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Vector Algebra — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsVector Algebra1 Mark
MCQ
Choose the correct answer
If $\theta$ is the angle between any two vectors $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}, \text{then}\ |\vec{\text{a}}\cdot\vec{\text{b}}|=|\vec{\text{a}}\times\vec{\text{b}}|\ \text{when}\ \theta$ is equal to
  • A
    0
  • ✓
    $\frac{\pi}{4}$
  • C
    $\frac{\pi}{2}$
  • D
    $\pi$
✓

Answer

Correct option: B.
$\frac{\pi}{4}$
Let $\theta$ be the angle between two vectors $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}.$
Then, without loss of generality, $\vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ are non-zero vectors, so that $\big|\vec{\text{a}}\big|\ \text{and}\ \Big|\vec{\text{b}}\Big|$ are position.
$\Big|\vec{\text{a}}\cdot\vec{\text{b}}\Big|=\Big|\vec{\text{a}}\times\vec{\text{b}}\Big|$
$\Rightarrow\big|\vec{\text{a}}\big|\Big|\vec{\text{b}}\Big|\cos\theta=\big|\vec{\text{a}}\big|\Big|\vec{\text{b}}\Big|\sin\theta$
$\Rightarrow\cos\theta=\sin\theta\ \ \ \Big[\big|\vec{\text{a}}\big|\ \text{and}\ \Big|\vec{\text{b}}\Big|\Big]\ \text{are positive}$
$\Rightarrow\tan\theta=1$
$\Rightarrow\theta=\frac{\pi}{4}$
Hence, $\Big|\vec{\text{a}}.\vec{\text{b}}\Big|=\Big|\vec{\text{a}}\times\vec{\text{b}}\Big|$ when $\theta$ is equal to $\frac{\pi}{4}.$
The correct answer is B.

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