The values of a, for which points A, B, C with position vectors 2i - j + k, i - 3j - 5k and ai - 3j

Q: Why can't we just use the distance formula and the Pythagorean theorem ($AC^2 + BC^2 = AB^2$)?

You absolutely could! However, in the JEE Advanced arena, time is your most precious currency. Using the distance formula involves calculating square roots and squaring binomials, which leads to a much more tedious algebraic expansion. The dot product method is the 'JEE-smart' way—it is elegant, direct, and minimizes the chance of calculation errors.

Q: Does it matter if I calculate $\vec{AC}$ instead of $\vec{CA}$?

That is a fantastic question! The answer is no, it does not matter. If $\vec{CA} \cdot \vec{CB} = 0$, then $(-\vec{AC}) \cdot (-\vec{BC}) = 0$. Since the negative signs cancel out, you are left with $\vec{AC} \cdot \vec{BC} = 0$. The result remains identical. Vector geometry is beautifully symmetric that way!

Q: Why did the $j$ and $k$ components vanish in the dot product calculation?

They didn't vanish by magic! When we computed $\vec{CA}$ and $\vec{CB}$, the subtraction resulted in zero coefficients for those specific components. When you perform the dot product, you multiply corresponding components. Since $2 \times 0 = 0$ and $0 \times (-6) = 0$, those terms effectively contributed nothing to the sum. It is a sign that you have simplified the geometry correctly.

Question

The values of $a$, for which points $A, B, C$ with position vectors $2\hat{i} - \hat{j} + \hat{k}, \hat{i} - 3\hat{j} - 5\hat{k}$ and $a\hat{i} - 3\hat{j} + \hat{k}$ respectively are the vertices of a right angled triangle with $C = \pi/2$ are

Options:

(A) 2 and 1

(B) $-2$ and $-1$

(C) $-2$ and 1

(D) 2 and $-1$

Sigma Percentile · Accepted Answer

Welcome, future engineers! Today, we are not just solving a problem; we are stepping into the elegant world of 3D vector geometry. Imagine you are standing in a vast, empty room with three points, $A$, $B$, and $C$, floating in space.

You are told that these points form a triangle, and specifically, that this triangle is right-angled at vertex $C$. This is not just a random shape; it is a rigid geometric constraint. Our mission is to find the value of the parameter $a$ that makes this specific configuration possible.

### The Geometry of Orthogonality

When we say a triangle is right-angled at $C$, we are essentially saying that the line segment $CA$ is perpendicular to the line segment $CB$. In the language of vectors, this means the vector $\vec{CA}$ and the vector $\vec{CB}$ are orthogonal.

The golden rule for orthogonality is the dot product. If two non-zero vectors are perpendicular, their dot product must be exactly zero. This is our master equation:

$$\vec{CA} \cdot \vec{CB} = 0$$

This single equation is the key that will unlock the entire problem.

### Defining the Vectors

Before we can perform the dot product, we must define our vectors. We are given the position vectors:
$\vec{A} = 2\hat{i} - \hat{j} + \hat{k}$
$\vec{B} = \hat{i} - 3\hat{j} - 5\hat{k}$
$\vec{C} = a\hat{i} - 3\hat{j} + \hat{k}$

To find the vector $\vec{CA}$, we subtract the position vector of $C$ from $A$. Think of this as finding the displacement from $C$ to $A$, where $\vec{CA} = \vec{A} - \vec{C}$.

Performing the subtraction component-wise:
The $i$ component is $(2 - a)$.
The $j$ component is $(-1 - (-3)) = 2$.
The $k$ component is $(1 - 1) = 0$.

Thus, we have:
$$\vec{CA} = (2 - a)\hat{i} + 2\hat{j} + 0\hat{k}$$

Now, let us do the same for $\vec{CB} = \vec{B} - \vec{C}$:
The $i$ component is $(1 - a)$.
The $j$ component is $(-3 - (-3)) = 0$.
The $k$ component is $(-5 - 1) = -6$.

So, we obtain:
$$\vec{CB} = (1 - a)\hat{i} + 0\hat{j} - 6\hat{k}$$

### The Power of the Dot Product

Now, we bring back our master equation: $\vec{CA} \cdot \vec{CB} = 0$. We substitute our vectors into this expression:

$$((2 - a)\hat{i} + 2\hat{j} + 0\hat{k}) \cdot ((1 - a)\hat{i} + 0\hat{j} - 6\hat{k}) = 0$$

The dot product is the sum of the products of the corresponding components: $(x_1x_2 + y_1y_2 + z_1z_2)$. Applying this, we get:

$$(2 - a)(1 - a) + (2)(0) + (0)(-6) = 0$$

Look at how the complexity melts away! The $j$ and $k$ terms vanish, leaving us with a beautifully simple quadratic equation:

$$(2 - a)(1 - a) = 0$$

### The Final Resolution

We have arrived at the final stage. We have a product of two factors equal to zero, which implies that either $(2 - a) = 0$ or $(1 - a) = 0$.

Solving these simple linear equations gives us the final values:

**$a = 2$ or $a = 1$**

These are the values that satisfy the condition of the right-angled triangle. Whenever you see "perpendicular" in a vector problem, let your mind immediately jump to the dot product; it is a powerful tool that simplifies the most complex 3D scenarios into manageable algebra.

Select Answer:

Visualized Solution

Introduction to the Problem

Defining Position Vectors

The Orthogonality Condition

Vector $C A$ Setup

Simplifying $C A$

Vector $C B$ Setup

Simplifying $C B$

Setting Up the Dot Product

Expanding the Dot Product

Solving for $a$

Final Result

The Sigma Insight: Scalar (Dot) Product

The Geometry of Orthogonality

Defining the Vectors

The Power of the Dot Product

The Final Resolution

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