NIMCET 2012 — Mathematics PYQ

• Given Information:

  $\vec{a},\vec{b},$ and $\vec{c}$ are unit vectors. Therefore:

  $$|\vec{a}|=1,|\vec{b}|=1,|\vec{c}|=1$$

  Also, the sum of the vectors is zero:

  $$\vec{a}+\vec{b}+\vec{c}=0$$

• Using the Square of the Sum Formula:

  We know the identity for the magnitude of the sum of three vectors:

  $$|\vec{a}+\vec{b}+\vec{c}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2+|\vec{c}{|}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})$$

• Substitute the known values:

  Since $\vec{a}+\vec{b}+\vec{c}=0$, its magnitude squared is also $0$:

  $$0=(1{)}^2+(1{)}^2+(1{)}^2+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})$$
  $$0=1+1+1+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})$$
  $$0=3+2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})$$

• Solve for the required expression:

  Rearrange the equation to isolate the dot product sum:

  $$2(\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a})=-3$$
  $$\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a}=-\frac{3}{2}$$

Conclusion:

The value of the expression is $-\frac{3}{2}$.

Correct Option: (d)