Vector Algebra
Vectors
star_batch_jee_advanced_2025
Grade 12
Question:
If non-zero vectors $\vec{a}, \vec{b}, \vec{c}$ and $\vec{d}$ satisfy $(\vec{a} \times \vec{b}) \cdot \vec{c} = |\vec{a}||\vec{b}||\vec{c}|$ holds then:
$\vec{a}\vec{b} = 0, \vec{b} \cdot \vec{c} = 0$
$\vec{b} \cdot \vec{c} = 0, \vec{c} \cdot \vec{a} = 0$
$\vec{c} \cdot \vec{a} = 0, \vec{a}\vec{b} = 0$
$\vec{a}\vec{b} = \vec{b} \times \vec{c} = \vec{c} \cdot \vec{a} = 0$
Step-by-Step Solution
Key Concept: The scalar triple product maximum occurs when vectors are mutually perpendicular, with magnitude equal to the product of individual magnitudes.
Use the scalar triple product formula: $(\vec{a} \cdot \vec{b}) \cdot \vec{c} = |\vec{a}||\vec{b}||\vec{c}|\sin(\vec{a}\times\vec{b}) \cos \beta$, where $\alpha$ is the angle between $\vec{a}$ and $\vec{b}$, and $\beta$ is the angle between $\vec{c}$ and the normal to $\vec{a} \times \vec{b}$. If $(\vec{a} \times \vec{b}) \cdot \vec{c} = |\vec{a}||\vec{b}||\vec{c}|$, then $\alpha = \frac{\pi}{2}$ and $\beta = 0$, indicating $\vec{c}$ is parallel to $\vec{a} \times \vec{b}$.
Correct Answer: 1,2,3,4