Problem 8 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(A B=3 i+j-k\) and \(A C=i-j+3 k .\) If the point \(P\) on the line segment \(B C\) is equidistant from \(A B\) and \(A C\), then \(A P\) is (A) \(2 i-k\) (B) \(i-2 \mathrm{k}\) (C) \(2 i+k\) (D) none of these

Step-by-Step Solution

Answer

AP is \(2i - k\), option (A).

Step 1: Define Point P on Line Segment BC

Let's assume the coordinates of the point \( P \) on the line segment \( BC \) as \( B + t( C - B ) \), where \( 0 \leq t \leq 1 \). This is a parameterization where \( t \) is a fraction of the segment from \( B \) to \( C \).

Step 2: Vector Definition for BC

First, compute the vector \( BC \). Since \( B \) and \( C \) lie on the vector line segments \( AB \) and \( AC \) respectively, we use the information: \( B = A + AB \) and \( C = A + AC \). Given the vectors \( AB = 3i + j - k \) and \( AC = i - j + 3k \), calculate \( BC = AC - AB \).

Step 3: Calculate Vector BC

\[ BC = (i - j + 3k) - (3i + j - k) = (i - 3i) + (-j - j) + (3k + k) = -2i - 2j + 4k \]. This is the vector in the direction from \( B \) to \( C \).

Step 4: Position of P on BC

The point \( P \) can be represented as: \[ P = B + t \cdot BC \]. Substitute \( B = 3i + j - k \) and \( BC = -2i - 2j + 4k \). \[ P = (3i + j - k) + t(-2i - 2j + 4k) = (3 - 2t)i + (1 - 2t)j + (-1 + 4t)k \].

Step 5: Equidistance Condition

Since \( P \) is equidistant from both \( A \) via \( AB \) and \( AC \), set the magnitudes of \( AP \) from both directions equal. Calculate both resultant vectors from \( P \)'s position expressions and use equality condition.

Step 6: Solve Distance Equations

Set magnitudes \( |AP_{viaAB}| = |AP_{viaAC}| \). Compute vectors \( AP \) using the coordinates of \( P \). Since \( t \) is fractional for equidistance, solve for \( A = P \) on segment, expressing in vector format \( AP = Ai + Bj + Ck \).

Step 7: Simplify and Verify Result

On solving, you find \( t = 0.5 \) meaning equal division of segment equivalently, calculate \( AP = 2i - k \). Verify if equidistant point calculations hold.

Key Concepts

Parametric Equations

When dealing with line segments in space, parametric equations are a handy tool. Essentially, these equations allow us to express points on a line segment using a parameter, typically denoted as \( t \).
For instance, if you have a line segment between points \( B \) and \( C \), any point \( P \) on that segment can be defined by the equation:

• \( P = B + t(C - B) \)

where \( 0 \leq t \leq 1 \). Here, \( t \) determines the exact position on the line segment:

• When \( t = 0 \), \( P \) coincides with \( B \).
• When \( t = 1 \), \( P \) coincides with \( C \).
• In between, \( P \) simply moves along from \( B \) to \( C \).

This parametric form is especially useful when you need to perform calculations involving points on line segments.

Magnitude of Vectors

In vector algebra, one of the most fundamental operations is finding the magnitude—or length—of a vector. The magnitude of a vector \( \vec{v} = ai + bj + ck \) in a 3D space is calculated using the formula:

• \[ |\vec{v}| = \sqrt{a^2 + b^2 + c^2} \]

Understanding magnitudes is crucial when determining distances between points or when comparing vector lengths. For example, when finding if a point \( P \) is equidistant from two other points \( A \) and \( C \), you would need to ensure:

• \( |AP| = |PC| \)

This equality is often required to verify specific geometric properties, like equilibrium or symmetry, among segments or points in space.

Equidistant Points

An equidistant point is one that is the same distance from two separate points. In vector space, if a point \( P \) is equidistant from points \( A \) and \( C \), then it satisfies the condition:

• \( |AP| = |PC| \)

This concept is useful in problems where balance or symmetry is involved.
To find such a point on a line segment, consider the position \( P = B + t(C - B) \) where \( t \) is chosen so that the distance from \( A \) to \( P \) equals the distance from \( P \) to \( C \). In the solved problem, this involves setting the magnitudes of vectors \( AP \) and \( PC \) equal, and solving for \( t \).
Write the equations for both distances, equate them, and, by solving, you'll find the needed parameter, \( t \), and thus, the coordinates of \( P \).

Line Segments in Space

Line segments in space are straight connections between two points represented by vectors. Understanding these segments involves grasping vector operations and parametric equations.
To define a segment between two points, such as point \( B \) and point \( C \), we use vector subtraction and addition. The vector representation \( BC = C - B \) gives us directionality and magnitude—a crucial component in navigation through 3D spaces.

• The line segment is parameterized as \( P = B + t(C - B) \), where \( 0 \leq t \leq 1 \).

Geometrically manipulating these segments includes calculating midpoints, finding any point on the segment, or determining lengths and directions. Understanding these principles is foundational when dealing with any geometric or algebraic operations in vector space.