Question #d7dc2

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Question #d7dc2

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Answer 1 · 2017-01-23T03:27:01

We know with reference to the given figure that by Law of parallelogram of vectors

$∣ ∣ ∣ \to R ∣ ∣ ∣ = \sqrt{{∣ ∣ ∣ \to P ∣ ∣ ∣}^{2} + {∣ ∣ ∣ \to Q ∣ ∣ ∣}^{2} + 2 ∣ ∣ ∣ \to P ∣ ∣ ∣ \cdot ∣ ∣ ∣ \to Q ∣ ∣ ∣ \cdot cos θ}$ $|vecR|=sqrt(|vecP|^2+|vecQ|^2+2|vecP|cdot|vecQ|cdot cos theta)$ .....(1)
and $α = {tan}^{- 1} ⎛ ⎜ ⎜ ⎜ ⎝ \frac{∣ ∣ ∣ \to Q ∣ ∣ ∣ sin θ}{∣ ∣ ∣ \to P ∣ ∣ ∣ + ∣ ∣ ∣ \to Q ∣ ∣ ∣ cos θ} ⎞ ⎟ ⎟ ⎟ ⎠$ $alpha=tan^-1((|vecQ|sin theta)/(|vecP|+|vecQ|cos theta))$ .....(2)

Inserting given values in equation (1) we get

$∣ ∣ ∣ \to R ∣ ∣ ∣ = \sqrt{{| 5 |}^{2} + {| - 3 |}^{2} + 2 | 5 | \cdot | - 3 | \cdot cos 60}$ $|vecR|=sqrt(|5|^2+|-3|^2+2|5|cdot|-3|cdot cos 60)$
$\Rightarrow ∣ ∣ ∣ \to R ∣ ∣ ∣ = \sqrt{23 + 9 + 2 \times 5 \times 3 \times 0.5}$ $=>|vecR|=sqrt(23+9+2xx5xx3xx 0.5)$
$\Rightarrow ∣ ∣ ∣ \to R ∣ ∣ ∣ = \sqrt{47}$ $=>|vecR|=sqrt47$
$\Rightarrow ∣ ∣ ∣ \to R ∣ ∣ ∣ = 6.86$ $=>|vecR|=6.86$ , rounded to two decimal places

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