( ) = ( ) ( ) = ( ) = + = = = ( ) Therefore: , where t. Note: If we start with the condition BM = tab, we will have BM = ( x + 2, y + 3, z 5)

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Marshall Garrett
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1 Chapter Exercise a) AB OB OA ( xb xa, yb ya, zb za),,, 0, b) AB OB OA ( xb xa, yb ya, zb za) ( ), ( ),, 0, c) AB OB OA x x, y y, z z (, ( ), ) (,, ) ( ) B A B A B A ( ) d) AB OB OA ( xb xa, yb ya, zb za) ( aa, a( a), a a) ( a, a, a) a) Given PQ OQ OP ( xq xp, yq yp, zq zp),, x Q yq zq,,,, Therefore: xq xq yq yq zq zq + So, Q,, b) Given PQ OQ OP ( xq xp, yq yp, zq zp),, x y z P, P, P,, Therefore: xp xp + yp yp + zp zp 0 So, P, 0, c) Given PQ OQ OP x x, y y, z z ( a, a, a) ( Q Q Q ) ( Q P Q P Q P) x a, y ( a), z a (, aa, ) Therefore: x a a x 0 Q Q y ( a) a y a a 4a Q Q zq a a zq a+ a a So, Q( 0, 4a, a) a) For points M, A and B to be collinear, it is sufficient to make AM parallel to AB If the two vectors are parallel, then one of them is a scalar multiple of the other, for example, AM tab AM ( xm xa, ym ya, zm za ) ( x 0, y 0, z ) AB x x, y y, z z ( 0, 0, 0 ) ( ) B A B A B A 70

2 ( ) ( ) Therefore: x, y, z t(,, ) x, y, z ( t, t, t) x t, y t, z t So, M( t, t, t), where t Note: If we start with the condition BM tab, we will have: BM x x, y y, z z x, y, z 0, AB (,, ) ( ) ( ) M B M B M b ( ) Therefore: x, y, z ( t, t, t) x + t, y + t, z t So, M( + t, + t, t), where t Both conditions describe the same set of points; for example, we can obtain point M(0, 0, ) by putting t 0 in M(t, t, t), or t in M( + t, + t, t); or M(,, ) by putting t in M(t, t, t), or t in M( + t, + t, t) b) For points M, A and B to be collinear, it is sufficient to make AM parallel to AB So, let s say, AM tab AM ( xm xa, ym ya, zm za ) ( x ( ), y 0, z ) AB ( xb xa, yb ya, zb za ) ( ( ), 0, ) Therefore: ( x +, y, z ) t( 4,, ) ( x +, y, z ) ( 4t, t, t) x + 4t, y t, z t So, M( + 4t, t, t), where t Note: If we start with the condition BM tab, we will have BM ( x, y, z + ) ; therefore, from BM tab, we will find x + 4t, y + t, z t c) For points M, A and B to be collinear, it is sufficient to make AM parallel to AB So, let s say, AM tab AM ( xm xa, ym ya, zm za ) ( x, y, z 4) AB x x, y y, z z (,, 4) ( ) B A B A B A Therefore: ( x, y, z 4) t( 4,, )( x, y, z 4) ( 4t, t,t) x 4t, y t, z 4 + t So, M( 4t, t, 4 + t), where t Note: If we start with the condition BM tab, we will have BM ( x +, y +, z ) ; therefore, from BM tab, we will find x 4t, y t, z + t 4 If C is the symmetric image of B with respect to A, points A, B and C are on the line and their positions are as shown below Their relationship can be expressed using different vector relationships; for example, AB CA, AB AC, CB AB, Here, we will use AB CA C A B a) For C( x, yz, ): AB (, 0( 4), 0) ( 44,, ), CA x, 4 y, 0 z Therefore: ( ) ( ) + x, 4 y, 0z ( 4, 4, ) 4 x, 4 4 yz, ; so, C( 7, 8, ) 7

