Question

Two vectors of magnitude 6 and 14 then their resultant will be

Answer 1

Explanation:

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Answer 2

Answer:

The resultant will either 8 units or 20 units or the value lies between 8 and 20.

Explanation:

Consider the magnitude of one vector (say),

$| \vec P| = 6$

Consider the magnitude of other vector (say),

$| \vec Q| = 14$

Also,

$P-Q=6-14$

          $=8$ (resultant cannot be negative)

$P+Q=6+14$

          $=20$

Then, the magnitude of resultant vector $| \vec R|$ is given by

$| \vec R| = \sqrt{P^2+Q^2+2PQcos \theta}$ ...... (1)

where $P$ and $Q$ are the two vectors and $\theta$ is the angle between the vectors $P$ and $Q$.

Since $-1 < cos \theta < 1$.

So, $|cos \theta|=1$.

Form (1), we get

⇒ $\sqrt{P^2+Q^2-2PQ} < | \vec R| < \sqrt{P^2+Q^2+2PQ}$ (Since cos$\theta$ = 1)

Using identities,

$(a+b)^2=a^2+b^2+2ab$ and $(a-b)^2=a^2+b^2-2ab$

⇒ $\sqrt{(P-Q)^2} \leq | \vec R| \leq \sqrt{(P+Q)^2}$

⇒ $P-Q \leq |\vec R| \leq P+Q$

Substitute the values of $P-Q$ and $P+Q$ as follows:

⇒ $8 \leq |\vec R| \leq 20$

Therefore, the resultant will either 8 units or 20 units or the value lies between 8 and 20.

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