Motion in a Plane Easy Questions and Answers

| A^{^→} × B^{^→} | = √3(A^{^→} . B^{^→})
⇒ AB sinθ = √3 AB cosθ
⇒ tan θ = √3 ⇒ θ = 60°

Correct Option: D

| A^{^→} × B^{^→} | = √3(A^{^→} . B^{^→})
⇒ AB sinθ = √3 AB cosθ
⇒ tan θ = √3 ⇒ θ = 60°

Using the law of vector addition, ( d^{^→} + e^{^→} ) is as shown in the fig.

∴ d^{^→} + e^{^→} = f^{^→}

Correct Option: C

Using the law of vector addition, ( d^{^→} + e^{^→} ) is as shown in the fig.

∴ d^{^→} + e^{^→} = f^{^→}

Vector triple product
A^{^→} × (B^{^→} × C^{^→}) = B^{^→}(A^{^→} . C^{^→}) - C^{^→}(A^{^→} . B^{^→}) = 0
⇒ A^{^→} || (B^{^→} × C^{^→})
[ ∵ A^{^→} . B^{^→} = 0 and A^{^→} . C^{^→} = 0 ]
(A^{^→} + B^{^→})² = (C^{^→})²
⇒ A^{^→}² + B^{^→}² + 2A^{^→} . B^{^→} = C²
⇒ 3² + 4² + 2A^{^→} . B^{^→} = 5²
⇒ 2A^{^→} . B^{^→} = 0
or ⇒ A^{^→} . B^{^→} = 0
∴ A^{^→} ⊥ B^{^→}
Hence A² + B² = C² . Hence A^{^→} ⊥ B^{^→}

Correct Option: D

V^{^→}_av =	∆r^{^→(displacement)}
V^{^→}_av =	∆t(time taken)

=	(13 - 2)î + (14 - 3)ĵ	=	11	(î + ĵ)
	5 - 0		5

Correct Option: D

V^{^→}_av =	∆r^{^→(displacement)}
V^{^→}_av =	∆t(time taken)

=	(13 - 2)î + (14 - 3)ĵ	=	11	(î + ĵ)
	5 - 0		5

|A^{^→} + B^{^→}| = |A^{^→} - B^{^→}|
Squaring on both sides
|A^{^→} + B^{^→}|² = |A^{^→} - B^{^→}|²
⇒ A^{^→} . A^{^→} + 2A^{^→} . B^{^→} + B^{^→} . B^{^→} = A^{^→} . A^{^→} - 2A^{^→} . B^{^→} + B^{^→} . B^{^→}
⇒ 4A^{^→} . B^{^→} = 0 ⇒ 4A^{^→} . B^{^→} cos θ = 0
⇒ cos θ = 0 ⇒ θ = 90°

Motion in a Plane

Motion in a Plane

Mechanics

Correct Option: D

Correct Option: C

Correct Option: D

Correct Option: D

Correct Option: B

	1	(13î + 14ĵ)
	5