Q: For any two points M and N in the X Y-plane, let M N denote the vector from M to N, and 0 denote the zero vector. Let P, Q and R be three distinct points in the X Y-plane. Let S be a point inside the

aligned & SP +5 SQ +6 SP = 0 \\ & (Let P.V. of P be p , Q be q and R be r ) \\ & aligned & ( p - s )+5( q - s )+6( r - s )= 0 \\ & S = p +5 q +6 r 12 \\ & E F = ( q - p 2 ) \\ & E S = ( 5 q -5 p 12 )= 5 12 ( q - p ) \\ & | E F | | E S | = 6 5 =1.2 aligned aligned

Q: Let w = i + j -2 k , and u and v be two vectors such that u v = w and v w = u . Let , , and t be real numbers such that u = i + j + k ,-t + + =0, -t + =0, and + -t =0. Match each entry in List-I to th

array lll u v = w & u v & u w \\ v w = u & v w & array aligned & w = i + j -2 k \\ & u = i + j + k aligned | array ccc - t & 1 & 1 \\ 1 & - t & 1 \\ 1 & 1 & - t array |=0 array ll u . w =0= + -2 & t=-1 or 2 \\ ( v . w ) v = u v & t=-1 \\ w | v |^2-0= w & + + =0 \\ | v |=1(P) & t=2 \\ & = = array aligned & Also, u v = w \\ & | u || v |=| w | \\ & | u |=| w |= 6 aligned Case-1: aligned & t =2 \\ & = = \\ & ^2+ ^2+ ^2=6 \\ & = 2 ,- 2 \\ & (S) t +3=5 aligned Case-2: aligned & t =-1 \\ & + + =0 \\ &

Q: Consider the vectors x = i +2 j +3 k , y =2 i +3 j + k , and z =3 i + j +2 k For two distinct positive real numbers and , define X = x + y - z , Y = y + z - x , and Z = z + x - y If the vectors X , Y

| array ccc & & -1 \\ -1 & & \\ & -1 & array | | array ccc 1 & 2 & 3 \\ 2 & 3 & 1 \\ 3 & 1 & 2 array | _ -0 =0 aligned & ( ^3+ ^3-1 )-(- - - )=0 \\ & ^3+ ^3+3 =1 \\ & ^3+ ^3+(-1)^3=3( )( )(-1) \\ & + -1=0 aligned So, + -3=-2

Q: Let O P = -1 i + j + k , O Q = i + -1 j + k and O R = i + j + 1 2 k be three vectors, where , R -\ 0\ and O denotes the origin. If ( O P O Q ) O R =0 and the point ( , , 2) lies on the plane 3 x+3 y-z

( OP OQ ) OR =0 | array ccc -1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 1 & 1 2 array |=0 + +1=0...(1) Also ( , , 2) lies on 3 x +3 y - z +l=0 3 +3 -2+l=0 l=2-3( + ) ...(2) use (1) in (2) l=5

Q: Let the position vectors of points P , Q , R and S be a → = i ^ + 2 j ^ - 5 k ^ , b → = 3 i ^ + 6 j ^ + 3 k ^ , c → = 17 5 i ^ + 16 5 j ^ + 7 k ^ and d → =

Given, P ( 1 , 2 , - 5 ) , Q ( 3 , 6 , 3 ) , R 17 5 , 16 5 , 7 & S ( 2 , 1 , 1 ) Now solving options, Option A checking points are coplanar, P Q → = 2 , 4 , 8 , P R → = 12 5 , 6 5 , 12 & P Q → = 1 , - 1 , 6 Now if points are coplanar then, P Q → P R → P S → = 0 Now solving L.H.S, we get, 2 4 8 12 5 6 5 12 1 - 1 6 = 2 5 1 2 4 12 6 60 1 - 1 6 = 2 5 96 - 24 - 72 = 0 Hence, they are co

Q: Let P be the plane 3 x + 2 y + 3 z = 16 and let S = α i ^ + β j ^ + γ k ^ : α 2 + β 2 + γ 2 = 1 and the distance of α , β , &#9

Given, Equation of plane, P : 3 x + 2 y + 3 z = 16 And S = α i ^ + β j ^ + γ k ^ : α 2 + β 2 + γ 2 = 1 which is equation of sphere, Also distance, d α , β , γ from P = 7 2 And relation between distinct vector is given by, u → - v → = v → - w → = w → - u → Now V : volume of parallelepiped by vectors u → , v → , w &#8

Q: Let i ^ , j ^ and k ^ be the unit vectors along the three positive coordinate axes. Let a → = 3 i ^ + j ^ - k ^ b → = i ^ + b 2 j ^ + b 3 k ^ , b 2 , b 3 ∈ ℝ c → =

