Let v = αi + 2j - 3k, w = 2αi + j - k, and u be a vector such that |u| = α 0. If the minimum value

Q: Why do we set the angle to 180 degrees?

The scalar triple product is a dot product between \vec{u} and the vector (\vec{v} \times \vec{w}). The dot product \vec{A} \cdot \vec{B} = |\vec{A}||\vec{B}| \cos \theta. To get the minimum value, we need \cos \theta to be -1, which happens at \theta = 180 degrees, meaning the vectors must be anti-parallel.

Q: Why is \alpha positive?

The problem explicitly states |\vec{u}| = \alpha > 0. This is a crucial constraint that allows us to divide by \alpha without worrying about division by zero or sign flips, simplifying our algebraic journey significantly.

Q: How do I know if m and n are coprime?

Check the prime factorization. We found m = 100 and n = 3401. 100 = 2^2 \times 5^2. 3401 is not divisible by 2 (it is odd) and not divisible by 5 (it does not end in 0 or 5). Thus, they share no common factors, confirming they are coprime.

Question

Let $\vec{v} = \alpha \hat{i} + 2\hat{j} - 3\hat{k}, \vec{w} = 2\alpha \hat{i} + \hat{j} - \hat{k}$, and $\vec{u}$ be a vector such that $|\vec{u}| = \alpha > 0$. If the minimum value of the scalar triple product $[\vec{u} \vec{v} \vec{w}]$ is $-\alpha \sqrt{3401}$, and $|\vec{u} \cdot \hat{i}|^2 = m/n$ where m and n are coprime natural numbers, then $m + n$ is equal to

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given three vectors:
$\vec{v} = \alpha \hat{i} + 2\hat{j} - 3\hat{k}$
$\vec{w} = 2\alpha \hat{i} + \hat{j} - \hat{k}$
$\vec{u}$ with magnitude $|\vec{u}| = \alpha$.

The scalar triple product is defined as $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} 	imes \vec{w})$. Geometrically, this represents the signed volume of the parallelepiped formed by these vectors.

### Calculating the Cross Product

To find the cross product $\vec{v} 	imes \vec{w}$, we evaluate the determinant:

$$
\vec{v} 	imes \vec{w} = 
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \
\alpha & 2 & -3 \
2\alpha & 1 & -1
\end{vmatrix}
$$

Expanding this determinant:
$\hat{i}$ component: $(2)(-1) - (-3)(1) = 1$
$\hat{j}$ component: $-(\alpha(-1) - (-3)(2\alpha)) = -(\alpha - 6\alpha) = -5\alpha$
$\hat{k}$ component: $(\alpha)(1) - (2)(2\alpha) = \alpha - 4\alpha = -3\alpha$

Thus, the cross product is:
$$\vec{v} 	imes \vec{w} = \hat{i} - 5\alpha\hat{j} - 3\alpha\hat{k}$$

### The Optimization Challenge

We aim to minimize the dot product $\vec{u} \cdot (\vec{v} 	imes \vec{w})$. Using the property $\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos 	heta$, the minimum value occurs when $\cos 	heta = -1$, implying $\vec{u}$ is anti-parallel to $(\vec{v} 	imes \vec{w})$.

The minimum value is given by:
$$	ext{Min} = -|\vec{u}| |\vec{v} 	imes \vec{w}|$$

Given $|\vec{u}| = \alpha$ and $|\vec{v} 	imes \vec{w}| = \sqrt{1^2 + (-5\alpha)^2 + (-3\alpha)^2} = \sqrt{1 + 34\alpha^2}$, we equate this to the provided minimum value $-\alpha \sqrt{3401}$:

$$-\alpha \sqrt{1 + 34\alpha^2} = -\alpha \sqrt{3401}$$

### Final Calculation

Squaring both sides, we obtain:
$$1 + 34\alpha^2 = 3401$$
$$34\alpha^2 = 3400$$
$$\alpha^2 = 100 \Rightarrow \alpha = 10$$

Since $\vec{u}$ is anti-parallel to $\vec{v} 	imes \vec{w}$, we have:
$$\vec{u} = -\alpha \frac{\vec{v} 	imes \vec{w}}{|\vec{v} 	imes \vec{w}|} = -10 \frac{\hat{i} - 50\hat{j} - 30\hat{k}}{\sqrt{3401}}$$

The square of the dot product of $\vec{u}$ and $\hat{i}$ is the square of the $x$-component of $\vec{u}$:
$$\left( \frac{-10}{\sqrt{3401}} ight)^2 = \frac{100}{3401}$$

Here, $m = 100$ and $n = 3401$. Since $m$ and $n$ are coprime, the final result is:
**$m + n = 100 + 3401 = 3501$**

Enter Numerical Value:

Visualized Solution

Identify Given Vectors

Define Scalar Triple Product

Calculate $v \times w$ (Setup)

Expand the Determinant

Simplify Cross Product

Calculate Magnitude of $v \times w$

Minimum Value Condition

Equate and Solve for $α$

Find the Value of $α$

Determine Vector $u$

Calculate $∣ u \cdot \hat{i} ∣^{2}$

Final Answer: $m + n$

The Sigma Insight: Scalar Triple Product

Analyzing the Setup

Calculating the Cross Product

The Optimization Challenge

Final Calculation

Similar Questions

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Enter Numerical Value:

Visualized Solution

Identify Given Vectors

Define Scalar Triple Product

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v×w (Setup)

Expand the Determinant

Simplify Cross Product

Calculate Magnitude of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } v×w

Minimum Value Condition

Equate and Solve for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } α

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

Determine Vector .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } u

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣u⋅i^∣2

Final Answer: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m+n