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Vector Algebra question

2025 · 22 Jan · Shift 2 · Q36
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Vector Algebra question

2025 · 22 Jan · Shift 2 · Q36

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a⃗\vec{a}a and b⃗\vec{b}b be two unit vectors such that the angle between them is π3\frac{\pi}{3}3π​. If λa⃗+2b⃗\lambda \vec{a}+2 \vec{b}λa+2b and 3a⃗−λb⃗3 \vec{a}-\lambda \vec{b}3a−λb are perpendicular to each other, then the number of values of λ\lambdaλ in [−1,3][-1,3][−1,3] is :
  1. A
    1
  2. B
    3
  3. C
    2
  4. D
    0
View written solutionFree

Correct answer: D

  1. Since the two vectors are perpendicular, their dot product must be zero:
(λa⃗+2b⃗)⋅(3a⃗−λb⃗)=0(\lambda \vec a + 2\vec b) \cdot (3\vec a - \lambda \vec b)=0(λa+2b)⋅(3a−λb)=0
  1. Use the given facts:
  • ∣a⃗∣=∣b⃗∣=1|\vec a|=|\vec b|=1∣a∣=∣b∣=1
  • angle between a⃗\vec aa and b⃗\vec bb is π3\frac{\pi}{3}3π​
  • so
a⃗⋅b⃗=∣a⃗∣∣b⃗∣cos⁡π3=1⋅1⋅12=12\vec a\cdot \vec b=|\vec a||\vec b|\cos\frac{\pi}{3}=1\cdot 1\cdot \frac12=\frac12a⋅b=∣a∣∣b∣cos3π​=1⋅1⋅21​=21​

Also,

a⃗⋅a⃗=1,b⃗⋅b⃗=1\vec a\cdot \vec a=1, \qquad \vec b\cdot \vec b=1a⋅a=1,b⋅b=1
  1. Expand the dot product:
(λa⃗+2b⃗)⋅(3a⃗−λb⃗)=λa⃗⋅3a⃗+λa⃗⋅(−λb⃗)+2b⃗⋅3a⃗+2b⃗⋅(−λb⃗)(\lambda \vec a + 2\vec b) \cdot (3\vec a - \lambda \vec b) = \lambda \vec a\cdot 3\vec a + \lambda \vec a\cdot (-\lambda \vec b) + 2\vec b\cdot 3\vec a + 2\vec b\cdot (-\lambda \vec b)(λa+2b)⋅(3a−λb)=λa⋅3a+λa⋅(−λb)+2b⋅3a+2b⋅(−λb) =3λ(a⃗⋅a⃗)−λ2(a⃗⋅b⃗)+6(b⃗⋅a⃗)−2λ(b⃗⋅b⃗)= 3\lambda (\vec a\cdot \vec a) - \lambda^2 (\vec a\cdot \vec b) + 6(\vec b\cdot \vec a) - 2\lambda (\vec b\cdot \vec b)=3λ(a⋅a)−λ2(a⋅b)+6(b⋅a)−2λ(b⋅b)

Substitute the values:

=3λ(1)−λ2(12)+6(12)−2λ(1)= 3\lambda(1) - \lambda^2\left(\frac12\right) + 6\left(\frac12\right) - 2\lambda(1)=3λ(1)−λ2(21​)+6(21​)−2λ(1) =3λ−λ22+3−2λ= 3\lambda - \frac{\lambda^2}{2} + 3 - 2\lambda=3λ−2λ2​+3−2λ =−λ22+λ+3= -\frac{\lambda^2}{2} + \lambda + 3=−2λ2​+λ+3

Since the vectors are perpendicular,

−λ22+λ+3=0-\frac{\lambda^2}{2} + \lambda + 3=0−2λ2​+λ+3=0

Multiply by −2-2−2:

λ2−2λ−6=0\lambda^2 - 2\lambda - 6=0λ2−2λ−6=0
  1. Solve the quadratic:
λ=2±4+242=2±282=1±7\lambda=\frac{2\pm\sqrt{4+24}}{2}=\frac{2\pm\sqrt{28}}{2}=1\pm \sqrt7λ=22±4+24​​=22±28​​=1±7​
  1. Check which values lie in [−1,3][-1,3][−1,3]:
  • 1+7≈1+2.646=3.6461+\sqrt7 \approx 1+2.646=3.6461+7​≈1+2.646=3.646, not in [−1,3][-1,3][−1,3]
  • 1−7≈1−2.646=−1.6461-\sqrt7 \approx 1-2.646=-1.6461−7​≈1−2.646=−1.646, not in [−1,3][-1,3][−1,3]

So, there are no values of λ\lambdaλ in the interval [−1,3][-1,3][−1,3].

  1. Hence the correct option is:
D: 0\boxed{\text{D: }0}D: 0​
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