Let a = i + 2j + 3k, b = 2i + 3j - 5k and c = 3i - j + λk be three vectors. Let r be a unit vector a

Q: Wait, why can't we just find $\vec{r}$ first?

Because $\vec{r}$ depends on $\lambda$ in the denominator, making it messy. It is much more elegant to keep it as a ratio $\frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|}$ and work with the dot product directly. This avoids unnecessary algebraic complexity.

Q: Why did the $\lambda^2$ term cancel out?

It is a linear equation in disguise! When you square both sides, you get $\lambda^2$ on both sides of the equation. Because they have the same coefficient, they subtract to zero. Don't panic when you see squares; they often vanish in these types of problems, leaving you with a simple linear equation.

Q: Is there a shortcut to finding $3\lambda$?

Yes! Notice that the final equation leads to $6\lambda = 50$. Instead of solving for $\lambda = \frac{50}{6}$ and then multiplying by $3$, just divide the entire equation by $2$. It saves time and reduces calculation errors—a classic JEE strategy.

Question

Let $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}$ and $\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$ be three vectors. Let $\vec{r}$ be a unit vector along $\vec{b} + \vec{c}$. If $\vec{r} \cdot \vec{a} = 3$, then $3\lambda$ is equal to:

Options:

(A) 21

(B) 30

(C) 25

(D) 27

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineers! Today, we are diving into a beautiful problem that tests your command over vector algebra. We are given three vectors: $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}$, and $\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$.

Our mission is to find the value of $3\lambda$ given a specific constraint involving a unit vector $\vec{r}$.

### The Vector Sum

Before we can tackle the unit vector $\vec{r}$, we must first understand the foundation it is built upon. The problem states that $\vec{r}$ is a unit vector along the direction of $\vec{b} + \vec{c}$.

We calculate this resultant vector by adding the corresponding components of $\vec{b}$ and $\vec{c}$:

$$\vec{b} + \vec{c} = (2+3)\hat{i} + (3-1)\hat{j} + (-5+\lambda)\hat{k} = 5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}$$

This resultant vector defines the direction we are interested in.

### Defining the Unit Vector

By definition, a unit vector along any vector $\vec{v}$ is given by $\frac{\vec{v}}{|\vec{v}|}$. It is a vector of length one that points in the same direction as $\vec{v}$.

Therefore, our unit vector $\vec{r}$ is defined as:

$$\vec{r} = \frac{\vec{b} + \vec{c}}{|\vec{b} + \vec{c}|}$$

This normalization process is crucial. It ensures that when we take the dot product later, we are essentially looking at the projection of $\vec{a}$ onto the direction of $\vec{b} + \vec{c}$.

### The Dot Product Condition

The problem provides the condition: $\vec{r} \cdot \vec{a} = 3$. Substituting our expression for $\vec{r}$, we obtain:

$$\frac{(\vec{b} + \vec{c}) \cdot \vec{a}}{|\vec{b} + \vec{c}|} = 3$$

This is our master equation. Let's evaluate the numerator and the denominator separately.

First, the numerator $(\vec{b} + \vec{c}) \cdot \vec{a}$:

$$(\vec{b} + \vec{c}) \cdot \vec{a} = 5(1) + 2(2) + (\lambda - 5)(3) = 5 + 4 + 3\lambda - 15 = 3\lambda - 6$$

Next, the denominator, which is the magnitude $|\vec{b} + \vec{c}|$:

$$|\vec{b} + \vec{c}| = \sqrt{5^2 + 2^2 + (\lambda - 5)^2} = \sqrt{29 + (\lambda - 5)^2}$$

### The Algebraic Dance

Now, we substitute these components back into our master equation:

$$\frac{3\lambda - 6}{\sqrt{29 + (\lambda - 5)^2}} = 3$$

Factoring out $3$ from the numerator gives $3(\lambda - 2)$. Dividing both sides by $3$ simplifies the expression:

$$\frac{\lambda - 2}{\sqrt{29 + (\lambda - 5)^2}} = 1$$

Cross-multiplying and squaring both sides to eliminate the radical yields:

$$(\lambda - 2)^2 = 29 + (\lambda - 5)^2$$

Expanding both sides using the identity $(a-b)^2 = a^2 - 2ab + b^2$:

$$\lambda^2 - 4\lambda + 4 = 29 + \lambda^2 - 10\lambda + 25$$

The $\lambda^2$ terms cancel out, leaving us with:

$$-4\lambda + 4 = 54 - 10\lambda$$
$$6\lambda = 50$$

### Final Calculation

We have determined that $6\lambda = 50$. The question asks for the value of $3\lambda$.

By dividing our result by $2$, we find:

**$3\lambda = 25$**

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{i} + 2 \hat{j} + 3 \hat{k}, b = 2 \hat{i} + 3 \hat{j} - 5 \hat{k}$ and $c = 3 \hat{i} - \hat{j} + λ \hat{k}$ be three vectors. Let $r$ be a unit vector along $b + c$ . If $r \cdot a = 3$ , then $3 λ$ is equal to:

Select Answer:

Visualized Solution

Introduction to the Problem

Summing Vectors $b$ and $c$

Defining the Unit Vector $r$

Applying the Dot Product Condition

Calculating the Dot Product Numerator

Calculating the Magnitude Denominator

Forming the Final Equation

Simplifying the Equation

Squaring Both Sides

Expanding the Squares

Simplifying and Solving for $λ$

Final Calculation of $3 λ$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Vector Sum

Defining the Unit Vector

The Dot Product Condition

The Algebraic Dance

Final Calculation

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