Let a=-5i+j-3k, b=i+2j-4k and c=(((axb)xi)xi)xi. Then ccdot(-i+j+k) is equal to:

Year	2026	2025	2024	Total
Questions	15	17	14	46

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Related Formula

Vector Triple Product Identity:

( u \times v) \times w = ( u \cdot w) v - ( v \cdot w) a

Core Logic

Given vectors $a$ $\vec{a}$ and $b$ $\vec{b}$: First, let's find the inner vector triple product $( a \times b) \times i$ :

( a \times b) \times i = ( a \cdot i) b - ( b \cdot i) a

From the given components:

$a \cdot i = -5$ $\vec{a} \cdot \hat{i} = -5$
$b \cdot i = 1$ $\vec{b} \cdot \hat{i} = 1$

( a \times b) \times i = -5 b - a

Step 1: Substitute Vector Coordinates

Substitute coordinates of $a$ $\vec{a}$ and $b$ $\vec{b}$ into the expression:

-5 b - a = -5( i + 2 j - 4 k) - (-5 i + j - 3 k)

= -5 i - 10 j + 20 k + 5 i - j + 3 k = -11 j + 23 k

Now, perform successive cross products with $i$ $\hat{i}$:

Next Step = (-11 j + 23 k) \times i = -11( j \times i) + 23( k \times i) = 11 k + 23 j

Final Vector c = (23 j + 11 k) \times i = 23( j \times i) + 11( k \times i) = -23 k + 11 j

Step 2: Calculate Dot Product

Now evaluate $c \cdot (- i + j + k)$ :

c = 11 j - 23 k

c \cdot (- i + j + k) = (0)( -1) + (11)(1) + (-23)(1) = 11 - 23 = -12

Pattern Recognition

Sees: Repeated vector cross products with standard unit vectors. Shortcut: Observe that crossing a vector in the $y-z$ $y-z$ plane with $i$ $\hat{i}$ rotates its components by $90°$ $90^{\circ}$ in that plane. Doing it twice returns it to the original plane with flipped coefficients and signs ( $v \times i \times i = - v$ ).

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 $a=-5 i+ j-3 k$ $\vec{a}=-5\hat{i}+\hat{j}-3\hat{k}$, $b= i+2 j-4 k$ $\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$ and $c=((( a\times b)\times i)\times i)\times i.$ Then $c\cdot(- i+ j+ k)$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Substitute Vector Coordinates

Step 2: Calculate Dot Product

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_01_february_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Let a=-5 i+ j-3 k$\vec{a}=-5\hat{i}+\hat{j}-3\hat{k}$, b= i+2 j-4 k$\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$ and c=((( a× b)× i)× i)× i.$\vec{c}=(((\vec{a}\times\vec{b})\times\hat{i})\times\hat{i})\times\hat{i}.$ Then c·(- i+ j+ k)$\vec{c}\cdot(-\hat{i}+\hat{j}+\hat{k})$ is equal to:

Solution & Explanation

Related Formula

Core Logic

Step 1: Substitute Vector Coordinates

Step 2: Calculate Dot Product

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_01_february_morning

Select Your Target Exam

Full Mock Test Hub

Topic-wise Practice Hub

Exam Analysis Tools

Chapter Weightage & Trends

Topic-wise Practice Hub

JEE Main Exam Papers

Chapter Weightage Board

Formula Recognition

Exam Questions

Select an Exam

Welcome back

Let $a=-5 i+ j-3 k$ $\vec{a}=-5\hat{i}+\hat{j}-3\hat{k}$, $b= i+2 j-4 k$ $\vec{b}=\hat{i}+2\hat{j}-4\hat{k}$ and $c=((( a\times b)\times i)\times i)\times i.$ Then $c\cdot(- i+ j+ k)$ is equal to: