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Mathematics · Vector Algebra

JEE Main 2024 — 9 April, Shift 2 — Question 10

Between the following two statements :

Statement-I : Let a→=i^+2j^−3k^\overrightarrow{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}} and b⃗=2i^+j^−k^\vec{b}=2 \hat{i}+\hat{j}-\hat{k}. Then the vector r⃗\vec{r} satisfying a→×r→=a→×b→\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}} and a→⋅r→=0\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{r}}=0 is of magnitude 10\sqrt{10}.

Statement-II : In a triangle ABC,cos⁡2 A+cos⁡2 B\mathrm{ABC}, \cos 2 \mathrm{~A}+\cos 2 \mathrm{~B} +cos⁡2C≥−32+\cos 2 C \geq-\frac{3}{2}.

In the light of the above statements, choose the correct answer from the options given below :

  1. Option A:

    Both Statement-I and Statement-II are incorrect

  2. Option B:

    Statement-I is incorrect but Statement-II is correct

    Correct
  3. Option C:

    Both Statement-I and Statement-II are correct

  4. Option D:

    Statement-I is correct but Statement-II is incorrect

Answer: B

Step-by-step solution

a‾=i^+2j^−3k^\quad \overline{\mathrm{a}}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-3 \hat{k}

b‾=2i^+j^−k^\overline{\mathrm{b}}=2 \hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}

a‾×r‾=a‾×b‾\overline{\mathrm{a}} \times \overline{\mathrm{r}}=\overline{\mathrm{a}} \times \overline{\mathrm{b}}

⇒a‾×(r‾−b‾)=0‾\Rightarrow \overline{\mathrm{a}} \times(\overline{\mathrm{r}}-\overline{\mathrm{b}})=\overline{0}

⇒a‾=λ(r‾−b‾)\Rightarrow \overline{\mathrm{a}}=\lambda(\overline{\mathrm{r}}-\overline{\mathrm{b}})

a‾.a‾=λ(a‾⋅r‾−a‾⋅b‾)\overline{\mathrm{a}} . \overline{\mathrm{a}}=\lambda(\overline{\mathrm{a}} \cdot \overline{\mathrm{r}}-\overline{\mathrm{a}} \cdot \overline{\mathrm{b}})

14=−7λ⇒λ=−214=-7 \lambda \Rightarrow \lambda=-2

−a‾2=r‾−b‾⇒r‾=b‾−a‾2\frac{-\overline{\mathrm{a}}}{2}=\overline{\mathrm{r}}-\overline{\mathrm{b}} \Rightarrow \overline{\mathrm{r}}=\overline{\mathrm{b}}-\frac{\overline{\mathrm{a}}}{2}

=2 b‾−a‾2=3i^+k^2=\frac{2 \overline{\mathrm{~b}}-\overline{\mathrm{a}}}{2}=\frac{3 \hat{\mathrm{i}}+\hat{\mathrm{k}}}{2}

Statement (I) is incorrect

cos⁡2 A+cos⁡2 B+cos⁡2c≥−32\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{c} \geq-\frac{3}{2}

2 A+2 B+2C=2π2 \mathrm{~A}+2 \mathrm{~B}+2 \mathrm{C}=2 \pi

cos⁡2 A+cos⁡2 B+cos⁡2C\cos 2 \mathrm{~A}+\cos 2 \mathrm{~B}+\cos 2 \mathrm{C}

=−1−4cos⁡ A⋅cos⁡ B⋅cos⁡C=-1-4 \cos \mathrm{~A} \cdot \cos \mathrm{~B} \cdot \cos \mathrm{C}

≥−1−4×12×12×12\geq-1-4 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}

=−32=-\frac{3}{2}

Statement (II) is correct.

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 2
Subject
Mathematics
Chapter
Vector Algebra
Topic
Vector or Cross Product of Two Vectors

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