Let α = 3i + j and β = 2i - j + 3k. If β = β1 - β2, where β1 is parallel to α and β2 is perpendicula

Q: Why do we write $\vec{\beta}_1 = \lambda \vec{\alpha}$? Can't we just use components?

Excellent question! Writing $\vec{\beta}_1 = \lambda \vec{\alpha}$ is the most powerful way to enforce the 'parallel' condition. It captures the direction of $\vec{\alpha}$ while leaving the magnitude flexible (the scalar $\lambda$). It turns a geometric constraint into a simple algebraic variable, which is exactly what we need to solve for the unknown.

Q: Wait, the problem says $\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2$. Why did we rearrange it to $\vec{\beta}_2 = \vec{\beta}_1 - \vec{\beta}$?

It's all about isolating the unknown. We know $\vec{\beta}$ and we have a form for $\vec{\beta}_1$. By rearranging, we express the 'unknown' $\vec{\beta}_2$ in terms of things we already know or can easily define. It's a strategic move to make the algebra work for us, not against us.

Q: Is there a shortcut for the cross product calculation?

While you could use the geometric definition $|\vec{\beta}_1| |\vec{\beta}_2| \sin \theta$, the determinant method is the 'gold standard' for JEE Advanced. It is systematic, less prone to sign errors, and works perfectly in 3D space. Stick to the determinant—it's your best friend for reliability.

Question

Let $\vec{\alpha} = 3\hat{i} + \hat{j}$ and $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}$. If $\vec{\beta} = \vec{\beta}_1 - \vec{\beta}_2$, where $\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$, then $\vec{\beta}_1 	imes \vec{\beta}_2$ is equal to

Options:

(A) $-3\hat{i} + 9\hat{j} + 5\hat{k}$

(B) $3\hat{i} - 9\hat{j} - 5\hat{k}$

(C) $\frac{1}{2}(-3\hat{i} + 9\hat{j} + 5\hat{k})$

(D) $\frac{1}{2}(3\hat{i} - 9\hat{j} + 5\hat{k})$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are dissecting the very anatomy of space.

We are given two vectors, $\vec{\alpha} = 3\hat{i} + \hat{j}$ and $\vec{\beta} = 2\hat{i} - \hat{j} + 3\hat{k}$. We are asked to decompose $\vec{\beta}$ into two components, $\vec{\beta}_1$ and $\vec{\beta}_2$.

We are told $\vec{\beta}_1$ is parallel to $\vec{\alpha}$ and $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$. This is the language of projections and rejections.

### Phase 1: The Parallel Constraint

When we say $\vec{\beta}_1$ is parallel to $\vec{\alpha}$, we are saying they share the same direction. Mathematically, this is the most elegant relationship in vector algebra:
$$\vec{\beta}_1 = \lambda \vec{\alpha}$$

Here, $\lambda$ is our unknown scalar, the 'scaling factor' that determines the magnitude of $\vec{\beta}_1$. By substituting $\vec{\alpha}$, we get:
$$\vec{\beta}_1 = \lambda(3\hat{i} + \hat{j}) = 3\lambda\hat{i} + \lambda\hat{j}$$

### Phase 2: The Orthogonality Filter

Now, we look at the relation $\vec{\beta} = \vec{\beta}_1 + \vec{\beta}_2$. We need to isolate $\vec{\beta}_2$. Rearranging gives us:
$$\vec{\beta}_2 = \vec{\beta} - \vec{\beta}_1$$

Substituting our expressions, we get $\vec{\beta}_2 = (2\hat{i} - \hat{j} + 3\hat{k}) - (3\lambda\hat{i} + \lambda\hat{j})$. Simplifying this, we find:
$$\vec{\beta}_2 = (2 - 3\lambda)\hat{i} + (-1 - \lambda)\hat{j} + 3\hat{k}$$

Now, the magic happens. We are told $\vec{\beta}_2$ is perpendicular to $\vec{\alpha}$. This is our 'golden key'. In the world of vectors, perpendicularity is synonymous with a dot product of zero:
$$\vec{\beta}_2 \cdot \vec{\alpha} = 0$$

When we compute $((2 - 3\lambda)\hat{i} + (-1 - \lambda)\hat{j} + 3\hat{k}) \cdot (3\hat{i} + 1\hat{j} + 0\hat{k}) = 0$, we are essentially filtering out the parallel component. The calculation yields:
$$(2 - 3\lambda)(3) + (-1 - \lambda)(1) + (3)(0) = 0$$

This simplifies to $6 - 9\lambda - 1 - \lambda = 0$, or $5 = 10\lambda$. Thus, $\lambda = \frac{1}{2}$.

### Phase 3: The Final Cross Product

With $\lambda = \frac{1}{2}$, we can define our vectors precisely:
$$\vec{\beta}_1 = \frac{3}{2}\hat{i} + \frac{1}{2}\hat{j}$$
$$\vec{\beta}_2 = (2 - \frac{3}{2})\hat{i} + (-1 - \frac{1}{2})\hat{j} + 3\hat{k} = \frac{1}{2}\hat{i} - \frac{3}{2}\hat{j} + 3\hat{k}$$

The final step is the cross product $\vec{\beta}_1 	imes \vec{\beta}_2$. We set up the determinant:
$$\vec{\beta}_1 	imes \vec{\beta}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ \frac{3}{2} & \frac{1}{2} & 0 \ \frac{1}{2} & -\frac{3}{2} & 3 \end{vmatrix}$$

Expanding this, the components are:
- $\hat{i}$ component: $(\frac{1}{2})(3) - (0)(-\frac{3}{2}) = \frac{3}{2}$
- $\hat{j}$ component: $-[(\frac{3}{2})(3) - (0)(\frac{1}{2})] = -\frac{9}{2}$
- $\hat{k}$ component: $(\frac{3}{2})(-\frac{3}{2}) - (\frac{1}{2})(\frac{1}{2}) = -\frac{9}{4} - \frac{1}{4} = -\frac{10}{4} = -\frac{5}{2}$

Our final result is:
$$\mathbf{\frac{1}{2}(3\hat{i} - 9\hat{j} - 5\hat{k})}$$

Animated Solution for Mathematics - Vector Algebra: Let $α = 3 \hat{i} + \hat{j}$ and $β = 2 \hat{i} - \hat{j} + 3 \hat{k}$ . If $β = β_{1} - β_{2}$ , where $β_{1}$ is parallel to $α$ and $β_{2}$ is perpendicular to $α$ , then $β_{1} \times β_{2}$ is equal to

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Visualized Solution

Visualizing the Vectors

Defining $β_{1}$ using Parallelism

Expressing $β_{2}$ in terms of $λ$

Substituting Vectors into $β_{2}$

Applying the Perpendicular Condition

Setting up the Dot Product

Solving for $λ$

Finding the Exact Vectors $β_{1}$ and $β_{2}$

Setting up the Cross Product

Expanding the Determinant: $\hat{i}$ and $\hat{j}$ components

Expanding the Determinant: $\hat{k}$ component

Final Simplification

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Phase 1

Phase 2

Phase 3

Similar Questions

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Visualized Solution

Visualizing the Vectors

Defining .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β​1​ using Parallelism

Expressing .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β​2​ in terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Substituting Vectors into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β​2​

Applying the Perpendicular Condition

Setting up the Dot Product

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Finding the Exact Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β​1​ and .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } β​2​

Setting up the Cross Product

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