Let a = i + 2j + k and b = 2i + j - k. Let c be a unit vector in the plane of the vectors a and b and be perpendicular…

Year	2026	2025	2024	Total
Questions	15	17	14	46

Q52 jee_main_2025_07_april_evening Vector Magnitude and Operations

Let

a

and

b

be the vectors of the same magnitude such that

| a + b| + | a - b|| a + b| - | a - b| = \sqrt(2) + 1

. Then

| a + b|²| a|²

is:

A. $2 + 4\sqrt(2)$ $2 + 4\sqrt{2}$
B. $1 + \sqrt(2)$ $1 + \sqrt{2}$
C. $2 + \sqrt(2)$ $2 + \sqrt{2}$
D. $4 + 2\sqrt(2)$ $4 + 2\sqrt{2}$

Solution

Related Formula

Componendo and Dividendo rule states that if $(x)/(y) = (p)/(q)$ $\frac{x}{y} = \frac{p}{q}$, then:

(x+y)/(x-y) = (p+q)/(p-q)

Core Logic

Given expression:

| a + b| + | a - b|| a + b| - | a - b| = \sqrt(2) + 11

Applying Componendo and Dividendo:

2| a + b|2| a - b| = (\sqrt(2) + 1) + 1(\sqrt(2) + 1) - 1 = \sqrt(2) + 2\sqrt(2) = 1 + \sqrt(2)

Squaring both sides:

| a + b|² = (1 + \sqrt(2))² | a - b|²

| a + b|² = (3 + 2\sqrt(2)) | a - b|²

Step 1: Vector Expansion

Expanding using dot products, keeping in mind that $| a| = | b|$ $|\vec{a}| = |\vec{b}|$:

| a|² + | b|² + 2 a\cdot b = (3 + 2\sqrt(2))(| a|² + | b|² - 2 a\cdot b)

2| a|² + 2 a\cdot b = (3 + 2\sqrt(2))(2| a|² - 2 a\cdot b)

2| a|² (1 - (3 + 2\sqrt(2))) = -2 a\cdot b (1 + 3 + 2\sqrt(2))

Simplifying directly leads to:

a\cdot b| a|² = 2 + 2\sqrt(2)4 + 2\sqrt(2) = 1\sqrt(2)

Step 2: Final Calculation

We need to find $| a + b|²| a|²$ :

| a + b|²| a|² = | a|² + | b|² + 2 a\cdot b| a|² = 1 + 1 + 2 a\cdot b| a|²

= 2 + 2( 1\sqrt(2)) = 2 + \sqrt(2)

Pattern Recognition

Whenever symmetric sums and differences like $| x|+| y|$ $|\vec{x}|+|\vec{y}|$ and $| x|-| y|$ $|\vec{x}|-|\vec{y}|$ occur in ratios, Componendo-Dividendo should be applied immediately to isolate the ratio of the individual magnitudes.

Chapter Mix

Class 12 Physics: Vector Algebra Class 12 Mathematics: Vector Algebra

Q56 jee_main_2025_24_jan_evening Centroid, Orthocenter, and Circumcenter

Let the position vectors of three vertices of a

be

4 p+ q-3 r

,

-5 p+ q+2 r

and

2 p- q+2 r

If the position vectors of the orthocenter and the circumcenter of the

are

p+ q+ r4

and

α p+β q+γ r

respectively, then

α+2β+5γ

is equal to:

A. $3$ $3$
B. $1$ $1$
C. $6$ $6$
D. $4$ $4$

Solution

Related Formula

Centroid ( $G$ $G$) of a $\triangle$ with vertices $A, B, C$ $A, B, C$ is given by:

G = A + B + C3

Euler's line property: The orthocenter ( $O$ $O$), centroid ( $G$ $G$), and circumcenter ( $C$ $C$) are collinear, and $G$ $G$ divides the segment $OC$ $OC$ internally in the ratio $2:1$ $2:1$.

Step 1: Compute the Centroid Vector

Sum the vectors of the three given vertices:

A = 4 p+ q-3 r

B = -5 p+ q+2 r

C = 2 p- q+2 r

G = (4 - 5 + 2) p + (1 + 1 - 1) q + (-3 + 2 + 2) r3 = p + q + r3

Step 2: Apply Euler Line Section Ratio

Using the section formula ratio $O-G-C$ $O-G-C$ as $2:1$ $2:1$:

Euler Line section diagram for Q56 - JEE Main 2025 Evening

G = 2 C + O3 \Rightarrow 3 G = 2 C + O

2 C = 3 G - O = 3( p + q + r3) - p + q + r4

2 C = ( p + q + r) - (1)/(4)( p + q + r) = (3)/(4)( p + q + r)

C = (3)/(8) p + (3)/(8) q + (3)/(8) r

Step 3: Coefficient Matching

Compare with the given circumcenter format $α p + β q + γ r$ :

α = (3)/(8), β = (3)/(8), γ = (3)/(8)

Calculate $α + 2β + 5γ$ $\alpha + 2\beta + 5\gamma$:

(3)/(8) + 2((3)/(8)) + 5((3)/(8)) = (3 + 6 + 15)/(8) = (24)/(8) = 3

Pattern Recognition

Euler line configuration is universally $O-G-C$ $O-G-C$ in $2:1$ $2:1$. Remember the mnemonic 'Oil-Gas-Company' or simply $3G = 2C + O$ $3G = 2C + O$ to prevent swapping structural coefficients under exam stress.

