Let a = i + 2j - 3k and b = 2i - 3j + 5k. If r x a = b x r, r . (αi + 2j + k) = 3 and r . (2i + 5j -

Q: Why does $\vec{r} \times \vec{a} = \vec{b} \times \vec{r}$ imply $\vec{r}$ is parallel to $\vec{a} + \vec{b}$?

Great question! When you move $\vec{b} \times \vec{r}$ to the left side, it becomes $\vec{r} \times \vec{a} + \vec{r} \times \vec{b} = \vec{0}$. By the distributive property, this is $\vec{r} \times (\vec{a} + \vec{b}) = \vec{0}$. Since the cross product of two non-zero vectors is zero if and only if they are parallel, $\vec{r}$ must be parallel to the sum vector.

Q: Can we just assume $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ and solve for $x, y, z$?

You absolutely could, but that is the 'brute force' path! It would lead to a system of three linear equations with four variables ($x, y, z, \lambda$), which is much more tedious. Using the parallelism property reduces the problem to a single scalar variable $\lambda$, which is the hallmark of an elegant JEE solution.

Q: Why did the $\lambda$ terms cancel out in the first dot product?

That is the beauty of the problem design! When you expand the dot product, you get terms like $3\lambda\alpha - 2\lambda + 2\lambda$. The $-2\lambda$ and $+2\lambda$ are additive inverses. This cancellation is a sign that you are on the right track—it simplifies the algebra significantly, allowing you to isolate $\lambda\alpha$.

Question

Let $\vec{a} = \hat{i} + 2\hat{j} - 3\hat{k}$ and $\vec{b} = 2\hat{i} - 3\hat{j} + 5\hat{k}$. If $\vec{r} 	imes \vec{a} = \vec{b} 	imes \vec{r}$, $\vec{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$ and $\vec{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$, $\alpha \in R$, then the value of $\alpha + |\vec{r}|^2$ is equal to :

Options:

(A) 9

(B) 15

(C) 13

(D) 11

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are choreographing a dance between vectors.

When you look at the equation $\vec{r} 	imes \vec{a} = \vec{b} 	imes \vec{r}$, do not see it as a dry algebraic constraint. See it as a geometric puzzle where the cross product measures perpendicularity and area.

To solve this, we bring the terms together. Recalling that the cross product is anti-commutative, we know that $\vec{b} 	imes \vec{r} = -(\vec{r} 	imes \vec{b})$.

Substituting this into our original equation, we transform it into:

$$\vec{r} 	imes \vec{a} + \vec{r} 	imes \vec{b} = \vec{0}$$

### The Power of Parallelism

Now, we apply the distributive property to factor out $\vec{r}$:

$$\vec{r} 	imes (\vec{a} + \vec{b}) = \vec{0}$$

This is the "Aha!" moment. In vector algebra, if the cross product of two vectors is the zero vector, they must be parallel. This implies that $\vec{r}$ is a scaled version of the sum of $\vec{a}$ and $\vec{b}$.

Let us calculate that sum:

$$\vec{a} + \vec{b} = (\hat{i} + 2\hat{j} - 3\hat{k}) + (2\hat{i} - 3\hat{j} + 5\hat{k}) = 3\hat{i} - \hat{j} + 2\hat{k}$$

We define our unknown vector as $\vec{r} = \lambda(3\hat{i} - \hat{j} + 2\hat{k})$. We have successfully reduced a complex vector problem into a single scalar variable, $\lambda$.

### The Algebraic Symphony

With $\vec{r} = 3\lambda\hat{i} - \lambda\hat{j} + 2\lambda\hat{k}$, we tackle the dot product conditions. The first condition, $\vec{r} \cdot (\alpha\hat{i} + 2\hat{j} + \hat{k}) = 3$, expands as follows:

$$3\lambda\alpha - 2\lambda + 2\lambda = 3$$

The terms $-2\lambda$ and $+2\lambda$ cancel out, leaving us with $3\lambda\alpha = 3$, or simply:

$$\lambda\alpha = 1$$

Next, we apply the second condition: $\vec{r} \cdot (2\hat{i} + 5\hat{j} - \alpha\hat{k}) = -1$. Expanding this gives:

$$6\lambda - 5\lambda - 2\lambda\alpha = -1$$

Simplifying this expression, we obtain $\lambda - 2\lambda\alpha = -1$. Substituting $\lambda\alpha = 1$ into this equation yields $\lambda - 2(1) = -1$, which leads us directly to **$\lambda = 1$**.

### Final Calculation

With $\lambda = 1$, it follows that $\alpha = 1$. Consequently, our vector is $\vec{r} = 3\hat{i} - \hat{j} + 2\hat{k}$.

The problem asks for the value of $\alpha + |\vec{r}|^2$. First, we calculate the magnitude squared:

$$|\vec{r}|^2 = 3^2 + (-1)^2 + 2^2 = 9 + 1 + 4 = 14$$

Adding $\alpha$ to this result, we get $1 + 14 = 15$.

The final answer is **$15$**.

Select Answer:

Visualized Solution

Analyze the Cross Product Condition

Rearrange to Find Parallelism

Calculate the Sum Vector $a + b$

Define $r$ using Scalar $λ$

Apply First Dot Product Condition

Simplify First Equation

Apply Second Dot Product Condition

Solve for $λ$

Find the Value of $α$

Calculate Magnitude Squared $∣ r ∣^{2}$

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of Parallelism

The Algebraic Symphony

Final Calculation

Similar Questions

Let $a = 2 \hat{i} - 3 \hat{j} + 4 \hat{k}$ and $b = 7 \hat{i} + \hat{j} - 6 \hat{k}$ . If $r \times a = r \times b$ , $r \cdot (\hat{i} + 2 \hat{j} + \hat{k}) = - 3$ , then $r \cdot (2 \hat{i} - 3 \hat{j} + \hat{k})$ is equal to

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Select Answer:

Visualized Solution

Analyze the Cross Product Condition

Rearrange to Find Parallelism

Calculate the Sum Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a+b

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r using Scalar .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Apply First Dot Product Condition

Simplify First Equation

Apply Second Dot Product Condition

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

Find the Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

Calculate Magnitude Squared .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣r∣2

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Power of Parallelism

The Algebraic Symphony

Final Calculation

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