Let a vector a has a magnitude 9. Let a vector b be such that for every (x, y) ∈ R × R - {(0, 0)}, t

Q: Wait, why can we just set the coefficients to zero? Is that always allowed?

That is a fantastic question! In this problem, we are told the condition holds for *every* $(x, y) \neq (0, 0)$. If a polynomial expression in $x$ and $y$ is zero for all values of $x$ and $y$, the only way that is possible is if the coefficient of each term is zero. It is like saying $Ax^2 + Bxy + Cy^2 = 0$ for all $x, y$; if you pick $(1, 0)$, you get $B=0$. If you pick $(0, 1)$, you get $C=0$. It is a fundamental property of identities!

Q: Why is $\vec{a} \cdot \vec{b} = 0$?

When we expand the dot product, we get terms involving $x^2(\vec{a} \cdot \vec{b})$ and $y^2(\vec{a} \cdot \vec{b})$. Since the equation must hold for all $x$ and $y$, the coefficient of these terms must be zero. Thus, $6(\vec{a} \cdot \vec{b}) = 0$, which directly implies $\vec{a} \cdot \vec{b} = 0$. This tells us that the vectors $\vec{a}$ and $\vec{b}$ are orthogonal.

Q: How do I know the cross product magnitude is just $|\vec{a}||\vec{b}|$?

Great observation! The magnitude of the cross product is defined as $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin(\theta)$, where $\theta$ is the angle between the vectors. Since we just proved that $\vec{a} \cdot \vec{b} = 0$, we know the angle $\theta$ is $90^{\circ}$. Since $\sin(90^{\circ}) = 1$, the expression simplifies beautifully to just $|\vec{a}||\vec{b}|$.

Question

Let a vector $\vec{a}$ has a magnitude 9. Let a vector $\vec{b}$ be such that for every $(x, y) \in \mathbb{R} 	imes \mathbb{R} - \{(0, 0)\}$, the vector $(x\vec{a} + y\vec{b})$ is perpendicular to the vector $(6y\vec{a} - 18x\vec{b})$. Then the value of $|\vec{a} 	imes \vec{b}|$ is equal to:

Options:

(A) $9\sqrt{3}$

(B) $27\sqrt{3}$

(C) 9

(D) 81

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional space. You have two vectors, $\vec{a}$ and $\vec{b}$, and a mysterious condition that binds them together.

The problem states that for any combination of scalars $x$ and $y$, the vector $(x\vec{a} + y\vec{b})$ is perpendicular to $(6y\vec{a} - 18x\vec{b})$. When two vectors are perpendicular, their dot product is zero.

### The Algebraic Expansion

We begin by setting the dot product of the two vectors to zero:

$$(x\vec{a} + y\vec{b}) \cdot (6y\vec{a} - 18x\vec{b}) = 0$$

Using the distributive property of the dot product, we expand the expression into four distinct terms:

$$6xy|\vec{a}|^2 - 18x^2(\vec{a} \cdot \vec{b}) + 6y^2(\vec{b} \cdot \vec{a}) - 18xy|\vec{b}|^2 = 0$$

Since the dot product is commutative, we know that $\vec{b} \cdot \vec{a} = \vec{a} \cdot \vec{b}$. Substituting this into our equation simplifies the expression significantly.

### The Identity Principle

Grouping the terms by their variable components, we obtain:

$$6xy(|\vec{a}|^2 - 3|\vec{b}|^2) + 6(\vec{a} \cdot \vec{b})(y^2 - 3x^2) = 0$$

This equation must hold for **all** values of $x$ and $y$. For a polynomial in $x$ and $y$ to be zero for all inputs, the coefficients of the independent terms must vanish.

This forces two critical conditions:

$$6(|\vec{a}|^2 - 3|\vec{b}|^2) = 0 \quad 	ext{and} \quad 6(\vec{a} \cdot \vec{b}) = 0$$

### The Geometric Revelation

From the second condition, we immediately conclude that $\vec{a} \cdot \vec{b} = 0$, which implies that $\vec{a}$ and $\vec{b}$ are **perpendicular**.

Now, we solve the first condition using the given magnitude $|\vec{a}| = 9$:

$$|\vec{a}|^2 - 3|\vec{b}|^2 = 0 \implies 81 - 3|\vec{b}|^2 = 0$$

Solving for $|\vec{b}|$, we find:

$$|\vec{b}|^2 = 27 \implies |\vec{b}| = \sqrt{27} = 3\sqrt{3}$$

### Final Calculation

We are tasked with finding the magnitude of the cross product $|\vec{a} 	imes \vec{b}|$. Since $\vec{a} \perp \vec{b}$, the angle between them is $90^{\circ}$, and the magnitude is given by:

$$|\vec{a} 	imes \vec{b}| = |\vec{a}||\vec{b}|\sin(90^{\circ}) = |\vec{a}||\vec{b}|$$

Substituting our known values:

$$|\vec{a} 	imes \vec{b}| = 9 	imes 3\sqrt{3} = 27\sqrt{3}$$

The final result is **$27\sqrt{3}$**.

Animated Solution for Mathematics - Vector Algebra: Let a vector $a$ has a magnitude 9. Let a vector $b$ be such that for every $(x, y) \in R \times R - {(0, 0)}$ , the vector $(x a + y b)$ is perpendicular to the vector $(6 y a - 18 x b)$ . Then the value of $∣ a \times b ∣$ is equal to:

Select Answer:

Visualized Solution

Understanding the Perpendicularity Condition

Setting up the Dot Product

Expanding the Dot Product

Simplifying the Terms

Grouping the Terms

Applying the Identity Condition

Deducing Perpendicularity

Calculating Magnitude of Vector $b$

Finding the Cross Product Magnitude

Final Calculation

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Algebraic Expansion

The Identity Principle

The Geometric Revelation

Final Calculation

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Select Answer:

Visualized Solution

Understanding the Perpendicularity Condition

Setting up the Dot Product

Expanding the Dot Product

Simplifying the Terms

Grouping the Terms

Applying the Identity Condition

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