Let O be the origin and the position vector of A and B be 2i+2j+k and 2i+4j+4k respectively. If the internal bisector…

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Q53 jee_main_2025_28_jan_evening Components of Vectors

If the components of

a=α i+β j+γ k

along and perpendicular to

b=3 i+ j- k

respectively, are

(16)/(11)(3 i+ j- k)

and

(1)/(11)(-4 i-5 j-17 k)

, then

α²+β²+γ²

is equal to:

A. $23$ $23$
B. $18$ $18$
C. $16$ $16$
D. $26$ $26$

Solution

Related Formula

Any vector $a$ $\vec{a}$ can be written as the sum of its component parallel to $b$ $\vec{b}$ (along $b$ $\vec{b}$) and its component perpendicular to $b$ $\vec{b}$:

a = a∥ + a⊥

Magnitude squared:

| a|² = α² + β² + γ²

Core Logic

Given:

a∥ = (16)/(11)(3 i+ j- k)

a⊥ = (1)/(11)(-4 i-5 j-17 k)

Step 1: Reconstruct Vector a

Add both components to find $a$ $\vec{a}$:

a = (16)/(11)(3 i+ j- k) + (1)/(11)(-4 i-5 j-17 k)

a = (1)/(11) [ (48 - 4) i + (16 - 5) j + (-16 - 17) k ]

a = (1)/(11) [ 44 i + 11 j - 33 k ] = 4 i + j - 3 k

Therefore, $α = 4$ $\alpha = 4$, $β = 1$ $\beta = 1$, $γ = -3$ $\gamma = -3$.

Step 2: Calculate Sum of Squares

α² + β² + γ² = 4² + 1² + (-3)² = 16 + 1 + 9 = 26

Pattern Recognition

Since parallel and perpendicular components are orthogonal vectors, you can also use directly $| a|² = | a∥|² + | a⊥|²$ to save algebra step: $| a∥|² = (16²)/(11²)(9+1+1) = (256)/(11)$ , $| a⊥|² = (1)/(11²)(16+25+289) = (330)/(121) = (30)/(11)$ . Total = $(286)/(11) = 26$ $\frac{286}{11} = 26$.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q55 jee_main_2025_29_jan_morning Vector Cross Product and Dot Product

Let

a = 2 i - j + 3 k

,

b = 3 i - 5 j + k

and

c

be a vector such that

a \times c = c \times b

and

( a + c) . ( b + c) = 168.

Then the maximum value of

| c|²

is:

A. 77
B. 462
C. 308
D. 154

Solution

Related Formula

u \times v = - v \times u

If u \times v = 0 u ∥ v u = λ v

Core Logic

Given $a \times c = c \times b a \times c + b \times c = 0$

( a + b) \times c = 0 c = λ( a + b)

Step 1: Compute a + b

a + b = (2+3) i + (-1-5) j + (3+1) k = 5 i - 6 j + 4 k

c = λ(5 i - 6 j + 4 k)

| c|² = λ²(25 + 36 + 16) = 77λ²

Step 2: Expand the Dot Product Condition

( a + c) \cdot ( b + c) = 168

a \cdot b + c \cdot ( a + b) + | c|² = 168

Evaluate $a \cdot b = (2)(3) + (-1)(-5) + (3)(1) = 6 + 5 + 3 = 14$ . Substitute $c \cdot ( a + b) = λ | a + b|² = 77λ$ :

14 + 77λ + 77λ² = 168 77λ² + 77λ - 154 = 0

λ² + λ - 2 = 0 λ = 1 or λ = -2

Step 3: Maximize $| c|²$ $|\vec{c}|^2$

Maximum value occurs when $λ = -2$ $\lambda = -2$:

| c|² = 77(-2)² = 77 \times 4 = 308

Pattern Recognition

Recognize the cross-product rule inversion immediately: $x \times y = y \times z ( x+ z) ∥ y$ . This linear reduction circumvents solving complex linear systems.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q8 jee_main_2024_01_february_morning Vector Triple Product

Let

a=-5 i+ j-3 k

,

b= i+2 j-4 k

and

c=((( a\times b)\times i)\times i)\times i.

Then

c\cdot(- i+ j+ k)

is equal to:

A. $-12$ $-12$
B. $-10$ $-10$
C. $-13$ $-13$
D. $-15$ $-15$

Solution

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

Q jee_main_2024_29_january_evening Area of Quadrilateral and Parallelogram

Let

OA = a, OB = 12 a + 4 b

and

OC = b

, where

O

is the origin. If

S

is the parallelogram with adjacent sides

OA

and

OC

, then the ratio of the area of the quadrilateral OABC to the area of

S

is equal to

A. 6
B. 10
C. 7
D. 8

Solution

Related Formula

Area of Parallelogram S = | a \times b|

Area of Quadrilateral OABC = Area(Δ OAB) + Area(Δ OBC)

Core Logic

Let us compute the component areas using vector cross products:

Area(Δ OAB) = (1)/(2) | OA \times OB| = (1)/(2) | a \times (12 a + 4 b)| = (1)/(2) |4( a \times b)| = 2| a \times b|

Area(Δ OBC) = (1)/(2) | OB \times OC| = (1)/(2) |(12 a + 4 b) \times b| = (1)/(2) |12( a \times b)| = 6| a \times b|

