(0,0,4)\cdot(0,-1,2)=8\), so \(-y+2z=8\). Check \(Q\): \(8\) ✓, \(R\): \(-4+12=8\) ✓.
   3. Area \(=\tfrac12\sqrt{144+576}=\tfrac12\sqrt{720}=\) 13.4 m\(^2\).
   4. The horizontal has normal \(\mathbf k\). \(\cos\theta=\dfrac{|\mathbf n\cdot\mathbf k|}{|\mathbf n|}=\dfrac2{\sqrt5}\), so \(\theta=\) 26.6\(^\circ\).

9. Answer to Question 9

   1. \(\mathbf r=(0,0,9)+\lambda(1,2,-3)\)
   2. Substitute: \(2\lambda+2\lambda+2(9-3\lambda)=12\), so \(4\lambda+18-6\lambda=12\), \(\lambda=3\). The point is \((3,6,0)\).
   3. \(\sin\varphi=\dfrac{|(1,2,-3)\cdot(2,1,2)|}{\sqrt{14}\times3}=\dfrac2{3\sqrt{14}}\), so \(\varphi=\) 10.3\(^\circ\) (a shallow angle).
   4. \((1,2,-2)\cdot(2,1,2)=2+2-4=0\), so the beam is parallel to the mirror and never meets it (the point \(A\) is not on the mirror since \(18\ne12\)).

10. Answer to Question 10

    1. Equating: \(1+\lambda=2+2\mu\), \(2-\lambda=\mu\), \(3+2\lambda=-1+\mu\). From the second, \(\mu=2-\lambda\); the first gives \(1+\lambda=6-2\lambda\), \(\lambda=\tfrac53\), \(\mu=\tfrac13\). The third: \(3+\tfrac{10}3=\tfrac{19}3\) but \(-1+\tfrac13=-\tfrac23\). Not equal, so the lines do not meet; they are skew.
    2. \(\mathbf d_1\times\mathbf d_2=(-3,3,3)\), or \((-1,1,1)\).
    3. \(\overrightarrow{AB}=(1,-2,-4)\) with \(A\) on \(l_1\) and \(B\) on \(l_2\). \(|\overrightarrow{AB}\cdot\mathbf n|/|\mathbf n|=\dfrac{|-3-6-12|}{\sqrt{27}}=\dfrac{21}{5.196}=\) 4.04 km.
    4. \(4.04\ge3\), so yes, the requirement is met.

11. Answer to Question 11

    Method: the scalar product is \(\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta\). Two non-zero vectors are perpendicular exactly when their scalar product is 0.

    1. Work done is the scalar product of force and displacement. \(12-2+10=\) 20 J So the crane does 20 J of work.
    2. Use Pythagoras in three dimensions. \(|\mathbf F|=\sqrt{16+4+25}=\sqrt{45}\approx6.71\) N; \(|\mathbf s|=\sqrt{9+1+4}=\sqrt{14}\approx3.74\) m.
    3. Use \(\cos\theta=\dfrac{\mathbf F\cdot\mathbf s}{|\mathbf F||\mathbf s|}\). \(\cos\theta=\dfrac{20}{\sqrt{45}\sqrt{14}}=\dfrac{20}{25.1}=0.7968\), so \(\theta=\) 37\(^\circ\) So the force acts at about \(37^\circ\) to the motion.
    4. The component of \(\mathbf F\) along \(\mathbf s\) is \(\mathbf F\cdot\hat{\mathbf s}=|\mathbf F|\cos\theta\). \(\dfrac{\mathbf F\cdot\mathbf s}{|\mathbf s|}=\dfrac{20}{\sqrt{14}}=\) 5.35 N Check: \(5.35\times\sqrt{14}=20\), the work done.

12. Answer to Question 12

    1. \(\mathbf r\times\mathbf F=\left(0\cdot(-10)-0.1\cdot30,\ 0.1\cdot0-0.4\cdot(-10),\ 0.4\cdot30-0\right)=\) \((-3,4,12)\) N m
    2. \(\sqrt{9+16+144}=\sqrt{169}=\) 13 N m
    3. The \(z\)-component is 12 N m, which turns the bolt.
    4. The moment has magnitude \(|\mathbf r||\mathbf F|\sin\theta\), so doubling \(|\mathbf r|\) doubles the turning effect for the same force.

13. Answer to Question 13

    Method: the vector product \(\mathbf a\times\mathbf b\) is perpendicular to both vectors and has magnitude \(|\mathbf a||\mathbf b|\sin\theta\), the area of the parallelogram with sides \(\mathbf a\) and \(\mathbf b\). Compute it as a \(3\times3\) determinant.

