The resultant of two vectors P- and Q- is R- If Q- is doubled then the new resultant vector is perpendicular to P- Then magnitude of R- is-

The correct option is B Q R⃗=P⃗+Q⃗,R⃗ '=P⃗+2Q⃗ ∵R⃗ '.P⃗=0 ∴ [ P⃗+2Q⃗ ].P⃗=0 ⇒ P^2+2Q⃗.P⃗=0 R^2=P^2+Q^2+2P⃗.Q⃗=P^2+Q^2 P^2=Q^2 ⇒ R=Q . ...

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Byju's Answer

Standard XII

Physics

Work Done as Dot Product

The resultant...

Question

The resultant of two vectors $\to P$ and $\to Q$ is $\to R$ . If $\to Q$ is doubled then the new resultant vector is perpendicular to $\to P$ . Then magnitude of $\to R$ is:

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

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Q

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\frac{P}{Q}

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

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Solution

The correct option is B $Q$
$\to R = \to P + \to Q, \to R^{'} = \to P + 2 \to Q$
$∵ \to R^{'} . \to P = 0 ∴ (\begin{matrix} \to P + 2 \to Q \end{matrix}) . \to P = 0 \Rightarrow P^{2} + 2 \to Q . \to P = 0$
$R^{2} = P^{2} + Q^{2} + 2 \to P . \to Q = P^{2} + Q^{2} - P^{2} = Q^{2} \Rightarrow R = Q$ .

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Similar questions

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. The resultant of two vectors

\to P

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\to Q

\to R

. If the magnitude of

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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

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. lf the magnitude of

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\to P

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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)

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and

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are coplanar
b) Angle between

\to P

and

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may be less than

90^{0}

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d)

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\to p, \to q, \to r

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The resultant of two vectors →P and →Q is →R. If →Q is doubled then the new resultant vector is perpendicular to →P. Then magnitude of →R is:

The correct option is B Q→R=→P+→Q,→R′=→P+2→Q∵→R′.→P=0∴(→P+2→Q).→P=0⇒P2+2→Q.→P=0R2=P2+Q2+2→P.→Q=P2+Q2−P2=Q2⇒R=Q.

The resultant of two vectors $\to P$ and $\to Q$ is $\to R$ . If $\to Q$ is doubled then the new resultant vector is perpendicular to $\to P$ . Then magnitude of $\to R$ is: