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Order of Magnitude

Three force v...

Question

Three force vectors $\to P, \to Q, \to R$ of 15kg. each act along AB, BC and CA respectively. The position vectors of $A, B$ and $C$ are given by $\to O A = 2 i - 4 j + 3 k$ , $\to O B = 5 i + 3 j - 2 k,$ $\to O C = - 2 i + 2 j + 3 k .$ Find the resultant force vector $\to S$ of the vectors $\to P, \to Q$ and $\to R .$

A

= 15 [\frac{3 i + 7 j + 5 k}{\sqrt{(83)}} + \frac{7 i - j + 5 k}{5 \sqrt{(3)}} + \frac{4 i - 6 j}{2 \sqrt{(13)}}]

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B

= 15 [\frac{3 i - 7 j - 5 k}{\sqrt{(83)}} + \frac{- 7 i - j - 5 k}{5 \sqrt{(3)}} + \frac{4 i + 6 j}{2 \sqrt{(13)}}]

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C

= 15 [\frac{3 i + 7 j - 5 k}{\sqrt{(83)}} + \frac{- 7 i - j + 5 k}{5 \sqrt{(3)}} + \frac{4 i - 6 j}{2 \sqrt{(13)}}]

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D

= 15 [\frac{3 i + 7 j + 5 k}{\sqrt{(83)}} + \frac{7 i + j + 5 k}{5 \sqrt{(3)}} + \frac{4 i + 6 j}{2 \sqrt{(13)}}]

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Solution

The correct option is C $= 15 [\frac{3 i + 7 j - 5 k}{\sqrt{(83)}} + \frac{- 7 i - j + 5 k}{5 \sqrt{(3)}} + \frac{4 i - 6 j}{2 \sqrt{(13)}}]$
$\to A B = \to O B - \to O A$
$= (5 i + 3 j - 2 k) - (2 i - 4 j + 3 k) = 3 i + 7 j - 5 k$
$\to B C = \to O C - \to O B$
$= (- 2 i + 2 j + 3 k) - (5 i + 3 j - 2 k) = - 7 i - j + 5 k$

$\to C A = \to O A - \to O C$
$= (- 2 i - 4 j + 3 k) - (- 2 i + 2 j + 3 k) = 4 i - 6 j$

$∴ \to S = \to P + \to Q + \to R = 15 ⎡ ⎢ ⎣ \frac{\to A B}{∣ ∣ \to A B ∣ ∣} + \frac{\to B C}{∣ ∣ \to B C ∣ ∣} + \frac{\to C A}{∣ ∣ \to C A ∣ ∣} ⎤ ⎥ ⎦$

$= 15 [\frac{3 i + 7 j - 5 k}{\sqrt{(83)}} + \frac{- 7 i - j + 5 k}{5 \sqrt{(3)}} + \frac{4 i - 6 j}{2 \sqrt{(13)}}]$

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