Let a = a1i + a2j + a3k, b = b1i + b2j + b3k and c = c1i + c2j + c3k be three non-zero vectors such

Q: Wait, why do we equate the determinant to the Scalar Triple Product? Is this always true?

Yes, it is a fundamental property of vector algebra. The determinant of a $3 \times 3$ matrix whose rows are the components of three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is exactly equal to the Scalar Triple Product $[\vec{a} \ \vec{b} \ \vec{c}]$. This is because the determinant calculates the signed volume of the parallelepiped defined by these vectors, which is the geometric definition of the Scalar Triple Product.

Q: If $\vec{c}$ is perpendicular to $\vec{a}$ and $\vec{b}$, does it have to be $\vec{a} \times \vec{b}$?

Not necessarily. The cross product $\vec{a} \times \vec{b}$ is a vector perpendicular to both $\vec{a}$ and $\vec{b}$, but its magnitude is $|\vec{a}| |\vec{b}| \sin(\theta)$. Since $\vec{c}$ is a unit vector, it is parallel to the direction of $\vec{a} \times \vec{b}$, but it might have a different magnitude (specifically, magnitude 1). Therefore, $\vec{c}$ is a unit vector along the same line as $\vec{a} \times \vec{b}$.

Q: Why does the answer involve the magnitudes squared? Where did the components go?

The components are hidden inside the definition of the magnitude squared. Recall that for any vector $\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}$, the magnitude squared is $|\vec{v}|^2 = v_1^2 + v_2^2 + v_3^2$. When we square our result, we naturally get the sum of the squares of the components of $\vec{a}$ and $\vec{b}$.

Question

Let $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ and $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$ be three non-zero vectors such that $\vec{c}$ is a unit vector perpendicular to both the vectors $\vec{a}$ and $\vec{b}$. If the angle between $\vec{a}$ and $\vec{b}$ is $\frac{\pi}{6}$, then $\begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix}^2$ is equal to

Options:

(A) 0

(B) 1

(C) $\frac{1}{4}(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)$

(D) $\frac{3}{4}(a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)(c_1^2 + c_2^2 + c_3^2)$

Sigma Percentile · Accepted Answer

### Analyzing the Geometric Setup

Imagine you are standing in a three-dimensional space, looking at three vectors: $\vec{a}$, $\vec{b}$, and $\vec{c}$. These are the edges of a parallelepiped, a slanted box that holds the secret to our problem.

### The Determinant as a Geometric Soul

When you see a $3 	imes 3$ determinant, do not just think of rows and columns. Think of volume.

The determinant of a matrix formed by three vectors is the **Scalar Triple Product**, denoted as $[\vec{a} \ \vec{b} \ \vec{c}]$. This product is defined as $(\vec{a} 	imes \vec{b}) \cdot \vec{c}$ and represents the volume of the parallelepiped spanned by these vectors.

### The Special Role of Vector $\vec{c}$

The problem provides a specific condition: $\vec{c}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$. This implies that $\vec{c}$ acts as the height of our parallelepiped.

Because $\vec{c}$ is perpendicular to the plane containing $\vec{a}$ and $\vec{b}$, the parallelepiped is a right prism. The cross product $\vec{a} 	imes \vec{b}$ yields a vector perpendicular to the base, with a magnitude equal to the area of the base parallelogram. Since $\vec{c}$ is also perpendicular to this base, $\vec{c}$ must be parallel to $\vec{a} 	imes \vec{b}$.

### The Final Calculation

The Scalar Triple Product is $(\vec{a} 	imes \vec{b}) \cdot \vec{c}$. Since $\vec{c}$ is parallel to $\vec{a} 	imes \vec{b}$, the angle $	heta$ between them is either $0$ or $\pi$.

Thus, the dot product is $\pm |\vec{a} 	imes \vec{b}| |\vec{c}|$. Given that $\vec{c}$ is a unit vector, $|\vec{c}| = 1$. The magnitude of the cross product is calculated as:

$$|\vec{a} 	imes \vec{b}| = |\vec{a}| |\vec{b}| \sin\left(\frac{\pi}{6}ight) = \frac{1}{2} |\vec{a}| |\vec{b}|$$

Squaring this result gives us $\frac{1}{4} |\vec{a}|^2 |\vec{b}|^2$. Substituting the component forms, we arrive at the elegant result:

$$\frac{1}{4} (a_1^2 + a_2^2 + a_3^2)(b_1^2 + b_2^2 + b_3^2)$$

This result is a testament to the beauty of vector geometry, where complex determinants simplify into the product of magnitudes and angles.

Select Answer:

Visualized Solution

Visualizing the Vector Setup

Connecting the Determinant to Vector Algebra

Geometric Interpretation of the Scalar Triple Product

Direction of $a \times b$ and $c$

Expanding the Dot Product

Calculating $∣ a \times b ∣$

Finding the Scalar Triple Product Value

Squaring the Scalar Triple Product

Final Expression and Verification

The Sigma Insight: Scalar Triple Product

Analyzing the Geometric Setup

The Determinant as a Geometric Soul

The Special Role of Vector $c$

The Final Calculation

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