Chapter Mix

Class 12 Mathematics: Vector Algebra Class 11 Mathematics: Properties of Triangles

Q64 jee_main_2025_24_jan_evening Vector Triple Product and Projection

Let

a=3 i- j+2 k,

b= a\times( i-2 k)

and

c= b\times k

. Then the projection of

c-2 j

on

a

is:

A. $3\sqrt(7)$ $3\sqrt{7}$
B. $\sqrt(14)$ $\sqrt{14}$
C. $2\sqrt(14)$ $2\sqrt{14}$
D. $2\sqrt(7)$ $2\sqrt{7}$

Solution

Related Formula

The scalar projection of vector $v$ $\vec{v}$ onto vector $w$ $\vec{w}$ is calculated as:

Projection = v \cdot w| w|

Step 1: Calculate $b$ $\vec{b}$

Compute the cross product using standard matrix expansion:

b = a \times ( i - 2 k) = vmatrix i & j & k 3 & -1 & 2 1 & 0 & -2 vmatrix

b = i(2 - 0) - j(-6 - 2) + k(0 - (-1)) = 2 i + 8 j + k

Step 2: Calculate $c$ $\vec{c}$ and $c - 2 j$ $\vec{c} - 2\hat{j}$

Perform the second cross product with unit vector $k$ $\hat{k}$:

c = b \times k = (2 i + 8 j + k) \times k = 2( i \times k) + 8( j \times k) + 0

c = 2(- j) + 8( i) = 8 i - 2 j

Subtract $2 j$ $2\hat{j}$:

c - 2 j = (8 i - 2 j) - 2 j = 8 i - 4 j

Step 3: Compute the projection onto $a$ $\vec{a}$

Using the dot product formula :

Projection = ( c - 2 j) \cdot a| a| = 8, -4, 0 \cdot 3, -1, 2 \sqrt(3² + (-1)² + 2²)

Projection = 24 + 4 + 0\sqrt(9 + 1 + 4) = 28\sqrt(14) = 2\sqrt(14)

Pattern Recognition

Keep cyclic unit cross products clear: $i \times k = - j$ $\hat{i} \times \hat{k} = -\hat{j}$ and $j \times k = i$ $\hat{j} \times \hat{k} = \hat{i}$. Missing a negative sign during basic cross multiplications ruins multi-step vector projections easily.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q51 jee_main_2025_24_jan_morning Coplanar Vectors and Vector Products

Let

a = i + 2 j + 3 k

,

b = 3 i + j - k

and \|\vec{\mathbf{c}}\| be three vectors such that

c

is coplanar with

a

and

b

. If the vector

c

is perpendicular to

b

and

a \cdot c = 5

, then

| c|

is equal to :

A. $13\sqrt(2)$ $\frac{1}{3\sqrt{2}}$
B. $18$ $18$
C. $16$ $16$
D. $\sqrt((11)/(6))$ $\sqrt{\frac{11}{6}}$

Solution

Related Formula

A vector $c$ $\vec{\mathbf{c}}$ coplanar with $a$ $\vec{\mathbf{a}}$ and $b$ $\vec{\mathbf{b}}$ and perpendicular to $b$ $\vec{\mathbf{b}}$ can be expressed using the vector triple product command template:

c = λ ( b \times ( a \times b)) = λ [( b \cdot b) a - ( a \cdot b) b]

Core Logic

First, find the dot products of the given vectors:

b \cdot b = 3² + 1² + (-1)² = 9 + 1 + 1 = 11

a \cdot b = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2

Substituting these values into the expression for $c$ $\vec{\mathbf{c}}$:

c = λ [11 a - 2 b]

c = λ [11( i + 2 j + 3 k) - 2(3 i + j - k)]

c = λ (5 i + 20 j + 35 k) = 5λ ( i + 4 j + 7 k)

Step 1: Determine Lambda

Using the given condition $a \cdot c = 5$ :

a \cdot [5λ ( i + 4 j + 7 k)] = 5

5λ (1 \cdot 1 + 2 \cdot 4 + 3 \cdot 7) = 5

5λ (1 + 8 + 21) = 5 30λ = 1 λ = (1)/(30)

Thus, the vector $c$ $\vec{\mathbf{c}}$ is:

c = (5)/(30)( i + 4 j + 7 k) = (1)/(6)( i + 4 j + 7 k)

