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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

p = K( a + λ b)

p \cdot a = 0

Core Logic

Formulate a coplanar parameterization vector, apply the zero dot-product geometric orthogonality constraint to pin down the linear parameter, and then normalize.

Step 1: Define Coplanar Structural Form

Let the targeting vector path be:

p = K( a + λ b) = K( (1+2λ) i + (2+λ) j + (1-λ) k )

Step 2: Force Orthogonality Constraint

Impose $p \cdot a = 0$ $\vec{p} \cdot \vec{a} = 0$:

1(1+2λ) + 2(2+λ) + 1(1-λ) = 0

1 + 2λ + 4 + 2λ + 1 - λ = 0 6 + 3λ = 0 λ = -2

Step 3: Substitute and Normalize

Substitute $λ = -2$ $\lambda = -2$ back into the base formulation:

p = K(-3 i + 3 k)

Normalizing to turn this vector into a proper unit scale form:

c = \pm - i + k\sqrt(2)

Pattern Recognition

Finding coplanar vectors orthogonal to one base component matches taking cross expansions like $( a \times b) \times a$ up to scalar metrics.

Chapter Mix

Class 12 Mathematics: 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 $a = i + 2 j + k$ 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:

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 10

Solution

Core Logic

Step 1: Compute Modulus of c

Step 2: Final Calculation

Chapter Mix

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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 10

Solution

Core Logic

Step 1: Compute Modulus of c

Step 2: Final Calculation

Chapter Mix

More Vector Algebra Questions — jee_main_2025_08_april_evening

Select Your Target Exam

Full Mock Test Hub

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Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

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Formula Recognition

Exam Questions

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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: