RI Vectors Lines and Planes 7B Solns

Page 1 RAFFLES INSTITUTION H2 Mathematics 9758 2023 Year 6 Term 3 Revision 7b (Tutorial) Topic: Vectors 3 (Lines and Planes) Summary for Lines and Planes [Refer to Revision 7a] Revision Tutorial Questions Source of Question: NJC Prelim 9758/2018/02/Q2 1 The planes 1p and 2 ,p have equations 2360xyz++= and 1 26 2   = r  respectively. (i) Find a vector equation of the line of intersection, l , between 1p and 2.p [2] The line m passes through the points ( )2 ,1 ,1A and ( )5 ,4 ,2 .B (ii) Verify that A lies on 2.p [1] (iii) Find the coordinates of the points on m that are equidistant from planes 1p and 2p . [5] Solution (i) 1 : 2 3 6 0p xyz ++= 2 : 226px y z++= Using GC, 18 6 12 2 , 01 λλ −−     = +∈       r  (ii) Since 2 2(1) 2(1) 6,++= or 21 1 2 2226 , 12   =++=  the point A lies on 2.p (iii) Let the point that is equidistant from both planes be C. 523 41 3 211  −=

Page 2 23 13 11 OC t    = +      for some t∈  Distance of C from 1p = Distance of C from 2p 222 222 23 2 1 23 0 2 13 1 2 13 0 3 11 2 10 6 122 236 3 6 2 46 39 66 37 11 13 21 37 77 39 63 tt tt tt ttt t t t tt tt +  +          +− ⋅ +− ⋅          ++      = ++ ++ ++ ++ ++ += += = + 77 39 63 or 77 39 63 140 39 or 14 39 39 39 or 140 14 t tt t tt tt = −− =+ = −= =−= 2 3 163 39 11 3 23140 1401 1 101 OC        = +− =            or 2 3 145 39 11 3 13114 141 1 53 OC        = +=             The two points are 163 23 101,,140 140 140   and 145 131 53,,14 14 14   . Source of Question: NJC JC2 Mid-Year CT 9758/2018/01/Q6 2 The equation of the plane p is given by 2 5 7,ax y z−−= where a is a real constant. (i) Given that the line l with equation 35 5 2 , 60 λλ      = +− ∈       r intersects the plane p exactly once, find the possible values of a. [1] Assume for the remainder of this question that a = 3. (ii) Find the coordinates of the point of intersection between l and p. [3] (iii) Find the acute angle between the line l and the plane p. [2] (iv) Find the vector equation of the line of reflection of the line l in the plane p. [5]

Page 3 Solution (i) Since 5 42 2 0 5 40 505 a aa    −⋅ − ≠⇒ +≠⇒≠ −   −  a can be any real number except 4 5− . (ii) Let N be the point of intersection between l and p. Since N lies on l, 35 52 6 λ λ + = −  ON for some .λ∈  Since N also lies on plane p, 3 27 5  ⋅− =−  ON 35 3 52 . 2 7 65 9 15 10 4 30 7 19 7 9 10 30 38 2 λ λ λλ λ λ +     − −=     −   + −+−= =−+ + = = Hence, 3 5(2) 13 5 2(2) 1 . 66 +   = −=     ON Therefore, coordinates of point of intersection are (13,1,6). (iii) Let θ be the acute angle required. 1 53 22 05sin 29 38 34.9 or 0.609 rad θ θ −    − ⋅−   − = =  (iv) The equation of the line 1l passing through (3, 5, 6).A and perpendicular to p has equation 33 5 2 , . 65 µµ      = +− ∈     −   r Let F be the foot of the perpendicular from the point A(3, 5, 6) onto p.

