Let a = -i + j + 2k, b = i - j - 3k, c = a x b and d = c x a. Then (a

Year	2026	2025	2024	Total
Questions	15	17	14	46

Year

2026

2025

2024

Total

Questions

Solution & Explanation

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

More Vector Algebra Previous-Year Questions — Page 10

Q28 jee_main_2024_31_jan_morning Vector Triple Product

Let

a

and

b

be two vectors such that

| a| = 1, | b| = 4

and

a \cdot b = 2

. If

c = (2 a \times b) - 3 b

and the angle between

b

and

c

α

, then

192 ²α

is equal to

Numerical Answer. Answer: 48 to 48

Solution

Core Logic

b \cdot c = b \cdot ((2 a \times b) - 3 b)

|b||c| α = 2( b \cdot ( a \times b)) - 3|b|²

Since $b \cdot ( a \times b) = 0$ , we have $|b||c| α = -3|b|²$ $|b||c|\cos\alpha = -3|b|^2$.

|c| α = -3|b| = -12 |c|² ² α = 144

Step 1: Compute Modulus of c

|c|² = |2 a \times b - 3 b|²

= 4| a \times b|² + 9| b|² - 12(( a \times b) \cdot b)

= 4| a \times b|² + 9| b|²

Given $a \cdot b = 2 |a||b| θ = 2 1 \cdot 4 θ = 2 θ = (π)/(3)$ .

| a \times b|² = |a|²|b|² ²θ = 1 \cdot 16 \cdot (3)/(4) = 12

|c|² = 4(12) + 9(16) = 48 + 144 = 192

Step 2: Final Calculation

We know $|c|² ² α = 144$ $|c|^2 \cos^2 \alpha = 144$.

192 ² α = 144

192(1 - ² α) = 144

192 ² α = 192 - 144 = 48

Chapter Mix

Class 12 Maths: Vector Algebra

Let $a = - i + j + 2 k$ , $b = i - j - 3 k$ , $c = a \times b$ $\vec{c} = \vec{a} \times \vec{b}$ and $d = c \times a$ $\vec{d} = \vec{c} \times \vec{a}$. Then $( a - b) \cdot d$ $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to:

Solution & Explanation

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

Reference Study Guides

More Vector Algebra Previous-Year Questions — Page 10

Solution

Core Logic

Step 1: Compute Modulus of c

Step 2: Final Calculation

Chapter Mix

More Vector Algebra Questions — jee_main_2026_23_january_morning

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Let a = - i + j + 2 k$\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, b = i - j - 3 k$\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, c = a × b$\vec{c} = \vec{a} \times \vec{b}$ and d = c × a$\vec{d} = \vec{c} \times \vec{a}$. Then ( a - b) · d$(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to:

Solution & Explanation

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

Reference Study Guides

More Vector Algebra Previous-Year Questions — Page 10

Solution

Core Logic

Step 1: Compute Modulus of c

Step 2: Final Calculation

Chapter Mix

More Vector Algebra Questions — jee_main_2026_23_january_morning

Select Your Target Exam

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Let $a = - i + j + 2 k$ , $b = i - j - 3 k$ , $c = a \times b$ $\vec{c} = \vec{a} \times \vec{b}$ and $d = c \times a$ $\vec{d} = \vec{c} \times \vec{a}$. Then $( a - b) \cdot d$ $(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to: