Let a = 3i + 2j + xk and b = i - j + k, for some real x. Then |a x b| = r is possible if : (1) 3√(3/

Q: Wait, why is the j-component $(x-3)$? I thought it was $(3-x)$ when expanding the determinant.

That is a classic trap! When expanding the determinant for the $\hat{j}$ component, we have a negative sign outside the minor: $-\hat{j} [ (3)(1) - (x)(1) ]$. This becomes $-(3-x)$, which simplifies to $(x-3)$. Always watch those signs!

Q: Why can't we just pick a value for $x$ to find $r$?

Because $x$ is a variable, not a constant. The problem asks for the *possible* values of $r$, which implies a range. We need to find the minimum possible value of $r$ and show that $r$ can take any value greater than or equal to that minimum.

Q: Is the cross product always perpendicular to both vectors?

Absolutely! That is the fundamental geometric property of the cross product. Whether $x$ is $1$, $10$, or $-5$, the resulting vector $\vec{a} \times \vec{b}$ will always be orthogonal to both $\vec{a}$ and $\vec{b}$. It is a beautiful invariant in a changing system.

Question

Let $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$, for some real x. Then $|\vec{a} 	imes \vec{b}| = r$ is possible if :

Options:

(A) $3\sqrt{\frac{3}{2}} < r \le 5\sqrt{\frac{3}{2}}$

(B) $0 < r \le \sqrt{\frac{3}{2}}$

(C) $\sqrt{\frac{3}{2}} < r \le 3\sqrt{\frac{3}{2}}$

(D) $r \ge 5\sqrt{\frac{3}{2}}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given two vectors in 3D space: $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. The vector $\vec{b}$ is fixed, while the vector $\vec{a}$ varies based on the parameter $x$.

Our goal is to determine the range of the magnitude of their cross product, defined as $r = |\vec{a} 	imes \vec{b}|$. Geometrically, this magnitude represents the area of the parallelogram formed by the two vectors.

### The Determinant Tool

To compute the cross product, we utilize the determinant method. We arrange the unit vectors and the components of $\vec{a}$ and $\vec{b}$ into a matrix:

$$\vec{a} 	imes \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & 2 & x \ 1 & -1 & 1 \end{vmatrix}$$

Expanding this determinant along the first row, we calculate the components:

$$\vec{a} 	imes \vec{b} = \hat{i}(2 - (-x)) - \hat{j}(3 - x) + \hat{k}(-3 - 2)$$

Simplifying these terms, we obtain the resultant vector:

$$\vec{a} 	imes \vec{b} = (x+2)\hat{i} + (x-3)\hat{j} - 5\hat{k}$$

### The Quadratic Heart

The magnitude $r$ is defined as the square root of the sum of the squares of the components. We express this as:

$$r = \sqrt{(x+2)^2 + (x-3)^2 + (-5)^2}$$

Expanding the squares, we get $(x^2 + 4x + 4) + (x^2 - 6x + 9) + 25$. Combining like terms, the expression inside the square root simplifies to:

$$r = \sqrt{2x^2 - 2x + 38}$$

This quadratic expression represents a parabola that opens upwards. Consequently, it possesses a distinct minimum value at its vertex.

### The Final Optimization

To find the minimum, we apply the vertex formula $x = -\frac{b}{2a}$. Given our quadratic $2x^2 - 2x + 38$, we have $a=2$ and $b=-2$:

$$x = -\frac{-2}{2(2)} = \frac{1}{2}$$

Substituting $x = \frac{1}{2}$ back into the expression for $r$, we calculate the minimum magnitude:

$$r_{min} = \sqrt{2\left(\frac{1}{4}ight) - 2\left(\frac{1}{2}ight) + 38} = \sqrt{0.5 - 1 + 38} = \sqrt{37.5}$$

Converting the decimal to a fraction, we find $r_{min} = \sqrt{\frac{75}{2}} = 5\sqrt{\frac{3}{2}}$. Therefore, the range of the magnitude is **$r \ge 5\sqrt{\frac{3}{2}}$**.

Animated Solution for Mathematics - Vector Algebra: Let $a = 3 \hat{i} + 2 \hat{j} + x \hat{k}$ and $b = \hat{i} - \hat{j} + \hat{k}$ , for some real x. Then $∣ a \times b ∣ = r$ is possible if :

Select Answer:

Visualized Solution

Defining Vectors $a$ and $b$

Cross Product Setup

Expanding the Determinant

The Resultant Vector $a \times b$

Defining the Magnitude $r$

Algebraic Expansion

Simplifying to a Quadratic Form

Quadratic Minimization Strategy

Finding the Vertex $x$

Calculating the Minimum Value of $r$

Simplifying the Final Bound

The Sigma Insight: Vector (Cross) Product

Analyzing the Setup

The Determinant Tool

The Quadratic Heart

The Final Optimization

Similar Questions

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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=3i^+2j^​+xk^ and b=i^−j^​+k^, for some real x. Then ∣a×b∣=r is possible if :

Select Answer:

Visualized Solution

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

Cross Product Setup

Expanding the Determinant

The Resultant Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a×b

Defining the Magnitude .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r

Algebraic Expansion

Simplifying to a Quadratic Form

Quadratic Minimization Strategy

Finding the Vertex .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x

Calculating the Minimum Value of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } r