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Vector Algebra question

2024 · 9 Apr · Shift 2 · Q81
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  5. /2024 · 9 Apr · Shift 2 · Q81

Vector Algebra question

2024 · 9 Apr · Shift 2 · Q81

JEE MainPhysicsVector AlgebraNumerical+4 / −1
The resultant of two vectors A⃗\vec{A}A and B⃗\vec{B}B is perpendicular to A⃗\vec{A}A and its magnitude is half that of B⃗\vec{B}B. The angle between vectors A⃗\vec{A}A and B⃗\vec{B}B is ‾\underline{\hspace{2cm}}​∘^\circ∘.
Numerical answer
View written solutionFree

Correct answer: 150

  1. Let the resultant be
R⃗=A⃗+B⃗.\vec{R}=\vec{A}+\vec{B}.R=A+B.
  1. Given that the resultant is perpendicular to A⃗\vec{A}A, so
R⃗⋅A⃗=0.\vec{R}\cdot \vec{A}=0.R⋅A=0.

Thus,

(A⃗+B⃗)⋅A⃗=0(\vec{A}+\vec{B})\cdot \vec{A}=0(A+B)⋅A=0 A⃗⋅A⃗+B⃗⋅A⃗=0\vec{A}\cdot\vec{A}+\vec{B}\cdot\vec{A}=0A⋅A+B⋅A=0 A2+ABcos⁡θ=0A^2+AB\cos\theta=0A2+ABcosθ=0

where θ\thetaθ is the angle between A⃗\vec{A}A and B⃗\vec{B}B.

So,

ABcos⁡θ=−A2AB\cos\theta=-A^2ABcosθ=−A2 Bcos⁡θ=−AB\cos\theta=-ABcosθ=−A cos⁡θ=−AB.(1)\cos\theta=-\frac{A}{B}. \qquad (1)cosθ=−BA​.(1)
  1. Also given that the magnitude of the resultant is half that of B⃗\vec{B}B:
∣R⃗∣=B2.|\vec{R}|=\frac{B}{2}.∣R∣=2B​.

Now,

R2=∣A⃗+B⃗∣2=A2+B2+2ABcos⁡θ.R^2=|\vec{A}+\vec{B}|^2=A^2+B^2+2AB\cos\theta.R2=∣A+B∣2=A2+B2+2ABcosθ.

Using R=B2R=\frac{B}{2}R=2B​,

(B2)2=A2+B2+2ABcos⁡θ.\left(\frac{B}{2}\right)^2=A^2+B^2+2AB\cos\theta.(2B​)2=A2+B2+2ABcosθ.

From (1),

ABcos⁡θ=−A2.AB\cos\theta=-A^2.ABcosθ=−A2.

Hence,

B24=A2+B2−2A2=B2−A2.\frac{B^2}{4}=A^2+B^2-2A^2=B^2-A^2.4B2​=A2+B2−2A2=B2−A2.

So,

A2=B2−B24=3B24.A^2=B^2-\frac{B^2}{4}=\frac{3B^2}{4}.A2=B2−4B2​=43B2​.

Thus,

A=32B.A=\frac{\sqrt{3}}{2}B.A=23​​B.
  1. Substitute into (1):
cos⁡θ=−AB=−32.\cos\theta=-\frac{A}{B}=-\frac{\sqrt{3}}{2}.cosθ=−BA​=−23​​.

Therefore,

θ=150∘.\theta=150^\circ.θ=150∘.
  1. Final integer answer:
150\boxed{150}150​
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