Let P be a point in the plane of the vectors AB = 3i + j - k and AC = i - j + 3k such that P is equi

Q: Wait, why do we reject the negative root for $\lambda$?

In this problem, we are looking for the vector $\vec{AP}$ that lies on the bisector. While $\lambda$ could technically be negative (pointing in the opposite direction), the problem asks for the area of the triangle formed by $A, B,$ and $P$. The area is a scalar quantity and will be the same regardless of the sign of $\lambda$, as the magnitude of the cross product involves the absolute value of the scalar.

Q: Why can't we just use the distance formula from a point to a line?

You absolutely could! You could define $P$ as $(x, y, z)$ and set the distance from $P$ to line $AB$ equal to the distance from $P$ to line $AC$. However, that would lead to a system of equations that is much more algebraically intensive. The geometric approach using the angle bisector is the 'JEE way'—it's faster, more elegant, and less prone to calculation errors.

Q: Why is the area of $\triangle ABP$ equal to $\frac{1}{2} |\vec{AB} \times \vec{AP}|$?

This is a fundamental result in vector geometry. The magnitude of the cross product of two vectors $\vec{u}$ and $\vec{v}$ gives the area of the parallelogram they span. Since a triangle is exactly half of that parallelogram, the area of $\triangle ABP$ is $\frac{1}{2} |\vec{AB} \times \vec{AP}|$. It's a beautiful, direct application of the cross product's geometric meaning.

Question

Let $P$ be a point in the plane of the vectors $\vec{AB} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{AC} = \hat{i} - \hat{j} + 3\hat{k}$ such that $P$ is equidistant from the lines AB and AC. If $|\vec{AP}| = \frac{\sqrt{5}}{2}$, then the area of the triangle ABP is:

Options:

(A) 2

(B) $\frac{\sqrt{30}}{4}$

(C) $\frac{3}{2}$

(D) $\frac{\sqrt{26}}{4}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two vectors, $\vec{AB} = 3\hat{i} + \hat{j} - \hat{k}$ and $\vec{AC} = \hat{i} - \hat{j} + 3\hat{k}$, originating from a common point $A$. We seek the area of $	riangle ABP$, where point $P$ lies on the plane defined by $A, B,$ and $C$.

The problem states that $P$ is equidistant from the lines $AB$ and $AC$. Geometrically, the locus of points equidistant from two intersecting lines is the **angle bisector** of the angle formed by those lines.

### Phase 1: The Geometric Insight

First, we calculate the magnitudes of the given vectors:
$|\vec{AB}| = \sqrt{3^2 + 1^2 + (-1)^2} = \sqrt{11}$
$|\vec{AC}| = \sqrt{1^2 + (-1)^2 + 3^2} = \sqrt{11}$

Since the magnitudes are equal, the angle bisector vector $\vec{d}$ is simply the sum of the two vectors:
$$\vec{d} = \vec{AB} + \vec{AC} = (3+1)\hat{i} + (1-1)\hat{j} + (-1+3)\hat{k} = 4\hat{i} + 2\hat{k}$$

We can simplify the direction vector to $\vec{v} = 2\hat{i} + \hat{k}$. Since $P$ lies on this bisector, the vector $\vec{AP}$ must be a scalar multiple of this direction: $\vec{AP} = \lambda(2\hat{i} + \hat{k})$.

### Phase 2: Determining Vector AP

We are given that $|\vec{AP}| = \frac{\sqrt{5}}{2}$. Substituting our expression for $\vec{AP}$ into this magnitude constraint:
$$|\lambda| \sqrt{2^2 + 0^2 + 1^2} = |\lambda| \sqrt{5} = \frac{\sqrt{5}}{2}$$

Solving for $\lambda$, we find $|\lambda| = \frac{1}{2}$. Taking the positive direction, we obtain:
$$\vec{AP} = \hat{i} + \frac{1}{2}\hat{k}$$

### Phase 3: Final Calculation

The area of $	riangle ABP$ is given by the formula $	ext{Area} = \frac{1}{2} |\vec{AB} 	imes \vec{AP}|$. We compute the cross product using the determinant form:
$$\vec{AB} 	imes \vec{AP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 1 & -1 \ 1 & 0 & \frac{1}{2} \end{vmatrix}$$

Expanding the determinant:
$$\vec{AB} 	imes \vec{AP} = \hat{i}\left(1 \cdot \frac{1}{2} - 0ight) - \hat{j}\left(3 \cdot \frac{1}{2} - (-1)ight) + \hat{k}(0 - 1) = \frac{1}{2}\hat{i} - \frac{5}{2}\hat{j} - \hat{k}$$

Now, we calculate the magnitude of this resulting vector:
$$|\vec{AB} 	imes \vec{AP}| = \sqrt{\left(\frac{1}{2}ight)^2 + \left(-\frac{5}{2}ight)^2 + (-1)^2} = \sqrt{\frac{1}{4} + \frac{25}{4} + 1} = \sqrt{\frac{30}{4}} = \frac{\sqrt{30}}{2}$$

Finally, the area of the triangle is:
$$	ext{Area} = \frac{1}{2} 	imes \frac{\sqrt{30}}{2} = \mathbf{\frac{\sqrt{30}}{4}}$$

Select Answer:

Visualized Solution

Visualizing Vectors $A B$ and $A C$

The Equidistant Property

Magnitudes of Vectors

Finding the Bisector Direction

Defining Vector $A P$

Solving for $λ$

Area of Triangle Formula

Setting up the Cross Product

Computing $A B \times A P$

Magnitude of the Cross Product

Final Area Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

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