Let P be a point in the plane of the vector AB = 3i + j - k and AC = i - j + 3k such that P is equidistant from the lines AB and AC. If |AP| = sqrt(5)2, then the area of the triangle ABP is:

Since P is equidistant from AB and AC, AP must lie on the angle bisector of BAC. Let the angle between vectors AB and AC be 2theta. We need theta to find the perpendicular distance from P to AB (which acts as the height of Delta ABP). Using half-angle formula 1 - 22theta = 2theta: The area of triangle ABP is (1/2) x base x height. Base = |AB| = sqrt(11). Height = |AP| = sqrt(5)2 sqrt((6/11)). The phrase "equidistant from two lines" defines the angle bisector. Instantly calculate half-angle trigonometric ratios from the standard dot-product full angle to unlock perpendicular triangle heights. Class 12 Maths: Vector Algebra

Let P be a point in the plane of the vector AB = 3i + j - k and AC = i - j + 3k such that P is equidistant from the lines AB and AC. If |AP| = sqrt(5)2, then the area of the triangle ABP is:

Since P is equidistant from AB and AC, AP must lie on the angle bisector of BAC. Let the angle between vectors AB and AC be 2theta. We need theta to find the perpendicular distance from P to AB (which acts as the height of Delta ABP). Using half-angle formula 1 - 22theta = 2theta: The area of triangle ABP is (1/2) x base x height. Base = |AB| = sqrt(11). Height = |AP| = sqrt(5)2 sqrt((6/11)). The phrase "equidistant from two lines" defines the angle bisector. Instantly calculate half-angle trigo

Let P be a point in the plane of the vector AB = 3i + j - k and AC = i - j + 3k such that P is equidistant from the lines AB and AC. If |AP| = sqrt(5)2, then the area of the triangle ABP is:

Since P is equidistant from AB and AC, AP must lie on the angle bisector of BAC. Let the angle between vectors AB and AC be 2theta. We need theta to find the perpendicular distance from P to AB (which acts as the height of Delta ABP). Using half-angle formula 1 - 22theta = 2theta: The area of triangle ABP is (1/2) x base x height. Base = |AB| = sqrt(11). Height = |AP| = sqrt(5)2 sqrt((6/11)). The phrase "equidistant from two lines" defines the angle bisector. Instantly calculate half-angle trigonometric ratios from the standard dot-product full angle to unlock perpendicular triangle heights. Class 12 Maths: Vector Algebra

Let P be a point in the plane of the vector AB = 3i + j - k and AC = i - j + 3k such that P is equidistant from the lines AB and AC. If |AP| = sqrt(5)2, then the area of the triangle ABP is:

Since P is equidistant from AB and AC, AP must lie on the angle bisector of BAC. Let the angle between vectors AB and AC be 2theta. We need theta to find the perpendicular distance from P to AB (which acts as the height of Delta ABP). Using half-angle formula 1 - 22theta = 2theta: The area of triangle ABP is (1/2) x base x height. Base = |AB| = sqrt(11). Height = |AP| = sqrt(5)2 sqrt((6/11)). The phrase "equidistant from two lines" defines the angle bisector. Instantly calculate half-angle trigo

Let P be a point in the plane of the vector AB = 3i + j - k and AC = i - j + 3k such that P is equidistant from the…

Year	2026	2025	2024	Total
Questions	15	17	14	46

Q8 jee_main_2024_30_january_evening Cross Product

Let

a

and

b

be two vectors such that

| b| = 1

and

| b \times a| = 2

. Then

|( b \times a) - b|²

is equal to

A. $3$ $3$
B. $5$ $5$
C. $1$ $1$
D. $4$ $4$

Solution

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

Q4 jee_main_2024_30_jan_morning Vector Cross Product

Let

a = a₁ i + a₂ j + a₃ k

and

b = b₁ i + b₂ j + b₃ k

be two vectors such that

| a| = 1

;

a\cdot b = 2

and

| b| = 4

. If

c = 2( a \times b) - 3 b

, then the angle between

b

and

c

is equal to:

A. $⁻¹( 2\sqrt(3))$
B. $⁻¹(- 1\sqrt(3))$
C. $⁻¹(- \sqrt(3)2)$
D. $⁻¹((2)/(3))$ $\cos^{-1}\left(\frac{2}{3}\right)$

Solution

Related Formula

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

θ = b \cdot c| b| | c|

Core Logic

Given $| a| = 1, | b| = 4, a \cdot b = 2$ . Let's evaluate $| a \times b|²$ $|\vec{a} \times \vec{b}|^2$ using Lagrange's identity:

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

| a \times b|² = (1)(16) - (2)² = 16 - 4 = 12

Step 1: Dot products with c

We are given:

c = 2( a \times b) - 3 b

To find the angle between $b$ $\vec{b}$ and $c$ $\vec{c}$, we need $b \cdot c$ $\vec{b} \cdot \vec{c}$ and $| c|$ $|\vec{c}|$. Taking the dot product with $b$ $\vec{b}$ on both sides:

b \cdot c = 2( b \cdot ( a \times b)) - 3( b \cdot b)

Since $b \cdot ( a \times b) = 0$ (scalar triple product with repeated vector):

b \cdot c = 0 - 3| b|² = -3(16) = -48 (1)

Step 2: Finding magnitude of c

Now, let's find $| c|²$ $|\vec{c}|^2$:

| c|² = c \cdot c = (2( a \times b) - 3 b) \cdot (2( a \times b) - 3 b)

Since $( a \times b) \cdot b = 0$ , the cross terms vanish:

| c|² = 4| a \times b|² + 9| b|²

| c|² = 4(12) + 9(16) = 48 + 144 = 192

| c| = \sqrt(192) = 8\sqrt(3)

Step 3: Calculating angle

Now apply the angle formula:

θ = b \cdot c| b| | c| = -48(4)(8\sqrt(3))

θ = -4832\sqrt(3) = -32\sqrt(3) = - \sqrt(3)2

θ = ⁻¹(- \sqrt(3)2)

Pattern Recognition

Always remember that the cross product vector $( a \times b)$ $(\vec{a} \times \vec{b})$ is orthogonal to both $a$ $\vec{a}$ and $b$ $\vec{b}$. This immediately eliminates cross terms when finding the magnitude of linear combinations like $x( a \times b) + y b$ $x(\vec{a} \times \vec{b}) + y\vec{b}$.

Chapter Mix

Class 12 Maths: Vector Algebra

Q18 jee_main_2024_30_jan_morning Cross Product

Let

A (2, 3, 5)

and

C(-3, 4, -2)

be opposite vertices of a parallelogram ABCD if the diagonal

BD = i + 2 j + 3 k

then the area of the parallelogram is equal to

A. $(1)/(2)\sqrt(410)$ $\frac{1}{2}\sqrt{410}$
B. $(1)/(2)\sqrt(474)$ $\frac{1}{2}\sqrt{474}$
C. $(1)/(2)\sqrt(586)$ $\frac{1}{2}\sqrt{586}$
D. $(1)/(2)\sqrt(306)$ $\frac{1}{2}\sqrt{306}$

Solution

Related Formula

Area of parallelogram = (1)/(2) | d₁ \times d₂|

Core Logic

The diagonals of the parallelogram are $AC$ $\overrightarrow{AC}$ and $BD$ $\overrightarrow{BD}$. First, calculate the diagonal vector $AC$ $\overrightarrow{AC}$:

AC = Position vector of C - Position vector of A

AC = (-3 - 2) i + (4 - 3) j + (-2 - 5) k = -5 i + 1 j - 7 k

Alternatively, taking $CA = 5 i - j + 7 k$ . Let's use $CA$ $\overrightarrow{CA}$ or $AC$ $\overrightarrow{AC}$, the magnitude of the cross product will be the same. The second diagonal is given:

BD = i + 2 j + 3 k

Step 1: Finding Cross Product

Area = (1)/(2) | AC \times BD|

AC \times BD = vmatrix i & j & k -5 & 1 & -7 1 & 2 & 3 vmatrix

= i(3 - (-14)) - j(-15 - (-7)) + k(-10 - 1)

= i(17) - j(-8) + k(-11)

= 17 i + 8 j - 11 k

Step 2: Calculating Magnitude

Area = (1)/(2) |17 i + 8 j - 11 k|

= (1)/(2) \sqrt(17² + 8² + (-11)²)

= (1)/(2) \sqrt(289 + 64 + 121)

= (1)/(2) \sqrt(474)

Pattern Recognition

When opposite vertices and one full diagonal vector are provided, immediately calculate the second diagonal vector via displacement and evaluate half the magnitude of their cross product.

