Given that P = 12, Q = 5 and R = 13, also P→ + Q→=R→, then the angle between P→ and Q→ is

P = 12, Q = 5, R= 13P¯+Q¯=R¯ (12)2+(5)2=132 i.e.We have P2+Q2=R2⇒P⊥Q∴θ=π2c

Given that P = 12, Q = 5 and R = 13, also P→ + Q→=R→, then the angle between P→ and Q→ is

P = 12, Q = 5, R= 13P¯+Q¯=R¯ (12)2+(5)2=132 i.e.We have P2+Q2=R2⇒P⊥Q∴θ=π2c

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Given that P = 12, Q = 5 and R = 13, also $\vec{P} + \vec{Q} = \vec{R},$ then the angle between $\vec{P} a n d \vec{Q}$ is

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$π$

$\frac{π}{4}$

$\frac{π}{2}$

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Detailed Solution

P = 12, Q = 5, R= 13

$\bar{P} + \bar{Q} = \bar{R}$

${(12)}^{2} + {(5)}^{2} = 13^{2}$ i.e.

We have $P^{2} + Q^{2} = R^{2} \Rightarrow P ⊥ Q$

$∴$ $θ = {\frac{π}{2}}^{c}$

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Given that P = 12, Q = 5 and R = 13, also P→ + Q→=R→, then the angle between P→ and Q→ is

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Given that P = 12, Q = 5 and R = 13, also $\vec{P} + \vec{Q} = \vec{R},$ then the angle between $\vec{P} a n d \vec{Q}$ is