NIMCET 2021 — Mathematics PYQ

1. Given Information:

• Angle between $\vec{a}$ and $\vec{b}$ is $120^{\circ }$, so $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos (120^{\circ })$.

• Since $\cos (120^{\circ })=-\frac{1}{2}$, we have:

  $$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\left(-\frac{1}{2}\right)$$

• We are given $|\vec{b}|=2|\vec{a}|$. Let $|\vec{a}|=k$, then $|\vec{b}|=2k$.

• Substituting these into the dot product:

  $$\vec{a}\cdot \vec{b}=(k)(2k)\left(-\frac{1}{2}\right)=-k^2$$

  Also, note that $|\vec{a}{|}^2=k^2$ and $|\vec{b}{|}^2=(2k{)}^2=4k^2$.

2. Condition for Perpendicularity:

The vectors $(\vec{a}+x\vec{b})$ and $(\vec{a}-\vec{b})$ are at right angles, so their dot product must be $0$:

$$(\vec{a}+x\vec{b})\cdot (\vec{a}-\vec{b})=0$$

3. Expanding the Dot Product:

$$\vec{a}\cdot \vec{a}-\vec{a}\cdot \vec{b}+x\vec{b}\cdot \vec{a}-x\vec{b}\cdot \vec{b}=0$$

$$|\vec{a}{|}^2-(\vec{a}\cdot \vec{b})+x(\vec{a}\cdot \vec{b})-x|\vec{b}{|}^2=0$$

4. Substituting the Values:

Substitute $|\vec{a}{|}^2=k^2$, $\vec{a}\cdot \vec{b}=-k^2$, and $|\vec{b}{|}^2=4k^2$:

$$k^2-(-k^2)+x(-k^2)-x(4k^2)=0$$

$$k^2+k^2-x{k}^2-4x{k}^2=0$$

$$2k^2-5x{k}^2=0$$

Dividing by $k^2$ (assuming $k\ne 0$):

$$2-5x=0$$

$$5x=2$$

$$x=\frac{2}{5}$$

Conclusion:

The value of $x$ is $\frac{2}{5}$. Therefore, the correct option is D.