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 Page 1


JEE Mains Previous Year Questions 
(2021-2024): Vector Algebra 
2024 
Q1: Let ??? = -?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? 
= ??ˆ + ?? ??ˆ - ?? ?? ˆ
 and 
?? = ( ( ??? × ?? 
)× ??ˆ)× ??ˆ)× ??ˆ. Then ??? · ( -??ˆ + ??ˆ + ?? ˆ
) is equal to : 
(A) -12 
(B) -10 
(C) -13 
(D) -15    [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans:  (a) 
?? = -5 · ??ˆ + ??ˆ - 3?? ˆ
,?? ? 
= ??ˆ + 2??ˆ - 4?? ˆ
?? = ( ( ( ?? × ?? ? 
)× ??ˆ)× ??ˆ)× ??ˆ
 = ( ( ( ?? · ??ˆ) ?? ? 
- ( ?? ? 
· ??ˆ) ?? )× ??ˆ)× ??ˆ
 = ( ( -5?? ? 
- ?? )× ??ˆ)× ??ˆ
 = ( ( -11??ˆ + 23?? ˆ
)× ??ˆ)× ??ˆ
 = ( 11?? ˆ
+ 23??ˆ)× ??ˆ
 = ( 11??ˆ - 23?? ˆ
)
?? · ( -??ˆ + ??ˆ + ?? ˆ
)= 0 + 11- 23
 = -12
 
Q2: Let ??? ? = ?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? ? ? 
= ?? ??ˆ + ??ˆ + ?? ?? ˆ
 and ??? = ??ˆ - ?? ??ˆ + ?? ?? ˆ
 be three vectors. If a vectors ??? ?  
satisfies ??? ? × ?? ? ? 
= ??? × ?? ? ? 
 and ??? ? · ??? ? = ?? , then ??? ? · ( ??ˆ - ??ˆ - ?? ˆ
) is equal to 
(A)24 
(B) 32 
(C) 36 
(D )28    [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (b) 
?? × ?? ? 
- ?? × ?? ? 
= 0
? 
 ( ?? - ?? )× ?? ? 
= 0
? 
?? - ?? = ?? ?? ? 
? ?? = ?? + ?? ?? ? 
 
Now, p ? · a ? = 0 (given) 
Page 2


JEE Mains Previous Year Questions 
(2021-2024): Vector Algebra 
2024 
Q1: Let ??? = -?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? 
= ??ˆ + ?? ??ˆ - ?? ?? ˆ
 and 
?? = ( ( ??? × ?? 
)× ??ˆ)× ??ˆ)× ??ˆ. Then ??? · ( -??ˆ + ??ˆ + ?? ˆ
) is equal to : 
(A) -12 
(B) -10 
(C) -13 
(D) -15    [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans:  (a) 
?? = -5 · ??ˆ + ??ˆ - 3?? ˆ
,?? ? 
= ??ˆ + 2??ˆ - 4?? ˆ
?? = ( ( ( ?? × ?? ? 
)× ??ˆ)× ??ˆ)× ??ˆ
 = ( ( ( ?? · ??ˆ) ?? ? 
- ( ?? ? 
· ??ˆ) ?? )× ??ˆ)× ??ˆ
 = ( ( -5?? ? 
- ?? )× ??ˆ)× ??ˆ
 = ( ( -11??ˆ + 23?? ˆ
)× ??ˆ)× ??ˆ
 = ( 11?? ˆ
+ 23??ˆ)× ??ˆ
 = ( 11??ˆ - 23?? ˆ
)
?? · ( -??ˆ + ??ˆ + ?? ˆ
)= 0 + 11- 23
 = -12
 
Q2: Let ??? ? = ?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? ? ? 
= ?? ??ˆ + ??ˆ + ?? ?? ˆ
 and ??? = ??ˆ - ?? ??ˆ + ?? ?? ˆ
 be three vectors. If a vectors ??? ?  
satisfies ??? ? × ?? ? ? 
= ??? × ?? ? ? 
 and ??? ? · ??? ? = ?? , then ??? ? · ( ??ˆ - ??ˆ - ?? ˆ
) is equal to 
(A)24 
(B) 32 
(C) 36 
(D )28    [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (b) 
?? × ?? ? 
- ?? × ?? ? 
= 0
? 
 ( ?? - ?? )× ?? ? 
= 0
? 
?? - ?? = ?? ?? ? 
? ?? = ?? + ?? ?? ? 
 
