Let a = i + 2j + 3k, b = i - j + 2k and c = 5i - 3j + 3k be there vectors. If r is a vector such tha

Q: Wait, why can't we just say $\vec{r} = \vec{c}$? Doesn't the cross product condition imply that?

That is a common trap! The condition $\vec{r} \times \vec{b} = \vec{c} \times \vec{b}$ only implies that $(\vec{r} - \vec{c}) \times \vec{b} = \vec{0}$. This means $(\vec{r} - \vec{c})$ is parallel to $\vec{b}$, not necessarily zero. So, $\vec{r}$ can be any vector on the line passing through $\vec{c}$ with direction $\vec{b}$.

Q: Why do we use $\lambda$ instead of solving for a specific vector directly?

Because the first condition defines a line in 3D space, not a single point. We introduce the scalar $\lambda$ to capture all possible vectors $\vec{r}$ that satisfy the cross product condition. The second condition, $\vec{r} \cdot \vec{a} = 0$, then acts as a constraint that selects the specific point on that line where the vector is perpendicular to $\vec{a}$.

Q: Is there a geometric interpretation of $\vec{r} \cdot \vec{a} = 0$?

Absolutely! The equation $\vec{r} \cdot \vec{a} = 0$ represents a plane passing through the origin with $\vec{a}$ as its normal vector. So, we are essentially finding the intersection of a line (defined by the cross product condition) and a plane (defined by the dot product condition).

Question

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{b} = \hat{i} - \hat{j} + 2\hat{k}$ and $\vec{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$ be there vectors. If $\vec{r}$ is a vector such that, $\vec{r} 	imes \vec{b} = \vec{c} 	imes \vec{b}$ and $\vec{r} \cdot \vec{a} = 0$. Then $25|\vec{r}|^2$ is equal to

Options:

(A) 449

(B) 336

(C) 339

(D) 560

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given three vectors:
$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$
$\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$
$\vec{c} = 5\hat{i} - 3\hat{j} + 3\hat{k}$

Our mission is to find a vector $\vec{r}$ that satisfies two specific conditions:
1. $\vec{r} 	imes \vec{b} = \vec{c} 	imes \vec{b}$
2. $\vec{r} \cdot \vec{a} = 0$

### Decoding the Cross Product

The first condition is $\vec{r} 	imes \vec{b} = \vec{c} 	imes \vec{b}$. By rearranging the terms, we obtain:
$$\vec{r} 	imes \vec{b} - \vec{c} 	imes \vec{b} = \vec{0}$$

Applying the distributive property of the cross product, this simplifies to:
$$(\vec{r} - \vec{c}) 	imes \vec{b} = \vec{0}$$

Since the cross product of two vectors is the zero vector, the vectors must be collinear. Therefore, $(\vec{r} - \vec{c})$ must be parallel to $\vec{b}$, which we express as:
$$\vec{r} - \vec{c} = \lambda \vec{b} \implies \vec{r} = \vec{c} + \lambda \vec{b}$$

### The Dot Product Constraint

The second condition states that $\vec{r} \cdot \vec{a} = 0$. Substituting our parametric form of $\vec{r}$ into this equation, we get:
$$(\vec{c} + \lambda \vec{b}) \cdot \vec{a} = 0$$

Expanding the dot product, we have:
$$\vec{c} \cdot \vec{a} + \lambda (\vec{b} \cdot \vec{a}) = 0$$

### The Calculation

First, we compute the necessary dot products:
$$\vec{c} \cdot \vec{a} = (5)(1) + (-3)(2) + (3)(3) = 5 - 6 + 9 = 8$$
$$\vec{b} \cdot \vec{a} = (1)(1) + (-1)(2) + (2)(3) = 1 - 2 + 6 = 5$$

Substituting these values into the linear equation $8 + 5\lambda = 0$, we find:
$$\lambda = -\frac{8}{5}$$

Now, we determine the vector $\vec{r}$:
$$\vec{r} = \vec{c} - \frac{8}{5}\vec{b} = (5\hat{i} - 3\hat{j} + 3\hat{k}) - \frac{8}{5}(\hat{i} - \hat{j} + 2\hat{k})$$
$$\vec{r} = \frac{1}{5} [ (25\hat{i} - 15\hat{j} + 15\hat{k}) - (8\hat{i} - 8\hat{j} + 16\hat{k}) ] = \frac{1}{5} (17\hat{i} - 7\hat{j} - \hat{k})$$

### Final Calculation

We are asked to find the value of $25|\vec{r}|^2$. First, calculate the magnitude squared:
$$|\vec{r}|^2 = \left(\frac{1}{5}ight)^2 [ 17^2 + (-7)^2 + (-1)^2 ]$$
$$|\vec{r}|^2 = \frac{1}{25} [ 289 + 49 + 1 ] = \frac{339}{25}$$

Finally, multiplying by 25:
$$25|\vec{r}|^2 = 25 	imes \frac{339}{25} = 339$$

The final result is **339**.

Select Answer:

Visualized Solution

Identify Given Vectors

Analyze the Cross Product Condition

Condition for Parallelism

Using the Dot Product Constraint

Calculate $c \cdot a$

Calculate $b \cdot a$

Solving for $λ$

Finding Vector $r$

Calculating $∣ r ∣^{2}$

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

Decoding the Cross Product

The Dot Product Constraint

The Calculation

Final Calculation

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Select Answer:

Visualized Solution

Identify Given Vectors

Analyze the Cross Product Condition

Condition for Parallelism

Using the Dot Product Constraint

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } c⋅a

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

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Final Calculation

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