If the resultant of 3 forces F1¯=pi¯+3j¯−k¯, F2¯=−5i¯+j¯+2k¯ and F3¯=6i¯−k¯ acting on a particle has magnitude equal to 5 units then absolute value of sum of values of P is_____________

Resultant force =F1¯+F2¯+F3¯=(P+1)i¯+4j¯+0k¯Given |(P+1)i¯+4j¯|=5(P+1)2+16=25(P+1)2=9⇒P+1=±3⇒P=3−1,−3−1P=2,−4Sum of values=2-4=-2Absolute value=2

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If the resultant of 3 forces $\bar{F_{1}} = p \bar{i} + 3 \bar{j} - \bar{k}$ , $\bar{F_{2}} = - 5 \bar{i} + \bar{j} + 2 \bar{k}$ and $\bar{F_{3}} = 6 \bar{i} - \bar{k}$ acting on a particle has magnitude equal to 5 units then absolute value of sum of values of P is_____________

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answer is 2.

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Detailed Solution

Resultant force $= \bar{F_{1}} + \bar{F_{2}} + \bar{F_{3}}$
$= (P + 1) \bar{i} + 4 \bar{j} + 0 \bar{k}$
Given $| (P+1) \bar{i} + 4 \bar{j} | = 5$
$\begin{array}{l} (P + 1)^{2} + 16 = 25 \\ (P + 1)^{2} = 9 \Rightarrow P + 1 = \pm 3 \\ \Rightarrow P = 3 - 1, - 3 - 1 \\ P = 2, - 4 \end{array}$
Sum of values=2-4=-2
Absolute value=2