Now, p ? · a ? = 0 (given) 
 So, c · a ? + ?? a ? · b
? 
= 0
 ( 3 - 3 - 8)+ ?? ( 12+ 1 - 14)= 0
?? = -8
p ? = c - 8b
? 
p ? = -31i ˆ - 11j ˆ - 52k
ˆ
 So, p ? · ( i ˆ - j ˆ - k
ˆ
)
 = -31+ 11+ 52
 = 32
 
Q3: The distance of the point ?? ( ?? ,?? ,-?? ) form the line passing through the point ?? ( ?? ,-?? ,?? ) and 
perpendicular to the lines ??? = ( -?? ??ˆ + ?? ?? ˆ
)+ ?? ( ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
) ,?? ? R and ??? = ( ??ˆ - ?? ??ˆ + ?? ˆ
)+ ?? ( -??ˆ +
?? ??ˆ + ?? ?? ˆ
) ,?? ? R is : 
(A) v???? 
(B) v???? 
(C) v???? 
(D) v????   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (a) 
A vector in the direction of the required line can be obtained by cross product of 
|
??ˆ ??ˆ ?? ˆ
2 3 5
-1 3 2
|
 = -9??ˆ - 9??ˆ + 9?? ˆ
 
Required line 
r = ( 5i ˆ - 4j ˆ + 3k
ˆ
)+ ?? '
( -9i ˆ - 9j ˆ + 9k
ˆ
)
r = ( 5i ˆ - 4j ˆ + 3k
ˆ
)+ ?? ( i ˆ + j ˆ - k
ˆ
)
 
Now distance of ( 0,2,-2) 
 
Page 3


JEE Mains Previous Year Questions 
(2021-2024): Vector Algebra 
2024 
Q1: Let ??? = -?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? 
= ??ˆ + ?? ??ˆ - ?? ?? ˆ
 and 
?? = ( ( ??? × ?? 
)× ??ˆ)× ??ˆ)× ??ˆ. Then ??? · ( -??ˆ + ??ˆ + ?? ˆ
) is equal to : 
(A) -12 
(B) -10 
(C) -13 
(D) -15    [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans:  (a) 
?? = -5 · ??ˆ + ??ˆ - 3?? ˆ
,?? ? 
= ??ˆ + 2??ˆ - 4?? ˆ
?? = ( ( ( ?? × ?? ? 
)× ??ˆ)× ??ˆ)× ??ˆ
 = ( ( ( ?? · ??ˆ) ?? ? 
- ( ?? ? 
· ??ˆ) ?? )× ??ˆ)× ??ˆ
 = ( ( -5?? ? 
- ?? )× ??ˆ)× ??ˆ
 = ( ( -11??ˆ + 23?? ˆ
)× ??ˆ)× ??ˆ
 = ( 11?? ˆ
+ 23??ˆ)× ??ˆ
 = ( 11??ˆ - 23?? ˆ
)
?? · ( -??ˆ + ??ˆ + ?? ˆ
)= 0 + 11- 23
 = -12
 
Q2: Let ??? ? = ?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? ? ? 
= ?? ??ˆ + ??ˆ + ?? ?? ˆ
 and ??? = ??ˆ - ?? ??ˆ + ?? ?? ˆ
 be three vectors. If a vectors ??? ?  
satisfies ??? ? × ?? ? ? 
= ??? × ?? ? ? 
 and ??? ? · ??? ? = ?? , then ??? ? · ( ??ˆ - ??ˆ - ?? ˆ
) is equal to 
(A)24 
(B) 32 
(C) 36 
(D )28    [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (b) 
?? × ?? ? 
- ?? × ?? ? 
= 0
? 
 ( ?? - ?? )× ?? ? 
= 0
? 
?? - ?? = ?? ?? ? 
? ?? = ?? + ?? ?? ? 
 