3 Chapter b) For C( x, yz, ): AB ( ), 0, 4,,, CA ( x, y, z ) Therefore: 4 0,,,,,, ( x y z) x y + z ; so, C 9,, c) For C( x, yz, ): AB ( a, a, b ( ) ), CA ( x, y, z ) Therefore: ( a, a, b+ ) ( x, y, z) x a+ a, y a+ 4 a, z b b ; so, C( a, 4 a, b) a) For G( x, yz, ): 0 GA + GB + GC x, y, z x, y, z x, y, z ( ) + ( ) + ( ) ( 000,, ) ( x, y, z) x, y, z So, G,, b) For G( x, yz, ): 0 GA + GB + GC ( x, y, z) + ( x, y,z) + ( 0 x, 0 y, z) ( 000,, ) ( x, y, z) x, y, z So, G,, c) For G( x, yz, ): 0 GA + GB + GC ( a x, a y, a z) + ( b x, b y, b z) + ( c x, c y, cz) ( 000,, ) ( a+ b+ c x, a+ b+ c y, abc z) a + b + c a+ b+ c x, y, z a + b + c So, G a + b + c a + b +, c, a+ b+ c The relationship between points A, B, C, and D of parallelogram ABCD can be D expressed using different vector relationships; for example, AB DC, AD BC, BA CD, Here, we will use AD BC a) For D(x, y, z): BC (,, 0) (,, ) AD x, y, z ( ) x, y, z ( ) ( + ) Therefore: x x, y y, z + z So, D(,, ) b) For D(x, y, z): BC (, ( ), ),, 8 AD x, y, z ( ) ( ) A C B 7

4 Therefore: x x 4, y y, z 8 z 7 ( ) So, D 4,, 7 c) For D(x, y, z): BC 7,, (,, 4) AD x + y z x y z +,, 0,, Therefore: x + x, y y, z 4 So, D,, 4 7 Two vectors v and w have the same direction if, for t > 0, v t w Therefore: ( m, m+ n, m+ n) t( 4,, ) m t m+ n 4t m+ n t m t + m+ n 4t m+ n t t + + n 4t n t m+ n t ( t + ) + ( t ) t t Therefore: m t +, n t 8 a) The length of the vector v i + j k is + + ( ) 9, so the unit vector is ( i + j k) i + j k b) The length of the vector v i 4j+ k is + ( 4) + 4, so the unit vector is 4 4 ( i j+ k) i j+ k c) The length of the vector v i j k is + ( ) + ( ) 9, so the unit vector is ( i j k) i j k 9 a) u+ v ( i + j k)+ ( i + j) i + 4j k u+ v i + 4j k ( ) 9 b) u + v i + j k + i + j + + ( ) ( ) + ( ) + c) u i + j k i 9j k ; v i + j i j d) e) u + v ( ) + ( 9) u u 4 u ( 4 i + j k) 4 i + 4 j 4 k u u 4 i+ 4 j 4 k

5 Chapter 0 a) Using B(x, y, z) for the terminal point and A(,, ) for the initial point: AB ( x ( ), y, z ( ) ) ( x +, y, z + ), AB w x + 4 Therefore: y x y z, 4, So, the terminal point is (, 4, ) z + b) Using B(x, y, z) for the terminal point and A(,, 4) for the initial point: AB x ( ), y, z 4 x, y, z 4, AB v ( ) ( + ) x + Therefore: y z 4 x y z 0,, So, the terminal point is (0,, ) a) A vector opposite in direction and a third the magnitude of u is u Therefore: u 4 4 b) A vector in the same direction as w and whose magnitude equals is times a unit vector in the direction of w Therefore, the vector is of the form: 4 w w ( i+ j k) ( i+ j k ) ( 4i+ jk) + + ( ) c) If vectors are parallel, then one can be represented as t times the other Therefore: xi + yj k t ( i 4j+ k) From the k-coordinate, we can find the value of t: t t So, x and y ( 4 ) 8, and the vector is: + 8 i j k Let u be the vector from the vertex A to the midpoint of side BC; so, u AB + BC Let v be the vector from the vertex B to the midpoint of side AC; so, v BA + AC Let w be the vector from the vertex C to the midpoint of side AB; so, w CA + AB Adding the vectors: u+ v + w AB + BC + BA + AC + CA + AB BC + AC + CA AB + BC + CA + AB BC + CA + AB 0 0 Exercise a) u v + ( ) ( )+ 4 ( ) θ + ( ) ( ) + ( ) 9 4 θ 7 b) u v ( )+ ( ) θ θ + ( ) + 0 ( ) + ()

