9 Mixed Exercise. vector equation is. 4 a

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1 9 Mixed Exercise a AB r i j k j k c OA AB 7 i j 7 k A7,, and B,,8 8 AB 6 A vector equation is 7 r x 7 y z (i j k) j k a x y z a a 7, Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

2 Use the equation r 7 When, r, so the point 7,, lies on l 9 so these two vectors are parallel So an alternative form of the equation is r 6 6 a r Set 7 a AB, AC r x y z x () y () z () () () : z y 7 yz Into (): yz 7 y z y z x 7 ( x ) y z y z x y z Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

3 8 Three points on the plane are A6,,, B8,, and C,, AB, AC 6 r 9 a ML 7 7 MN 6 ML. MN 7 9 ML MN a 9 6 a,, c p 6 q 6 9 a 6 Equation of l: 9 r t Since C lies on l, 9 p t q 9 t t 6 t So p t 6 and q t Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

4 c cos OC. AB OC AB OC. AB OC AB 6 66 ( ) cos ( d.p.) d If OD and AB are perpendicular, d. ( a) = 9 t Since d lies on AB, use d t t 9 t t t (9 t) ( t) ( t) 7 9t 8 6t t t t d i j k a AB r Set Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

5 c Let a, a 8 a cos 8 o. (.d.p.) Let G e a general point on l d CG When C is closest to l, CG is perpendicular to l. CG 6 6 CG Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

6 a Line l: r λ 9 Line l: r μ Using the direction vectors: Since the scalar product is zero, the directions are perpendicular. λ 9 μ At an intersection point: λ μ λ μ 9 ( ) Adding: Intersection point: Position vector of A is a = i k. c Position vector of B: jk For l, to give zero as the x component, =. r So B lies on l. For l, to give as the z component, = 9. 9 r 9 So B does not lie on l. So Only one of the sumarines passes through B. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 6

7 d AB ( ) ( ) [ ( )] ( ) ( ) Since unit represents m, the distance AB is m. km. Let A and B e general points on l and l respectively. 6 t AB t s t AB is perpendicular to l 6 t t s t t s AB is perpendicular to l 6 t t s t 6t s Solving these simultaneous equations: s, t AB AB 8. ( s. f.) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 7

8 Let A and B e general points on l and l respectively. 6ts AB t s s AB is perpendicular to l 6 ts t s s ts AB is perpendicular to l 6 ts t s s t s Solving these simultaneous equations: 9 s, t 7 7 AB AB ( s. f.) a r and When, When, r r a So the position vectors, and all lie on the plane Suppose the plane has equation ax y cz Sustituting each of the coordinates into this equation gives: Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 8

9 a c a c a c Solving these equations simultaneously gives: 9 a,, c Therefore the equation of the plane is 9 x y z or x 9y z So the required equation is r. 9 r and When, When, r r a So the position vectors, and all lie on the plane Suppose the plane has equation ax y cz Sustituting each of the coordinates into this equation gives: a c a c a c Solving these equations simultaneously gives: a,, c Therefore the equation of the plane is x y z Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 9

10 or x y z So the required equation is r. c r and When, When, r r 7 a So the position vectors, and 8 all lie on the plane 6 Suppose the plane has equation ax y cz Sustituting each of the coordinates into this equation gives: a c a c 7a 8 6c Solving these equations simultaneously gives: 8 a,, c Therefore the equation of the plane is 8 x y z or 8x y z 8 So the required equation is r. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

11 6 The line is in the direction i j k. This lies in the plane. (,, ) is a point on the line. This also lies in the plane, as does the point (,, ). First otain the equation of the is a direction in the plane. plane in the form, rn an, then convert to Cartesian form. a Let n e perpendicular to the plane c a a So. and. c c Therefore a c and a Choosing a gives and c 8 Therefore a normal vector is given y 8 The equation of the plane is r (i j8 k) ( i j k) (i j8 k ) i.e: x+ y - 8z This is a Cartesian equation of the plane. 7 a AB and AC a Let n e perpendicular to the plane c Then n is perpendicular to oth and a a So. and. c c Therefore a c and a c Choosing a gives c and c Solving simultaneously gives and c Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

