If the resultant of three forces F1 - p - - 3- - k- F2 - -5 - - - - 2 k- and F3 - 6 - - k- acting on a particle have magnitude equal to 5 units- then the values of p is are

Resultant of all three vectors are, R = F_1 + F_2 + F_3 R = (p î+ 3ĵ k̂) + ( 5î+ ĵ +2k̂) + (6 î k̂) R = (p +1)î+ 4ĵ + 0 k̂ |R| = √((p +1)^2 + 4^2 +0^2) 5 = ...

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Byju's Answer

Standard XI

Physics

Vector Addition

If the result...

Question

If the resultant of three forces $_{1} = p^i + 3^j -^k,_{2} = - 5^i +^j + 2^k$ and $_{3} = 6^i -^k$ acting on a particle have magnitude equal to 5 units, then the value(s) of $p$ is (are)

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Solution

The correct option is C 2
Resultant of all three vectors are,
$\to R =_{1} +_{2} +_{3}$
$\to R = (p^i + 3^j -^k) + (- 5^i +^j + 2^k) + (6^i -^k)$
$\to R = (p + 1)^i + 4^j + 0^k$
$| \to R | = \sqrt{(p + 1)^{2} + 4^{2} + 0^{2}}$
$5 = \sqrt{(p + 1)^{2} + 4^{2}} p + 1 = \pm 3$
$\Rightarrow p = 2, - 4$

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If the resultant of three forces −→F1=p^i+3^j−^k,−→F2=−5^i+^j+2^k and −→F3=6^i−^k acting on a particle have magnitude equal to 5 units, then the value(s) of p is (are)

The correct option is C 2Resultant of all three vectors are, →R=−→F1+−→F2+−→F3 →R=(p^i+3^j−^k)+(−5^i+^j+2^k)+(6^i−^k) →R=(p+1)^i+4^j+0^k |→R|=√(p+1)2+42+02 5=√(p+1)2+42p+1=±3 ⇒p=2,−4

If the resultant of three forces $_{1} = p^i + 3^j -^k,_{2} = - 5^i +^j + 2^k$ and $_{3} = 6^i -^k$ acting on a particle have magnitude equal to 5 units, then the value(s) of $p$ is (are)

The correct option is C 2
Resultant of all three vectors are,
$\to R =_{1} +_{2} +_{3}$
$\to R = (p^i + 3^j -^k) + (- 5^i +^j + 2^k) + (6^i -^k)$
$\to R = (p + 1)^i + 4^j + 0^k$
$| \to R | = \sqrt{(p + 1)^{2} + 4^{2} + 0^{2}}$
$5 = \sqrt{(p + 1)^{2} + 4^{2}} p + 1 = \pm 3$
$\Rightarrow p = 2, - 4$