Let veca, vecb and vecc are three non-collinear vectors in a plane such that | veca|=2,| vecb|=5 and | vecc|=√29. If the angle between veca and vecc is θ1 and the angle between vecb and vecc is θ2, where θ1, θ2 ∈[(π/2), π], then the value of θ1+θ2 is equal to

Question

Let veca, vecb and vecc are three non-collinear vectors in a plane such that | veca|=2,| vecb|=5 and | vecc|=√29. If the angle between veca and vecc is θ1 and the angle between vecb and vecc is θ2, where θ1, θ2 ∈[(π/2), π], then the value of θ1+θ2 is equal to (A) (7 π /6) (B) (4 π /6) (C) (3 π /2) (D) (7 π /4)

Tardigrade Admin · Accepted Answer

$\frac{7 π}{6}$

$\frac{4 π}{6}$

$\frac{3 π}{2}$

$\frac{7 π}{4}$

$∣ a ∣^{2} + ∣ ∣ b ∣ ∣^{2} = ∣ c ∣^{2}$ (property of triangles)

From the diagram,
$π - θ_{1} + π - θ_{2} = \frac{π}{2}$
$\Rightarrow θ_{1} + θ_{2} = \frac{3 π}{2}$

Q. Let a,b and c are three non-collinear vectors in a plane such that ∣a∣=2,∣b∣=5 and ∣c∣=29​. If the angle between a and c is θ1​ and the angle between b and c is θ2​, where θ1​,θ2​∈[2π​,π], then the value of θ1​+θ2​ is equal to

67π​

64π​

23π​

47π​

∣a∣2+∣∣​b∣∣​2=∣c∣2 (property of triangles) From the diagram, π−θ1​+π−θ2​=2π​ ⇒θ1​+θ2​=23π​

$\frac{7 π}{6}$

$\frac{4 π}{6}$

$\frac{3 π}{2}$

$\frac{7 π}{4}$

$∣ a ∣^{2} + ∣ ∣ b ∣ ∣^{2} = ∣ c ∣^{2}$ (property of triangles)

From the diagram,
$π - θ_{1} + π - θ_{2} = \frac{π}{2}$
$\Rightarrow θ_{1} + θ_{2} = \frac{3 π}{2}$