6 c) u v + ( ) θ θ ( ) d) u v 0+ ( ) + 0 θ θ 74 + ( ) ( ) 0 9 π e) u v uvθ 4 4 ; angle is θ 0 π f) u v uvθ 4 4 ; angle is θ 0 a) u v ( )+ ( ) The dot product is zero; hence, the vectors are orthogonal b) u v + ( 7) The dot product is positive; hence, the angle is acute c) u v 0+ ( ) + 0 The dot product is zero; hence, the vectors are orthogonal a) v u y x + x y 0 The dot product is zero; hence, v is orthogonal to u w u y x + ( x) y 0 The dot product is zero; hence, w is orthogonal to u Pay attention to the relationship between the coordinates of a two-dimensional vector and a vector that is perpendicular to it b) The vectors perpendicular to u i j are i + j and i j Unit vectors in the direction of those vectors are: v i j + y ( ) and v i + j ( ) v v u x 4 a) i) v + ( ) + 4 i v + 0 ( )+ 0 α 4 j v 0 + ( )+ 0 β 4 k v ( )+ γ 4 4 ii) α + β + γ iii) α , β γ b) i) v + ( ) + i v + 0 ( )+ 0 α, 7

7 Chapter j v 0 + ( )+ 0 β k v ( )+ γ 4 ii) α + β + γ iii) α 90, β γ 90, c) i) v + ( ) + 4 i v + 0 ( )+ 0 α 4 j v 0 + ( )+ 0 β 4 k v ( )+ γ 4 9 ii) α + β + γ iii) α , β 4 γ , d) i) v ( 4) i v ( 4) α j v ( 4) 0 β 0 k v ( 4) 4 4 γ 4 9 ii) α + β + γ 0 + iii) α 0, β 0 90, γ a) u v ( m )+ ( m+ ) m Vectors are perpendicular if their dot product is zero; therefore: 8m+ 9 0 m 8 b) u v ( m) ( m)+ ( m) m+ ( m+ ) ( m) 4m m Vectors are perpendicular if their dot product is zero; therefore: 4m m 0 m, m 4 u w u ( u+ mv) u u+ mu v ( ) m (( ) + + ) 4 + m Vectors are orthogonal if their dot product is zero; therefore: 4 + m 0 m 4 7

8 7 a) u v ( ) + ( )+ 4 0 θ uv ( ) ( ) 7 θ b) u+ v ( 4,, )+(, 0, ) ( 44,, ) u ( u+ v) θ ) uu+ v θ 449 c) v ( u+ v) θ 4 + ) vu+ v θ AB (,, ( ) ) (,, ) AC ( m,, 0 m ( ) ) ( m,, 0 m + ) a) The points A, B and C are collinear if AC is parallel to AB Given this collinearity, AC tab(,, ) t ( m,, 0m+ ) From the second set of coordinates, we can determine t: t t Hence: ( m) m Checking with the third set of coordinates: 0 + ; so, it fits and for m the points are collinear b) AC AB ( m )+ ( )+ ( 0m+ ) 8m Vectors are perpendicular if their dot product is zero; therefore: AC AB 0 8m 0 m 4 9 The vector equation of the line is an equation of the form: r r0 + t v, where r 0 is the position vector of any point on the line and the direction vector v is a vector parallel to the line For the median through A, we can take the position vector of point A for r 0 and the vector from A to the midpoint of BC for v So: r 0 ( 4,, ), m,,,, BC, v 4 ( ) ( ),, 0,, Therefore: m : r (,, A 4 )+ m, 0, For the median through B, we can take the position vector of point B for r and the vector from B 0 to the midpoint of AC for v So: r 0 (,, ), m AC 4 7,,,,, v 7 ( ) ( ) 9,,,, Therefore: m : r (,, )+ B n, 9,