Chapter Mix

Class 12 Maths: Vector Algebra

Q26 jee_main_2024_31_jan_evening Vector Triple Product

Let

a = 3 i + 2 j + k

,

b = 2 i - j + 3 k

and

c

be a vector such that

( a + b) \times c = 2( a \times b) + 24 j - 6 k

and

( a - b + i) \cdot c = -3

. Then

| c|²

is equal to

Numerical Answer. Answer: 38 to 38

Solution

Core Logic

Evaluate base vectors:

a + b = (5, 1, 4)

a \times b = vmatrix i & j & k 3 & 2 & 1 2 & -1 & 3 vmatrix = (7, -7, -7)

Substitute into given cross product equation, letting $c = (x, y, z)$ $\vec{c} = (x, y, z)$:

(5 i + j + 4 k) \times (x i + y j + z k) = 2(7 i - 7 j - 7 k) + 24 j - 6 k

vmatrix i & j & k 5 & 1 & 4 x & y & z vmatrix = (14, -14, -14) + (0, 24, -6)

(z-4y) i - (5z-4x) j + (5y-x) k = (14, 10, -20)

Equating components:

z - 4y = 14 z = 4y + 14

$4x - 5z = 10$ $4x - 5z = 10$

5y - x = -20 x = 5y + 20

Now use the dot product constraint: $a - b + i = (3-2+1) i + (2+1) j + (1-3) k = (2, 3, -2)$ .

( a - b + i) \cdot c = -3

2x + 3y - 2z = -3

Substitute $x = 5y + 20$ $x = 5y + 20$ and $z = 4y + 14$ $z = 4y + 14$:

2(5y + 20) + 3y - 2(4y + 14) = -3

10y + 40 + 3y - 8y - 28 = -3

5y + 12 = -3 5y = -15 y = -3

Calculate $x$ $x$ and $z$ $z$:

x = 5(-3) + 20 = 5

z = 4(-3) + 14 = 2

So, $c = (5, -3, 2)$ $\vec{c} = (5, -3, 2)$.

| c|² = 5² + (-3)² + 2² = 25 + 9 + 4 = 38

Chapter Mix

Class 12 Maths: Vector Algebra

Q12 jee_main_2024_31_jan_morning Cross and Dot Product Operations

Let

a = 3 i + j - 2 k, b = 4 i + j + 7 k

and

c = i - 3 j + 4 k

be three vectors. If a vector

p

satisfies

p \times b = c \times b

and

p \cdot a = 0

, then

p \cdot ( i - j - k)

is equal to

A. $24$ $24$
B. $36$ $36$
C. $28$ $28$
D. $32$ $32$

Solution

Core Logic

Given $p \times b = c \times b$ .

( p - c) \times b = 0

Thus, $p - c = λ b p = c + λ b$ .

Step 1: Utilize Dot Product Condition

Given $p \cdot a = 0$ $\vec{p} \cdot \vec{a} = 0$.

( c + λ b) \cdot a = 0

c \cdot a + λ ( b \cdot a) = 0

Calculate $c \cdot a$ $\vec{c} \cdot \vec{a}$: $(1)(3) + (-3)(1) + (4)(-2) = 3 - 3 - 8 = -8$ . Calculate $b \cdot a$ $\vec{b} \cdot \vec{a}$: $(4)(3) + (1)(1) + (7)(-2) = 12 + 1 - 14 = -1$ .

-8 + λ(-1) = 0 λ = -8

Step 2: Substitute and Solve

Substitute $λ$ $\lambda$ back to find $p$ $\vec{p}$:

p = c - 8 b = ( i - 3 j + 4 k) - 8(4 i + j + 7 k)

p = -31 i - 11 j - 52 k

Now compute $p \cdot ( i - j - k)$ :

= (-31)(1) + (-11)(-1) + (-52)(-1) = -31 + 11 + 52 = 32

Chapter Mix

Class 12 Maths: Vector Algebra

Solution & Explanation

Core Logic

Execution

Pattern Recognition

Chapter Mix

Reference Study Guides

More Vector Algebra Previous-Year Questions — Page 9

Solution

Related Formula

Core Logic

Step 1: Simplify using Vector Properties

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Dot products with c

Step 2: Finding magnitude of c

Step 3: Calculating angle

Pattern Recognition

Chapter Mix

Solution

Related Formula

Core Logic

Step 1: Finding Cross Product

Step 2: Calculating Magnitude

Pattern Recognition

Chapter Mix

Solution

Core Logic

Chapter Mix

Solution

Core Logic

Step 1: Utilize Dot Product Condition

Step 2: Substitute and Solve

Chapter Mix

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