Page 4 Since F lies on 1,l 33 52 65 µ µ µ + = − −  OF for some .µ∈  Since F also lies on p, 3 27 5  ⋅− =−  OF 33 3 52 2 7 65 5 9 9 10 4 30 25 7 38 7 9 10 30 38 1. µ µ µ µµ µ µ µ +     − ⋅− =    −−   + −+ −+ = =−+ + = = So 3 3 6 6 13 7 5 2 3 3 1 2 65 1 1 6 5 +−         =−= ⇒ = − =        −−       OF NF Let A′ be the point of reflection of A about p. By Ratio Theorem, 2 63 13 5 216 63 9 23 5 1 16 4 ′+=     ′⇒= +          ′⇒= −=    −          OA OAOF OA OA 9 13 4 11 0 4 6 10 −     ′⇒= −=     −−     NA Therefore, equation of the line of reflection is 13 2 1 0 , . 65 νν      = +∈       r N(13, 1, 6) p (side view)

Page 5 Source of Question: YJC Prelim 9758/2018/01/Q9 3 The two lines 1l and 2l have equations 2, 02 zxy= −=− and 10 0 22 qµ    = +    r respectively, where µ∈  and q is a constant. The point A has coordinates ( )4 , 3 , k where k is a constant and the two planes 1p and 2p have equations 48 2x yz= −+ − and 2y tz t+= respectively, where t is a constant. (i) Find a vector equation of 1l and hence show that 1l lies in 1p . [3] (ii) Find the coordinates of the foot of perpendicular from A to 1p . Express your answer in terms of k. [3] (iii) Given that the angle between 1l and 2l is 60°, find the possible values of q . [2] (iv) Given that 1q= and that the point ( )1 , 2 , 2B is equidistant from 2l and 2p , find the possible values of t. [6] Solution (i) 1 : 2 2 Let 2 2 2 0 2 zlx zx x y z λ λ λ += − =+= − = −+ = =− Hence, vector equation of 1l is 21 = 0 0 , 02 λλ −   +∈   −  r  Note that 2 14 0 01 0 22 84 4 8 for all values of . λ λλ λ −      +−     −   = −+ − =−  11 11 Hence all points on lies in . Hence, lies in . lp lp

Page 6 (ii) Let the foot of perpendicular from A to be N. Equation of the line that passes through A and perpendicular to is 44 : = 3 1 , 2 ANl k αα      +− ∈       r . Since N lies on lAN, 44 = 3 , for some 2 ON k α αα α +  −∈ +   44 4 3 18 22 16 16 3 2 4 8 21 21 2 21 21 k k k k α α α αα α α α +     − •− = −    +   + −++ + = − = −− = −− 8844 21 21 22=31 4 21 21 4 1722 21 21 kk kkON kkk   −− −        ∴ ++ = +        −− −+      Hence, N is the point 8 2 17,4 , 221 21 21k kk−+ − . (iii) Direction vector of l2 is 0 2 q    . 2 2 2 2 01 0 22cos60 45 14 2 5 20 5 20 8 44 5 44 = 2.97 (3 s.f.)5 o q q q q q q    •   − = + = + += = = ±± 1p 1p

Page 7 (iv) 0 2 0 CB  =    Shortest distance from B to the line 2l : = 22 00 4 21 0 02 0 4 512 12    ×    = = ++ Let D be any point on plane 2p . 0 0 2 OD  =   1 2 0 DB OB OD  =−=        Shortest distance from B to plane 2p 22 10 21 0 2 11 t tt             = = ++  B(1,2,2) D B(1,2,2) C(1,0,2)

Page 8 Since point ( )1 ,2 ,2B is equidistant from 2l and 2p , 2 2 2 42 5 1 51 2 1 4 1 2 t t t t = + += = =± Source of Question: SAJC JC2 Mid-Year CT 9758/2018/01/Q8 4 (a) Given that 23= +−vi j k , find the direction cosines of v. Hence prove that the sum of the squares of the direction cosines is 1. [3] (b) The planes 12 and ∏∏ have equations 3 6 and 3 10 0x my z mx y z+ + = + ++ = respectively. The point P has coordinates (‒3, ‒3, 5) and O is the origin. (i) Given that point P is on 1∏ , find m. Verify that point P is also on 2∏ . [2] (ii) Hence find a vector equation of the line of intersection between the planes 12 and ∏∏ . [2] (iii) Point Q with coordinates (‒3, 0, 3) is a point on 1∏ . Find the shortest distance from point Q to 2∏ . [3] (iv) Suggest in parametric form, the equation of a plane 3∏ that contains points P and Q and is perpendicular to 2∏ . [1] Solution (a) Given . Let , and αβ γ be the angles formed by v with