Question

Three vectors \(\vec{a},\vec{b},\vec{c}\) satisfy the condition \[ \vec{a}+\vec{b}+\vec{c}=\vec{0}. \] If \[ |\vec{a}|=1,\quad |\vec{b}|=3,\quad |\vec{c}|=4, \] then \[ \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} = \] is:

• A. \(12\)
• B. \(-12\)
• C. \(-13\)
• D. \(13\)

Hint:

If \[ \vec{a}+\vec{b}+\vec{c}=\vec{0}, \] then squaring both sides gives a direct relation between magnitudes and dot products.

Answer 1

The Correct Option is C

Step 1: Use the given vector condition.
Given, \[ \vec{a}+\vec{b}+\vec{c}=\vec{0} \] Taking magnitude squared on both sides, \[ |\vec{a}+\vec{b}+\vec{c}|^2=|\vec{0}|^2 \] \[ |\vec{a}+\vec{b}+\vec{c}|^2=0 \]

Step 2: Expand the square of vector sum.
We know that \[ |\vec{a}+\vec{b}+\vec{c}|^2 = |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 +2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) \] Therefore, \[ |\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2 +2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) =0 \]

Step 3: Substitute the given magnitudes.
Given, \[ |\vec{a}|=1,\quad |\vec{b}|=3,\quad |\vec{c}|=4 \] So, \[ 1^2+3^2+4^2 +2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) =0 \] \[ 1+9+16 +2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) =0 \] \[ 26+2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) =0 \] \[ 2(\vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a}) =-26 \] \[ \vec{a}\cdot\vec{b}+\vec{b}\cdot\vec{c}+\vec{c}\cdot\vec{a} =-13 \]

Step 4: Final conclusion.
Hence, \[ \boxed{-13} \]