JEE Exam > JEE Notes > Mathematics (Maths) Main & Advanced > JEE Main Previous Year Questions (2025): Vector Algebra
JEE Main Previous Year Questions (2025): Vector Algebra
| Download, print and study this document offline |
Please wait while the PDF view is loading
Page 1 JEE Main Previous Year Questions (2025): Vector Algebra Q1: Let ??? be the projection vector of ?? ? ? = ?? ??ˆ + ?? ?? ˆ ,?? > ?? , on the vector ??? ? = ??ˆ + ?? ??ˆ + ?? ?? ˆ . If |??? ? + ??? | = ?? , then the area of the parallelogram formed by the vectors ?? ? ? and ??? is ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 16 Solution: To find the projection vector ?? of ?? ? = ?? ??ˆ + 4?? ˆ (where ?? > 0 ) onto vector ?? = ??ˆ + 2??ˆ + 2?? ˆ , we use the formula for the projection of a vector: ?? = ( ?? ? · ?? |?? | 2 )?? Calculate the dot product ?? ? · ?? : ?? ? · ?? = (?? ??ˆ + 4?? ˆ )· (??ˆ + 2??ˆ + 2?? ˆ )= ?? · 1 + 4 · 2 = ?? + 8 Calculate the magnitude of ?? : |?? | = v1 2 + 2 2 + 2 2 = v9 = 3 Now substitute to find ?? : ?? = ( ?? + 8 9 )(??ˆ + 2??ˆ + 2?? ˆ ) We know that the magnitude |?? + ?? | = 7. Thus, substituting ?? in this equation, we resolve it to find ?? : |?? + ?? | = 7 ? (?? = 4) This indicates that ?? has a value of 4 . Next, calculating the area of the parallelogram formed by vectors ?? ? and ?? involves finding the cross product |?? ? × ?? | : ?? ? = 4??ˆ + 4?? ˆ ,?? = ( 12 9 )(??ˆ + 2??ˆ + 2?? ˆ )= ( 4 3 )??ˆ + ( 8 3 )??ˆ + ( 8 3 )?? ˆ The cross product is: ?? ? × ?? = ??ˆ ??ˆ ?? ˆ 4 3 8 3 8 3 4 0 4 | Solving the determinant: ?? ? × ?? = (( 8 3 × 4 - 0 × 8 3 )??ˆ - ( 4 3 × 4 - 4 × 8 3 )??ˆ + ( 4 3 × 0 - 8 3 × 4)?? ˆ ) = ( 32 3 ??ˆ + 0??ˆ + (- 32 3 )?? ˆ ) Then, the magnitude is calculated as: Page 2 JEE Main Previous Year Questions (2025): Vector Algebra Q1: Let ??? be the projection vector of ?? ? ? = ?? ??ˆ + ?? ?? ˆ ,?? > ?? , on the vector ??? ? = ??ˆ + ?? ??ˆ + ?? ?? ˆ . If |??? ? + ??? | = ?? , then the area of the parallelogram formed by the vectors ?? ? ? and ??? is ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 16 Solution: To find the projection vector ?? of ?? ? = ?? ??ˆ + 4?? ˆ (where ?? > 0 ) onto vector ?? = ??ˆ + 2??ˆ + 2?? ˆ , we use the formula for the projection of a vector: ?? = ( ?? ? · ?? |?? | 2 )?? Calculate the dot product ?? ? · ?? : ?? ? · ?? = (?? ??ˆ + 4?? ˆ )· (??ˆ + 2??ˆ + 2?? ˆ )= ?? · 1 + 4 · 2 = ?? + 8 Calculate the magnitude of ?? : |?? | = v1 2 + 2 2 + 2 2 = v9 = 3 Now substitute to find ?? : ?? = ( ?? + 8 9 )(??ˆ + 2??ˆ + 2?? ˆ ) We know that the magnitude |?? + ?? | = 7. Thus, substituting ?? in this equation, we resolve it to find ?? : |?? + ?? | = 7 ? (?? = 4) This indicates that ?? has a value of 4 . Next, calculating the area of the parallelogram formed by vectors ?? ? and ?? involves finding the cross product |?? ? × ?? | : ?? ? = 4??ˆ + 4?? ˆ ,?? = ( 12 9 )(??