Step 1: Finding Total Ratio

Adding both triangles to find the total area of the quadrilateral OABC:

Area(OABC) = 2| a \times b| + 6| a \times b| = 8| a \times b|

Area of Quadrilateral and Parallelogram diagram for Q8 - JEE Main 2024 Evening

Dividing this total by the area of the baseline parallelogram $S = | a \times b|$ $S = |\vec{a} \times \vec{b}|$:

Ratio = 8| a \times b|| a \times b| = 8

Pattern Recognition

Since $a \times a = 0$ $\vec{a} \times \vec{a} = 0$ and $b \times b = 0$ $\vec{b} \times \vec{b} = 0$, cross product distributions yield terms with purely $a \times b$ $\vec{a} \times \vec{b}$, ensuring absolute scalability independent of vectors chosen.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Q jee_main_2024_29_january_evening Vector Operations and Angles

Let a unit vector

u = x i + y j + z k

make angles

(π)/(2)

,

(π)/(3)

, and

(2π)/(3)

with the vectors

1\sqrt(2) i + 1\sqrt(2) k

,

1\sqrt(2) j + 1\sqrt(2) k

, and

1\sqrt(2) i + 1\sqrt(2) j

respectively. If

v = 1\sqrt(2) i + 1\sqrt(2) j + 1\sqrt(2) k

, then

| u - v|²

is equal to

A. $(11)/(2)$ $\frac{11}{2}$
B. $(5)/(2)$ $\frac{5}{2}$
C. 9
D. 7

Solution

Related Formula

A \cdot B = | A| | B| φ

Core Logic

Let the given baseline vectors be $p₁, p₂, p₃$ $\vec{p}_1, \vec{p}_2, \vec{p}_3$. Note that $| p₁| = | p₂| = | p₃| = 1$ .

Angle with $p₁$ $\vec{p}_1$ is $(π)/(2)$ $\frac{\pi}{2}$:

u \cdot p₁ = 0 x\sqrt(2) + z\sqrt(2) = 0 x + z = 0 (i)

Angle with $p₂$ $\vec{p}_2$ is $(π)/(3)$ $\frac{\pi}{3}$:

u \cdot p₂ = (π)/(3) y\sqrt(2) + z\sqrt(2) = (1)/(2) y + z = 1\sqrt(2) (ii)

Angle with $p₃$ $\vec{p}_3$ is $(2π)/(3)$ $\frac{2\pi}{3}$:

u \cdot p₃ = (2π)/(3) x\sqrt(2) + y\sqrt(2) = -(1)/(2) x + y = - 1\sqrt(2) (iii)

Step 1: Finding Vector Coordinates

Subtracting (ii) from (iii):

(x + y) - (y + z) = - 1\sqrt(2) - 1\sqrt(2) x - z = -\sqrt(2) (iv)

Solving (i) and (iv):

2x = -\sqrt(2) x = - 1\sqrt(2)

z = 1\sqrt(2)

From (ii): $y = 1\sqrt(2) - 1\sqrt(2) = 0$ . So, $u = - 1\sqrt(2) i + 0 j + 1\sqrt(2) k$ .

Step 2: Evaluating the Norm Difference

Given $v = 1\sqrt(2) i + 1\sqrt(2) j + 1\sqrt(2) k$ :

u - v = (- 1\sqrt(2) - 1\sqrt(2)) i + (0 - 1\sqrt(2)) j + ( 1\sqrt(2) - 1\sqrt(2)) k

u - v = -\sqrt(2) i - 1\sqrt(2) j + 0 k

Evaluating the squared magnitude:

| u - v|² = (-\sqrt(2))² + (- 1\sqrt(2))² = 2 + (1)/(2) = (5)/(2)

Pattern Recognition

Setting up dot products systematically transforms descriptive geometric angles into solvable sets of linear equations.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Let O be the origin and the position vector of A and B be 2 i+2 j+ k$2\hat{i}+2\hat{j}+\hat{k}$ and 2 i+4 j+4 k$2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of ∠ AOB$\angle AOB$ meets the line AB at C, then the length of OC is

Solution & Explanation

Related Formula

Core Logic

Step 1: Apply Section Formula

Step 2: Compute Length of OC

Pattern Recognition

Chapter Mix

Reference Study Guides

More Vector Algebra Previous-Year Questions — Page 7

Solution

Related Formula

Core Logic

Step 1: Reconstruct Vector a

Step 2: Calculate Sum of Squares

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Compute a + b

Step 2: Expand the Dot Product Condition

Step 3: Maximize | c|²$|\vec{c}|^2$

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Substitute Vector Coordinates

Step 2: Calculate Dot Product

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Total Ratio

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Vector Coordinates

Step 2: Evaluating the Norm Difference

Pattern Recognition

Chapter Mix

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Let O be the origin and the position vector of A and B be $2 i+2 j+ k$ $2\hat{i}+2\hat{j}+\hat{k}$ and $2 i+4 j+4 k$ $2\hat{i}+4\hat{j}+4\hat{k}$ respectively. If the internal bisector of $∠ AOB$ $\angle AOB$ meets the line AB at C, then the length of OC is

Step 3: Maximize $| c|²$ $|\vec{c}|^2$