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JEE Main 09-Apr-2024 Paper - Shift 2 — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEEJEE Main 09-Apr-2024 Paper - Shift 24 Marks
MCQ
Between the following two statements :
Statement $-I$ : Let $\overrightarrow{ a }=\hat{ i }+2 \hat{ j }-3 \hat{ k }$ and $\vec{b}=2 \hat{i}+\hat{j}-\hat{k}$. Then the vector $\vec{r}$ satisfying $\overrightarrow{ a } \times \overrightarrow{ r }=\overrightarrow{ a } \times \overrightarrow{ b }$ and $\overrightarrow{ a } \cdot \overrightarrow{ r }=0$ is of magnitude $\sqrt{10}$
Statement $-II$ : In a triangle $\text{ABC} , \cos 2 A+\cos 2 B +\cos 2 C \geq-\frac{3}{2}$
  • A
    Both Statement-I and Statement-II are incorrect
  • B
    Statement-I is incorrect but Statement-II is correct
  • C
    Both Statement-I and Statement-II are correct
  • D
    Statement-I is correct but Statement-II is incorrect
✓

Answer

$\overline{a}=\hat{i}+2 \hat{j}-3 \hat{k}$
$\overline{a}=2 \hat{i}+\hat{j}-\hat{k}$
$\overline{a} \times \overline{r}=\overline{a} \times \overline{b} ; \overline{a} \cdot \overline{r}=0$
$\Rightarrow \overline{a} \times(\overline{r}-\overline{b})=\overline{o}$
$\Rightarrow \overline{a}=\lambda(\overline{r}-\overline{b})$
$\overline{a} \cdot \overline{a}=\lambda(\overline{a} \cdot \overline{r}-\overline{a} \cdot \overline{b})$
$14=-7 \lambda $
$\Rightarrow \lambda=-2$
$\frac{-\overline{a}}{2}=\overline{r}-\overline{b} $
$\Rightarrow \overline{r}=\overline{b}-\frac{\overline{a}}{2}$
$=\frac{2 \overline{b}-\overline{a}}{2}=\frac{3 \hat{i}+\hat{k}}{2}$
Statement $(I)$ is incorrect
$\cos 2 A+\cos 2 B+\cos 2 c \geq-\frac{3}{2}$
$2 A+2 B+2 C=2 \pi$
$\cos 2 A+\cos 2 B+\cos 2 C$
$=-1-4 \cos A \cdot \cos B \cdot \cos C$
$\geq-1-4 \times \frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}$
$=-\frac{3}{2}$
Statement $(II)$ is correct.

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