ˆ + 2??ˆ + 2?? ˆ )= ( 4 3 )??ˆ + ( 8 3 )??ˆ + ( 8 3 )?? ˆ The cross product is: ?? ? × ?? = ??ˆ ??ˆ ?? ˆ 4 3 8 3 8 3 4 0 4 | Solving the determinant: ?? ? × ?? = (( 8 3 × 4 - 0 × 8 3 )??ˆ - ( 4 3 × 4 - 4 × 8 3 )??ˆ + ( 4 3 × 0 - 8 3 × 4)?? ˆ ) = ( 32 3 ??ˆ + 0??ˆ + (- 32 3 )?? ˆ ) Then, the magnitude is calculated as: |?? ? × ?? | = v ( 32 3 ) 2 + 0 + (- 32 3 ) 2 = v ( 32 3 ) 2 + ( 32 3 ) 2 = v 2048 9 = v2048 3 = 16 3 × 3 ˜ 16 Therefore, the area of the parallelogram formed by ?? ? and ?? is 16 . Q2: Let ??? ? = ??ˆ + ??ˆ + ?? ˆ ,?? = ?? ??ˆ + ?? ??ˆ + ?? ˆ and ?? = ??? ? × ?? . If ?? is a vector such that ??? ? · ?? = |?? |,|?? - ?? ??? ? | ?? = ?? and the angle between ?? and ?? is ?? ?? , then |???? - ?? ?? ??? · ?? |+| ?? × ?? ?? is equal to ____ . JEE Main 2025 (Online) 28th January Morning Shift Ans: 6 Solution: ?? = ??ˆ + ??ˆ + ?? ˆ b ? = 2i ˆ + 2j ˆ + k ˆ d ? = a ? × b ? = -i ˆ + j ˆ |c - 2a ? | 2 = 8 |c| 2 + 4|a| 2 - 4(a· c)= 8 c 2 + 12- 4c = 8 ?? 2 - 4?? + 4 = 0 |c| = 2 d ? = a ? × b ? d ? × c = (a ??? × b ? ) 2 × c (|d||c|sin ?? 4 ) 2 = ((a· c)· b - (b · c)· a) 2 4 = 4 b 2 + (b · c) 2 2(a) 2 - 2( b · c)(a· b) Let b. c = x 4 = 36+ 3?? 2 - 20?? 3?? 2 - 20?? + 32 = 0 3?? 2 - 12?? - 8?? + 32 = 0 ?? = 8 3 ,4 b.c = 8 3 ,4 b.c = 8 3 Now | 10- 3 b. c|+|d × c| 2 |10- 8| + (2) 2 ? 6 Ans. Page 3 JEE Main Previous Year Questions (2025): Vector Algebra Q1: Let ??? be the projection vector of ?? ? ? = ?? ??ˆ + ?? ?? ˆ ,?? > ?? , on the vector ??? ? = ??ˆ + ?? ??ˆ + ?? ?? ˆ . If |??? ? + ??? | = ?? , then the area of the parallelogram formed by the vectors ?? ? ? and ??? is ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 16 Solution: To find the projection vector ?? of ?? ? = ?? ??ˆ + 4?? ˆ (where ?? > 0 ) onto vector ?? = ??ˆ + 2??ˆ + 2?? ˆ , we use the formula for the projection of a vector: ?? = ( ?? ? · ?? |?? | 2 )?? Calculate the dot product ?? ? · ?? : ?? ? · ?? = (?? ??ˆ + 4?? ˆ )· (??ˆ + 2??ˆ + 2?? ˆ )= ?? · 1 + 4 · 2 = ?? + 8 Calculate the magnitude of ?? : |?? | = v1 2 + 2 2 + 2 2 = v9 = 3 Now substitute to find ?? : ?? = ( ?? + 8 9 )(??ˆ + 2??ˆ + 2?? ˆ ) We know that the magnitude |?? + ?? | = 7. Thus, substituting ?? in this equation, we resolve it to find ?? : |?? + ?? | = 7 ? (?? = 4) This indicates that ?? has a value of 4 . Next, calculating the area of the parallelogram formed by vectors ?? ? and ?? involves finding the cross product |?? ? × ?? | : ?? ? = 4??ˆ + 4?? ˆ ,?? = ( 12 9 )(??ˆ + 2??ˆ + 2?? ˆ )= ( 4 3 )??ˆ + ( 8 3 )??ˆ + ( 8 3 )?? ˆ The cross product is: ?? ? × ?? = ??ˆ ??ˆ ?? ˆ 4 3 8 3 8 3 4 0 4 | Solving the determinant: ?? ? × ?? = (( 8 3 × 4 - 0 × 8 3 )??ˆ - ( 4 3 × 4 - 4 × 8 3 )??ˆ + ( 4 3 × 0 - 8 3 × 4)?? ˆ ) = ( 32 3 ??ˆ + 0??ˆ + (- 32 3 )?? ˆ ) Then, the magnitude is calculated as: |?? ? × ?? | = v ( 32 3 ) 2 + 0 + (- 32 3 ) 2 = v ( 32 3 ) 2 + ( 32 3 ) 2 = v 2048 9 = v2048 3 = 16 3 × 3 ˜ 16 Therefore, the area of the parallelogram formed by ?? ? and ?? is 16 . Q2: Let ??? ? = ??