Match List I with List II and select...

MathematicsScalar Triple ProductJEE Advanced 2013Moderate

Match List I with List II:

List-I

(P)

Volume of parallelepiped determined by vectors

a, b

and

c

is 2. Then the volume of the parallelepiped determined by vectors

2 (a \times b), 3 (b \times c)

and

2 (c \times a)

(Q)

Volume of parallelepiped determined by vectors

a, b

and

c

is 5. Then the volume of the parallelepiped determined by vectors

3 (a + b), 3 (b + c)

and

2 (c + a)

(R)

Area of a triangle with adjacent sides determined by vectors

a

and

b

is 20. Then the area of the triangle with adjacent sides determined by vectors

(2 a + 3 b)

and

(a - b)

(S)

Area of a parallelogram with adjacent sides determined by vectors

a

and

b

is 30. Then the area of the parallelogram with adjacent sides determined by vectors

(a + b)

and

a

List-II

(1)

100

(2)

(3)

(4)

Answer:P → 3, Q → 4, R → 1, S → 2

Visualized Solution (Hindi)

Introduction to Vector Geometry

We need to match the geometric properties in List I with numerical values in List II.
Key concepts involved:
1. Volume of Parallelepiped: $V = [a b c]$ (Scalar Triple Product).
2. Area of Triangle: $A = \frac{1}{2} ∣ a \times b ∣$ .
3. Area of Parallelogram: $A = ∣ a \times b ∣$ .
4. Triple Product Properties: $[a \times b, b \times c, c \times a] = [a b c]^{2}$ .

Part (P): Volume of Cross Products

Part (P): Given $V = [a b c] = 2$ .
We need to find $V^{'} = [2 (a \times b) 3 (b \times c) (c \times a)]$ .
Using the property: $[k_{1} u k_{2} v k_{3} w] = k_{1} k_{2} k_{3} [u v w]$ .
And the identity: $[a \times b b \times c c \times a] = [a b c]^{2}$ .

Calculating Part (P)

$V^{'} = (2 \times 3 \times 1) \times [a \times b b \times c c \times a]$
$V^{'} = 6 \times [a b c]^{2}$
Substitute $[a b c] = 2$ :
$V^{'} = 6 \times (2)^{2} = 6 \times 4 = 24$ .
Match: (P) $\to$ 24 (List II index 2)

Part (Q): Volume of Sum of Vectors

Part (Q): Given $V = [a b c] = 5$ .
We need to find $V^{'} = [3 (a + b) (b + c) 2 (c + a)]$ .
Using the identity: $[a + b b + c c + a] = 2 [a b c]$ .

Calculating Part (Q)

$V^{'} = (3 \times 1 \times 2) \times [a + b b + c c + a]$
$V^{'} = 6 \times 2 [a b c] = 12 [a b c]$
Substitute $[a b c] = 5$ :
$V^{'} = 12 \times 5 = 60$ .
Match: (Q) $\to$ 60 (List II index 3)

Part (R): Area of Triangle

Part (R): Given Area $= \frac{1}{2} ∣ a \times b ∣ = 20 \Rightarrow ∣ a \times b ∣ = 40$ .
New sides: $u = 2 a + 3 b$ and $v = a - b$ .
New Area $A^{'} = \frac{1}{2} ∣ (2 a + 3 b) \times (a - b) ∣$ .

Calculating Part (R)

Expand the cross product:
$(2 a + 3 b) \times (a - b) = 2 (a \times a) - 2 (a \times b) + 3 (b \times a) - 3 (b \times b)$
Since $a \times a = 0$ and $b \times a = - (a \times b)$ :
$= 0 - 2 (a \times b) - 3 (a \times b) - 0 = - 5 (a \times b)$
New Area $A^{'} = \frac{1}{2} ∣ - 5 (a \times b) ∣ = \frac{5}{2} ∣ a \times b ∣ = \frac{5}{2} \times 40 = 100$ .
Match: (R) $\to$ 100 (List II index 0)

Part (S): Area of Parallelogram

Part (S): Given $∣ a \times b ∣ = 30$ .
New sides: $u = a + b$ and $v = a$ .
New Area $A^{'} = ∣ (a + b) \times a ∣ = ∣ a \times a + b \times a ∣$ .
Since $a \times a = 0$ , $A^{'} = ∣ b \times a ∣ = ∣ a \times b ∣ = 30$ .
Match: (S) $\to$ 30 (List II index 1)

Final Matching Result

Final Matching Summary:
- (P) $\to$ 24 (Index 2)
- (Q) $\to$ 60 (Index 3)
- (R) $\to$ 100 (Index 0)
- (S) $\to$ 30 (Index 1)
The correct sequence is [[2], [3], [0], [1]].

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