The resultant of two vectors P- and Q- is R- lf the magnitude of Q- is doubled- the new resultant becomes perpendicular to P- then magnitude of R- is -

The correct option is A Q Resultant of two vector P⃗ and Q⃗ is R⃗=√(P^2+Q^2+2PQ cosθ) . #160;Where θ is the angle between the two ...

You visited us 3 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Vector Component

The resultant...

Question

The resultant of two vectors $\to P$ and $\to Q$ is $\to R$ . lf the magnitude of $\to Q$ is doubled, the new resultant becomes perpendicular to $\to P$ , then magnitude of $\to R$ is :

\frac{P^{2} - Q^{2}}{2 P Q}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{P + Q}{P - Q}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Q

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

\frac{P}{Q}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $Q$
Resultant of two vector $\to P$ and $\to Q$ is $\to R = \sqrt{P^{2} + Q^{2} + 2 P Q cos θ}$ .
Where $θ$ is the angle between the two vectors.
As magnitude of $\to Q$ is doubled the angle between the new resultant is $90^{o}$
Thus we get $t a n 90^{o} = \frac{2 Q s i n θ}{P + 2 Q c o s θ}$
As $t a n 90^{o} = \infty$ we have $P + 2 Q c o s θ = 0$

$P = - 2 Q c o s θ$

Now, $R^{2} = P^{2} + Q^{2} + 2 P Q \frac{- P}{2 Q}$

$R = Q$

Suggest Corrections

Similar questions

Q. The resultant of two vectors

\to P

and

\to Q

\to R

. If the magnitude of

\to Q

is doubled, the new resultant vector becomes perpendicular to

\to P

. Then, the magnitude of

\to R

is equal to

Q. The resultant of two vectors

\to P

and

\to Q

\to R

. If

\to Q

is doubled then the new resultant vector is perpendicular to

\to P

. Then magnitude of

\to R

is:

The resultant of two vectors $\to P$ and $\to Q$ is $\to R$ . If Q is doubled, the new resultant is perpendicular to P. Then R equals

Q. lf

\to P \times \to Q = \to R; \to Q \times \to R = \to P

and

\to R \times \to P = \to Q

are 3 non-zero vectors, then,
a)

\to P, \to Q

and

\to R

are coplanar
b) Angle between

\to P

and

\to Q

may be less than

90^{0}

\to P + \to Q + \to R

cannot be equal to zero
d)

\to P, \to Q

and

\to R

are mutually perpendicular
( Given

P \neq Q \neq R \neq 0

)

Q. Let

\to p, \to q, \to r

be three mutually perpendicular vectors of the same magnitude. lf a vector

\to x

satisfies the equation

\to p \times {(\to x - \to q) \times \to p} + \to q \times {(\to x - \to r) \times \to q} + \to r \times {(\to x - \to p) \times \to r}

is given by

Join BYJU'S Learning Program

The resultant of two vectors →P and →Q is →R. lf the magnitude of →Q is doubled, the new resultant becomes perpendicular to →P, then magnitude of →R is :

The resultant of two vectors $\to P$ and $\to Q$ is $\to R$ . lf the magnitude of $\to Q$ is doubled, the new resultant becomes perpendicular to $\to P$ , then magnitude of $\to R$ is :