ˆ + ??ˆ + ?? ˆ ,?? = ?? ??ˆ + ?? ??ˆ + ?? ˆ and ?? = ??? ? × ?? . If ?? is a vector such that ??? ? · ?? = |?? |,|?? - ?? ??? ? | ?? = ?? and the angle between ?? and ?? is ?? ?? , then |???? - ?? ?? ??? · ?? |+| ?? × ?? ?? is equal to ____ . JEE Main 2025 (Online) 28th January Morning Shift Ans: 6 Solution: ?? = ??ˆ + ??ˆ + ?? ˆ b ? = 2i ˆ + 2j ˆ + k ˆ d ? = a ? × b ? = -i ˆ + j ˆ |c - 2a ? | 2 = 8 |c| 2 + 4|a| 2 - 4(a· c)= 8 c 2 + 12- 4c = 8 ?? 2 - 4?? + 4 = 0 |c| = 2 d ? = a ? × b ? d ? × c = (a ??? × b ? ) 2 × c (|d||c|sin ?? 4 ) 2 = ((a· c)· b - (b · c)· a) 2 4 = 4 b 2 + (b · c) 2 2(a) 2 - 2( b · c)(a· b) Let b. c = x 4 = 36+ 3?? 2 - 20?? 3?? 2 - 20?? + 32 = 0 3?? 2 - 12?? - 8?? + 32 = 0 ?? = 8 3 ,4 b.c = 8 3 ,4 b.c = 8 3 Now | 10- 3 b. c|+|d × c| 2 |10- 8| + (2) 2 ? 6 Ans. Q3: Let ??? ? = ??ˆ + ??ˆ + ?? ˆ ,?? ? ? = ?? ??ˆ + ?? ??ˆ - ?? ˆ ,??? = ?? ??ˆ + ?? ?? ˆ and ?? ˆ be a unit vector such that ??? ? × ?? ˆ = ?? ? ? × ?? ˆ and ??? · ?? ˆ = ?? . If ??? is perpendicular to ??? ? , then |?? ?? ?? ˆ + ?? ??? | ?? is equal to ____ JEE Main 2025 (Online) 3rd April Morning Shift Ans: 5 Solution: Given the vectors ?? = ??ˆ + ??ˆ + ?? ˆ and ?? ? = 3??ˆ + 2??ˆ - ?? ˆ , along with ?? = ?? ??ˆ + ?? ?? ˆ , and ?? ˆ being a unit vector such that ?? × ?? ˆ = ?? ? × ?? ˆ and ?? · ?? ˆ = 1, we proceed as follows: Determine ?? ˆ : Since ?? × ?? ˆ = ?? ? × ?? ˆ , we have: (?? - ?? ? )× ?? ˆ = 0 Thus, ?? ˆ is parallel to ?? - ?? ? . Calculate: ?? - ?? ? = (-2??ˆ + ??ˆ + 2?? ˆ ) Therefore, we can express ?? ˆ as: ?? ˆ = -2 3 ??ˆ - 1 3 ??ˆ + 2 3 ?? ˆ Solve for ?? and ?? : Using the condition ?? · ?? ˆ = 1 : -?? 3 + 2?? 3 = 1 Simplify to: -?? + 2?? = 3 (?? ) Since ?? is perpendicular to ?? : ?? · ?? = 0 Which gives: ?? + ?? = 0 ? ?? = -?? Substitute ?? = -?? into equation (i): ?? + 2?? = 3 ? 3?? = 3 ? ?? = 1 Therefore, ?? = -1. Find |3?? ?? ˆ + ?? ?? | 2 : Calculate: |3?? ?? ˆ + ?? ?? | 2 = 9?? 2 |?? ˆ | 2 + ?? 2 |?? | 2 + 2· 3 · ?? · ?? · ?? · ?? Substituting the determined values: |?? ˆ | 2 = 1 (since ?? ˆ is a unit vector) ?? · ?? = 1 Thus: Page 4 JEE Main Previous Year Questions (2025): Vector Algebra Q1: Let ??? be the projection vector of ?? ? ? = ?? ??ˆ + ?? ?? ˆ ,?? > ?? , on the vector ??? ? = ??ˆ + ?? ??ˆ + ?? ?? ˆ . If |??? ? + ??? | = ?? , then the area of the parallelogram formed by the vectors ?? ? ? and ??? is ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 16 Solution: To find the projection vector ?? of ?? ? = ?? ??ˆ + 4?? ˆ (where ?? > 0 ) onto vector ?? = ??ˆ + 2??ˆ + 2?? ˆ , we use the formula for the projection of a vector: ?? = ( ?? ? · ?? |?? | 2 )?? Calculate the dot product ?? ? · ?? : ?? ? · ?? = (?? ??ˆ + 4?? ˆ )· (??ˆ + 2??