    1. \(\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\-1&3&2\\0&-2&4\end{vmatrix}=\mathbf i(12+4)-\mathbf j(-4-0)+\mathbf k(2-0)=\) \((16,4,2)\)
    2. This scalar triple product gives six times the volume of the tetrahedron. \(\overrightarrow{OA}\cdot(16,4,2)=32+4+0=36\). \(V=\dfrac{36}6=\) 6 m\(^3\). So the tent holds 6 m³ of air.
    3. The area of the triangle is half the magnitude of the vector product. \(\tfrac12\sqrt{256+16+4}=\tfrac12\sqrt{276}=\) 8.31 m\(^2\).
    4. Use the volume formula for a pyramid with base \(OBC\). \(V=\tfrac13\times\text{base}\times h\), so \(h=\dfrac{3V}{\text{area}}=\dfrac{18}{8.31}=\) 2.17 m. (Directly: \(\dfrac{36}{\sqrt{276}}=2.17\).) So \(A\) is about 2.17 m from the plane \(OBC\).

14. Answer to Question 14

    1. For A: \(1+2\lambda=3\), \(2-\lambda=1\), \(3+\lambda=4\) all give \(\lambda=1\). For B: \(4-\mu=3\), \(-1+2\mu=1\), \(2+2\mu=4\) all give \(\mu=1\). So the paths cross at \((3,1,4)\).
    2. Both submarines reach that point when \(t=1\), because the parameters are times. So yes, they collide at \(t=1\) hour.
    3. \(\cos\theta=\dfrac{|(2,-1,1)\cdot(-1,2,2)|}{\sqrt6\times3}=\dfrac{|-2-2+2|}{3\sqrt6}=\dfrac2{3\sqrt6}\), so \(\theta=\) 74.2\(^\circ\)
    4. A: \(\sqrt{4+1+1}=\sqrt6\approx\) 2.45 km h\(^{-1}\); B: \(\sqrt{1+4+4}=\) 3 km h\(^{-1}\).

15. Answer to Question 15

    1. \((1+\lambda)+2\lambda-(2+\lambda)=4\) gives \(2\lambda-1=4\), \(\lambda=\tfrac52\). The point is \(\left(\tfrac72,\tfrac52,\tfrac92\right)\).
    2. \(\sin\varphi=\dfrac{|1+2-1|}{\sqrt3\sqrt6}=\dfrac2{\sqrt{18}}=0.4714\), so \(\varphi=\) 28.1\(^\circ\)
    3. At \(P\): \(1+0-2=-1\ne4\). Distance \(=\dfrac{|-1-4|}{\sqrt6}=\dfrac5{\sqrt6}\approx\) 2.04.
    4. The normal \((1,2,-1)\) points towards the plane (the value \(-5\) is negative, so \(P\) is on the side where \(x+2y-z<4\)). The image is \(P+2\cdot\dfrac56(1,2,-1)=\) \(\left(\tfrac83,\tfrac{10}3,\tfrac13\right)\). Check: the midpoint of \(P\) and its image is \(\left(\tfrac{11}6,\tfrac53,\tfrac76\right)\), which satisfies \(x+2y-z=\tfrac{11}6+\tfrac{20}6-\tfrac76=4\). ✓

16. Answer to Question 16

    1. \(\mathbf w=(2,-1,3)-(1,1,0)=\) \((1,-2,3)\)
    2. Normal \(=\mathbf d\times\mathbf w=(1,2,1)\times(1,-2,3)=(8,-2,-4)\), or \((4,-1,-2)\). With the point \((1,1,0)\): \(4-1=3\). So \(4x-y-2z=3\). Check \((2,-1,3)\): \(8+1-6=3\) ✓.
    3. \(\dfrac{|\mathbf w\times\mathbf d|}{|\mathbf d|}=\dfrac{\sqrt{64+4+16}}{\sqrt6}=\dfrac{\sqrt{84}}{\sqrt6}=\) \(\sqrt{14}\approx3.74\)
    4. Sliding a point along a line changes \(\mathbf w\) by a multiple of \(\mathbf d\), and \(\mathbf d\times\mathbf d=\mathbf 0\), so the cross product, and hence the distance, is unchanged.

17. Answer to Question 17

    1. \(\overrightarrow{AB}=(1+2\mu-\lambda,\ \lambda+\mu-2,\ \mu-2\lambda-4)\)
    2. \(\mathbf d_1\): \((1+2\mu-\lambda)-(\lambda+\mu-2)+2(\mu-2\lambda-4)=0\) gives \(-6\lambda+3\mu-5=0\). \(\mathbf d_2\): \(2(1+2\mu-\lambda)+(\lambda+\mu-2)+(\mu-2\lambda-4)=0\) gives \(-3\lambda+6\mu-4=0\). Solving: from the first \(\mu=\dfrac{6\lambda+5}3\); substituting, \(-3\lambda+12\lambda+10-4=0\), \(\lambda=-\tfrac23\) and \(\mu=\tfrac13\).
    3. \(A=\left(\tfrac13,\tfrac83,\tfrac53\right)\), \(B=\left(\tfrac83,\tfrac13,-\tfrac23\right)\). \(\overrightarrow{AB}=\left(\tfrac73,-\tfrac73,-\tfrac73\right)\) and \(|AB|=\tfrac73\sqrt3=\) 4.04.
    4. \(\dfrac{7\sqrt3}3=4.04\) matches \(\dfrac{21}{\sqrt{27}}=\dfrac{21}{3\sqrt3}=\dfrac{7}{\sqrt3}\). ✓ Also \(\overrightarrow{AB}\parallel(-1,1,1)\), the common perpendicular.

