Let a vector a = sqrt(2)i - j + lambdak, lambda > 0, make an obtuse angle with the vector b = -lambda2i + 4sqrt(2)j +…

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More Vector Algebra Previous-Year Questions — Page 2

Q3 jee_main_2026_23_january_morning Vector Triple Product

Let

a = - i + j + 2 k

b = i - j - 3 k

c = a \times b

and

d = c \times a

. Then

( a - b) \cdot d

is equal to:

A. $4$ $4$
B. $-4$ $-4$
C. $-2$ $-2$
D. $2$ $2$

Solution

Related Formula

x \times ( y \times z) = ( x \cdot z) y - ( x \cdot y) z

Core Logic

Given $c = a \times b$ $\vec{c} = \vec{a} \times \vec{b}$ and $d = c \times a$ $\vec{d} = \vec{c} \times \vec{a}$. Substitute $c$ $\vec{c}$ into the expression for $d$ $\vec{d}$:

d = ( a \times b) \times a

Using the vector triple product expansion:

d = ( a \cdot a) b - ( a \cdot b) a = a² b - ( a \cdot b) a

Step 1: Calculate Magnitudes and Dot Products

For $a = - i + j + 2 k$ and $b = i - j - 3 k$ :

a² = | a|² = (-1)² + (1)² + (2)² = 1 + 1 + 4 = 6

b² = | b|² = (1)² + (-1)² + (-3)² = 1 + 1 + 9 = 11

a \cdot b = (-1)(1) + (1)(-1) + (2)(-3) = -1 - 1 - 6 = -8

Step 2: Express vector d

Substitute values into $d$ $\vec{d}$:

d = 6 b - (-8) a = 6 b + 8 a

Step 3: Evaluate Final Expression

Now compute $( a - b) \cdot d$ $(\vec{a} - \vec{b}) \cdot \vec{d}$:

( a - b) \cdot (8 a + 6 b)

= 8( a \cdot a) + 6( a \cdot b) - 8( b \cdot a) - 6( b \cdot b)

= 8a² - 2( a \cdot b) - 6b²

Substitute the known values:

= 8(6) - 2(-8) - 6(11) = 48 + 16 - 66 = 64 - 66 = -2

Pattern Recognition

Instead of computing cross products sequentially (which is tedious and error-prone), immediately expand nested cross products using the standard vector triple product identity $BAC-CAB$ $BAC-CAB$.

Chapter Mix

Class 12 Maths: Vector Algebra

Q3 jee_main_2026_23_january_evening Cross Product and Projection

Let

a= i-2 j+3 k, b=2 i+ j- k, c=λ i+ j+ k

and

v= a\times b

. If

v\cdot c=11

and the length of the projection of

b

c

p

, then

9p²

is equal to:

A. $9$ $9$
B. $6$ $6$
C. $4$ $4$
D. $12$ $12$

Solution

Related Formula

Length of projection of b on c = | b\cdot c|| c|

Core Logic

First, find $v = a \times b$ $\vec{v} = \vec{a} \times \vec{b}$.

a = i - 2 j + 3 k

b = 2 i + j - k

v = vmatrix i & j & k 1 & -2 & 3 2 & 1 & -1 vmatrix = i(2-3) - j(-1-6) + k(1+4) = - i + 7 j + 5 k

Given $v\cdot c = 11$ $\vec{v}\cdot\vec{c} = 11$ and $c = λ i + j + k$ :

(- i + 7 j + 5 k) \cdot (λ i + j + k) = 11

-λ + 7 + 5 = 11 λ = 1

Step 1: Projection Calculation

Now, $c = i + j + k$ $\vec{c} = \hat{i} + \hat{j} + \hat{k}$. The length of the projection of $b$ $\vec{b}$ on $c$ $\vec{c}$ is $p$ $p$:

p = | (2 i + j - k) \cdot ( i + j + k)\sqrt(1² + 1² + 1²) | = 2(1) + 1(1) - 1(1)\sqrt(3) = 2\sqrt(3)

Calculate $9p²$ $9p^2$:

9p² = 9((4)/(3)) = 12

Pattern Recognition

Standard sequence: compute cross product to find normal vector $v$ $\vec{v}$, take dot product to deduce missing parameter $λ$ $\lambda$, then substitute into scalar projection formula.

Chapter Mix

Class 12 Maths: Vector Algebra

Q10 jee_main_2026_23_january_evening Vector Product

Let

a, b, c

be three vectors such that

a\times b=2( a\times c)

. If

| a|=1, | b|=4, | c|=2

, and the angle between

b

and

c

60°

, then

| a\cdot c|

is:

A. $2$ $2$
B. $4$ $4$
C. $0$ $0$
D. $1$ $1$

Solution

Related Formula

u \times v = 0 u and v are parallel ($ u = λ v $)

| x + y|² = | x|² + | y|² + 2 x\cdot y

Core Logic

a \times b - 2( a \times c) = 0

a \times ( b - 2 c) = 0

This implies that $( b - 2 c)$ $(\vec{b} - 2\vec{c})$ is collinear with $a$ $\vec{a}$. So, $b - 2 c = λ a$ $\vec{b} - 2\vec{c} = \lambda \vec{a}$ for some scalar $λ$ $\lambda$.

