Question

The resultant of two vectors is perpendicular to first vector
of magnitude 6. If the resultant has magnitude 6√3, then
magnitude of second vector is
6√2 N
12 N
9√2N
6√3 N​

Answer 1

Given :

Angle b/w both vectors = 90°

Magnitude of reultant vector = 6√3 N

Magnitude of first vector = 6 N

To Find :

Magnitude of second vector$.$

Solution :

❒ By triangle law or parallelogram law of vector addition, the magnitude of resultant R at two vectors A and B inclined to each other at angle θ, is given by

• R² = A² + B² + 2AB cosθ

By substituting the given values;

➠ (6√3)² = (6)² + B² + 2(6)(B) cos90°

• cos 90° = 0

➠ 108 = 36 + B² + 0

➠ B² = 108 - 36

➠ B² = 72

➠ B = √72

➠ B = √(36 × 2)

➠ B = 6√2 N

Additional Information :

• The latin word vector means carrier.
• The physical quantities which have no specified direction and have different values in different directions are called tensors. For example, moment of inertia.
• A vector whose initial point is fixed is called a localised vector and whose initial point is not fixed is called non-localised vector.

Answer 2

${ \huge{ \bf{AnSwEr :}}}$

The resultant of two vectors is perpendicular

Magnitude of first vector is 6.

Magnitude of reultant vector is 6√3

To Find Magnitude of second vector

using parallelogram law,

${ \to{ \bf {R = \sqrt{A^2 + B^2 + 2AB cos \theta}}}}$

${ \to{ \sf {6 \sqrt{3} = \sqrt{6^2 + B^2 + 2(6)(B) cos90}}}}$

${ \to{ \sf {6 \sqrt{3} = \sqrt{36 + B^2 + 12B(0)}}}}$

${ \to{ \sf {6 \sqrt{3} = \sqrt{36 + B^2 +0}}}}$

${ \to{ \sf {(6 \sqrt{3})^{2} = \big (\sqrt{36 + B^2 } \: \big)^{2} }}}$

${ \to{ \sf {36 \times 3 = 36 + B^2 }}}$

${ \to{ \sf{108 = 36 +B^2 }}}$

${ \to{ \sf{108 - 36 = B^2 }}}$

${ \to{ \sf{72 = B^2 }}}$

${ \to{ \sf{B = \sqrt{72} }}}$

${ \to{ \sf{B = 6 \sqrt{2} } }}$

hence, option a, 6√2 is ur answer