ˆ + 2?? ˆ )= ?? · 1 + 4 · 2 = ?? + 8 Calculate the magnitude of ?? : |?? | = v1 2 + 2 2 + 2 2 = v9 = 3 Now substitute to find ?? : ?? = ( ?? + 8 9 )(??ˆ + 2??ˆ + 2?? ˆ ) We know that the magnitude |?? + ?? | = 7. Thus, substituting ?? in this equation, we resolve it to find ?? : |?? + ?? | = 7 ? (?? = 4) This indicates that ?? has a value of 4 . Next, calculating the area of the parallelogram formed by vectors ?? ? and ?? involves finding the cross product |?? ? × ?? | : ?? ? = 4??ˆ + 4?? ˆ ,?? = ( 12 9 )(??ˆ + 2??ˆ + 2?? ˆ )= ( 4 3 )??ˆ + ( 8 3 )??ˆ + ( 8 3 )?? ˆ The cross product is: ?? ? × ?? = ??ˆ ??ˆ ?? ˆ 4 3 8 3 8 3 4 0 4 | Solving the determinant: ?? ? × ?? = (( 8 3 × 4 - 0 × 8 3 )??ˆ - ( 4 3 × 4 - 4 × 8 3 )??ˆ + ( 4 3 × 0 - 8 3 × 4)?? ˆ ) = ( 32 3 ??ˆ + 0??ˆ + (- 32 3 )?? ˆ ) Then, the magnitude is calculated as: |?? ? × ?? | = v ( 32 3 ) 2 + 0 + (- 32 3 ) 2 = v ( 32 3 ) 2 + ( 32 3 ) 2 = v 2048 9 = v2048 3 = 16 3 × 3 ˜ 16 Therefore, the area of the parallelogram formed by ?? ? and ?? is 16 . Q2: Let ??? ? = ??ˆ + ??ˆ + ?? ˆ ,?? = ?? ??ˆ + ?? ??ˆ + ?? ˆ and ?? = ??? ? × ?? . If ?? is a vector such that ??? ? · ?? = |?? |,|?? - ?? ??? ? | ?? = ?? and the angle between ?? and ?? is ?? ?? , then |???? - ?? ?? ??? · ?? |+| ?? × ?? ?? is equal to ____ . JEE Main 2025 (Online) 28th January Morning Shift Ans: 6 Solution: ?? = ??ˆ + ??ˆ + ?? ˆ b ? = 2i ˆ + 2j ˆ + k ˆ d ? = a ? × b ? = -i ˆ + j ˆ |c - 2a ? | 2 = 8 |c| 2 + 4|a| 2 - 4(a· c)= 8 c 2 + 12- 4c = 8 ?? 2 - 4?? + 4 = 0 |c| = 2 d ? = a ? × b ? d ? × c = (a ??? × b ? ) 2 × c (|d||c|sin ?? 4 ) 2 = ((a· c)· b - (b · c)· a) 2 4 = 4 b 2 + (b · c) 2 2(a) 2 - 2( b · c)(a· b) Let b. c = x 4 = 36+ 3?? 2 - 20?? 3?? 2 - 20?? + 32 = 0 3?? 2 - 12?? - 8?? + 32 = 0 ?? = 8 3 ,4 b.c = 8 3 ,4 b.c = 8 3 Now | 10- 3 b. c|+|d × c| 2 |10- 8| + (2) 2 ? 6 Ans. Q3: Let ??? ? = ??ˆ + ??ˆ + ?? ˆ ,?? ? ? = ?? ??ˆ + ?? ??ˆ - ?? ˆ ,??? = ?? ??ˆ + ?? ?? ˆ and ?? ˆ be a unit vector such that ??? ? × ?? ˆ = ?? ? ? × ?? ˆ and ??? · ?? ˆ = ?? . If ??? is perpendicular to ??? ? , then |?? ?? ?? ˆ + ?? ??? | ?? is equal to ____ JEE Main 2025 (Online) 3rd April Morning Shift Ans: 5 Solution: Given the vectors ?? = ??ˆ + ??ˆ + ?? ˆ and ?? ? = 3??ˆ + 2??ˆ - ?? ˆ , along with ?? = ?? ??ˆ + ?? ?? ˆ , and ?? ˆ being a unit vector such that ?? × ?? ˆ = ?? ? × ?? ˆ and ?? · ?? ˆ = 1, we proceed as follows: Determine ?? ˆ : Since ?? × ?? ˆ = ?? ? × ?? ˆ , we have: (?? - ?? ? )× ?? ˆ = 0 Thus, ?? ˆ is parallel to ?? - ?? ? . Calculate: ?? - ?? ? = (-2??ˆ + ??ˆ + 2?? ˆ ) Therefore, we can express ?? ˆ as: ?? ˆ = -2 3 ??ˆ - 1 3 ??ˆ + 2 3 ?? ˆ Solve for ?? and ?? : Using the condition ?? · ?? ˆ = 1 : -?? 3 + 2?? 3 = 1 Simplify to: -?? + 2?? = 3 (?? ) Since ?? is perpendicular to ?? : ?? · ?? = 0 Which gives: ?? + ?? = 0 ? ?? = -?? Substitute ?? = -?? into equation (i): ?? + 2?? = 3 ? 3?? = 3 ? ?? = 1 Therefore, ?? = -1. Find |3?? ?? ˆ + ?? ?? | 2 : Calculate: |3?? ?? ˆ + ?? ?? | 2 = 9?? 