Now, p ? · a ? = 0 (given) 
 So, c · a ? + ?? a ? · b
? 
= 0
 ( 3 - 3 - 8)+ ?? ( 12+ 1 - 14)= 0
?? = -8
p ? = c - 8b
? 
p ? = -31i ˆ - 11j ˆ - 52k
ˆ
 So, p ? · ( i ˆ - j ˆ - k
ˆ
)
 = -31+ 11+ 52
 = 32
 
Q3: The distance of the point ?? ( ?? ,?? ,-?? ) form the line passing through the point ?? ( ?? ,-?? ,?? ) and 
perpendicular to the lines ??? = ( -?? ??ˆ + ?? ?? ˆ
)+ ?? ( ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
) ,?? ? R and ??? = ( ??ˆ - ?? ??ˆ + ?? ˆ
)+ ?? ( -??ˆ +
?? ??ˆ + ?? ?? ˆ
) ,?? ? R is : 
(A) v???? 
(B) v???? 
(C) v???? 
(D) v????   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (a) 
A vector in the direction of the required line can be obtained by cross product of 
|
??ˆ ??ˆ ?? ˆ
2 3 5
-1 3 2
|
 = -9??ˆ - 9??ˆ + 9?? ˆ
 
Required line 
r = ( 5i ˆ - 4j ˆ + 3k
ˆ
)+ ?? '
( -9i ˆ - 9j ˆ + 9k
ˆ
)
r = ( 5i ˆ - 4j ˆ + 3k
ˆ
)+ ?? ( i ˆ + j ˆ - k
ˆ
)
 
Now distance of ( 0,2,-2) 
 
 P.V. of P = ( 5 + ?? ) ??ˆ + ( ?? - 4) ??ˆ + ( 3 - ?? ) ?? ˆ
AP
????? 
= ( 5 + ?? ) ??ˆ + ( ?? - 6) ??ˆ + ( 5 - ?? ) ?? ˆ
AP
????? 
· ( i ˆ + j ˆ - k
ˆ
)= 0
5 + ?? + ?? - 6 - 5 + ?? = 0
?? = 2
 |AP
????? 
| = v49 + 16 + 9
 |AP
????? 
| = v74
 
Q4: Let ?? ?? :??? = ( ??ˆ - ??ˆ + ?? ?? ˆ
)+ ?? ( ??ˆ - ??ˆ + ?? ?? ˆ
) ,?? ? R, 
?? ?? :??? = ( ??ˆ - ?? ˆ
)+ ?? ( ?? ??ˆ + ??ˆ + ?? ?? ˆ
) ,?? ? R, and ?? ?? :??? = ?? ( ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
) ,?? ? R 
be three lines such that ?? ?? is perpendicular to ?? ?? and ?? ?? is perpendicular to both ?? ?? and ?? ?? . Then, the 
point which lies on ?? ?? is 
(A) ( ?? ,?? ,-?? ) 
(B) ( ?? ,-?? ,?? ) 
(C) ( -?? ,?? ,?? ) 
(D) ( -,?? - ?? ,?? )   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (c) 
L
1
? L
2
  L
3
? L
1
, L
2
3 - 1 + 2P = 0
P = -1
|
i ˆ j ˆ k
ˆ
1 -1 2
3 1 -1
| = -i ˆ + 7j ˆ + 4k
ˆ
 ? ( -?? ,7?? ,4?? ) will lie on L
3
 
For ?? = 1 the point will be ( -1,7,4) 
Q5: Let ??? ? = ??ˆ + ?? ??ˆ + ?? ?? ˆ
,?? ,?? ? R. Let a vector ?? ? ? 
 be such that the angle between ??? ?  and ?? ? ? 
 is 
?? ?? and 
|?? ? ? 
|
?? = ?? . If ??? ? · ?? ? ? 
= ?? v?? , then the value of ( ?? ?? + ?? ?? ) |??? ? × ?? ? ? 
|
?? is equal to 
A. 85 
B. 90 
C. 75 
D. 95   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (b) 
 |b
? 
|
2
= 6;|a ? ||b
? 
|cos ?? = 3v2
 |a ? |
2
|b
? 
|
2
cos
2
 ?? = 18
 |a ? |
2
= 6
 
