The vectors a and b are not perpendicular and c and d are two vectors satisfying b × c = b × d and a

Q: Wait, why can't we just cancel $\vec{b}$ from both sides of $\vec{b} \times \vec{c} = \vec{b} \times \vec{d}$?

That is a classic trap! In vector algebra, the cross product is not an invertible operation like multiplication in scalar algebra. You cannot simply 'divide' by a vector or cancel it out. Instead, you must use the distributive property to factor the expression, which leads to $\vec{b} \times (\vec{c} - \vec{d}) = \vec{0}$. This reveals that $(\vec{c} - \vec{d})$ is parallel to $\vec{b}$, not necessarily equal to $\vec{0}$.

Q: Why is the right side $\vec{0}$ and not just $0$?

This is a vital distinction. The cross product of two vectors is always a vector. Therefore, the result of $\vec{b} \times \vec{c} - \vec{b} \times \vec{d}$ must be the zero vector, denoted as $\vec{0}$. Treating it as a scalar zero is a common mistake that can lead to dimensional inconsistencies in your derivation.

Q: How do we know for sure that $\vec{a} \cdot \vec{b} \neq 0$?

Great observation! The problem explicitly states that $\vec{a}$ and $\vec{b}$ are not perpendicular. By definition, the dot product of two non-zero vectors is zero if and only if they are perpendicular. Since they are not, their dot product $\vec{a} \cdot \vec{b}$ is guaranteed to be non-zero, allowing us to safely divide by it to isolate $\lambda$.

Question

The vectors $\vec{a}$ and $\vec{b}$ are not perpendicular and $\vec{c}$ and $\vec{d}$ are two vectors satisfying $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{d}$ and $\vec{a} \cdot \vec{d} = 0$. Then the vector $\vec{d}$ is equal to

Options:

(A) $\vec{c} + \left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}ight)\vec{b}$

(B) $\vec{b} + \left(\frac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}ight)\vec{c}$

(C) $\vec{c} - \left(\frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}ight)\vec{b}$

(D) $\vec{b} - \left(\frac{\vec{b} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}ight)\vec{c}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a three-dimensional coordinate system, watching vectors $\vec{a}$, $\vec{b}$, $\vec{c}$, and $\vec{d}$ dance in space. This problem is defined by two powerful geometric clues: a cross product equality and a dot product orthogonality.

Our mission is to determine the identity of $\vec{d}$ based on these constraints.

### The Cross Product Mystery

We begin with the condition $\vec{b} 	imes \vec{c} = \vec{b} 	imes \vec{d}$. Rather than attempting to cancel $\vec{b}$, we must bring all terms to one side:

$$\vec{b} 	imes \vec{c} - \vec{b} 	imes \vec{d} = \vec{0}$$

By applying the distributive property, we transform this into:

$$\vec{b} 	imes (\vec{c} - \vec{d}) = \vec{0}$$

When the cross product of two vectors is the zero vector, it implies they are collinear. Thus, the vector $(\vec{c} - \vec{d})$ must be parallel to $\vec{b}$.

Mathematically, we express this as $\vec{c} - \vec{d} = \lambda \vec{b}$, where $\lambda$ is an unknown scalar. Rearranging this gives us our bridge to the solution:

$$\vec{d} = \vec{c} - \lambda \vec{b}$$

### The Dot Product Constraint

Now, we introduce the second condition: $\vec{a} \cdot \vec{d} = 0$. This indicates that $\vec{a}$ is perpendicular to $\vec{d}$.

We substitute our expression for $\vec{d}$ into this dot product:

$$\vec{a} \cdot (\vec{c} - \lambda \vec{b}) = 0$$

Using the distributive property of the dot product, we expand this to:

$$\vec{a} \cdot \vec{c} - \lambda (\vec{a} \cdot \vec{b}) = 0$$

We isolate $\lambda$ by rearranging the scalar terms:

$$\vec{a} \cdot \vec{c} = \lambda (\vec{a} \cdot \vec{b})$$

$$\lambda = \frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}}$$

Since the problem guarantees that $\vec{a}$ and $\vec{b}$ are not perpendicular, we know $\vec{a} \cdot \vec{b} 
eq 0$, making this division valid.

### Final Synthesis

We have found the missing piece of the puzzle. Substituting our value of $\lambda$ back into our expression for $\vec{d}$, we arrive at the final result:

$$\vec{d} = \vec{c} - \left( \frac{\vec{a} \cdot \vec{c}}{\vec{a} \cdot \vec{b}} ight) \vec{b}$$

This is the **elegant result** we were seeking. It demonstrates how vector conditions constrain the orientation and magnitude of vectors in space, turning abstract conditions into a concrete, logical conclusion.

Animated Solution for Mathematics - Vector Algebra: The vectors $a$ and $b$ are not perpendicular and $c$ and $d$ are two vectors satisfying $b \times c = b \times d$ and $a \cdot d = 0$ . Then the vector $d$ is equal to

Select Answer:

Visualized Solution

Problem Setup

The Cross Product Equation

Rearranging Terms

Distributive Property of Cross Product

Collinearity Condition

Expressing Parallel Vectors

Isolating $d$

The Dot Product Condition

Substituting $d$

Expanding the Equation

Finding the Scalar $λ$

Final Vector $d$

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Cross Product Mystery

The Dot Product Constraint

Final Synthesis

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