Given, a → = 3 i ^ + j ^ - k ^ b → = i ^ + b 2 j ^ + b 3 k ^ c → = c 1 i ^ + c 2 j ^ + c 3 k ^ 0 - c 3 c 2 c 3 0 - c 1 - c 2 c 1 0 1 b 2 b 3 = 3 - c 1 1 - c 2 - 1 - c 3 Now multiply and compare we get, b 2 c 3 - b 3 c 2 = c 1 - 3 . . . . . . . 1 c 3 - b 3 c 1 = 1 - c 2 . . . . . . . . . . . 2 c 2 - b 2 c 1 = 1 + c 3 . . . . . . . . . . . 3 Now applying operation equation 1 i ^ - 2 j ^ + 3 k ^ we get, i ^ b 2 c 3 - c 2 b 3 - j ^ c 3 - b 3 c 1 + k ^ c 2 -

Q: Let O be the origin and O A → = 2 i ^ + 2 j ^ + k ^ , O B → = i ^ - 2 j ^ + 2 k ^ and O C → = 1 2 O B → - λ O A → fo

O A → = 2 i ^ + 2 j ^ + k ^ O B → = i ^ − 2 j ^ + 2 k ^ O C → = 1 2 O B → − λ O A → ; λ > 0 O B → × O C → = 9 2 ⇒ O B → × 1 2 O B → − λ O A → = 9 2 ⇒ 1 2 O B → × O B → − λ O A → = 9 2 ⇒ 1 2 λ O A → × O B → = 9 2 ⇒ λ 2 O A → × O B → = 9 2 . . . 1 now

Question 1

Let a , b be two vectors, and let P, Q and R be the points with position vectors a , b and a + b , respectively, with respect to the origin O. If | a + b | = 21 , | a - b | = 3, and a and ( a - b ) ar

Accepted Answer

Given | a + b | = 21 and | a - b | = 3. Squaring both equations: | a |^2 + | b |^2 + 2 a b = 21 | a |^2 + | b |^2 - 2 a b = 9 Subtracting the second equation from the first gives: 4 a b = 12 a b = 3 Adding the two equations gives: 2(| a |^2 + | b |^2) = 30 | a |^2 + | b |^2 = 15 Since a and ( a - b ) are perpendicular, their dot product is zero: a ( a - b ) = 0 | a |^2 - a b = 0 | a |^2 = a b = 3 Substituting | a |^2 = 3 into | a |^2 + | b |^2 = 15 gives: 3 + | b |^2 = 15 | b |^2 = 12 Using Lagr

Question 2

For any two points M and N in the X Y-plane, let M N denote the vector from M to N, and 0 denote the zero vector. Let P, Q and R be three distinct points in the X Y-plane. Let S be a point inside the

Accepted Answer

aligned & SP +5 SQ +6 SP = 0 \ & (Let P.V. of P be p , Q be q and R be r ) \ & aligned & ( p - s )+5( q - s )+6( r - s )= 0 \ & S = p +5 q +6 r 12 \ & E F = ( q - p 2 ) \ & E S = ( 5 q -5 p 12 )= 5 12 ( q - p ) \ & | E F | | E S | = 6 5 =1.2 aligned aligned

Question 3

Let w = i + j -2 k , and u and v be two vectors such that u v = w and v w = u . Let , , and t be real numbers such that u = i + j + k ,-t + + =0, -t + =0, and + -t =0. Match each entry in List-I to th

Accepted Answer

array lll u v = w & u v & u w \ v w = u & v w & array aligned & w = i + j -2 k \ & u = i + j + k aligned | array ccc - t & 1 & 1 \ 1 & - t & 1 \ 1 & 1 & - t array |=0 array ll u . w =0= + -2 & t=-1 or 2 \ ( v . w ) v = u v & t=-1 \ w | v |^2-0= w & + + =0 \ | v |=1(P) & t=2 \ & = = array aligned & Also, u v = w \ & | u || v |=| w | \ & | u |=| w |= 6 aligned Case-1: aligned & t =2 \ & = = \ & ^2+ ^2+ ^2=6 \ & = 2 ,- 2 \ & (S) t +3=5 aligned Case-2: aligned & t =-1 \ & + + =0 \ &

Question 4

Consider the vectors x = i +2 j +3 k , y =2 i +3 j + k , and z =3 i + j +2 k For two distinct positive real numbers and , define X = x + y - z , Y = y + z - x , and Z = z + x - y If the vectors X , Y