2 |?? ˆ | 2 + ?? 2 |?? | 2 + 2· 3 · ?? · ?? · ?? · ?? Substituting the determined values: |?? ˆ | 2 = 1 (since ?? ˆ is a unit vector) ?? · ?? = 1 Thus: = 9(-1) 2 · 1 + 1 2 · (1 2 + 1 2 )+ 2· 3 · (-1)· 1 · 1 = 9 + 2- 6 = 5 Hence, |3?? ?? ˆ + ?? ?? | 2 = 5. Q4: Let ??? ? = ??ˆ + ?? ??ˆ + ?? ˆ ,?? ? ? = ?? ??ˆ - ?? ??ˆ + ?? ?? ˆ ,??? = ?? ??ˆ - ??ˆ + ?? ?? ˆ and ?? ? ? be a vector such that ?? ? ? × ?? ? ? = ??? × ?? ? ? and ??? ? · ?? ? ? = ?? . Then |(??? ? × ?? ? ? )| ?? is equal to ____ ? JEE Main 2025 (Online) 3rd April Evening Shift Ans: 128 Solution: Given Conditions : The condition ?? ? × ?? = ?? × ?? implies that (?? ? - ?? )× ?? = 0. This indicates that ?? is parallel to ?? ? - ?? . Expressing ?? ? - ?? : Calculate ?? ? - ?? = (3??ˆ - 3??ˆ + 3?? ˆ )- (2??ˆ - ??ˆ + 2?? ˆ )= ??ˆ - 2??ˆ + ?? ˆ . Express ?? : Since ?? is parallel to ?? ? - ?? , let ?? = ?? (??ˆ - 2??ˆ + ?? ˆ ) . Using the condition ?? · ?? = 4 : Compute ?? · ?? = (??ˆ + 2??ˆ + ?? ˆ )· (?? (??ˆ - 2??ˆ + ?? ˆ )) . This simplifies to ?? (1 · 1 + 2· (-2)+ 1 · 1)= ?? (1 - 4 + 1)= -2?? . Set -2?? = 4, which gives ?? = -2. Determine ?? using ?? : Therefore, ?? = -2(??ˆ - 2??ˆ + ?? ˆ )= -2??ˆ + 4??ˆ - 2?? ˆ . Calculate ?? × ?? : Use the determinant form for the cross product: ?? × ?? = ??ˆ ??ˆ ?? ˆ 1 2 1 -2 4 -2 | Calculate the determinant: For ??ˆ:(2× -2)- (1 × 4)= -4- 4 = -8 For ??ˆ:(-(1 × -2)- (1 × -2))= 2- 2 = 0 For ?? ˆ :(1 × 4)- (2× -2)= 4 + 4 = 8 Thus, ?? × ?? = -8??ˆ + 0??ˆ + 8?? ˆ . Calculate the magnitude squared: Page 5 JEE Main Previous Year Questions (2025): Vector Algebra Q1: Let ??? be the projection vector of ?? ? ? = ?? ??ˆ + ?? ?? ˆ ,?? > ?? , on the vector ??? ? = ??ˆ + ?? ??ˆ + ?? ?? ˆ . If |??? ? + ??? | = ?? , then the area of the parallelogram formed by the vectors ?? ? ? and ??? is ____ . JEE Main 2025 (Online) 22nd January Morning Shift Ans: 16 Solution: To find the projection vector ?? of ?? ? = ?? ??ˆ + 4?? ˆ (where ?? > 0 ) onto vector ?? = ??ˆ + 2??ˆ + 2?? ˆ , we use the formula for the projection of a vector: ?? = ( ?? ? · ?? |?? | 2 )?? Calculate the dot product ?? ? · ?? : ?? ? · ?? = (?? ??ˆ + 4?? ˆ )· (??ˆ + 2??ˆ + 2?? ˆ )= ?? · 1 + 4 · 2 = ?? + 8 Calculate the magnitude of ?? : |?? | = v1 2 + 2 2 + 2 2 = v9 = 3 Now substitute to find ?? : ?? = ( ?? + 8 9 )(??ˆ + 2??ˆ + 2?? ˆ ) We know that the magnitude |?? + ?? | = 7. Thus, substituting ?? in this equation, we resolve it to find ?? : |?? + ?? | = 7 ? (?? = 4) This indicates that ?? has a value of 4 . Next, calculating the area of the parallelogram formed by vectors ?? ? and ?? involves finding the cross product |?? ? × ?? | : ?? ? = 4??ˆ + 4?? ˆ ,?? = ( 12 9 )(??ˆ + 2??ˆ + 2?? ˆ )= ( 4 3 )??ˆ + ( 8 3 )??ˆ + ( 8 3 )?? ˆ The cross product is: ?? ? × ?? = ??ˆ ??ˆ ?? ˆ 4 3 8 3 8 3 4 0 4 | Solving the determinant: ?? ? × ?? = (( 8 3 × 4 - 0 × 8 3 )??