JEE Mains · Maths · STD 12 - 10. vector algebra
If four distinct points with position vectors \(\vec{a}, \vec{b}, \vec{c}\) and \(\vec{d}\) are coplanar; then \([\vec{a} \vec{b} \vec{c}]\) is equal to
- A \([\overrightarrow{ d } \overrightarrow{ c } \overrightarrow{ a }]+[\overrightarrow{ b } \overrightarrow{ d } \overrightarrow{ a }]+[\overrightarrow{ c } \overrightarrow{ d } \overrightarrow{ b }]\)
- B \([\overrightarrow{ d } \overrightarrow{ b } \overrightarrow{ d }]+[\overrightarrow{ a } \overrightarrow{ c } \overrightarrow{ d }]+[\overrightarrow{ d } \overrightarrow{ b } \overrightarrow{ c }]\)
- C \([\overrightarrow{ a } \overrightarrow{ d } \overrightarrow{ b }]+[\overrightarrow{ d } \overrightarrow{ c } \overrightarrow{ a }]+[\overrightarrow{ d } \overrightarrow{ b } \overrightarrow{ c }]\)
- D \([\overrightarrow{ b } \overrightarrow{ c } \overrightarrow{ d }]+[\overrightarrow{ d } \overrightarrow{ a } \overrightarrow{ c }]+[\overrightarrow{ d } \overrightarrow{ b } \overrightarrow{ a }]\)
Answer & Solution
Correct Answer
(A) \([\overrightarrow{ d } \overrightarrow{ c } \overrightarrow{ a }]+[\overrightarrow{ b } \overrightarrow{ d } \overrightarrow{ a }]+[\overrightarrow{ c } \overrightarrow{ d } \overrightarrow{ b }]\)
Step-by-step Solution
Detailed explanation
\(\vec{a}, \vec{b}, \vec{c}, \vec{d}\) are coplanar points. \(\vec{b}-\vec{a}, \vec{c}-\vec{a}, \vec{d}-\vec{a}\) are coplanar vectors. So, \([\overrightarrow{ b }-\overrightarrow{ a } \overrightarrow{ c }-\overrightarrow{ a } \overrightarrow{ d }-\overrightarrow{ a }]=0\)…
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