Find the magnitude of each of the two vectors a→ and b→, having the same magnitude such that the angle between them is 600 and their scalar product is 92.

Assume that a→and b→be two such that vectors.Now, a→⋅b→=|a→||b→|cos⁡θ .....(1)It is given that |a→|=|b→|,θ=60∘ and a→⋅b→=92∴Putting these values in equation (i) 92 =|a→||a→|cos⁡60∘⇒ 92 =|a→|2⋅12⇒ |a→|2 =9⇒ |a→| =3 ∴ |a→| =|b→|=3Therefore, the magnitude is 3 of each of the two vectorsa→ and b→, having the same magnitude such that the angle between them is 60° and their scalar product is 92.

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Find the magnitude of each of the two vectors $\vec{a}$ and $\vec{b}$ , having the same magnitude such that the angle between them is $60^{0}$ and their scalar product is $\frac{9}{2} .$

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Detailed Solution

Assume that $\vec{a}$ and $\vec{b}$ be two such that vectors.
Now, $\vec{a} \cdot \vec{b} = | \vec{a} | | \vec{b} | \cos θ . . . . . (1)$
It is given that $| \vec{a} | = | \vec{b} |, θ = 60^{\circ} and \vec{a} \cdot \vec{b} = \frac{9}{2}$
$∴$ Putting these values in equation (i)
$\begin{array}{l} \frac{9}{2} = | \vec{a} | | \vec{a} | \cos 60^{\circ} \\ \Rightarrow \frac{9}{2} = | \vec{a} |^{2} \cdot \frac{1}{2} \\ \Rightarrow | \vec{a} |^{2} = 9 \\ \Rightarrow | \vec{a} | = 3 \\ ∴ | \vec{a} | = | \vec{b} | = 3 \end{array}$
Therefore, the magnitude is 3 of each of the two vectors $\vec{a}$ and $\vec{b}$ , having the same magnitude such that the angle between them is 60° and their scalar product is $\frac{9}{2}$ .