ˆ - ( 4 3 × 4 - 4 × 8 3 )??ˆ + ( 4 3 × 0 - 8 3 × 4)?? ˆ ) = ( 32 3 ??ˆ + 0??ˆ + (- 32 3 )?? ˆ ) Then, the magnitude is calculated as: |?? ? × ?? | = v ( 32 3 ) 2 + 0 + (- 32 3 ) 2 = v ( 32 3 ) 2 + ( 32 3 ) 2 = v 2048 9 = v2048 3 = 16 3 × 3 ˜ 16 Therefore, the area of the parallelogram formed by ?? ? and ?? is 16 . Q2: Let ??? ? = ??ˆ + ??ˆ + ?? ˆ ,?? = ?? ??ˆ + ?? ??ˆ + ?? ˆ and ?? = ??? ? × ?? . If ?? is a vector such that ??? ? · ?? = |?? |,|?? - ?? ??? ? | ?? = ?? and the angle between ?? and ?? is ?? ?? , then |???? - ?? ?? ??? · ?? |+| ?? × ?? ?? is equal to ____ . JEE Main 2025 (Online) 28th January Morning Shift Ans: 6 Solution: ?? = ??ˆ + ??ˆ + ?? ˆ b ? = 2i ˆ + 2j ˆ + k ˆ d ? = a ? × b ? = -i ˆ + j ˆ |c - 2a ? | 2 = 8 |c| 2 + 4|a| 2 - 4(a· c)= 8 c 2 + 12- 4c = 8 ?? 2 - 4?? + 4 = 0 |c| = 2 d ? = a ? × b ? d ? × c = (a ??? × b ? ) 2 × c (|d||c|sin ?? 4 ) 2 = ((a· c)· b - (b · c)· a) 2 4 = 4 b 2 + (b · c) 2 2(a) 2 - 2( b · c)(a· b) Let b. c = x 4 = 36+ 3?? 2 - 20?? 3?? 2 - 20?? + 32 = 0 3?? 2 - 12?? - 8?? + 32 = 0 ?? = 8 3 ,4 b.c = 8 3 ,4 b.c = 8 3 Now | 10- 3 b. c|+|d × c| 2 |10- 8| + (2) 2 ? 6 Ans. Q3: Let ??? ? = ??ˆ + ??ˆ + ?? ˆ ,?? ? ? = ?? ??ˆ + ?? ??ˆ - ?? ˆ ,??? = ?? ??ˆ + ?? ?? ˆ and ?? ˆ be a unit vector such that ??? ? × ?? ˆ = ?? ? ? × ?? ˆ and ??? · ?? ˆ = ?? . If ??? is perpendicular to ??? ? , then |?? ?? ?? ˆ + ?? ??? | ?? is equal to ____ JEE Main 2025 (Online) 3rd April Morning Shift Ans: 5 Solution: Given the vectors ?? = ??ˆ + ??ˆ + ?? ˆ and ?? ? = 3??ˆ + 2??ˆ - ?? ˆ , along with ?? = ?? ??ˆ + ?? ?? ˆ , and ?? ˆ being a unit vector such that ?? × ?? ˆ = ?? ? × ?? ˆ and ?? · ?? ˆ = 1, we proceed as follows: Determine ?? ˆ : Since ?? × ?? ˆ = ?? ? × ?? ˆ , we have: (?? - ?? ? )× ?? ˆ = 0 Thus, ?? ˆ is parallel to ?? - ?? ? . Calculate: ?? - ?? ? = (-2??ˆ + ??ˆ + 2?? ˆ ) Therefore, we can express ?? ˆ as: ?? ˆ = -2 3 ??ˆ - 1 3 ??ˆ + 2 3 ?? ˆ Solve for ?? and ?? : Using the condition ?? · ?? ˆ = 1 : -?? 3 + 2?? 3 = 1 Simplify to: -?? + 2?? = 3 (?? ) Since ?? is perpendicular to ?? : ?? · ?? = 0 Which gives: ?? + ?? = 0 ? ?? = -?? Substitute ?? = -?? into equation (i): ?? + 2?? = 3 ? 3?? = 3 ? ?? = 1 Therefore, ?? = -1. Find |3?? ?? ˆ + ?? ?? | 2 : Calculate: |3?? ?? ˆ + ?? ?? | 2 = 9?? 2 |?? ˆ | 2 + ?? 2 |?? | 2 + 2· 3 · ?? · ?? · ?? · ?? Substituting the determined values: |?? ˆ | 2 = 1 (since ?? ˆ is a unit vector) ?? · ?? = 1 Thus: = 9(-1) 2 · 1 + 1 2 · (1 2 + 1 2 )+ 2· 3 · (-1)· 1 · 1 = 9 + 2- 6 = 5 Hence, |3?? ?? ˆ + ?? ?? | 2 = 5. Q4: Let ??? ? = ??ˆ + ?? ??ˆ + ?? ˆ ,?? ? ? = ?? ??ˆ - ?? ??ˆ + ?? ?? ˆ ,??? = ?? ??ˆ - ??ˆ + ?? ?? ˆ and ?? ? ? be a vector such that ?? ? ? × ?? ? ? = ??? × ?? ? ? and ??? ? · ?? ? ? = ?? . Then |(??? ? × ?? ? ? )| ?? is equal to ____ ? JEE Main 2025 (Online) 3rd April Evening Shift Ans: 128 Solution: Given Conditions : The condition ?? ? × ?? = ?? × ?? implies that (?? ? - ?? )× ?? = 0. This indicates that ?? is parallel to ?? ? - ?? . Expressing ?? ? - ?? : Calculate ?? ? - ?? = (3??