Match List I with List II and select the correct answer using the code given below the lists :
List

Q: Wait, why does the volume of the sum vectors $[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]$ result in $2[\vec{a}, \vec{b}, \vec{c}]$? It feels like it should be more complex.

That is a fantastic observation! When you expand the scalar triple product $[\vec{a}+\vec{b}, \vec{b}+\vec{c}, \vec{c}+\vec{a}]$, you get a sum of eight triple products. However, any triple product containing repeated vectors, such as $[\vec{a}, \vec{b}, \vec{a}]$, is zero because the volume of a parallelepiped with two identical vectors is zero (it collapses into a plane). After canceling all those zero terms, you are left with only $[\vec{a}, \vec{b}, \vec{c}] + [\vec{b}, \vec{c}, \vec{a}]$, which simplifies to $2[\vec{a}, \vec{b}, \vec{c}]$. It is a beautiful example of how symmetry simplifies complex algebra.

Q: In the area calculation, why does the negative sign disappear when we get $-5(\vec{a} \times \vec{b})$?

Remember that the area of a triangle or parallelogram is defined by the magnitude of the cross product, which is always a non-negative scalar. When we calculate the magnitude $|-5(\vec{a} \times \vec{b})|$, we are taking the absolute value of the scalar $-5$, which is $5$. Thus, the negative sign is absorbed by the magnitude operator. Always keep in mind that geometric quantities like area and volume must be positive!

Q: Is there a geometric intuition for why $[\vec{a} \times \vec{b}, \vec{b} \times \vec{c}, \vec{c} \times \vec{a}] = [\vec{a}, \vec{b}, \vec{c}]^2$?

Yes, there is! This identity relates the volume of the original parallelepiped to the volume of the parallelepiped formed by the normal vectors of its faces. These normal vectors are the cross products. This identity is essentially a statement about the relationship between a basis and its reciprocal basis in 3D space. While the algebraic proof is straightforward, visualizing it as the volume of the 'dual' parallelepiped is a powerful way to remember it for the exam.

Question

Match List I with List II:

List-I:

P. Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is 2. Then the volume of the parallelepiped determined by vectors $2(\vec{a} 	imes \vec{b}), 3(\vec{b} 	imes \vec{c})$ and $2(\vec{c} 	imes \vec{a})$ is

Q. Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is 5. Then the volume of the parallelepiped determined by vectors $3(\vec{a} + \vec{b}), 3(\vec{b} + \vec{c})$ and $2(\vec{c} + \vec{a})$ is

R. Area of a triangle with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is 20. Then the area of the triangle with adjacent sides determined by vectors $(2\vec{a} + 3\vec{b})$ and $(\vec{a} - \vec{b})$ is

S. Area of a parallelogram with adjacent sides determined by vectors $\vec{a}$ and $\vec{b}$ is 30. Then the area of the parallelogram with adjacent sides determined by vectors $(\vec{a} + \vec{b})$ and $\vec{a}$ is

List-II:

1. 100

2. 30

3. 24

4. 60

Sigma Percentile · Accepted Answer

### Analyzing the Scalar Triple Product Identity

The volume of a parallelepiped defined by vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ is given by the scalar triple product $[\vec{a}, \vec{b}, \vec{c}]$. A fundamental identity in vector algebra states that the scalar triple product of the cross products of these vectors is equal to the square of the original scalar triple product:

$$[\vec{a} 	imes \vec{b}, \vec{b} 	imes \vec{c}, \vec{c} 	imes \vec{a}] = [\vec{a}, \vec{b}, \vec{c}]^2$$

Given the volume of the original parallelepiped is $2$, we have $[\vec{a}, \vec{b}, \vec{c}] = 2$. Consequently, the scalar triple product of the cross products is $2^2 = 4$.

### Calculating the First Volume

We are asked to find the volume of the parallelepiped defined by $2(\vec{a} 	imes \vec{b})$, $3(\vec{b} 	imes \vec{c})$, and $1(\vec{c} 	imes \vec{a})$. By the linearity property of the scalar triple product, we can factor out the constants:

$$	ext{Volume} = |2 \cdot 3 \cdot 1| \cdot [\vec{a} 	imes \vec{b}, \vec{b} 	imes \vec{c}, \vec{c} 	imes \vec{a}]$$

Substituting the known values:

$$	ext{Volume} = 6 \cdot 4 = \mathbf{24}$$

### Evaluating Sums of Vectors

Next, we consider the volume of a parallelepiped defined by $3(\vec{a} + \vec{b})$, $(\vec{b} + \vec{c})$, and $2(\vec{c} + \vec{a})$, given $[\vec{a}, \vec{b}, \vec{c}] = 5$. First, we factor out the constants $3$, $1$, and $2$:

$$	ext{Volume} = |3 \cdot 1 \cdot 2| \cdot [\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}]$$

Using the expansion property of the scalar triple product for sums of vectors, we know that:

$$[\vec{a} + \vec{b}, \vec{b} + \vec{c}, \vec{c} + \vec{a}] = 2[\vec{a}, \vec{b}, \vec{c}]$$

Substituting the given volume of $5$:

$$	ext{Volume} = 6 \cdot (2 \cdot 5) = \mathbf{60}$$

### Geometric Transformation of Areas

The area of a triangle with sides $\vec{a}$ and $\vec{b}$ is given by $A = \frac{1}{2}|\vec{a} 	imes \vec{b}|$. We transform the sides to $2\vec{a} + 3\vec{b}$ and $\vec{a} - \vec{b}$. The new area $A'$ is:

$$A' = \frac{1}{2} |(2\vec{a} + 3\vec{b}) 	imes (\vec{a} - \vec{b})|$$

Expanding the cross product using the distributive property and the fact that $\vec{a} 	imes \vec{a} = 0$ and $\vec{b} 	imes \vec{a} = -(\vec{a} 	imes \vec{b})$:

$$(2\vec{a} + 3\vec{b}) 	imes (\vec{a} - \vec{b}) = -2(\vec{a} 	imes \vec{b}) + 3(\vec{b} 	imes \vec{a}) = -5(\vec{a} 	imes \vec{b})$$