Also 1 + ?? 2
+ ?? 2
= 6 
?? 2
+ ?? 2
= 5 
to find 
Page 4


JEE Mains Previous Year Questions 
(2021-2024): Vector Algebra 
2024 
Q1: Let ??? = -?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? 
= ??ˆ + ?? ??ˆ - ?? ?? ˆ
 and 
?? = ( ( ??? × ?? 
)× ??ˆ)× ??ˆ)× ??ˆ. Then ??? · ( -??ˆ + ??ˆ + ?? ˆ
) is equal to : 
(A) -12 
(B) -10 
(C) -13 
(D) -15    [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans:  (a) 
?? = -5 · ??ˆ + ??ˆ - 3?? ˆ
,?? ? 
= ??ˆ + 2??ˆ - 4?? ˆ
?? = ( ( ( ?? × ?? ? 
)× ??ˆ)× ??ˆ)× ??ˆ
 = ( ( ( ?? · ??ˆ) ?? ? 
- ( ?? ? 
· ??ˆ) ?? )× ??ˆ)× ??ˆ
 = ( ( -5?? ? 
- ?? )× ??ˆ)× ??ˆ
 = ( ( -11??ˆ + 23?? ˆ
)× ??ˆ)× ??ˆ
 = ( 11?? ˆ
+ 23??ˆ)× ??ˆ
 = ( 11??ˆ - 23?? ˆ
)
?? · ( -??ˆ + ??ˆ + ?? ˆ
)= 0 + 11- 23
 = -12
 
Q2: Let ??? ? = ?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? ? ? 
= ?? ??ˆ + ??ˆ + ?? ?? ˆ
 and ??? = ??ˆ - ?? ??ˆ + ?? ?? ˆ
 be three vectors. If a vectors ??? ?  
satisfies ??? ? × ?? ? ? 
= ??? × ?? ? ? 
 and ??? ? · ??? ? = ?? , then ??? ? · ( ??ˆ - ??ˆ - ?? ˆ
) is equal to 
(A)24 
(B) 32 
(C) 36 
(D )28    [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (b) 
?? × ?? ? 
- ?? × ?? ? 
= 0
? 
 ( ?? - ?? )× ?? ? 
= 0
? 
?? - ?? = ?? ?? ? 
? ?? = ?? + ?? ?? ? 
 
Now, p ? · a ? = 0 (given) 
 So, c · a ? + ?? a ? · b
? 
= 0
 ( 3 - 3 - 8)+ ?? ( 12+ 1 - 14)= 0
?? = -8
p ? = c - 8b
? 
p ? = -31i ˆ - 11j ˆ - 52k
ˆ
 So, p ? · ( i ˆ - j ˆ - k
ˆ
)
 = -31+ 11+ 52
 = 32
 
Q3: The distance of the point ?? ( ?? ,?? ,-?? ) form the line passing through the point ?? ( ?? ,-?? ,?? ) and 
perpendicular to the lines ??? = ( -?? ??ˆ + ?? ?? ˆ
)+ ?? ( ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
) ,?? ? R and ??? = ( ??ˆ - ?? ??ˆ + ?? ˆ
)+ ?? ( -??ˆ +
?? ??ˆ + ?? ?? ˆ
) ,?? ? R is : 
(A) v???? 
(B) v???? 
(C) v???? 
(D) v????   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (a) 
A vector in the direction of the required line can be obtained by cross product of 
|
??ˆ ??ˆ ?? ˆ
2 3 5
-1 3 2
|
 = -9??ˆ - 9??ˆ + 9?? ˆ
 
Required line 
r = ( 5i ˆ - 4j ˆ + 3k
ˆ
)+ ?? '
( -9i ˆ - 9j ˆ + 9k
ˆ
)
r = ( 5i ˆ - 4j ˆ + 3k
ˆ
)+ ?? ( i ˆ + j ˆ - k
ˆ
)
 
Now distance of ( 0,2,-2) 
 
 P.V. of P = ( 5 + ?? ) ??ˆ + ( ?? - 4) ??ˆ + ( 3 - ?? ) ?? ˆ
AP
????? 
= ( 5 + ?? ) ??ˆ + ( ?? - 6) ??ˆ + ( 5 - ?? ) ?? ˆ
AP
????? 
· ( i ˆ + j ˆ - k
ˆ
)= 0
5 + ?? + ?? - 6 - 5 + ?? = 0
?? = 2
 |AP
????? 
| = v49 + 16 + 9
 |AP
????? 
| = v74
 