ˆ - 3??ˆ + 3?? ˆ )- (2??ˆ - ??ˆ + 2?? ˆ )= ??ˆ - 2??ˆ + ?? ˆ . Express ?? : Since ?? is parallel to ?? ? - ?? , let ?? = ?? (??ˆ - 2??ˆ + ?? ˆ ) . Using the condition ?? · ?? = 4 : Compute ?? · ?? = (??ˆ + 2??ˆ + ?? ˆ )· (?? (??ˆ - 2??ˆ + ?? ˆ )) . This simplifies to ?? (1 · 1 + 2· (-2)+ 1 · 1)= ?? (1 - 4 + 1)= -2?? . Set -2?? = 4, which gives ?? = -2. Determine ?? using ?? : Therefore, ?? = -2(??ˆ - 2??ˆ + ?? ˆ )= -2??ˆ + 4??ˆ - 2?? ˆ . Calculate ?? × ?? : Use the determinant form for the cross product: ?? × ?? = ??ˆ ??ˆ ?? ˆ 1 2 1 -2 4 -2 | Calculate the determinant: For ??ˆ:(2× -2)- (1 × 4)= -4- 4 = -8 For ??ˆ:(-(1 × -2)- (1 × -2))= 2- 2 = 0 For ?? ˆ :(1 × 4)- (2× -2)= 4 + 4 = 8 Thus, ?? × ?? = -8??ˆ + 0??ˆ + 8?? ˆ . Calculate the magnitude squared: Find |?? × ?? | 2 = (-8) 2 + 0 2 + 8 2 = 64+ 0 + 64 = 128. Therefore, |(?? × ?? )| 2 = 128. Q5: Let the three sides of a triangle ABC be given by the vectors ?? ??ˆ - ??ˆ + ?? ˆ ,??ˆ - ?? ??ˆ - ?? ?? ˆ and ?? ??ˆ - ?? ??ˆ - ?? ?? ˆ . Let ?? be the centroid of the triangle ?????? . Then ?? (|???? ????? | ?? + |???? ????? | ?? + |???? ????? | ?? ) is equal to ____ . JEE Main 2025 (Online) 4th April Evening Shift Ans: 164 Solution: Assuming Vertex ?? to be origin ?? = ?? 1 = 0 ? ?? ? = ?? 1 + ??? = ??? = 2??ˆ - ??ˆ + ?? ˆ ?? = ?? 1 + ?? = ?? = 3??ˆ - 4??ˆ - 4?? ˆ One solving ?? = 0 ? ,?? ? = 2??ˆ - ??ˆ + ?? ˆ and ?? = 3??ˆ - 4??ˆ - 4?? ˆ , are the position vector of vertices ???? and ?? respectively. ?? = 1 3 (?? + ?? ? + ?? )= 1 3 (0 ? + ?? ? + ?? )= 1 3 (?? ? + ?? ) ? ?? = 5 3 ??ˆ - 5 3 ??ˆ - ?? ˆ ???? ????? = ?? - ?? = ?? |???? ????? | 2 = ( 5 3 ) 2 + ( 5 3 ) 2 + (1) 2 = 25 9 + 25 9 + 1 = 50 9 + 1 = 59 9 ???? ????? = ?? - ?? ? ?? ? = 2??ˆ - ??ˆ + ?? ˆ |???? ????? | 2 = ( 1 3 ) 3 + ( 2 3 ) 2 + 4 = 1 9 + 4 9 + 4 = 5 9 + 4 = 41 9 ???? ????? = ?? - ?? ?? = 3??ˆ - 4??ˆ - 4?? ˆ ???? ????? | 2 = ( 4 3 ) 2 + ( 7 3 ) 2 + 9 = 16 9 + 49 9 + 9 = 65 9 + 9 = 65 9 + 81 9 = 146 9 6(|???? ????? | 2 + |???? ????? | 2 + |???? ????? | 2 )= 6 · ( 59 9 + 41 9 + 146 9 ) = 6 · 246 9 = 164 Q6: Let ??? ? and ?? ? ? be two unit vectors such that the angle between them is ?? ?? . If ?? ??? ? + ?? ?? ? ? and ?? ??? ? - ?? ?? ? ? are perpendicular to each other, then the number of values of ?? in [-?? ,?? ] is :Read More
FAQs on JEE Main Previous Year Questions (2025): Vector Algebra
| 1. What is vector addition, and how is it represented geometrically? |  |
Ans.Vector addition involves combining two or more vectors to produce a resultant vector. Geometrically, this is represented using the head-to-tail method, where the tail of one vector is placed at the head of another. The resultant vector is drawn from the tail of the first vector to the head of the last vector.
