Mathematics · Vector Algebra

JEE Main 2024 — 4 April, Shift 1 — Question 11

Let a unit vector which makes an angle of 60∘60^{\circ} with 2i^+2j^−k^2 \hat{i}+2 \hat{j}-\hat{k} and an angle of 45∘45^{\circ} with i^−k^\hat{i}-\hat{k} be C⃗\vec{C}. Then C→+(−12i^+132j^−23k^)\overrightarrow{\mathrm{C}}+\left(-\frac{1}{2} \hat{\mathrm{i}}+\frac{1}{3 \sqrt{2}} \hat{\mathrm{j}}-\frac{\sqrt{2}}{3} \hat{\mathrm{k}}\right) is :

  1. Option A:

    −23i^+23j^+(12+223)k^-\frac{\sqrt{2}}{3} \hat{\mathrm{i}}+\frac{\sqrt{2}}{3} \hat{\mathrm{j}}+\left(\frac{1}{2}+\frac{2 \sqrt{2}}{3}\right) \hat{\mathrm{k}}

  2. Option B:

    23i^+132j^−12k^\frac{\sqrt{2}}{3} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{1}{2} \hat{k}

  3. Option C:

    (13+12)i^+(13−132)j^+(13+23)k^\left(\frac{1}{\sqrt{3}}+\frac{1}{2}\right) \hat{\mathrm{i}}+\left(\frac{1}{\sqrt{3}}-\frac{1}{3 \sqrt{2}}\right) \hat{\mathrm{j}}+\left(\frac{1}{\sqrt{3}}+\frac{\sqrt{2}}{3}\right) \hat{\mathrm{k}}

  4. Option D:

    23i^−12k^\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k}

    Correct

Answer: D

Step-by-step solution

C→=C1i^+C2j^+C3k^\overrightarrow{\mathrm{C}}=\mathrm{C}_{1} \hat{i}+\mathrm{C}_{2} \hat{\mathrm{j}}+\mathrm{C}_{3} \hat{k}

C12+C22+C32=1\mathrm{C}_{1}^{2}+\mathrm{C}_{2}^{2}+\mathrm{C}_{3}^{2}=1

C→⋅(2i^+2j^−k^)=∣C∣9cos⁡60∘\overrightarrow{\mathrm{C}} \cdot(2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\hat{\mathrm{k}})=|C| \sqrt{9} \cos 60^{\circ}

2C1+2C2−C3=322 \mathrm{C}_{1}+2 \mathrm{C}_{2}-\mathrm{C}_{3}=\frac{3}{2}

C1−C3=1\mathrm{C}_{1}-\mathrm{C}_{3}=1

C1+2C2=12\mathrm{C}_{1}+2 \mathrm{C}_{2}=\frac{1}{2}

C1=23+12\mathrm{C}_{1}=\frac{\sqrt{2}}{3}+\frac{1}{2}

C2=−132\mathrm{C}_{2}=\frac{-1}{3 \sqrt{2}}

C3=23−12\mathrm{C}_{3}=\frac{\sqrt{2}}{3}-\frac{1}{2}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Subject
Mathematics
Chapter
Vector Algebra
Topic
Scalar or Dot Product of Two Vectors