A vector A- of length 10 units makes an angle of 60- with a vector B- of length 6 units- Find the magnitude of the vector difference A-B- the angle it makes with vector A-

The correct option is C 2√(19), tan^ 1[ 3√(3)/7 ] A⃗ B⃗=√(A^2+B^2 2AB cosθ) #160; #160; #160; #160; #160; #160; #160; #160; ...

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Standard XII

Mathematics

Integration of Piecewise Continuous Functions

A vector A⃗...

Question

A vector $\to A$ of length $10$ units makes an angle of $60^{\circ}$ with a vector $\to B$ of length $6$ units. Find the magnitude of the vector difference $\to A - \to B$ & the angle it makes with vector $\to A$ .

2 \sqrt{19}, t a n^{- 1} (\begin{matrix} \frac{3 \sqrt{3}}{7} \end{matrix})

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\sqrt{19}, t a n^{- 1} (\begin{matrix} \frac{3 \sqrt{3}}{7} \end{matrix})

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2 \sqrt{19}, t a n^{- 1} (\begin{matrix} \frac{\sqrt{3}}{7} \end{matrix})

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\sqrt{19}, t a n^{- 1} (\begin{matrix} \frac{\sqrt{3}}{7} \end{matrix})

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Solution

The correct option is C $2 \sqrt{19}, t a n^{- 1} (\begin{matrix} \frac{3 \sqrt{3}}{7} \end{matrix})$
$∣ ∣ \begin{matrix} \to A - \to B \end{matrix} ∣ ∣ = \sqrt{A^{2} + B^{2} - 2 A B c o s θ}$
$= \sqrt{(10)^{2} + (6)^{2} - 2 (10) (6) c o s 60^{\circ}} = 2 \sqrt{19}$
$t a n α = \frac{6 s i n 60^{\circ}}{10 - 6 c o s 60^{\circ}} = \frac{6 \times \sqrt{3} / 2}{10 - 3} = \frac{3 \sqrt{3}}{7}$
$\Rightarrow α = t a n^{- 1} (\begin{matrix} \frac{3 \sqrt{3}}{7} \end{matrix})$

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A vector →A of length 10 units makes an angle of 60∘ with a vector →B of length 6 units. Find the magnitude of the vector difference →A−→B & the angle it makes with vector →A.

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