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JEE Main 20244 AprilEvening Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Vector Algebra

JEE Main 2024 · 4 April · Evening Shift · Mathematics · Vector Algebra

For λ>0 , let θ be the angle between the vectors a = i +λ j -3 k...

For λ>0\lambda>0, let θ\theta be the angle between the vectors a⃗=i^+λj^−3k^\vec{a}=\hat{i}+\lambda \hat{j}-3 \hat{k} and b⃗=3i^−j^+2k^\vec{b}=3 \hat{i}-\hat{j}+2 \hat{k}. If the vectors a⃗+b⃗\vec{a}+\vec{b} and a⃗−b⃗\vec{a}-\vec{b} are mutually perpendicular, then the value of (14 cos θ)2\theta)^2 is equal to

Options

  1. A50
  2. B40
  3. C25
  4. D20

Correct answer

C. 25

Step-by-step solution

(a⃗+b⃗)⋅(a⃗−b⃗)=0,λ>0∣a⃗∣2−∣b⃗∣2=0→1+λ2+9=9+1+4∴λ=2,cos⁡θ=a⃗−b⃗∣a⃗∣⋅∣b⃗∣=3−λ−614⋅1414cos⁡θ=3−8=−5∴(14cos⁡θ)2=25\begin{aligned} & (\vec{a}+\vec{b}) \cdot(\vec{a}-\vec{b})=0, \lambda>0 \\ & |\vec{a}|^2-|\vec{b}|^2=0 \rightarrow 1+\lambda^2+9=9+1+4 \\ & \therefore \lambda=2, \cos \theta=\frac{\vec{a}-\vec{b}}{|\vec{a}| \cdot|\vec{b}|}=\frac{3-\lambda-6}{\sqrt{14} \cdot \sqrt{14}} \\ & 14 \cos \theta=3-8=-5 \\ & \therefore(14 \cos \theta)^2=25\end{aligned}
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