Question

If \(\overset{⃗}{a},\overset{⃗}{b},\overset{⃗}{c}\) are three non-coplanar vectors and \(\overset{⃗}{p},\overset{⃗}{q},\overset{⃗}{r}\) are defined as \(\overset{⃗}{p} = \frac{\overset{⃗}{b}\times \overset{⃗}{c}}{[\overset{⃗}{a} \overset{⃗}{b} \overset{⃗}{c}]},\overset{⃗}{q} = \frac{\overset{⃗}{c}\times \overset{⃗}{a}}{[\overset{⃗}{a} \overset{⃗}{b} \overset{⃗}{c}]},\overset{⃗}{r} = \frac{\overset{⃗}{a}\times \overset{⃗}{b}}{[\overset{⃗}{a} \overset{⃗}{b} \overset{⃗}{c}]}\), then \([(\overset{⃗}{a}+\overset{⃗}{b})\cdot \overset{⃗}{p}+(\overset{⃗}{b}+\overset{⃗}{c})\cdot \overset{⃗}{q}+(\overset{⃗}{c}+\overset{⃗}{a})\cdot \overset{⃗}{r}]\) is equal to

• A. \(0\)
• B. \(1\)
• C. \(2\)
• D. \(3\)

Hint:

The vectors p, q, r form the reciprocal system, so a dot p equals 1 and b dot p equals 0.

Answer 1

The Correct Option is D

Step 1: Approach
Compute one term explicitly and use cyclic symmetry.

Step 2: First term
$(\vec a+\vec b)\cdot\vec p=\dfrac{(\vec a+\vec b)\cdot(\vec b\times\vec c)}{[\vec a\,\vec b\,\vec c]}=\dfrac{[\vec a\,\vec b\,\vec c]+[\vec b\,\vec b\,\vec c]}{[\vec a\,\vec b\,\vec c]}=1+0=1$.

Step 3: Other terms
The other two terms follow by cyclic change $a\to b\to c\to a$, each equal to 1.

Step 4: Total
$3$, option (D).

Final Answer:
Each bracket equals 1 for the reciprocal system, so the sum is 3, option (D). \[ \boxed{3} \]