Step 1: Finding Lambda

Square both sides of the relation:

|λ a|² = | b - 2 c|²

λ² | a|² = | b|² + 4| c|² - 4( b\cdot c)

Given $| a| = 1, | b| = 4, | c| = 2$ and the angle between $b$ $\vec{b}$ and $c$ $\vec{c}$ is $60^°$ $60^\circ$:

b\cdot c = | b|| c| 60^° = (4)(2)((1)/(2)) = 4

Substitute the values:

λ²(1)² = 4² + 4(2)² - 4(4)

λ² = 16 + 16 - 16 = 16

$λ = \pm 4$ $\lambda = \pm 4$

Step 2: Calculating Dot Product

We have $b - 2 c = \pm 4 a$ $\vec{b} - 2\vec{c} = \pm 4\vec{a}$. Take the dot product with $c$ $\vec{c}$ on both sides:

( b - 2 c) \cdot c = (\pm 4 a) \cdot c

b\cdot c - 2| c|² = \pm 4( a\cdot c)

4 - 2(2)² = \pm 4( a\cdot c)

4 - 8 = \pm 4( a\cdot c)

-4 = \pm 4( a\cdot c)

| a\cdot c| = 1

Pattern Recognition

A cross-product equation structured as $a\times( X)=0$ $\vec{a}\times(\vec{X})=0$ immediately gives $X = λ a$ $\vec{X} = \lambda\vec{a}$. Expanding the magnitude squared is the standard method to expose the dot products and solve for $λ$ $\lambda$.

Chapter Mix

Class 12 Maths: Vector Algebra

Q5 jee_main_2026_24_january_morning Cross and Dot Products

Let

a = 2 i + j - 2 k

b = i + j

and

c = a \times b

. Let

d

be a vector such that

| d - a| = \sqrt(11)

| c \times d| = 3

and the angle between

c

and

d

(π)/(4)

. Then

a \cdot d

is equal to

A. $11$ $11$
B. $3$ $3$
C. $0$ $0$
D. $1$ $1$

Solution

Related Formula

| x \times y| = | x| | y| θ

| x - y|² = | x|² + | y|² - 2( x \cdot y)

Step 1: Finding Vector c

c = a \times b = vmatrix i & j & k 2 & 1 & -2 1 & 1 & 0 vmatrix

= i(0 - (-2)) - j(0 - (-2)) + k(2 - 1) = 2 i - 2 j + k

However, evaluating carefully from the pdf steps, wait, $1 \times -2$ $1 \times -2$ is $-2$ $-2$, so $- j(0 - (-2))$ $-\hat{j}(0 - (-2))$ is $-2 j$ $-2\hat{j}$. The PDF indicates $c = 2 i - 2 j + k$ (magnitude is $3$ $3$). Let's use $| c| = \sqrt(4 + 4 + 1) = 3$ $|\vec{c}| = \sqrt{4 + 4 + 1} = 3$.

Step 2: Finding magnitude of d

| c \times d| = 3

| c| | d| (π)/(4) = 3

3 | d| 1\sqrt(2) = 3 \Rightarrow | d| = \sqrt(2)

Step 3: Calculating dot product

Given $| d - a| = \sqrt(11)$ $|\vec{d} - \vec{a}| = \sqrt{11}$. Squaring both sides:

| d|² + | a|² - 2( a \cdot d) = 11

We have $| a|² = 2² + 1² + (-2)² = 9$ $|\vec{a}|^2 = 2^2 + 1^2 + (-2)^2 = 9$.

(\sqrt(2))² + 9 - 2( a \cdot d) = 11

2 + 9 - 2( a \cdot d) = 11

11 - 2( a \cdot d) = 11 \Rightarrow 2( a \cdot d) = 0

a \cdot d = 0

Pattern Recognition

Isolate the unknown dot product by exploiting vector magnitude identities (difference square expansion). The cross-product magnitude uniquely isolates the remaining magnitude variable.

Chapter Mix

Class 12 Maths: Vector Algebra

Solution & Explanation

Related Formula

Core Logic

Step 1: Obtuse Angle Condition

Step 2: Combine Intervals

Pattern Recognition

Chapter Mix

Reference Study Guides

More Vector Algebra Previous-Year Questions — Page 2

Solution

Related Formula

Core Logic

Step 1: Calculate Magnitudes and Dot Products

Step 2: Express vector d

Step 3: Evaluate Final Expression

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Projection Calculation

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Lambda

Step 2: Calculating Dot Product

Pattern Recognition

Chapter Mix

Solution

Related Formula

Step 1: Finding Vector c

Step 2: Finding magnitude of d

Step 3: Calculating dot product

Pattern Recognition

Chapter Mix

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