9 Chapter For the median through C, we can take the position vector of point C for r 0 and the vector from C to the midpoint of AB for v So: + + ( ) + ( ) r 0 (,, ), m AB 4 7 7,,,,, v 7 7 9,,,, Therefore: mc : r (,, )+ k 9,, The centroid is the point where all the medians meet We will find the intersection point of two lines, and then check that this point is also on the third line If m A and m B intersect, then: (,, )+ m,,,,,, ( )+ n Therefore, we have: 4 m + n m n n m + m + n Putting m n we can see that it fits the equation, so the point of intersection of m A and m B is: (,, )+,,, 0, Now, we have to check that 0, 0, is on the third line as 0 well: 9, 0,,,,, ( )+ Hence, we can see that the centroid is 0, 0, 0 4 Note: For the centroid, it holds that:, 0,,, The formula xa + xb + xc ya + yb + yc za + zb + zc,, holds, in general, for a triangle with vertices A, B, C 0 AB (,, ) ( 40,, ) ; AB 0 AC (, 4, ) ( 0,, 0) ; AC AD (,, ) ( 0,, ) ; AD 40 BC BD ( ( ), 4, ) ( 4,, ) ; BC ( ( ),, ) ( 0,, 4) ; BD CD (, ( 4), ) (,, ) ; CD 7 We will calculate the angles by finding the dot product and using the ine angle formula: 78

10 AB AC 0 angle 90 AB AD angle AB BD angle AB BC angle AC AD 0 angle 90 AC BC - 99 angle 7 AC CD angle AD BD angle 8 AD CD angle 4 BC BD - 77 angle 7 BC CD angle BD CD angle Total surface area consists of triangles ABC, ABD, ACD, and BCD: A ABC ABAC 90 sin 0 44 (Note: AB and AC are perpendicular) AABD ABADsin θ 0 40 sin (Note: θ AB, AD 8 ) A ACD ACCD 7 sin θ sin (Note: θ is the angle between the vectors CA, CD 80 AC, CD 47 ) A BCD BCCD 7 sin θ sin 704 (Note: θ is the angle between the vectors CB, CD 80 AC, CD ) Therefore, A 7 CD (,, ) DC (,, ) i DC j DC - 7 k DC

11 Chapter Given AD ( 0,, ) DA ( 0,, ), BD ( 0,, 4) DB (, 04, ), AC ( 0,, 0) : DA DB AC (( 0,, )( 04,, )) ( 0,, 0) ( 40,, ) ( 0,, 0) 0 4 π k k θ + k k 9+ k k k + 0 Therefore: + k k k k 4k+ 4 0 k k + 0 x x y 0 + y 4 x 8 x y 0 + y Hence, we have to solve the system of equations: + x y 0 8 x + y 0 Adding the equations: 4 + y 0 y 4 and x 0 Two vectors are parallel if t exists such that u t v Therefore: x x x t( x) x t + x ; and x t( + x) + x + x + x t( + x) 4 From the last two equations, we can see that + x x x ; hence, + t( + ) t Verifying using the first equation, we can see that x is the solution 7 ABC BABC, So: BA OA OB 4 4 ABC ABC AC AB + BC BA + BC BAC ABAC,