Q4: Let ?? ?? :??? = ( ??ˆ - ??ˆ + ?? ?? ˆ
)+ ?? ( ??ˆ - ??ˆ + ?? ?? ˆ
) ,?? ? R, 
?? ?? :??? = ( ??ˆ - ?? ˆ
)+ ?? ( ?? ??ˆ + ??ˆ + ?? ?? ˆ
) ,?? ? R, and ?? ?? :??? = ?? ( ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
) ,?? ? R 
be three lines such that ?? ?? is perpendicular to ?? ?? and ?? ?? is perpendicular to both ?? ?? and ?? ?? . Then, the 
point which lies on ?? ?? is 
(A) ( ?? ,?? ,-?? ) 
(B) ( ?? ,-?? ,?? ) 
(C) ( -?? ,?? ,?? ) 
(D) ( -,?? - ?? ,?? )   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (c) 
L
1
? L
2
  L
3
? L
1
, L
2
3 - 1 + 2P = 0
P = -1
|
i ˆ j ˆ k
ˆ
1 -1 2
3 1 -1
| = -i ˆ + 7j ˆ + 4k
ˆ
 ? ( -?? ,7?? ,4?? ) will lie on L
3
 
For ?? = 1 the point will be ( -1,7,4) 
Q5: Let ??? ? = ??ˆ + ?? ??ˆ + ?? ?? ˆ
,?? ,?? ? R. Let a vector ?? ? ? 
 be such that the angle between ??? ?  and ?? ? ? 
 is 
?? ?? and 
|?? ? ? 
|
?? = ?? . If ??? ? · ?? ? ? 
= ?? v?? , then the value of ( ?? ?? + ?? ?? ) |??? ? × ?? ? ? 
|
?? is equal to 
A. 85 
B. 90 
C. 75 
D. 95   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (b) 
 |b
? 
|
2
= 6;|a ? ||b
? 
|cos ?? = 3v2
 |a ? |
2
|b
? 
|
2
cos
2
 ?? = 18
 |a ? |
2
= 6
 
Also 1 + ?? 2
+ ?? 2
= 6 
?? 2
+ ?? 2
= 5 
to find 
( ?? 2
+ ?? 2
) |?? |
2
|?? ? 
|
2
sin
2
 ?? = ( 5) ( 6) ( 6)(
1
2
)
 = 90
 
Q6: Let ??  and ?? ? 
 be two vectors such that |?? ? 
| = 1 and |?? ? 
× ?? | = 2. Then |( ?? ? 
× ?? )- ?? ? 
|
2
 is equal to 
(A) 1 
(B) 3 
(C) 5 
(D) 4    [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (c) 
 |?? ? 
| = 1&|?? ? 
× ?? | = 2
 ( ?? ? 
× ?? )· ?? ? 
= ?? ? 
· ( ?? ? 
× ?? )= 0
 |( ?? ? 
× ?? )- ?? ? 
|
2
= |?? ? 
× ?? |
2
+ |?? ? 
|
2
 = 4 + 1 = 5
 
Q7: Let ??? = ?? ?? ??ˆ + ?? ?? ??ˆ + ?? ?? ?? ˆ
 and ?? 
= ?? ?? ??ˆ
ˆ
+ ?? ?? ??ˆ + ?? ?? ?? ˆ
 be two vectors such that |??? | = ?? ,??? ? · ?? ? ? 
= ?? 
and |?? ? ? 
| = ?? . If ??? = ?? ( ??? ? × ?? ? ? 
)- ?? ?? ? ? 
, then the angle between ?? ? ? 
 and ???  is equal to: 
(A) ?????? -?? (-
?? v?? ) 
(B) ?????? -?? (
?? ?? ) 
(C) ?????? -?? (
?? v?? ) 
(D) ?????? -?? ( -
v?? ?? )    [JEE Main 2024 (Online) 30th January Morning Shift] 
Ans: (d) 
Given |?? | = 1,|?? ? 
| = 4,?? · ?? ? 
= 2 
?? = 2( ?? × ?? ? 
)- 3?? ? 
 
Dot product with a ?  on both sides 
?? . ?? ?? = -6… 
Dot product with ?? ? 
 on both sides 
bc
???? 
= -48 …( 2)
c · c = 4a ? × b
? 
|
2
+ 9|b
? 
|
2
 |c |
2
= 4[|a|
2
| b|
2
- ( a · b
? 
)
2
] + 9|b
? 
|
2
 |c |
2
= 4[( 1) ( 4)
2
- ( 4) ] + 9( 16)
 |c |
2
= 4[12] + 144
 |c |
2
= 48 + 144
 |c |
2
= 192
 
