Let a = j - k and c = i - j - k. Then the vector b satisfying a × b + c = 0 and a · b = 3 (a) 2i - j

Q: Wait, why is $\vec{b} \cdot \vec{c} = 0$? I thought the cross product was just a vector.

That is a great question! Remember the fundamental property of the cross product: $\vec{a} \times \vec{b}$ is always perpendicular to both $\vec{a}$ and $\vec{b}$. Since the problem states $\vec{a} \times \vec{b} = -\vec{c}$, the vector $-\vec{c}$ must be perpendicular to $\vec{b}$. If a vector is perpendicular to $\vec{b}$, then its dot product with $\vec{b}$ is zero. Thus, $\vec{b} \cdot (-\vec{c}) = 0$, which simplifies to $\vec{b} \cdot \vec{c} = 0$.

Q: Can we use the vector triple product formula to solve this?

You certainly could! You could take the cross product of $\vec{a}$ with both sides of $\vec{a} \times \vec{b} = -\vec{c}$, leading to $\vec{a} \times (\vec{a} \times \vec{b}) = \vec{a} \times (-\vec{c})$. While this is mathematically valid, it introduces more complex vector identities. In a competitive exam like JEE, the component method we used is often faster and less prone to algebraic errors.

Q: Why did we get a family of vectors instead of a single unique solution?

The two conditions provided ($\vec{a} \times \vec{b} = -\vec{c}$ and $\vec{a} \cdot \vec{b} = 3$) are not enough to fix a single point in 3D space; they actually define a line of possible vectors. Any vector on this line satisfies the given conditions. The options provided in the question are simply specific points on this line, and our job was to verify which one of those points matches our derived expression.

Question

Let $\vec{a} = \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$. Then the vector $\vec{b}$ satisfying $\vec{a} 	imes \vec{b} + \vec{c} = \vec{0}$ and $\vec{a} \cdot \vec{b} = 3$

Options:

(A) $2\hat{i} - \hat{j} + 2\hat{k}$

(B) $\hat{i} - \hat{j} - 2\hat{k}$

(C) $\hat{i} + \hat{j} - 2\hat{k}$

(D) $-\hat{i} + \hat{j} - 2\hat{k}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system. You have two fixed vectors, $\vec{a} = \hat{j} - \hat{k}$ and $\vec{c} = \hat{i} - \hat{j} - \hat{k}$, acting as anchors in space.

Your goal is to find a third, elusive vector $\vec{b}$ that satisfies two specific constraints. This is a puzzle of orientation and magnitude.

### The Geometric Insight

The first condition, $\vec{a} 	imes \vec{b} + \vec{c} = \vec{0}$, is our gateway. By rearranging this, we find:

$$\vec{a} 	imes \vec{b} = -\vec{c}$$

The cross product of two vectors, $\vec{a}$ and $\vec{b}$, creates a new vector that is perfectly perpendicular to the plane containing $\vec{a}$ and $\vec{b}$. This means the resulting vector, $-\vec{c}$, is orthogonal to $\vec{b}$.

If $-\vec{c}$ is perpendicular to $\vec{b}$, then $\vec{c}$ is also perpendicular to $\vec{b}$. Mathematically, this is our first golden key:

$$\vec{b} \cdot \vec{c} = 0$$

### The Algebraic Bridge

Let us define our mystery vector as $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$. We utilize the orthogonality condition $\vec{b} \cdot \vec{c} = 0$:

$$(b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) \cdot (\hat{i} - \hat{j} - \hat{k}) = 0$$

Expanding this, we obtain $b_1(1) + b_2(-1) + b_3(-1) = 0$, which simplifies to:

$$b_1 = b_2 + b_3$$

Next, we turn to the second condition: $\vec{a} \cdot \vec{b} = 3$. Substituting $\vec{a} = 0\hat{i} + 1\hat{j} - 1\hat{k}$, we get:

$$(0\hat{i} + 1\hat{j} - 1\hat{k}) \cdot (b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) = 3$$

The $b_1$ term vanishes, leaving us with $b_2 - b_3 = 3$, or:

$$b_2 = 3 + b_3$$

### The Final Synthesis

We now have a system of two equations: $b_1 = b_2 + b_3$ and $b_2 = 3 + b_3$. By substituting the second into the first, we find:

$$b_1 = (3 + b_3) + b_3 = 3 + 2b_3$$

Our mystery vector $\vec{b}$ is now expressed in terms of a single parameter, $b_3$:

$$\vec{b} = (3 + 2b_3)\hat{i} + (3 + b_3)\hat{j} + b_3\hat{k}$$

This represents a family of vectors. Testing the specific case where $b_3 = -2$, we calculate:

$$b_1 = 3 + 2(-2) = -1$$
$$b_2 = 3 + (-2) = 1$$

The resulting vector is **$-\hat{i} + \hat{j} - 2\hat{k}$**, which satisfies all given constraints.

Animated Solution for Mathematics - Vector Algebra: Let $a = \hat{j} - \hat{k}$ and $c = \hat{i} - \hat{j} - \hat{k}$ . Then the vector $b$ satisfying $a \times b + c = 0$ and $a \cdot b = 3$

Select Answer:

Visualized Solution

Visualizing the Given Vectors

Analyzing the Cross Product Condition

The Dot Product Consequence

Setting up the First Equation

Expanding the Dot Product

Utilizing the Second Condition

Setting up the Second Equation

Expanding the Second Dot Product

Expressing $b_{1}$ in terms of $b_{3}$

Constructing the General Vector $b$

Testing the Options

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Geometric Insight

The Algebraic Bridge

The Final Synthesis

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Animated Solution for Mathematics - Vector Algebra: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a=j^​−k^ and c=i^−j^​−k^. Then the vector b satisfying a×b+c=0 and a⋅b=3