Given the original area $A = 20$, we have $\frac{1}{2}|\vec{a} 	imes \vec{b}| = 20$, which implies $|\vec{a} 	imes \vec{b}| = 40$. Therefore, the new area is:

$$A' = \frac{1}{2} \cdot 5 \cdot 40 = \mathbf{100}$$

Animated Solution for Mathematics - Vector Algebra: Match List I with List II:

List-I

List-II

Select Matching Pairs:

Visualized Solution

Introduction to Vector Geometry

Part (P): Volume of Cross Products

Part (P): Applying the Triple Product Identity

Part (P): Final Calculation

Part (Q): Volume of Sum of Vectors

Part (Q): Applying the Sum Identity

Part (R): Area of Triangle

Part (R): Cross Product Expansion

Part (S): Area of Parallelogram

Final Matching Result

The Sigma Insight: Scalar Triple Product

Analyzing the Scalar Triple Product Identity

Calculating the First Volume

Evaluating Sums of Vectors

Geometric Transformation of Areas

Similar Questions

Let the vectors $a, b, c$ represent three coterminous edges of a parallelopiped of volume $V$ . Then the volume of the parallelopiped, whose coterminous edges are represented by $a, b + c$ and $a + 2 b + 3 c$ is equal to

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

Let $a, b, c$ be three non-coplanar vectors and $p, q, r$ are vectors defined by the relations $p = \frac{b \times c}{[ a b c ]}, q = \frac{c \times a}{[ a b c ]}, r = \frac{a \times b}{[ a b c ]}$ then the value of the expression $(a + b) \cdot p + (b + c) \cdot q + (c + a) \cdot r$ is equal to

The volume of the parallelopiped whose sides are given by $O A = 2 \hat{i} - 2 \hat{j}, O B = \hat{i} + \hat{j} - \hat{k}, O C = 3 \hat{i} - \hat{k}$ , is

Match the statements / expressions given in Column-I with the values given in Column-II.

List-I

List-II

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = 3 (\hat{i} - \hat{j} + \hat{k})$ . Let $c$ be the vector such that $a \times c = b$ and $a \cdot c = 3$ . Then $a \cdot ((c \times b) - b - c)$ is equal to :

If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

Let $a$ and $b$ be two vectors. Let $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ , then the value of $b \cdot c$ is

Animated Solution for Mathematics - Vector Algebra: Match List I with List II:

List-I

List-II

Select Matching Pairs:

Visualized Solution

Introduction to Vector Geometry

Part (P): Volume of Cross Products

Part (P): Applying the Triple Product Identity

Part (P): Final Calculation

Part (Q): Volume of Sum of Vectors

Part (Q): Applying the Sum Identity

Part (R): Area of Triangle

Part (R): Cross Product Expansion

Part (S): Area of Parallelogram

Final Matching Result

The Sigma Insight: Scalar Triple Product

Analyzing the Scalar Triple Product Identity

Calculating the First Volume

Evaluating Sums of Vectors

Geometric Transformation of Areas

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=αi^+βj^​+3k^, b=βi^−αj^​−k^ and c=i^−2j^​−k^ such that a⋅b=1 and b⋅c=−3, then 31​((a×b)⋅c) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The volume of the parallelopiped whose sides are given by OA=2i^−2j^​,OB=i^+j^​−k^,OC=3i^−k^, is

Match the statements / expressions given in Column-I with the values given in Column-II.

List-I

List-II

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If a=2i^+j^​+3k^,b=3i^+3j^​+k^ and c=c1​i^+c2​j^​+c3​k^ are coplanar vectors and a⋅c=5,b⊥c, then 122(c1​+c2​+c3​) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a and b be two vectors. Let ∣a∣=1,∣b∣=4 and a⋅b=2. If c=(2a×b)−3b, then the value of b⋅c is

If $a = α \hat{i} + β \hat{j} + 3 \hat{k}$ , $b = β \hat{i} - α \hat{j} - \hat{k}$ and $c = \hat{i} - 2 \hat{j} - \hat{k}$ such that $a \cdot b = 1$ and $b \cdot c = - 3$ , then $\frac{1}{3} ((a \times b) \cdot c)$ is equal to

The volume of the parallelopiped whose sides are given by $O A = 2 \hat{i} - 2 \hat{j}, O B = \hat{i} + \hat{j} - \hat{k}, O C = 3 \hat{i} - \hat{k}$ , is

Let $a = \hat{i} + 2 \hat{j} + \hat{k}$ , $b = 3 (\hat{i} - \hat{j} + \hat{k})$ . Let $c$ be the vector such that $a \times c = b$ and $a \cdot c = 3$ . Then $a \cdot ((c \times b) - b - c)$ is equal to :

If $a = 2 \hat{i} + \hat{j} + 3 \hat{k}, b = 3 \hat{i} + 3 \hat{j} + \hat{k}$ and $c = c_{1} \hat{i} + c_{2} \hat{j} + c_{3} \hat{k}$ are coplanar vectors and $a \cdot c = 5, b ⊥ c$ , then $122 (c_{1} + c_{2} + c_{3})$ is equal to

Let $a$ and $b$ be two vectors. Let $∣ a ∣ = 1, ∣ b ∣ = 4$ and $a \cdot b = 2$ . If $c = (2 a \times b) - 3 b$ , then the value of $b \cdot c$ is