Page 5


JEE Mains Previous Year Questions 
(2021-2024): Vector Algebra 
2024 
Q1: Let ??? = -?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? 
= ??ˆ + ?? ??ˆ - ?? ?? ˆ
 and 
?? = ( ( ??? × ?? 
)× ??ˆ)× ??ˆ)× ??ˆ. Then ??? · ( -??ˆ + ??ˆ + ?? ˆ
) is equal to : 
(A) -12 
(B) -10 
(C) -13 
(D) -15    [JEE Main 2024 (Online) 1st February Morning Shift] 
Ans:  (a) 
?? = -5 · ??ˆ + ??ˆ - 3?? ˆ
,?? ? 
= ??ˆ + 2??ˆ - 4?? ˆ
?? = ( ( ( ?? × ?? ? 
)× ??ˆ)× ??ˆ)× ??ˆ
 = ( ( ( ?? · ??ˆ) ?? ? 
- ( ?? ? 
· ??ˆ) ?? )× ??ˆ)× ??ˆ
 = ( ( -5?? ? 
- ?? )× ??ˆ)× ??ˆ
 = ( ( -11??ˆ + 23?? ˆ
)× ??ˆ)× ??ˆ
 = ( 11?? ˆ
+ 23??ˆ)× ??ˆ
 = ( 11??ˆ - 23?? ˆ
)
?? · ( -??ˆ + ??ˆ + ?? ˆ
)= 0 + 11- 23
 = -12
 
Q2: Let ??? ? = ?? ??ˆ + ??ˆ - ?? ?? ˆ
,?? ? ? 
= ?? ??ˆ + ??ˆ + ?? ?? ˆ
 and ??? = ??ˆ - ?? ??ˆ + ?? ?? ˆ
 be three vectors. If a vectors ??? ?  
satisfies ??? ? × ?? ? ? 
= ??? × ?? ? ? 
 and ??? ? · ??? ? = ?? , then ??? ? · ( ??ˆ - ??ˆ - ?? ˆ
) is equal to 
(A)24 
(B) 32 
(C) 36 
(D )28    [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (b) 
?? × ?? ? 
- ?? × ?? ? 
= 0
? 
 ( ?? - ?? )× ?? ? 
= 0
? 
?? - ?? = ?? ?? ? 
? ?? = ?? + ?? ?? ? 
 
Now, p ? · a ? = 0 (given) 
 So, c · a ? + ?? a ? · b
? 
= 0
 ( 3 - 3 - 8)+ ?? ( 12+ 1 - 14)= 0
?? = -8
p ? = c - 8b
? 
p ? = -31i ˆ - 11j ˆ - 52k
ˆ
 So, p ? · ( i ˆ - j ˆ - k
ˆ
)
 = -31+ 11+ 52
 = 32
 
Q3: The distance of the point ?? ( ?? ,?? ,-?? ) form the line passing through the point ?? ( ?? ,-?? ,?? ) and 
perpendicular to the lines ??? = ( -?? ??ˆ + ?? ?? ˆ
)+ ?? ( ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
) ,?? ? R and ??? = ( ??ˆ - ?? ??ˆ + ?? ˆ
)+ ?? ( -??ˆ +
?? ??ˆ + ?? ?? ˆ
) ,?? ? R is : 
(A) v???? 
(B) v???? 
(C) v???? 
(D) v????   [JEE Main 2024 (Online) 31st January Morning Shift] 
Ans: (a) 
A vector in the direction of the required line can be obtained by cross product of 
|
??ˆ ??ˆ ?? ˆ
2 3 5
-1 3 2
|
 = -9??ˆ - 9??ˆ + 9?? ˆ
 
Required line 
r = ( 5i ˆ - 4j ˆ + 3k
ˆ
)+ ?? '
( -9i ˆ - 9j ˆ + 9k
ˆ
)
r = ( 5i ˆ - 4j ˆ + 3k
ˆ
)+ ?? ( i ˆ + j ˆ - k
ˆ
)
 
Now distance of ( 0,2,-2) 
 
 P.V. of P = ( 5 + ?? ) ??ˆ + ( ?? - 4) ??ˆ + ( 3 - ?? ) ?? ˆ
AP
????? 
= ( 5 + ?? ) ??ˆ + ( ?? - 6) ??ˆ + ( 5 - ?? ) ?? ˆ
AP
????? 
· ( i ˆ + j ˆ - k
ˆ
)= 0
5 + ?? + ?? - 6 - 5 + ?? = 0
?? = 2
 |AP
????? 
| = v49 + 16 + 9
 |AP
????? 
| = v74
 