Accepted Answer

| array ccc & & -1 \ -1 & & \ & -1 & array | | array ccc 1 & 2 & 3 \ 2 & 3 & 1 \ 3 & 1 & 2 array | _ -0 =0 aligned & ( ^3+ ^3-1 )-(- - - )=0 \ & ^3+ ^3+3 =1 \ & ^3+ ^3+(-1)^3=3( )( )(-1) \ & + -1=0 aligned So, + -3=-2

Question 5

Let p =2 i + j +3 k and q = i - j + k . If for some real numbers , , and , we have 15 i +10 j +6 k = (2 p + q )+ ( p -2 q )+ ( p q ) then the value of is

Accepted Answer

15 i +10 j +6 k = (2 p + q )+ ( p -2 q )+ ( p q ) taking dot with ( p q ) | array ccc 15 & 10 & 6 \ 2 & 1 & 3 \ 1 & -1 & 1 array |=0+0+ ( p ^2 q ^2-( p q )^2 ) [ ( p q )^2+( p q )^2= p ^2 q ^2 ] aligned & 52=26 \ & =2 aligned

Question 6

Let O P = -1 i + j + k , O Q = i + -1 j + k and O R = i + j + 1 2 k be three vectors, where , R -\ 0\ and O denotes the origin. If ( O P O Q ) O R =0 and the point ( , , 2) lies on the plane 3 x+3 y-z

Accepted Answer

( OP OQ ) OR =0 | array ccc -1 & 1 & 1 \ 1 & -1 & 1 \ 1 & 1 & 1 2 array |=0 + +1=0...(1) Also ( , , 2) lies on 3 x +3 y - z +l=0 3 +3 -2+l=0 l=2-3( + ) ...(2) use (1) in (2) l=5

Question 7

Let the position vectors of points P , &#160; Q , &#160; R and S  be a &#8594; = i ^ + 2 j ^ - 5 k ^ , b &#8594; = 3 i ^ + 6 j ^ + 3 k ^ , c &#8594; = 17 5 i ^ + 16 5 j ^ + 7 k ^ and d &#8594; =

Accepted Answer

Given, P ( 1 , 2 , - 5 ) , &#160; Q ( 3 , 6 , 3 ) , &#160; R 17 5 , 16 5 , 7 &#160; &#38; &#160; S ( 2 , 1 , 1 ) Now solving options, Option  A  checking points are coplanar, P Q &#8594; = 2 , 4 , 8 , &#160; P R &#8594; = 12 5 , 6 5 , 12 &#160; &#38; &#160; P Q &#8594; = 1 , - 1 , 6 Now if points are coplanar then,  P Q &#8594; P R &#8594; P S &#8594; = 0 Now solving L.H.S, we get, 2 4 8 12 5 6 5 12 1 - 1 6 = 2 5 1 2 4 12 6 60 1 - 1 6 = 2 5 96 - 24 - 72 = 0 Hence, they are co

Question 8

Let P be the plane 3 x + 2 y + 3 z = 16  and let S = &#945; i ^ + &#946; j ^ + &#947; k ^ : &#160; &#945; 2 + &#946; 2 + &#947; 2 = 1  and the distance of &#945; , &#160; &#946; , &#160; &#9

Accepted Answer

Given, Equation of plane, P : 3 x + 2 y + 3 z = 16 And  S = &#945; i ^ + &#946; j ^ + &#947; k ^ &#160; : &#160; &#945; 2 + &#946; 2 + &#947; 2 = 1  which is equation of sphere, Also distance,  d &#945; , &#160; &#946; , &#160; &#947;  from  P = 7 2 And relation between distinct vector is given by, u &#8594; - v &#8594; = v &#8594; - w &#8594; = w &#8594; - u &#8594; Now   V : volume of parallelepiped by vectors  u &#8594; , &#160; v &#8594; , &#160; w &#8

Question 9

Let i ^ , j ^ and k ^ be the unit vectors along the three positive coordinate axes. Let a &#8594; = 3 i ^ + j ^ - k ^ b &#8594; = i ^ + b 2 j ^ + b 3 k ^ , &#160; b 2 , b 3 &#8712; &#8477; c &#8594; =

Accepted Answer

Given, a &#8594; = 3 i ^ + j ^ - k ^ b &#8594; = i ^ + b 2 j ^ + b 3 k ^ c &#8594; = c 1 i ^ + c 2 j ^ + c 3 k ^ 0 - c 3 c 2 c 3 0 - c 1 - c 2 c 1 0 1 b 2 b 3 = 3 - c 1 1 - c 2 - 1 - c 3 Now multiply and compare we get, b 2 c 3 - b 3 c 2 = c 1 - 3 &#160; . . . . . . . 1 c 3 - b 3 c 1 = 1 - c 2 . . . . . . . . . . . 2 c 2 - b 2 c 1 = 1 + c 3 . . . . . . . . . . . 3 Now applying operation  equation&#160; 1 i ^ - 2 j ^ + 3 k ^  we get, i ^ b 2 c 3 - c 2 b 3 - j ^ c 3 - b 3 c 1 + k ^ c 2 -

