Let AB = 2i + 4j - 5k and AD = i + 2j + λk, λ ∈ R. Let the projection of the vector v = i + j + k on

Q: Why do we use the absolute value in the projection formula? Can't we just take the dot product?

That is a brilliant question. The dot product $\vec{a} \cdot \vec{b}$ can be negative if the angle between the vectors is obtuse (greater than 90 degrees). However, a 'length' or 'magnitude' must always be a non-negative scalar. The absolute value ensures that we are calculating the physical distance of the shadow cast by the vector, regardless of whether it points in the same general direction or the opposite direction of the reference vector.

Q: I tried to find the unit vector of $\vec{AC}$ first and then dot it with $\vec{v}$. Is that wrong?

Not at all! That is actually a very elegant way to think about it. The projection of $\vec{v}$ onto $\vec{AC}$ is simply $\vec{v} \cdot \hat{AC}$, where $\hat{AC}$ is the unit vector. Since $\hat{AC} = \frac{\vec{AC}}{|\vec{AC}|}$, you get $\vec{v} \cdot \frac{\vec{AC}}{|\vec{AC}|}$, which is exactly the same as the formula we used. You are thinking like a physicist—always look for the most intuitive path!

Q: Why did $\lambda^2$ cancel out when solving for $\lambda$? Did I make a mistake?

That is actually a sign that you are on the right track! In many JEE problems, complex-looking quadratic terms often simplify into linear equations. If the $\lambda^2$ didn't cancel, you might have had a much harder time. Trust your algebra; when the math simplifies, it is usually a sign that you have set up the problem correctly.

Question

Let $\vec{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{AD} = \hat{i} + 2\hat{j} + \lambda\hat{k}, \lambda \in \mathbb{R}$. Let the projection of the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$ on the diagonal $\vec{AC}$ of the parallelogram ABCD be of length one unit. If $\alpha, \beta$, where $\alpha > \beta$, be the roots of the equation $\lambda^2x^2 - 6\lambda x + 5 = 0$, then $2\alpha - \beta$ is equal to

Options:

(A) 3

(B) 4

(C) 1

(D) 6

Sigma Percentile · Accepted Answer

### The Geometry of Vectors: A Journey into Parallelograms

Welcome, student. Today, we are not just solving a problem; we are exploring the elegant architecture of 3D space. When you look at a problem involving a parallelogram like $ABCD$, I want you to stop seeing it as a flat shape on your paper.

Imagine it floating in three-dimensional space. We are given two vectors, $\vec{AB} = 2\hat{i} + 4\hat{j} - 5\hat{k}$ and $\vec{AD} = \hat{i} + 2\hat{j} + \lambda\hat{k}$. These are the building blocks of our structure.

### Phase 1: The Diagonal as a Resultant

In the world of vectors, the parallelogram law is our most trusted companion. When we define a parallelogram by two adjacent vectors originating from a common vertex, the diagonal originating from that same vertex is simply the vector sum of those two sides.

Thus, we define our diagonal $\vec{AC}$ as:

$$\vec{AC} = \vec{AB} + \vec{AD}$$

By performing the vector addition, we combine the components: $(2+1)\hat{i} + (4+2)\hat{j} + (-5+\lambda)\hat{k}$. This yields our diagonal vector:

$$\vec{AC} = 3\hat{i} + 6\hat{j} + (\lambda-5)\hat{k}$$

This vector is the backbone of our problem. It contains the unknown parameter $\lambda$, which we must now hunt down.

### Phase 2: The Shadow of a Vector

Now, let us introduce the vector $\vec{v} = \hat{i} + \hat{j} + \hat{k}$. The problem asks us to consider the projection of $\vec{v}$ onto $\vec{AC}$. Imagine you are standing in a room, and there is a light source directly above $\vec{v}$, shining perpendicular to the line of $\vec{AC}$.

The shadow that $\vec{v}$ casts on $\vec{AC}$ is exactly what we call the projection. The length of this shadow is given by the formula:

$$	ext{Projection} = \frac{|\vec{v} \cdot \vec{AC}|}{|\vec{AC}|}$$

We are told this length is exactly $1$. This is our constraint, and the key that unlocks the door to $\lambda$.

### Phase 3: The Algebraic Grind

Let us calculate the components of our equation. First, the dot product $\vec{v} \cdot \vec{AC}$:

$$\vec{v} \cdot \vec{AC} = (1)(3) + (1)(6) + (1)(\lambda-5) = 3 + 6 + \lambda - 5 = \lambda + 4$$

Next, the magnitude $|\vec{AC}|$, which is the square root of the sum of the squares of its components:

$$|\vec{AC}| = \sqrt{3^2 + 6^2 + (\lambda-5)^2} = \sqrt{9 + 36 + (\lambda-5)^2} = \sqrt{45 + (\lambda-5)^2}$$

Now, we equate the projection to $1$:

$$\frac{|\lambda + 4|}{\sqrt{45 + (\lambda-5)^2}} = 1$$

To solve this, we square both sides. This is a crucial step—it removes the absolute value and the square root, transforming a radical equation into a manageable algebraic one:

$$(\lambda + 4)^2 = 45 + (\lambda - 5)^2$$

Expanding both sides, we get:

$$\lambda^2 + 8\lambda + 16 = 45 + \lambda^2 - 10\lambda + 25$$

Observe the beauty of the cancellation! The $\lambda^2$ terms vanish, leaving us with a simple linear equation:

$$8\lambda + 16 = -10\lambda + 70$$

$$18\lambda = 54 \implies \lambda = 3$$

### Phase 4: The Quadratic Finale

We have found $\lambda = 3$. Now, we turn to the quadratic equation provided: $\lambda^2x^2 - 6\lambda x + 5 = 0$. Substituting $\lambda = 3$, we get:

$$9x^2 - 18x + 5 = 0$$

To find the roots $\alpha$ and $\beta$, we factorize. We look for two numbers that multiply to $9 	imes 5 = 45$ and add to $-18$. These are $-15$ and $-3$:

$$9x^2 - 15x - 3x + 5 = 0$$

$$3x(3x - 5) - 1(3x - 5) = 0$$

$$(3x - 5)(3x - 1) = 0$$

Thus, the roots are $x = \frac{5}{3}$ and $x = \frac{1}{3}$. Since the problem specifies $\alpha > \beta$, we assign $\alpha = \frac{5}{3}$ and $\beta = \frac{1}{3}$.

### The Final Result

Finally, we calculate $2\alpha - \beta$:

$$2\left(\frac{5}{3}ight) - \frac{1}{3} = \frac{10}{3} - \frac{1}{3} = \frac{9}{3} = 3$$

And there we have it. Through geometry, projection, and algebra, we have arrived at the answer: **3**. Remember, every step in this process was logical. When you face such problems in the exam, do not panic; break them down, visualize the vectors, and trust the math.

Select Answer:

Visualized Solution

Visualizing the Parallelogram $A B C D$

The Diagonal $A C$

Computing $A C$ Components

Introducing Vector $v$ and its Projection

The Projection Formula

Calculating the Dot Product

Calculating the Magnitude of $A C$

Solving for $λ$

Expanding and Finding $λ$

The Quadratic Equation

Finding Roots $α$ and $β$

Assigning $α$ and $β$

Final Calculation: $2 α - β$

The Sigma Insight: Scalar (Dot) Product

The Geometry of Vectors

Phase 1

Phase 2

Phase 3

Phase 4

The Final Result

Similar Questions

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Select Answer:

Visualized Solution

Visualizing the Parallelogram .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ABCD

The Diagonal .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC

Computing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } AC Components

Introducing Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v and its Projection