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Mathematics · Vector Algebra

JEE Main 2025 — 4 April, Morning Shift — Question 24

Consider two vectors u⃗=3i^−j^\vec{u}=3 \hat{i}-\hat{j} and v⃗=2i^+j^−λk^,λ>0\vec{v}=2 \hat{i}+\hat{j}-\lambda \hat{k}, \lambda>0. The angle between them is given by

cos⁡−1(527)\cos ^{-1}\left(\frac{\sqrt{5}}{2 \sqrt{7}}\right). Let v⃗=v⃗1+v⃗2\vec{v}=\vec{v}_{1}+\vec{v}_{2}, where v⃗1\vec{v}_{1} is parallel to u⃗\vec{u} and v⃗2\vec{v}_{2} is perpendicular to u⃗\vec{u}. Then the value ∣v⃗1∣2+∣v⃗2∣2\left|\vec{v}_{1}\right|^{2}+\left|\vec{v}_{2}\right|^{2} is equal to

  1. Option A:

    232\frac{23}{2}

  2. Option B:

    252\frac{25}{2}

  3. Option C:

    14

    Correct
  4. Option D:

    10

Answer: C

Step-by-step solution

u⃗⋅v⃗=∣u∣⋅∣v∣⋅cos⁡θ\vec{u} \cdot \vec{v}=|u| \cdot|v| \cdot \cos \theta

⇒6−1=10⋅5+λ2⋅527⇒1=2⋅5+λ2⋅127⇒14=5+λ2⇒λ2=9⇒λ=3v1=ku⃗v⃗=v⃗1+v⃗2⇒v⃗=ku⃗+v⃗2v⃗u⃗=k⋅∣u⃗∣2⇒5=k⋅10⇒k=12∴v⃗1=u⃗2=3i^2−j2\begin{aligned} \Rightarrow & 6-1=\sqrt{10} \cdot \sqrt{5+\lambda^{2}} \cdot \frac{\sqrt{5}}{2 \sqrt{7}} \\& \Rightarrow 1=\sqrt{2} \cdot \sqrt{5+\lambda^{2}} \cdot \frac{1}{2 \sqrt{7}} \\& \Rightarrow 14=5+\lambda^{2} \\& \Rightarrow \lambda^{2}=9 \\& \Rightarrow \lambda=3 \\& v_{1}=k \vec{u} & \vec{v}=\vec{v}_{1}+\vec{v}_{2} \\& \Rightarrow \vec{v}=k \vec{u}+\vec{v}_{2} & \vec{v} \vec{u}=k \cdot|\vec{u}|^{2} \\& \Rightarrow 5=k \cdot 10 \Rightarrow k=\frac{1}{2} \\& \therefore \vec{v}_{1}=\frac{\vec{u}}{2}=\frac{3 \hat{i}}{2}-\frac{j}{2} \end{aligned}

∣v⃗1∣2=104\left|\vec{v}_{1}\right|^{2}=\frac{10}{4}

v⃗2=v⃗−v⃗1\vec{v}_{2}=\vec{v}-\vec{v}_{1}

=12i^+3j^2−3k^=\frac{1}{2} \hat{i}+\frac{3 \hat{j}}{2}-3 \hat{k}

∣v⃗2∣2=104+9\left|\vec{v}_{2}\right|^{2}=\frac{10}{4}+9

∣v⃗1∣2+∣v⃗2∣2=104+104+9=14\left|\vec{v}_{1}\right|^{2}+\left|\vec{v}_{2}\right|^{2}=\frac{10}{4}+\frac{10}{4}+9=14

Answer key and solution verified before publishing.

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Exam
JEE Main 2025
Paper
4 April, Morning Shift
Subject
Mathematics
Chapter
Vector Algebra
Topic
Applications of Vectors