12 Therefore a normal vector is given y, or Therefore a unit vector normal to the plane is ( i j k ) (i j k ) The equation of the plane may e written as r (i j k) ( i j k) (i j k) ie.. x y z c The perpendicular distance from the origin to the plane is. 8 a The plane with vector equation r i sj t( i k ) is perpendicular to i k, as ( i k) j and ( i k) ( i k ) The plane also has equation r ( i k) i ( i k ), as i is the position vector of a point on the plane. i.e. r ( i k ) The perpendicular distance from the origin to this plane is The Cartesian form of the equation of the plane is xz. or.77 ( s.f.) 9 a AB OB OA (i jk) ( i j k ) ij AC OC OA (i j 6 k) ( i j k) i j k AB and AC Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

13 a Let n e perpendicular to the plane c a a So. and. c c Therefore a and a c Choosing a gives and c Therefore a normal vector is given y Or any multiple of i jk An equation of the plane containing A, B and C is r (i j k) ( i j k) (i j k ) i.e. x y z a (i j k) ( i j k ) 9 (,, ) lies on the plane ( i j k) ( i j k ) the normal to plane is perpendicular to the normal to plane. is perpendicular to. c r i j k ( i j k ) d This line meets the plane when ( ) i ( ) j ( ) k ( i j k ) i.e. i.e. 69 Sustitute into the equation of the line: then i.e. The line meets at the point,, e The distance required is r i j k ( ( )) 9.67 ( s.f.) Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

14 a l : r and l : r If l and l intersect, then () () Equation () gives Sustituting into () gives 7, so Check for consistency: 6 and 6, so these equations are consistent. Therefore l and l intersect. 7 Sustitute into line l, so r 6 7 Therefore the position vector of their point of intersection is 6 c The cosine of the acute angle etween the lines is the cosine of the acute angle etween their respective direction vectors.. 8 So cos Therefore and 8 8 cos, as required. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

15 a c 6 8 a a Position vector of P is 8 9 P,9, 9 a 6 AB 6 6 r Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

16 c CP The position vector of P is 6 9 P, 9, or P 7.,., a 8 jand k components, 6 Check i component: () 7 6 So the equations are consistent Therefore the lines meet, and they meet at 7 A7,, Let a and a a 6 6 cos 6 o 8. ( d.p.) c Set Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 6

17 d The shortest distance of B to the line l is given y d BA sin 7 OA and OB 7 BA OA OB BA 6 So d 6 sin8.. a AP a n an a 8 n 7 sin o to the nearest degree () Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 7

18 6 a Let l denote the path of the first aeroplane, and l the second. 8 l : r 8 8 l : r i and k components, Check j component: 8 So the equations are consistent Therefore the paths intersect, and they intersect at ,, The planes pass through the same point, ut not necessarily at the same time. Challenge a ( x + y z) = ( ) ( ) gives x y + 6z = matrix A is singular if det A = (c + ) + ( c + a) + 6( a) = 6a 6a + + c c = c i a = n, = n, c = where n R, n ii a = 6, = and c = 9 Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 8

19 Challenge Coordinates A,,, B,, and C,, lie on the circumference of the circle. Let O e the centre of the circle. Then A, B, C and O lie on the same plane. Without loss of generality, suppose the plane has equation ax y cz Then sustituting point A gives: a c () sustituting point B gives: c () sustituting point C gives: c () From equation (), c Sustituting into () gives Sustituting and c in equation () gives a a a So the equation of the plane is x y z Or x y z () The centre of the circle lies on the intersection of any two perpendicular isectors chosen from AB, BC or AC. Each perpendicular isector must also lie on the plane x y z To find the equation of the perpendicular isector of AB: The mid-point of AB is M,, M,, Therefore the equation of the line MO may e written as r AB 8 and AB is perpendicular to the direction of MO, so AB i.e. 8, so 8 () The line r lies on the plane x y z for all values of. Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free. 9

20 Choosing, you have Challenge continued r Solving r and x y z simultaneously gives, or (6) Equation (6) is equivalent to Adding this to equation () gives, so and Therefore r Now to find the equation of the perpendicular isector of BC: The mid-point of BC is M,, M,, Therefore the equation of the line MO may e written as r BC and BC is perpendicular to the direction of MO, so BC i.e., so, or (7) The line r lies on the plane x y z for all values of. Choosing, you have r Solving r and x y z simultaneously gives, or (8) Solving equations (7) and (8) simultaneously gives Therefore r Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.

21 Challenge continued Now consider the equations of the two perpendicular isectors: r and r These meet at the centre of the circle, O. Equating the j components gives, so Therefore the two isectors must meet at the position vector The centre of the circle is therefore O,, The radius is the distance etween O and any of the points A, B or C C,, Consider the point Then OC Therefore the circle has radius r Pearson Education Ltd 7. Copying permitted for purchasing institution only. This material is not copyright free.