| 2. How do you calculate the dot product of two vectors, and what does it represent? | |
Ans.The dot product of two vectors A and B is calculated using the formula A · B = |A| |B| cos(θ), where θ is the angle between the two vectors. The dot product represents the magnitude of one vector in the direction of another and can be used to determine if the vectors are orthogonal (if A · B = 0).
| 3. What is the significance of the cross product in vector algebra? | |
Ans.The cross product of two vectors A and B, represented as A × B, results in a third vector that is perpendicular to both A and B. The magnitude of the cross product is given by |A| |B| sin(θ), where θ is the angle between the vectors. The cross product is significant in physics for calculating torque and angular momentum.
| 4. How can vectors be expressed in component form, and why is this useful? | |
Ans.Vectors can be expressed in component form as A = ai + bj + ck, where a, b, and c are the scalar components along the x, y, and z axes, respectively. This representation is useful because it allows for easier calculations of vector operations such as addition, subtraction, and dot and cross products.
| 5. What are unit vectors, and how are they used in vector algebra? | |
Ans.Unit vectors are vectors that have a magnitude of 1 and are used to indicate direction. They are often denoted as i, j, and k in three-dimensional space, representing the directions along the x, y, and z axes, respectively. Unit vectors are used in vector algebra to simplify expressions and calculations, as they provide a clear way to express direction without concern for magnitude.
About this Document
4.72/5 Rating
Apr 25, 2026 Last updated
Related Exams
Document Description: JEE Main Previous Year Questions (2025): Vector Algebra for JEE 2026 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The notes and questions for JEE Main Previous Year Questions (2025): Vector Algebra have been prepared according to the JEE exam syllabus. Information about JEE Main Previous Year Questions (2025): Vector Algebra covers topics like and JEE Main Previous Year Questions (2025): Vector Algebra Example, for JEE 2026 Exam. Find important definitions, questions, notes, meanings, examples, exercises and tests below for JEE Main Previous Year Questions (2025): Vector Algebra.
Introduction of JEE Main Previous Year Questions (2025): Vector Algebra in English is available as part of our Mathematics (Maths) for JEE Main & Advanced for JEE & JEE Main Previous Year Questions (2025): Vector Algebra in Hindi for Mathematics (Maths) for JEE Main & Advanced course. Download more important topics related with notes, lectures and mock test series for JEE Exam by signing up for free. JEE: JEE Main Previous Year Questions (2025): Vector Algebra
Description
JEE Main Previous Year Questions (2025): Vector Algebra of Maths is important for the exam. Get access to all the questions with solutions asked in past years of JEE exam. Download it from EduRev.
Information about JEE Main Previous Year Questions (2025): Vector Algebra
In this doc you can find the meaning of JEE Main Previous Year Questions (2025): Vector Algebra defined & explained in the simplest way possible. Besides explaining types of JEE Main Previous Year Questions (2025): Vector Algebra theory, EduRev gives you an ample number of questions to practice JEE Main Previous Year Questions (2025): Vector Algebra tests, examples and also practice JEE tests
Related Searches
Extra Questions, JEE Main Previous Year Questions (2025): Vector Algebra, JEE Main Previous Year Questions (2025): Vector Algebra, past year papers, Semester Notes, Sample Paper, ppt, MCQs, study material, Objective type Questions, mock tests for examination, video lectures, JEE Main Previous Year Questions (2025): Vector Algebra, Important questions, Summary, Free, pdf , Previous Year Questions with Solutions, shortcuts and tricks, practice quizzes, Exam, Viva Questions;