12 BAC BAC a) ( b,, ) (, b, ) b+ b+ 4b Vectors are orthogonal if their dot product is zero; therefore: 4b+ 0 b b) ( 4,, 7) ( b, b, 0) 4b b b b Vectors are orthogonal if their dot product is zero; therefore: b b b, For b 0, the vector ( b, b, 0) is a zero vector A zero vector has no direction; therefore, it is not orthogonal to any vector Hence, the vectors are only orthogonal for b 9 To determine the angle between two vectors, we are going to find their dot product: ( )( ) p+ q p q p q Since, for any vector: v v v 0 v v we have: p+ q p q p q p q therefore, the vectors are perpendicular ( )( ) 0 We can find the z-component by transforming 00 m/min into km/h: 00 m/min 0 km/min 0 0 km/h 8 km/h The velocity vector in the xy-plane is parallel to the vector (, ) The unit vector in this direction is (, ) Since the airspeed is 00 km/h, and its vertical component is 8 km/h, then the velocity of the xy-component is: km/h So, the velocity vector in the xy-plane is: 9 7 (, ) 9 88 (, ) ( 40, ) Hence, the velocity vector is ( 40, 8 40, 88) 0; Exercise a) In the equation r r0 + t u, vector r 0 0, and u, so the vector equation of the line is 4 r 0 + t The parametric equations are 4 z b) Substituting r 0 and u r + t The parametric equations are: z c) Substituting r 0 and u x y + t t 4t into r r0 + t u, we get the vector equation of the line: x y + t t + t into r r u + 0 t, we get the vector equation of the line: 8

13 Chapter x + t r + t The parametric equations are: y t + z t 7+ 8 a) In the equation r r 0 + tu, vector r0 r 4 A, and u AB 4, so the vector 0 8 equation of the line is: r 4 + t Note: For r 0 we can use r B and for u we can use BA, or any vector parallel to this vector Therefore, it is possible to find different, but correct, equations of the line b) Substituting r0 r A and u AB into r r0 + t u, we get the vector 4 4 equation of the line: r + t 4 4 Note: Since u is parallel to AB, we can use u AB 4 4 line would be r + s, and the vector equation of the c) Substituting r0 r A and u AB + 4 into r r 0 + tu, we get the vector 4 equation of the line: r + t a) In the equation r a + tb, we substitute a ra and u AB + equation of the line is r + t b) In the equation r a + tb, we substitute a ra and u AB equation of the line is r + t, so the, so the 4 Method I: To determine the equation of the line in the required form, we need to find two points on the line r (, ) + t (, ) One point is (, ) Another point we can find by letting, for example, t ; therefore, the point is (, ) The equation of the line through those two points is: y y x x y x ( ) ( ) + 7

14 Method II: We can write the equation of the line r (, ) + t (, ) in parametric form: x t y + x t From the first row: x + t t Substituting t into the second row, we get: x y t ( y ) ( x ) x + y 7 Method III: We can write the equation of the line r (, ) + t (, ) in parametric form: x t y + t Transforming and adding the equations: x + t / y t / x 4+ t + y t x + y 7 In the equation r r 0 + tu, the vector r 0 r A i j and u can be the same as the direction vector of the given line; therefore, u 4i j So: r i j+ λ 4ij ( ) In the equation r r0 + t u, the vector r0 ra ( 4,, ) and u (, 4, 7) So, we have: r ( 4,, )+ t (, 47, ) Solution Paper type 7 a The lines are not parallel since the direction vectors (,, ) and (, 4, ) are not a scalar multiple of each other For lines to intersect, there should be some point (x 0, y 0, z 0,) which satisfies the equations of both lines, r (,, )+ t (,, ) and r ( 4,, )+ s (, 4, ), for some values of z and s (Note: We have to change the parameter in one of the equations so that they are not the same) So: x 0 + t + s y 0 + t + 4s z 0 + t 4 + s From the first equation, we see that t s Substituting into the second equation: + t + 4t t t s Finally, substituting these values into the third equation: 4 Hence, the lines intersect, and the point of intersection is: (,, )+( )(,, ) (,, ) 8

Chapter Solution Paper type 7 a We can solve this system using matrices Firstly, transform the system of equations: + t + s t s 0 + t + 4s t 4s + t 4 + s t s and then use a GDC: From the result

