If (2,3, —1) is the foot of the perpendicular from (4, 2, 1) to a plane, then the equation of the plane is © 2x + y4+22- @ 2x -y+22=0 O 2x +yt2 woe (FB (4,2,°) AWA BR sev se aa a (2,327 & Ai aaa Gi Atago\ Fl — Pusves = D Since, the line joining the two points is perpendicular to the plane, it's DR's will given the normal to the plane. DR's of the normal = (4 — 2,2 — 3,1 +1) = (2,-1,2) Hence, a = 2,b = —Lande Since, the plane passes through (2, 3, —1). d =2(2) +3(-1) —1(2) d=-1 Hence, the required equation of plane is 2a — y + 22 =Wa=2+2j+3k, b=—i+2j+ hand c= 3i + jare the vectors such that @ + Ab is perpendicular to ¢, then value of Nis 4 = 2 wf Fe oteai ed, B= -ir2atk ote a mga 2H ark aaa 2 aad Rh AS oR Sr Ancwex 5 SB Sitio a a+ Ab = 21+ 27+ 3k + A(-1+ 2] +k) = (2—A)i+ (2+2A)j + (34 A)k (a+Ab)-¢=0 => (2 — A)(3) + (2+ 2A)(1) + (3 + A)(0) = 0 =>8-A=0 =>rA=8if [a|=3,]b] =4, then the value of 4 for which a+Ab is perpendicular to a— Ab, is * 3 A) 76 ®) 4 a © > x »¢ 3 z aR HSK \al= . Bleu J a wi _ RAL cid 2 iy A HA tn Pes = optim © Solwtt a+Ab is perpendicular to a—Ab (a+Ab)-(a-2b)=The angle between the straight lines sol _ y2 _ edig 1. 45° Z 80" 3. 60° Direction ratios of lines are ay = 2, by = 5, cy = 4 and ap =1, by =2, cp =-3 As we know, The angle between the lines is given by ae ayaytbibycyey (is) (vs) 2x1 +5%244x—3 (varee)-(VireCa) => cos0 = 1. 0= 90°The lengths of the intercepts on the co-ordinate axes made by the plane 5x + 2y+z-13=0are A 5.2, Tunit Q 2B, B, 13unit € Sd, g unit P 1,2, 5unit as aad = Sn+2yt+Z-'3=0 GTX Ata vay RK aS Ae SOT At Pmswer + ops @) Sol Equation of plane 5x + 2y+z-13=0 = 5x+2y+z=13 bx +2y+z _ > 13 ae >#$#+H4#+4=1 Is Is Is 3 2 1 .. Lengths of intercepts are XB, 8, 13 unitIf @ be the angle between any two vectors g and §, then |a-b|=|axb, when @ is equal to 1 1 (A) 0 (B) 7 OF (D) x 4 2 EHR SD ae a IA, be B= fal [b|coso =| |al[b|sin 0 tan@ =1 6-2. 4‘The area of the quadrilateral ABCD when A(0,4, 1), B(2, 3, —1),O(4, 5, 0) and D(2, 6, 2) is equal to sq units — © 18 squnis @ 2 saunis © 21 squnits KC0,4,1), B(213-1),¢ C4,6,0) STR D¢2,62) tar jE Abc HT fmvwey = seria D (2, 6, 2) C (4,5, 0) A (0, 4, 1) B(2,3,-1)If sum of two unit vectors is itself a unit vector, then the magnitude of their difference is AZ BB © (D) 2 wm Gy cork aiea FT UT 2% cai afta, % a sy ay Wer ater ae Gi aru em — Let 7 and § be two unit vectors. Jato 2+2cos@=1 cose ==! 2 Now, |a—bf [af + [of -2a-6a meaning ful ere ate 2 Cra a aldal Auswes = Open (8) (A) (a-b)x(e-4) (2) (axb)-(exd) (C) (a-b)-€ (0) a-(b-e) a-(b-) is meaningless. Here, @ is a vector and b-@ is a scalar. Dot product of vector and scalar is not possible. (a-b)x(-d) is meaningless. Here, g.pand ¢-q both are scalars. Cross product of two scalars is not possible. (axb)-(e%d) is meaningful Here, @xb and Tx both are vectors. Dot product of two vectors is possible. It will give a scalar. (a-b)-T is meaningless. Here, q-b is a scalar and @ is a vector. Dot product of a scalar and a vector is not possible, It will give a vector.Let the vectors, b,& be such that ja| = 2, |b] = 4 and el = 4. Ifthe projection of bon ais equal to the projection of on a and bis perpendicular to ¢ ea Se He Galet, (el=4 - a ey pie ra a wy 243 6 yO Sy 2h afta t afer 2 ta we e ay AR am - Prswera 6 2, [8] = 4, |2] =4 According to the given condition, Projection of b on @ = projection of Zona Now, |a + 6 — @| = y/|a+b— a2 jal? + |b—2|2+ 2a. (6-2) = y (2)? + |b- e+ 2(0) -... [From (i)] 4+ |b]? + |e? — 2(6- 2) 4+ 42+ 42—2(0) ... [bis perpendicular to ¢] = V36 =6