Q4: Let ?? ?? :??? = ( ??ˆ - ??ˆ + ?? ?? ˆ
)+ ?? ( ??ˆ - ??ˆ + ?? ?? ˆ
) ,?? ? R, 
?? ?? :??? = ( ??ˆ - ?? ˆ
)+ ?? ( ?? ??ˆ + ??ˆ + ?? ?? ˆ
) ,?? ? R, and ?? ?? :??? = ?? ( ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
) ,?? ? R 
be three lines such that ?? ?? is perpendicular to ?? ?? and ?? ?? is perpendicular to both ?? ?? and ?? ?? . Then, the 
point which lies on ?? ?? is 
(A) ( ?? ,?? ,-?? ) 
(B) ( ?? ,-?? ,?? ) 
(C) ( -?? ,?? ,?? ) 
(D) ( -,?? - ?? ,?? )   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (c) 
L
1
? L
2
  L
3
? L
1
, L
2
3 - 1 + 2P = 0
P = -1
|
i ˆ j ˆ k
ˆ
1 -1 2
3 1 -1
| = -i ˆ + 7j ˆ + 4k
ˆ
 ? ( -?? ,7?? ,4?? ) will lie on L
3
 
For ?? = 1 the point will be ( -1,7,4) 
Q5: Let ??? ? = ??ˆ + ?? ??ˆ + ?? ?? ˆ
,?? ,?? ? R. Let a vector ?? ? ? 
 be such that the angle between ??? ?  and ?? ? ? 
 is 
?? ?? and 
|?? ? ? 
|
?? = ?? . If ??? ? · ?? ? ? 
= ?? v?? , then the value of ( ?? ?? + ?? ?? ) |??? ? × ?? ? ? 
|
?? is equal to 
A. 85 
B. 90 
C. 75 
D. 95   [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (b) 
 |b
? 
|
2
= 6;|a ? ||b
? 
|cos ?? = 3v2
 |a ? |
2
|b
? 
|
2
cos
2
 ?? = 18
 |a ? |
2
= 6
 
Also 1 + ?? 2
+ ?? 2
= 6 
?? 2
+ ?? 2
= 5 
to find 
( ?? 2
+ ?? 2
) |?? |
2
|?? ? 
|
2
sin
2
 ?? = ( 5) ( 6) ( 6)(
1
2
)
 = 90
 
Q6: Let ??  and ?? ? 
 be two vectors such that |?? ? 
| = 1 and |?? ? 
× ?? | = 2. Then |( ?? ? 
× ?? )- ?? ? 
|
2
 is equal to 
(A) 1 
(B) 3 
(C) 5 
(D) 4    [JEE Main 2024 (Online) 30th January Evening Shift] 
Ans: (c) 
 |?? ? 
| = 1&|?? ? 
× ?? | = 2
 ( ?? ? 
× ?? )· ?? ? 
= ?? ? 
· ( ?? ? 
× ?? )= 0
 |( ?? ? 
× ?? )- ?? ? 
|
2
= |?? ? 
× ?? |
2
+ |?? ? 
|
2
 = 4 + 1 = 5
 
Q7: Let ??? = ?? ?? ??ˆ + ?? ?? ??ˆ + ?? ?? ?? ˆ
 and ?? 
= ?? ?? ??ˆ
ˆ
+ ?? ?? ??ˆ + ?? ?? ?? ˆ
 be two vectors such that |??? | = ?? ,??? ? · ?? ? ? 
= ?? 
and |?? ? ? 
| = ?? . If ??? = ?? ( ??? ? × ?? ? ? 
)- ?? ?? ? ? 
, then the angle between ?? ? ? 
 and ???  is equal to: 
(A) ?????? -?? (-
?? v?? ) 
(B) ?????? -?? (
?? ?? ) 
(C) ?????? -?? (
?? v?? ) 
(D) ?????? -?? ( -
v?? ?? )    [JEE Main 2024 (Online) 30th January Morning Shift] 
Ans: (d) 
Given |?? | = 1,|?? ? 
| = 4,?? · ?? ? 
= 2 
?? = 2( ?? × ?? ? 
)- 3?? ? 
 
