Let $\vec{\alpha} . \vec{\beta} , \vec{\gamma}$ be three unit vectors such that $\vec{\alpha} . \vec{\beta} = \vec{\alpha} . \vec{\gamma} = 0 $ and the angle between $\vec{\beta}$ and $\vec{\gamma}$ is $30^{\circ}$ . Then $\vec{\alpha} $ is
- $2 ( \vec{\beta} \times \vec{\gamma})$
- $ - 2 ( \vec{\beta} \times \vec{\gamma})$
- $\pm 2 ( \vec{\beta} \times \vec{\gamma})$
- $ ( \vec{\beta} \times \vec{\gamma})$
The Correct Option is C
Solution and Explanation
$\Rightarrow |\vec{\alpha}|=|\lambda|\left(|\beta||\vec{\gamma}| \cdot \frac{1}{2}\right)$
$\Rightarrow 1=|\lambda| \cdot 1 \cdot 1 \cdot \frac{1}{2}$
$\Rightarrow |\lambda|=2$
$\Rightarrow \lambda=\pm 2$
$\therefore \vec{\alpha}=\pm 2(\vec{\beta} \times \vec{\gamma})$
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Concepts Used:
Multiplication of a Vector by a Scalar
When a vector is multiplied by a scalar quantity, the magnitude of the vector changes in proportion to the scalar magnitude, but the direction of the vector remains the same.
Properties of Scalar Multiplication:
The Magnitude of Vector:
In contrast, the scalar has only magnitude, and the vectors have both magnitude and direction. To determine the magnitude of a vector, we must first find the length of the vector. The magnitude of a vector formula denoted as 'v', is used to compute the length of a given vector ‘v’. So, in essence, this variable is the distance between the vector's initial point and to the endpoint.