Step 2: Calculate Magnitude

The magnitude of $c$ $\vec{\mathbf{c}}$ is evaluated as:

| c| = \sqrt(1² + 4² + 7²)6 = \sqrt(1 + 16 + 49)6 = \sqrt(66)6 = \sqrt((66)/(36)) = \sqrt((11)/(6))

Pattern Recognition

Whenever a vector is specified to be coplanar with $a, b$ $\vec{\mathbf{a}}, \vec{\mathbf{b}}$ and perpendicular to $b$ $\vec{\mathbf{b}}$, direct setup with the cross-product template $b \times ( a \times b)$ circumvents solving cumbersome linear systems of scalar variables.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q52 jee_main_2025_28_jan_evening Vector Operations and Components

Let A, B, C be three points in xy-plane, whose position vector are given by

\sqrt(3) i+ j

,

i+\sqrt(3) j

and

a i+(1-a) j

respectively with respect to the origin O. If the distance of the point C from the line bisecting the angle between the vectors

OA

and

OB

is

9\sqrt(2)

then the sum of all the possible values of a is:

A. $1$ $1$
B. $9/2$ $9/2$
C. $0$ $0$
D. $2$ $2$

Solution

Related Formula

The line bisecting the angle between two symmetric vectors passing through the origin in the first quadrant is given by $y = x$ $y = x$ or $x - y = 0$ $x - y = 0$.

Distance from point $(x₁, y₁)$ $(x_1, y_1)$ to line $Ax + By + C = 0$ $Ax + By + C = 0$ is:

d = |Ax₁ + By₁ + C|\sqrt(A² + B²)

Core Logic

Vectors $OA = \sqrt(3) i+ j$ and $OB = i+\sqrt(3) j$ are symmetric about the line $y = x$ $y = x$. Therefore, the angle bisector of $OA$ $\overline{OA}$ and $OB$ $\overline{OB}$ is the line $x - y = 0$ $x - y = 0$.

Point C has coordinates $(a, 1 - a)$ $(a, 1 - a)$.

Step 1: Calculate Distance and Solve for a

The perpendicular distance from $C(a, 1 - a)$ $C(a, 1 - a)$ to $x - y = 0$ $x - y = 0$ is:

d = |a - (1 - a)|\sqrt(1² + (-1)²) = |2a - 1|\sqrt(2)

Given that this distance is $9\sqrt(2)$ $\frac{9}{\sqrt{2}}$:

|2a - 1|\sqrt(2) = 9\sqrt(2) |2a - 1| = 9

This gives two solutions:

$2a - 1 = 9 2a = 10 a = 5$
$2a - 1 = -9 2a = -8 a = -4$

Step 2: Find the Sum of Values

Sum of all possible values of $a$ $a$:

Sum = 5 + (-4) = 1

Pattern Recognition

Notice that $OA$ $\overline{OA}$ and $OB$ $\overline{OB}$ have swapped coordinates, meaning they are symmetric with respect to $y=x$ $y=x$. Thus, the angle bisector equation is immediate ( $x-y=0$ $x-y=0$), simplifying the problem to a standard point-to-line distance calculation.

Chapter Mix

Class 11 Mathematics: Straight Lines Class 12 Mathematics: Vector Algebra

Let a = i + 2 j + k$\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and b = 2 i + j - k$\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$. Let c$\hat{c}$ be a unit vector in the plane of the vectors a$\vec{a}$ and b$\vec{b}$ and be perpendicular to a$\vec{a}$. Then such a vector c$\hat{c}$ is:

Solution & Explanation

Related Formula

Core Logic

Step 1: Define Coplanar Structural Form

Step 2: Force Orthogonality Constraint

Step 3: Substitute and Normalize

Pattern Recognition

Chapter Mix

Reference Study Guides

More Vector Algebra Previous-Year Questions — Page 6

Solution

Related Formula

Core Logic

Step 1: Vector Expansion

Step 2: Final Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Compute the Centroid Vector

Step 2: Apply Euler Line Section Ratio

Step 3: Coefficient Matching

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Calculate b$\vec{b}$

Step 2: Calculate c$\vec{c}$ and c - 2 j$\vec{c} - 2\hat{j}$

Step 3: Compute the projection onto a$\vec{a}$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Determine Lambda

Step 2: Calculate Magnitude

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Calculate Distance and Solve for a

Step 2: Find the Sum of Values

Pattern Recognition

Chapter Mix

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Let $a = i + 2 j + k$ and $b = 2 i + j - k$ . Let $c$ $\hat{c}$ be a unit vector in the plane of the vectors $a$ $\vec{a}$ and $b$ $\vec{b}$ and be perpendicular to $a$ $\vec{a}$. Then such a vector $c$ $\hat{c}$ is:

Step 1: Calculate $b$ $\vec{b}$

Step 2: Calculate $c$ $\vec{c}$ and $c - 2 j$ $\vec{c} - 2\hat{j}$

Step 3: Compute the projection onto $a$ $\vec{a}$