A vector of magnitude 5 $\sqrt{3}$ units and another vector of magnitude 10 units are inclined to each other at an angle of $30^{\circ}$ . The magnitude of their vector product is

\sqrt{3}

units

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\sqrt{3}

units

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10 units

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50 units

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Solution

The correct option is B 25 $\sqrt{3}$ units
If two vector are given $\to a, \to b$
Their vector product & given by:
$∣ ∣ ∣ \to a \times \to b ∣ ∣ ∣ = | a | | b | sin θ$
where is angle between them in the given case,
magnitude of vector $= | A | | B | sin θ$
$\begin{matrix} = 5 \sqrt{3} \times 10 sin 30 = 25 \sqrt{3} \end{matrix}$

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A vector of magnitude 5√3 units and another vector of magnitude 10 units are inclined to each other at an angle of 30∘ . The magnitude of their vector product is

The correct option is B 25√3 unitsIf two vector are given →a,→bTheir vector product & given by:∣∣∣→a×→b∣∣∣=|a||b|sinθwhere is angle between them in the given case,magnitude of vector =|A||B|sinθ=5√3×10sin30=25√3

A vector of magnitude 5 $\sqrt{3}$ units and another vector of magnitude 10 units are inclined to each other at an angle of $30^{\circ}$ . The magnitude of their vector product is