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

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

Since $P$ $P$ is equidistant from $AB$ $AB$ and $AC$ $AC$, $AP$ $AP$ must lie on the angle bisector of $∠ BAC$ $\angle BAC$. Let the angle between vectors $AB$ $\overrightarrow{AB}$ and $AC$ $\overrightarrow{AC}$ be $2θ$ $2\theta$.

2θ = AB \cdot AC| AB| | AC| = 3(1) + 1(-1) + (-1)(3)\sqrt(9+1+1) \sqrt(1+1+9) = 3-1-3\sqrt(11)\sqrt(11) = -(1)/(11)

Execution

We need $θ$ $\sin \theta$ to find the perpendicular distance from $P$ $P$ to $AB$ $AB$ (which acts as the height of $Δ ABP$ $\Delta ABP$). Using half-angle formula $1 - 2 ²θ = 2θ$ $1 - 2\sin^2\theta = \cos 2\theta$:

1 - 2 ²θ = -(1)/(11) \Rightarrow 2 ²θ = (12)/(11) \Rightarrow θ = \sqrt((6)/(11))

Angle bisector in triangle vector representation

The area of triangle $ABP$ $ABP$ is $(1)/(2) \times base \times height$ . Base $= | AB| = \sqrt(11)$ $= |\overrightarrow{AB}| = \sqrt{11}$. Height $= | AP| θ = \sqrt(5)2 \sqrt((6)/(11))$ .

Area(Δ ABP) = (1)/(2) \times \sqrt(11) \times \sqrt(5)2 \times \sqrt((6)/(11)) = \sqrt(30)4

Pattern Recognition

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.

Chapter Mix

Class 12 Maths: 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 $P$ $P$ be a point in the plane of the vector $AB = 3 i + j - k$ and $AC = i - j + 3 k$ such that $P$ $P$ is equidistant from the lines $AB$ $AB$ and $AC$ $AC$. If $| AP| = \sqrt(5)2$ , then the area of the triangle $ABP$ $ABP$ is:

Solution & Explanation

Core Logic

Execution

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_2026_28_january_evening

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Solution & Explanation

Core Logic

Execution

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_2026_28_january_evening

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