18. Answer to Question 18

    Method: a plane with normal \(\mathbf n\) through the point \(A\) has equation \(\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n\), that is \(n_1x+n_2y+n_3z=d\). A normal to a plane containing two directions is their vector product.

    1. Intercepts 2, 3, 6: \(\dfrac x2+\dfrac y3+\dfrac z6=1\), or \(3x+2y+z=6\). Check via \(\overrightarrow{AB}\times\overrightarrow{AC}=(18,12,6)\parallel(3,2,1)\).
    2. \(V=\tfrac16\times2\times3\times6=\) 6.
    3. \(\tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|=\tfrac12\sqrt{324+144+36}=\tfrac12\sqrt{504}=\) 11.2
    4. \(V=\tfrac13\times11.22\times h=6\) gives \(h=\dfrac{18}{11.22}=1.60\). Formula: \(\dfrac{|0-6|}{\sqrt{9+4+1}}=\dfrac6{\sqrt{14}}=1.60\). ✓

19. Answer to Question 19

    Method: a plane with normal \(\mathbf n\) through the point \(A\) has equation \(\mathbf r\cdot\mathbf n=\mathbf a\cdot\mathbf n\), that is \(n_1x+n_2y+n_3z=d\). A normal to a plane containing two directions is their vector product.

    1. A normal is perpendicular to both direction vectors of the plane, so take their vector product. \(\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\1&1&0\\0&1&1\end{vmatrix}=\mathbf i(1)-\mathbf j(1)+\mathbf k(1)=\) \((1,-1,1)\) Check: \((1,-1,1)\cdot(1,1,0)=0\) and \((1,-1,1)\cdot(0,1,1)=0\).
    2. Use the point \((1,0,2)\) to find \(d\). \(\mathbf r\cdot\mathbf n=(1,0,2)\cdot(1,-1,1)=3\): \(x-y+z=3\)
    3. Substitute into the Cartesian equation, then compare components with the vector equation. \(3-1+1=3\) ✓. Equate: \(1+\lambda=3\) gives \(\lambda=2\); \(\lambda+\mu=1\) gives \(\mu=-1\); \(2+\mu=1\) ✓. So \(\lambda=2\), \(\mu=-1\). All three equations are consistent, so the point lies in the plane.
    4. Use: the distance from the point \((x_1,y_1,z_1)\) to the plane \(ax+by+cz=d\) is \(\dfrac{|ax_1+by_1+cz_1-d|}{\sqrt{a^2+b^2+c^2}}\). \(\dfrac{|0-3|}{\sqrt3}=\dfrac3{\sqrt3}=\) \(\sqrt3\approx1.73\)

20. Answer to Question 20

    1. Dividing by \(|\mathbf a|^2|\mathbf b|^2\): \(\dfrac{|\mathbf a\times\mathbf b|^2}{|\mathbf a|^2|\mathbf b|^2}=1-\cos^2\theta=\sin^2\theta\). Taking positive square roots (since \(0\le\theta\le\pi\)) gives \(|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta\).
    2. \(|\mathbf a\times\mathbf b|^2=230\). \(|\mathbf a|^2|\mathbf b|^2-(\mathbf a\cdot\mathbf b)^2=14\times21-64=294-64=230\). ✓
    3. The left side is \((a_2b_3-a_3b_2)^2+(a_3b_1-a_1b_3)^2+(a_1b_2-a_2b_1)^2\). Expanding gives the sum of terms \(a_i^2b_j^2\) with \(i\ne j\), minus \(2a_ia_jb_ib_j\) over the three pairs. The right side is \((a_1^2+a_2^2+a_3^2)(b_1^2+b_2^2+b_3^2)-(a_1b_1+a_2b_2+a_3b_3)^2\), and the terms \(a_i^2b_i^2\) cancel, leaving the same expression. So the two sides are equal.
    4. A parallelogram with sides \(\mathbf a\) and \(\mathbf b\) has base \(|\mathbf a|\) and height \(|\mathbf b|\sin\theta\), so its area is \(|\mathbf a||\mathbf b|\sin\theta=|\mathbf a\times\mathbf b|\).