Question 10

Let  O  be the origin and  O A &#8594; = 2 i ^ + 2 j ^ + k ^ , &#160;&#160; O B &#8594; = i ^ - 2 j ^ + 2 k ^  and  O C &#8594; = 1 2 O B &#8594; - &#955; O A &#8594;  fo

Accepted Answer

O A &#8594; = 2 i ^ + 2 j ^ + k ^ O B &#8594; = i ^ &#8722; 2 j ^ + 2 k ^ O C &#8594; = 1 2 O B &#8594; &#8722; &#955; O A &#8594; ; &#955; &#62; 0 O B &#8594; &#215; O C &#8594; = 9 2 &#8658; O B &#8594; &#215; 1 2 O B &#8594; &#8722; &#955; O A &#8594; = 9 2 &#8658; 1 2 O B &#8594; &#215; O B &#8594; &#8722; &#955; O A &#8594; = 9 2 &#8658; 1 2 &#955; O A &#8594; &#215; O B &#8594; = 9 2 &#8658; &#955; 2 O A &#8594; &#215; O B &#8594; = 9 2 &#160; &#160; &#160; &#160; &#160; &#160; . . . 1 now

Question 11

Let u &#8594; , v &#8594; and w &#8594; be vectors in three-dimensional space, where u &#8594; and v &#8594; are unit vectors which are not perpendicular to each other and  u &#8594; &#183; w &#8

Accepted Answer

Volume of the parallelopiped is  u &#8594; v &#8594; w &#8594; = 2 &#8658; u &#8594; v &#8594; w &#8594; 2 = 2 &#8658; u &#8594; &#183; u &#8594; u &#8594; &#183; v u &#8594; &#183; w &#8594; v &#8594; &#183; u &#8594; v &#8594; &#183; v &#8594; v &#8594; &#183; w &#8594; w &#8594; &#183; u &#8594; w &#8594; &#183; v &#8594; w &#8594; &#183; w &#8594; = 2 Using  u &#8594; &#183; w &#8594; = 1 , &#160; v &#8594; &#183; w &#8594; = 1 , &#160; w &#8594; &#183; w &#8594; = 4 , we get

Question 12

Let a and b be positive real numbers. Suppose P Q &#8594; = a i ^ + b j ^ and P S &#8594; = a i ^ - b j ^ are adjacent sides of a parallelogram P Q R S . Let u &#8594; and v &#8594; be the projection

Accepted Answer

Projection of a → a l o n g b → = a → . b → b → · b ^ u → = i ^ + j ^ · a i ^ + b j ^ a 2 + b 2 a i ^ + b j ^ a 2 + b 2 = a + b a 2 + b 2 · a i ^ + b j ^ a 2 + b 2 v → = i ^ + j ^ · a i ^ - b j ^ a 2 + b 2 a i ^ - b j ^ a 2 + b 2 = a - b a 2 + b 2 · a i ^ - b j ^ a 2 + b 2 given i j k a b 0 a - b 0 = 8 k ^ - a b - a b = 8 a b = 4 Given u → + | v → | = w → a + b a 2 + b 2 a - b a 2 + b 2 = 2 For a > b , 2 a = 2 a 2 + b 2 2 a 2 = a 2 + b 2 a 2 = b 2 ⇒ a = b = 2 Similarly, for b > a , a = b = 2 a +

Question 13

In a triangle P Q R , let a &#8594; = Q R &#8594; , &#160; b &#8594; = R P &#8594; and c &#8594; = P Q &#8594; . If a &#8594; = 3 , &#160; b &#8594; = 4 and a &#8594; . c &#8594; &#8722; b &#8594; c &

Accepted Answer

a &#8594; + b &#8594; + c &#8594; = 0 &#8757; a &#8594; . c &#8594; &#8722; b &#8594; c &#8594; . a &#8594; &#8722; b &#8594; = a &#8594; a &#8594; + b &#8594; &#8658; &#8722; b &#8594; + c &#8594; &#8901; c &#8594; &#8722; b &#8594; &#8722; a &#8594; + b &#8594; &#8901; a &#8594; &#8722; b &#8594; = a &#8594; a &#8594; + b &#8594; &#8658; c &#8594; 2 &#8722; b &#8594; 2 a &#8594; 2 &#8722; b &#8594; 2 = 3 3 + 4 = 3 7 &#8658; c &#8594; 2 = 13 Now,  a &#8594; + b &#8594; = &#8722; c &#8594;

Vector Algebra — Previous Year Questions & Solutions