Between the following two statements : Statement-I : Let {a} = {i} + 2 {j} - 3 {k} and {b} = 2 {i} + {j} - {k}. Then the vector {r} satisfying {a} × {r} = {a} × {b} and {a} · {r} = 0 is of magnitude {10}. Statement-II : In a triangle ABC, 2A + 2B + 2C ≥ -3/2. [9 Apr 2024 Shift 2]

Correct Answer is : Statement-I is incorrect but Statement-II is correct

Both Statement-I and Statement-II are incorrect

Statement-I is incorrect but Statement-II is correct

Both Statement-I and Statement-II are correct

Statement-I is correct but Statement-II is incorrect

Correct Answer is : Statement-I is incorrect but Statement-II is correct

Test Index

Vector Algebra Part 3

Section: Mathematics

Question : 43 of 72

Marks: +1, -0

\vec{a} = \hat{i} + 2 \hat{j} - 3 \hat{k}

\vec{b} = 2 \hat{i} + \hat{j} - \hat{k}

\vec{r}

\vec{a} \times \vec{r} = \vec{a} \times \vec{b}

\vec{a} \cdot \vec{r} = 0

\sqrt{10}

ABC, \cos 2A + \cos 2B

+ \cos 2C \ge -\frac{3}{2}.

Solution: 👈: Video Solution

\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{0}

\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{\hat{3i + \hat{k}}}{2}

Statement (I) is incorrect

\cos 2A + \cos 2B + \cos 2c \ge -\frac{3}{2}

2A + 2B + 2C = 2\pi

\cos 2A + \cos 2B + \cos 2C

= -1 - 4 \cos A \cdot \cos B \cdot \cos C

\ge -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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