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Two Vectors Have Magnitudes 2 M and 3m. the Angle Between Them is 60°. Find (A) the Scalar Product of the Two Vectors, (B) the Magnitude of Their Vector Product. - Physics

Answer in Brief

Two vectors have magnitudes 2 m and 3m. The angle between them is 60°. Find (a) the scalar product of the two vectors, (b) the magnitude of their vector product.

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Solution

Let the two vectors be \[\left| \vec{a} \right| = 2 m\text { and } \left| \vec{b} \right| = 3 m\].

Angle between the vectors, θ = 60°
(a) The scalar product of two vectors is given by \[\vec{a} . \vec{b} = \left| \vec{a} \right| . \left| \vec{b} \right| \cos\theta^\circ\]

∴ \[\vec{a} . \vec{b} = \left| \vec{a} \right| . \left| \vec{b} \right| \cos 60^\circ\]

\[= 2 \times 3 \times \frac{1}{2} = 3 m^2\]

(b) The vector product of two vectors is given by \[\left| \vec{a} \times \vec{b} \right| = \left| \vec{a} \right| \left| \vec{a} \right| \sin\theta^\circ\].

∴ \[\left| \vec{a} \times \vec{b} \right| = \left| \vec{a} \right| \left| \vec{a} \right| \sin 60^\circ\]

\[= 2 \times 3 \times \frac{\sqrt{3}}{2}\]

\[ = 3\sqrt{3} m^2\]

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APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 2 Physics and Mathematics
Exercise | Q 11 | Page 29
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