If the resultant of three forces F→$$1=pi^+3j^$$−$$k^$$,F→$$2=$$−$$5i^+j^+2k^$$ and F→$$3=6i^$$−$$k^$$ acting on a particle has a magnitude equal to $$5$$ units, then what is difference in the values of p?

Let R→ be the resultant. ThenR→$$=F$$→$$1+F$$→$$2+F$$→$$3=(p+1)i^+4j^$$ Given, |R→|$$=5$$, therefore $$R2=25$$∵ $$(p+1)2+16=25$$ or $$p+1=$$±$$3$$ or $$p=2$$,−$$4$$

Algebra of vectors

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Question

If the resultant of three forces ${\vec{F}}_{1} = p \hat{i} + 3 \hat{j} - \hat{k}, {\vec{F}}_{2} = - 5 \hat{i} + \hat{j} + 2 \hat{k} and {\vec{F}}_{3} = 6 \hat{i} - \hat{k}$ acting on a particle has a magnitude equal to 5 units, then what is difference in the values of p?

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Solution

Let $\vec{R}$ be the resultant. Then
$\vec{R} = {\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} = (p + 1) \hat{i} + 4 \hat{j}$
$Given, | \vec{R} | = 5, therefore R^{2} = 25$
$\begin{array}{ll} ∵ (p + 1)^{2} + 16 = 25 or p + 1 = \pm 3 \\ or p = 2, - 4 \end{array}$

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