Match List-I with List - II and select the correct answer using the code given below the lists List-I List-II A. Volume of parallelepiped determined by vectors a → , b → and c → is 2 . Then the volume

Question

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

	List-I		List-II
A.	Volume of parallelepiped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $2$ . Then the volume of the parallelepiped determined by vectors $2 (\vec{a} \times \vec{b}), 3 (\vec{b} \times \vec{c})$ and $(\vec{c} \times \vec{a})$ is	P.	$100$
B.	Volume of parallel piped determined by vectors $\vec{a}, \vec{b}$ and $\vec{c}$ is $5$ . Then the volume of the parallelepiped determined by vectors $3 (\vec{a} + \vec{b}), (\vec{b} + \vec{c})$ and $2 (\vec{c} + \vec{a})$ is	Q.	$30$
C	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	R.	$24$
D	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	S.	$60$

TestHub · Accepted Answer

(A) Given, a → b → c → = 2 . . . 1 The volume of the parallelepiped determined by vectors 2 a → × b → , 3 b → × c → , c → × a → is the scalar tripple product of given three vectors is ⇒ V = 2 a → × b → 3 b → × c → c → × a → . . . 2 ⇒ V = 6 a → × b → b → × c → c → × a → Using this result , a → × b → b → × c → c → × a → = a → b → c → 2 . . . 3 ⇒ V = 6 a → b → c → 2 Again putting the value of a → b → c → = 2 We get the volume of the parallelepiped, V = 6 × 4 = 24 . A→R (B) Given, a → b → c → = 5 . . . 1 V = 3 a → + b → b → + c → 2 c → + a → ⇒ V = 6 a → + b → b → + c → c → + a → . . . 2 Using results, a → + b → b → + c → c → + a → = 2 a → b → c → . . . 3 ⇒ V = 6 × 2 a → b → c → ⇒ V = 12 a → b → c → Putting the value a → b → c → from equation 1 , we get the volume of the parallelepiped , ⇒ V = 1 2 × 5 = 6 0 . B⟶S (C)) Given,Area of triangle whose adjacents sides are a → a n d b → , ⇒ ∆ = 1 2 a → × b → = 2 0 . . . 1 Δ 1 = 1 2 2 a → + 3 b → × a → - b → ⇒ ∆ 1 = 1 2 - 2 a → × b → - 3 a → × b → ⇒ ∆ 1 = 5 1 2 a → × b → Using the results from equation 1 ⇒ ∆ 1 = 5 × 2 0 = 1 0 0 C⟶P (D) Given, The area of the parallelogram with adjacent sides determined by vectors a → and b → a → × b → = 3 0 . . . 1 The area of the parallelogram with adjacent sides determined by vectors a → + b → and a → ⇒ Area A = a → + b → × a → = b → × a → , Because a → × a → = 0 ⇒ Area A = b → × a → Using results from 1 ,we get ⇒ Area A = b → × a → = 30 D⟶Q

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