Question

If four distinct points with position vectors \(\overrightarrow a,\overrightarrow b,\overrightarrow c\)  and \(\overrightarrow d\)  are coplanar, then \([\overrightarrow a \overrightarrow b \overrightarrow c]\) is equal to

• A. \([\overrightarrow a \overrightarrow d \overrightarrow b]+[\overrightarrow d \overrightarrow c \overrightarrow a]+[\overrightarrow d\overrightarrow b\overrightarrow c] \)
• B. \([\overrightarrow b \overrightarrow c \overrightarrow d]+[\overrightarrow d \overrightarrow a \overrightarrow c]+[\overrightarrow d\overrightarrow b\overrightarrow a]\)
• C. \([\overrightarrow d \overrightarrow b \overrightarrow a]+[\overrightarrow a \overrightarrow c \overrightarrow c]+[\overrightarrow d\overrightarrow b\overrightarrow c]\)
• D. \([\overrightarrow d \overrightarrow c \overrightarrow a]+[\overrightarrow b \overrightarrow d \overrightarrow a]+[\overrightarrow c\overrightarrow d\overrightarrow b]\)

Answer 1

The Correct Option is D

1. Represent the condition of coplanarity: 
   Since the vectors ⃗b − ⃗a, ⃗c − ⃗b, and ⃗d − ⃗c are coplanar, we have: 
   [⃗b − ⃗a, ⃗c − ⃗b, ⃗d − ⃗c] = 0
2. Expand the determinant:
   Expanding the determinant using the scalar triple product: 
   (⃗b − ⃗a) · ((⃗c − ⃗b) × (⃗d − ⃗c)) = 0
3. Simplify the expression:
   Using the distributive property: 
   (⃗b − ⃗a) · (⃗c × ⃗b − ⃗c × ⃗a − ⃗a × ⃗d) = 0
4. Represent as scalar triple products:
   The scalar triple products can be written as: 
   [⃗b ⃗c ⃗d] − [⃗b ⃗c ⃗a] − [⃗b ⃗a ⃗d] − [⃗a ⃗c ⃗d] = 0
5. Rearrange to find [⃗a ⃗b ⃗c]:
   Finally, rearranging the terms gives: 
   [⃗a ⃗b ⃗c] = [⃗d ⃗c ⃗a] + [⃗b ⃗d ⃗a] + [⃗c ⃗d ⃗b]

Final Answer:

[⃗a ⃗b ⃗c] = [⃗d ⃗c ⃗a] + [⃗b ⃗d ⃗a] + [⃗c ⃗d ⃗b]