If the components of a = αi + βj + γk along and perpendicular to b = 3i + j - k respectively, are 16

Q: Wait, why are we just adding them? Shouldn't we use the projection formula $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}$?

That is a brilliant question! You absolutely could use that formula if you were only given $\vec{a}$ and $\vec{b}$. However, in this specific problem, the question has already provided the parallel and perpendicular components explicitly. Using the projection formula would be like trying to build a house from scratch when the architect has already handed you the finished blueprints. Trust the information given; the vector sum principle is the most direct path to the solution.

Q: Is it possible that the perpendicular component is not orthogonal to $\vec{b}$?

In the context of vector decomposition, by definition, the perpendicular component $\vec{a}_{\perp}$ must be orthogonal to $\vec{b}$. If you ever doubt this, take the dot product of the given $\vec{a}_{\perp}$ and $\vec{b}$. If the result is zero, you have confirmed the orthogonality. It is a great way to verify your understanding!

Q: Why do we square the components at the end? Is there a geometric significance?

Yes, there is! The expression $\alpha^2 + \beta^2 + \gamma^2$ is simply the square of the magnitude of the vector $\vec{a}$, denoted as $|\vec{a}|^2$. By finding $\alpha$, $\beta$, and $\gamma$, you are essentially reconstructing the vector in its Cartesian form, and the final step is just calculating the squared length of that vector.

Question

If the components of $\vec{a}=\alpha\hat{i}+\beta\hat{j}+\gamma\hat{k}$ along and perpendicular to $\vec{b}=3\hat{i}+\hat{j}-\hat{k}$ respectively, are $\frac{16}{11}(3\hat{i}+\hat{j}-\hat{k})$ and $\frac{1}{11}(-4\hat{i}-5\hat{j}-17\hat{k})$, then $\alpha^{2}+\beta^{2}+\gamma^{2}$ is equal to :

Options:

(A) 26

(B) 18

(C) 23

(D) 16

Sigma Percentile · Accepted Answer

### The Geometry of Decomposition: Unlocking the Vector Mystery

Welcome, future engineer. Today, we are going to peel back the layers of a classic JEE Advanced problem.

At first glance, you might see a vector $\vec{a} = \alpha\hat{i} + \beta\hat{j} + \gamma\hat{k}$ and feel the urge to dive into complex dot products or projection formulas. But pause for a moment.

The beauty of physics and mathematics lies not in the complexity of the tools we use, but in the elegance of the principles we apply. This problem is a masterclass in the principle of superposition.

### The Principle of Reconstruction

Imagine you are standing on a flat plane. You have a vector $\vec{a}$ pointing somewhere in space. Now, imagine a reference vector $\vec{b}$.

We can break $\vec{a}$ down into two distinct parts: one part that is perfectly parallel to $\vec{b}$ (let us call it $\vec{a}_{\parallel}$) and one part that is perfectly perpendicular to $\vec{b}$ (let us call it $\vec{a}_{\perp}$).

The fundamental truth here is that these two components are not just random vectors; they are the building blocks of $\vec{a}$. If you place them head-to-tail, they reconstruct the original vector exactly. Mathematically, this is expressed as:

$$\vec{a} = \vec{a}_{\parallel} + \vec{a}_{\perp}$$

This is the heartbeat of the entire problem.

### The Art of Substitution

The question provides us with the parallel component $\vec{a}_{\parallel} = \frac{16}{11}(3\hat{i} + \hat{j} - \hat{k})$ and the perpendicular component $\vec{a}_{\perp} = \frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k})$. Instead of getting intimidated by the fractions, let us embrace them.

We substitute these into our reconstruction equation:

$$\vec{a} = \frac{16}{11}(3\hat{i} + \hat{j} - \hat{k}) + \frac{1}{11}(-4\hat{i} - 5\hat{j} - 17\hat{k})$$

This is where the magic happens. We are not doing calculus; we are doing bookkeeping. We are simply grouping the terms by their unit vectors.

### The Arithmetic of Precision

Let us focus on the $\hat{i}$ component first. We have:

$$\alpha = \frac{16 	imes 3}{11} + \frac{-4}{11} = \frac{48 - 4}{11} = \frac{44}{11} = 4$$

Next, the $\hat{j}$ component:

$$\beta = \frac{16 	imes 1}{11} + \frac{-5}{11} = \frac{16 - 5}{11} = \frac{11}{11} = 1$$

Finally, the $\hat{k}$ component:

$$\gamma = \frac{16 	imes (-1)}{11} + \frac{-17}{11} = \frac{-16 - 17}{11} = \frac{-33}{11} = -3$$

See how the fractions vanished? That is the reward for staying calm and methodical.

### The Final Victory

We have successfully reconstructed our vector: $\vec{a} = 4\hat{i} + \hat{j} - 3\hat{k}$. The question asks for the sum of the squares of these components: $\alpha^2 + \beta^2 + \gamma^2$.

Plugging in our values, we get:

$$4^2 + 1^2 + (-3)^2 = 16 + 1 + 9 = 26$$

You have just navigated a problem that tests your conceptual clarity and your ability to execute arithmetic with precision. Remember, in the JEE Advanced exam, the most complex-looking problems often yield to the simplest fundamental principles. Keep this clarity, keep this focus, and you will conquer any challenge.

Select Answer:

Visualized Solution

Understanding Vector Decomposition

The Vector Sum Principle

Raw Setup (Substitution)

Combining the $\hat{i}$ Components

Combining the $\hat{j}$ Components

Combining the $\hat{k}$ Components

Finding the Magnitude Squared

The Sigma Insight: Components of a Vector

The Geometry of Decomposition

The Principle of Reconstruction

The Art of Substitution

The Arithmetic of Precision

The Final Victory

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Select Answer:

Visualized Solution

Understanding Vector Decomposition

The Vector Sum Principle

Raw Setup (Substitution)

Combining the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } i^ Components

Combining the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } j^​ Components

Combining the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } k^ Components