15 Chapter Solution Paper type 7 a We can solve this system using matrices Firstly, transform the system of equations: + t + s t s 0 + t + 4s t 4s + t 4 + s t s and then use a GDC: From the result screen, we can see that an intersection exists and that t and s We can also find the solutions of the system using the Apps menu Of course, a GDC will only solve for the parameters to find the point of intersection, we have to substitute the values of the parameters into the equations of the lines Solution Paper type b The lines are not parallel since the direction vectors (4,, 0) and (,, ) are not a scalar multiple of each other For lines to intersect, there should be some point (x 0, y 0, z 0 ) which satisfies the equations of both lines, r (,, ) + t (4,, 0) and r (,, ) + s (,, ), for some values of z and s (Note: We have to change the parameter in one of the equations so that they are not the same) So: x 0 + 4t + s y 0 + t + s z t + s From the last equation, we see that s Substituting into the second equation: + + t 4 t Finally, substituting these values into the first equation: Hence, the lines intersect, and the point of intersection is: r (,, )+( 4)( 40,, ) ( 7,, ) Solution Paper type b We can solve this system using matrices Firstly, transform the system of equations: + 4t + s 4t s + t + s t s and then use a GDC: + 0 t + s s From the result screen, we can see that an intersection exists, and that t 4 and s Of course, a GDC will only solve for the parameters to find the point of intersection we have to substitute the values of the parameters into the equations of the lines 84

16 Solution Paper type 7 c The lines are not parallel since the direction vectors (7,, ) and (, 0, ) are not a scalar multiple of each other For lines to intersect, there should be some point (x 0, y 0, z 0 ) which satisfies the equations of both lines, r (,, ) + t (7,, ) and r (4,, 7) + s (, 0, ), for some values of z and s (Note: We have to change the parameter in one of the equations so that they are not the same) So: x 0 + 7t 4 s y 0 + t + 0 s z 0 t 7+ s From the second equation, we can see that t Substituting into the first equation: + 4 s s 8 Finally, substituting these values into the last equation: 7+ ( 8)4 9 Hence, the lines do not intersect; they are skew Solution Paper type c + 7t 4 s 7t + s + t + 0 s t + 0 s t 7+ s t s From the result screen, we can see that an intersection does not exist since we interpret the screen as: t 0, s 0 and 0 Alternatively, solving for parameters using Apps: From the result screen, we can see that there is no solution Solution Paper type d 4 4 The lines have parallel direction vectors and, since To check whether the lines coincide, we examine the point (, 4, ), which is on the first line x x 4 y 4 z + t, and see whether it lies on the second line, y z + s 7 4 4s s So: 4 + s + s s 7 4 s 7 We can see that the point is not on the other line, so the lines do not coincide; therefore, the lines are parallel 8

17 Chapter Solution Paper type d t 4 4s t + 4s 4 + t + s t s t 7s t + s From the result screen, we can see that an intersection does not exist and the lines are parallel We arrive at this conclusion by interpreting the screen as: t s 0, 0 and 0 0 Alternatively, solving for parameters using Apps: From the result screen, we can see that there is no solution and that the lines are parallel 8 a) A direction vector is: (, ( ) ) (, ); hence, the equations are: x t r (, )+ t (, ) and y + + t 8 b) We know a point and a direction vector, so the equations are: x t r (, )+ t ( 7, ) and y + 7t c) For the direction vector, we can use any vector perpendicular to 7 So, we use vector 7 7 as the direction vector of the line, since Therefore, the equations x t are: r (, )+ t ( 7, ) and y t d) We know a point and a direction vector, so the equations are: x t r ( 0, )+ t (, 4) and y 4t 9 a) Substituting the point 0 x 9,, into the equation y t 4 + : z 0 0 t t t 9 t From the first equation, we can see that t Checking, using the second equation: 4 +, and the third equation: So, the point is on the line when t

