If vectors P- Q and R have magnitude 5- 12 and 13 units and P- -Q-R- the angle between Q and R is

|P⃗|=5, |Q⃗|=12 and |R⃗|=13 cos θ =Q/R=12/13 ∴θ =cos^ 1 ( 12/13 ) ...

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Byju's Answer

Standard XI

Physics

Vector Addition

If vectors P,...

Question

If vectors P, Q and R have magnitude 5, 12 and 13 units and $\to P + \to Q = \to R$ , the angle between Q and R is

$c o s^{- 1} \frac{5}{12}$

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$c o s^{- 1} \frac{5}{13}$

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$c o s^{- 1} \frac{12}{13}$

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$c o s^{- 1} \frac{7}{13}$

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Solution

The correct option is C
$c o s^{- 1} \frac{12}{13}$

$| \to P | = 5, | \to Q | = 12 a n d | \to R | = 13 c o s θ = \frac{Q}{R} = \frac{12}{13} ∴ θ = c o s^{- 1} (\frac{12}{13})$

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Similar questions

Q. If vectors P. Q and R have magnitude 5, 12 and 13 units

\to P + \to Q = \to R

, the angle between Q and R is

Q. Given that

P = 12, Q = 5

and

R = 13

also

\to P + \to Q = \to R,

then the angle between

\to P

and

\to Q

will be:

Q. Two vectors

\to P

and

\to Q

have magnitudes

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units and

4

units respectively.

\to R

is the resultant

\to P

and

\to Q

. Find the magnitude and direction of

\to R

when the angle

θ

between

\to P

and

\to Q

30^{o}

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

)

If vectors P, Q and R have magnitude 5, 12 and 13 units and $\to P + \to Q = \to R$ , the angle between Q and R is

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If vectors P, Q and R have magnitude 5, 12 and 13 units and →P+→Q=→R, the angle between Q and R is

The correct option is C cos−11213 |→P|=5,|→Q|=12and|→R|=13cosθ=QR=1213∴θ=cos−1(1213)

If vectors P, Q and R have magnitude 5, 12 and 13 units and $\to P + \to Q = \to R$ , the angle between Q and R is

The correct option is C
$c o s^{- 1} \frac{12}{13}$

$| \to P | = 5, | \to Q | = 12 a n d | \to R | = 13 c o s θ = \frac{Q}{R} = \frac{12}{13} ∴ θ = c o s^{- 1} (\frac{12}{13})$