Let a and b be two vectors such that |b| = 1 and |b x a| = 2 . Then |(b x a)

Year	2026	2025	2024	Total
Questions	15	17	14	46

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

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

( b \times a) \cdot b = 0 (Scalar triple product with repeated vectors is zero)

Core Logic

Given $| b| = 1$ $|\vec{b}| = 1$ and $| b \times a| = 2$ $|\vec{b} \times \vec{a}| = 2$. Expand the requested expression:

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

Step 1: Simplify using Vector Properties

The cross product $( b \times a)$ $(\vec{b} \times \vec{a})$ produces a vector orthogonal to both $b$ $\vec{b}$ and $a$ $\vec{a}$. Therefore, $( b \times a) \cdot b = 0$ .

Substituting this back:

|( b \times a) - b|² = | b \times a|² + | b|² - 0

|( b \times a) - b|² = (2)² + (1)²

|( b \times a) - b|² = 4 + 1 = 5

Pattern Recognition

A cross product $u \times v$ $\vec{u} \times \vec{v}$ is inherently perpendicular to $u$ $\vec{u}$. Any length squared involving $( u \times v) \pm u$ $(\vec{u} \times \vec{v}) \pm \vec{u}$ resolves simply via Pythagorean sum.

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 $a$ $\vec{a}$ and $b$ $\vec{b}$ be two vectors such that $| b| = 1$ $|\vec{b}| = 1$ and $| b \times a| = 2$ $|\vec{b} \times \vec{a}| = 2$ . Then $|( b \times a) - b|²$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify using Vector Properties

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_2024_30_january_evening

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Let a$\vec{a}$ and b$\vec{b}$ be two vectors such that | b| = 1$|\vec{b}| = 1$ and | b × a| = 2$|\vec{b} \times \vec{a}| = 2$ . Then |( b × a) - b|²$\left|\left(\vec{b} \times \vec{a}\right) - \vec{b}\right|^2$ is equal to

Solution & Explanation

Related Formula

Core Logic

Step 1: Simplify using Vector Properties

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_2024_30_january_evening

Select Your Target Exam

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Let $a$ $\vec{a}$ and $b$ $\vec{b}$ be two vectors such that $| b| = 1$ $|\vec{b}| = 1$ and $| b \times a| = 2$ $|\vec{b} \times \vec{a}| = 2$ . Then $|( b \times a) - b|²$ is equal to