Dot product with a ?  on both sides 
?? . ?? ?? = -6… 
Dot product with ?? ? 
 on both sides 
bc
???? 
= -48 …( 2)
c · c = 4a ? × b
? 
|
2
+ 9|b
? 
|
2
 |c |
2
= 4[|a|
2
| b|
2
- ( a · b
? 
)
2
] + 9|b
? 
|
2
 |c |
2
= 4[( 1) ( 4)
2
- ( 4) ] + 9( 16)
 |c |
2
= 4[12] + 144
 |c |
2
= 48 + 144
 |c |
2
= 192
 
 ? cos ?? =
b
? 
· c 
|b
? 
c |
??????? 
 ? cos ?? =
-48
v192· 4
 ? cos ?? =
-48
8v3 · 4
 ? cos ?? =
-3
2v3
 ? cos ?? =
-v3
2
? ?? = cos
-1
 (
-v3
2
)
 
Q8: Let a unit vector ?? ˆ = ?? ??ˆ + ?? ??ˆ + ?? ?? ˆ
 make angles 
?? ?? ,
?? ?? and 
?? ?? ?? with the vectors 
?? v?? ??ˆ +
?? v?? ?? ˆ
,
?? v?? ??ˆ +
?? v?? ?? ˆ
 
and 
?? v?? ??ˆ +
?? v?? ??ˆ respectively. If ??? ? =
?? v?? ??ˆ +
?? v?? ??ˆ +
?? v?? ?? ˆ
 then |??ˆ - ??? ? |
?? is equal to 
A. 
????
?? 
B. 
?? ?? 
C. 7 
D. 9    [JEE Main 2024 (Online) 29th January Evening Shift] 
Ans: (b) 
Unit vector u ˆ = xi ˆ + yj ˆ + zk 
p
1
???? =
1
v2
i ˆ +
1
v2
k
ˆ
,p ? 2 =
1
v2
j ˆ +
1
v2
k
ˆ
p ? 3 =
1
v2
i ˆ +
1
v2
j ˆ
 
Now angle between u ˆ and p ? 
1
=
?? 2
 
u ˆ · p ? 
1
= 0 ?
x
v2
+
z
v2
= 0
 ? x + z = 0…… (i) 
 
Angle between u ˆ and p ? 
2
=
?? 3
 
u ˆ · p
2
???? = |u ˆ|· |p
2
???? |cos 
?? 3
 ?
?? v2
+
?? v2
=
1
2
? ?? + ?? =
1
v2
 
Angle between u ˆ and p ? 
3
=
2?? 3
 
u ˆ · p ? 
3
= |u ˆ|· |P
? ? 
3|cos 
2?? 3
 ?
?? v2
+
4
v2
=
-1
2
? ?? + ?? =
-1
v2
…..
 
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FAQs on Vector Algebra & Three Dimensional Geometry: JEE Mains Previous Year Questions (2021-2024) - Mathematics (Maths) for JEE Main & Advanced

1. How to find the angle between two vectors in three-dimensional space?
Ans. To find the angle between two vectors in three-dimensional space, you can use the formula: cosθ = (a · b) / (|a||b|), where a and b are the two vectors, · denotes the dot product, and |a| and |b| are the magnitudes of the vectors.
2. How to determine if two vectors are orthogonal in three-dimensional space?
Ans. Two vectors are orthogonal in three-dimensional space if their dot product is equal to zero. In other words, if a · b = 0, where a and b are the vectors, then they are orthogonal.
3. How to find the distance between a point and a plane in three-dimensional space?
Ans. To find the distance between a point and a plane in three-dimensional space, you can use the formula: d = |ax + by + cz + d| / √(a^2 + b^2 + c^2), where (a, b, c) is the normal vector to the plane, (x, y, z) is the coordinates of the point, and d is the constant term in the plane equation ax + by + cz + d = 0.
4. How to determine if three points are collinear in three-dimensional space?
Ans. Three points are collinear in three-dimensional space if the vectors formed by connecting them are linearly dependent. This means that the determinant of the matrix formed by the coordinates of the points should be zero.
5. How to find the equation of a plane passing through a point and perpendicular to a given vector in three-dimensional space?
Ans. To find the equation of a plane passing through a point and perpendicular to a given vector in three-dimensional space, you can use the formula: a(x - x₁) + b(y - y₁) + c(z - z₁) = 0, where (a, b, c) is the given vector, and (x₁, y₁, z₁) is the point the plane passes through.
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