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Vector Algebra appeared 46 times across 3 years — 5.3% of Mathematics. This question is from Projection Of A Vector On Another Vector.

Year 2026 2025 2024 Total
Questions 15 17 14 46

Let AB=2 i+4 j-5 k and AD= i+2 j+λ k, λin R. Let the projection of the vector v= i+ j+ k on the diagonal AC of the parallelogram ABCD be of length one unit. If α,β, where α>β, be the roots of the equation λ²x²-6λ x+5=0, then 2α-β is equal to

Solution & Explanation

Related Formula
Projection of v on a = v · a| a| AC = AB + AD
Core Logic

Using the parallelogram law of vector addition, the diagonal AC is given by:

AC = AB + AD = (2 i+4 j-5 k) + ( i+2 j+λ k) AC = 3 i + 6 j + (λ - 5) k
Step 1: Applying the Projection Condition

The projection of v = i+ j+ k on AC is 1. Therefore:

v · AC| AC| = 1 3(1) + 6(1) + (λ - 5)(1)√(3² + 6² + (λ - 5)²) = 1 3 + 6 + λ - 5 = √(9 + 36 + (λ - 5)²) λ + 4 = √(45 + (λ - 5)²)

Squaring both sides:

(λ + 4)² = 45 + (λ - 5)² λ² + 8λ + 16 = λ² - 10λ + 25 + 45 18λ = 54 λ = 3
Step 2: Solving the Quadratic Equation

Substitute λ = 3 into the given quadratic equation λ²x² - 6λ x + 5 = 0:

9x² - 18x + 5 = 0 (3x - 1)(3x - 5) = 0 x = (1)/(3), (5)/(3)

Given α > β, we have α = (5)/(3) and β = (1)/(3).

Calculate 2α - β:

2((5)/(3)) - (1)/(3) = (10 - 1)/(3) = 3

Projection of A Vector diagram for Q1 - JEE Main 2026 Morning
Projection of A Vector diagram for Q1 - JEE Main 2026 Morning

Pattern Recognition

Projection questions in 3D geometry often couple a simple dot product identity with an unknown parameter. Squaring a linear equation derived from vector magnitudes often cancels the squared terms, making it easily solvable.

Chapter Mix

Class 12 Maths: Vector Algebra Class 11 Maths: Quadratic Equations

Reference Study Guides

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 · b = 2. If c = (2 a × b) - 3 b and the angle between b and c is α, then 192 ²α is equal to
Numerical Answer. Answer: 48 to 48

Solution

Core Logic
b · c = b · ((2 a × b) - 3 b) |b||c| α = 2( b · ( a × b)) - 3|b|²

Since b · ( a × b) = 0, we have |b||c| α = -3|b|².

|c| α = -3|b| = -12 |c|² ² α = 144
Step 1: Compute Modulus of c
|c|² = |2 a × b - 3 b|² = 4| a × b|² + 9| b|² - 12(( a × b) · b) = 4| a × b|² + 9| b|²

Given a · b = 2 |a||b| θ = 2 1 · 4 θ = 2 θ = (π)/(3).

| a × b|² = |a|²|b|² ²θ = 1 · 16 · (3)/(4) = 12 |c|² = 4(12) + 9(16) = 48 + 144 = 192
Step 2: Final Calculation

We know |c|² ² α = 144.

192 ² α = 144 192(1 - ² α) = 144 192 ² α = 192 - 144 = 48
Chapter Mix

Class 12 Maths: Vector Algebra

More Vector Algebra Questions — jee_main_2026_22_january_morning

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