18 b) To check whether the point is on the line, we have to find whether the system of equations has a solution: 4 4 t + t 4 4+ t t From the last two equations, we can see that t 0, but this will not satisfy the first equation; hence, there is no solution to the system and the point does not lie on the line c) We have to solve the system of equations: m m 4 + t m t + m 4 t t 7 From the last equation, we can see that t Substituting into the second equation: m 7 m 7 Checking, using the first equation: Therefore, the point will be on the line when m 7 0 a) i) The starting position is when t 0, so the point is (, 4) ii) The velocity vector is v 7 4 iii) The speed is v b) i) The starting position is when t 0, so the point is (, ) ii) The velocity vector is v iii) The speed is v + ( ) c) i) The starting position is when t 0, so the point is (, ) ii) The velocity vector is v ( 4, 7) iii) The speed is v 4 + ( 7) a) The direction of the velocity vector is given by the unit vector: So, the velocity vector is: b) The direction of the velocity vector is given by the unit vector: So, the velocity vector is: ( ) ( ) 87

Chapter a) The car is travelling from the point (, ) to (7, ), so the direction vector of the velocity vector is given by the unit vector: v, where v v ( 7, ) ( 4, ) Therefore, the unit vector is: (

19 Chapter a) The car is travelling from the point (, ) to (7, ), so the direction vector of the velocity vector is given by the unit vector: v, where v v ( 7, ) ( 4, ) Therefore, the unit vector is: ( 4) ( 4 4 ) +,, ; and the velocity vector is: 0 ( 4, ) ( 4, 8 ) b) The starting point is (, ) and the direction vector of the line is (4, 8), so the equation of the position of the car after t hours is r (, ) + t (4, 8) c) We have to determine the parameter of the point (7, ) ( 7, ) (, )+ t ( 4, 8) ( 4, ) t( 4, 8) t Therefore, in of an hour, ie 0 minutes, the car will reach the traffic light a) To be perpendicular to the vectors, both dot products have to be zero a a b 0 + a a b 0 + b b a+ b So, we have to solve the system: a, b a b v w b) θ v w + ( ) ( ) + + ( ) 4 c) Using the Pythagorean identity for sine, sin θ θ, and the fact that sine is positive for angles from 0 80 we have: sinθ + θ Area of triangle OPQ is: A OPOQ sin POQ v w sin ( v, w ) 7 7 So, A 4 4 a) Firstly, we have to determine vectors AB and AC: ( ) 0 AB 0( ), AC θ ( ) + ( ) 7 Therefore, θ 4 θ

20 b) The area of the triangle is: A ABAC sin θ 4 sin θ 9 c) i) Line L goes through the point (,, 0) and its direction vector is AB 0 0, so its equation is: r + t 0 Line L goes through the point (,, ) and its direction vector is AC, so its equation is: r + s ii) We have to solve the system of equations: 0 + s + t + s + t s 0 t s From the first equation, we have s, and from the second t 7 Substituting these values into the third equation: ( 7) 4 So, there is no point of intersection a) Let the direction vector be a vector parallel to AB : AB 7 0 8( 7) Thus, we can use the vector as the direction vector Therefore, the parametric equations of the x + t line are: y t z 7 + t + t b) If point P is on the line, then vector OP t If OP is perpendicular to the line, then 7 + t OP and the direction vector of the line are perpendicular, and their dot product is zero 0 OP + t t 7 + t + t t + 8+ t 0 t 90 t + So, OP and P ( 4,, ) 89

chapter 1 vector geometry solutions V Consider the parallelogram shown alongside. Which of the following statements are true?

chapter 1 vector geometry solutions V Consider the parallelogram shown alongside. Which of the following statements are true? chapter vector geometry solutions V. Exercise A. For the shape shown, find a single vector which is equal to a)!!! " AB + BC AC b)! AD!!! " + DB AB c)! AC + CD AD d)! BC + CD!!! " + DA BA e) CD!!! " "