NIMCET 2021 — Mathematics PYQ

We are given the system of equations:

1. $2\mathbf{a}+\mathbf{b}={\mathbf{e}}_1$

2. $\mathbf{a}+2\mathbf{b}={\mathbf{e}}_2$

To isolate $\mathbf{a}$, multiply equation (2) by 2 and subtract equation (1):

Similarly, to isolate $\mathbf{a}$:

2. Calculating the components of a and b

Given ${\mathbf{e}}_1=(1,1,1)$ and ${\mathbf{e}}_2=(1,1,-1)$:

For vector a:

For vector b:

3. Finding the Angle ($\theta$)

The formula for the angle between two vectors is:

• Dot Product ($\mathbf{a}\cdot \mathbf{b}$):

  $$\mathbf{a}\cdot \mathbf{b}=\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)+\left(\frac{1}{3}\right)\left(\frac{1}{3}\right)+(1)(-1)=\frac{1}{9}+\frac{1}{9}-1=\frac{2}{9}-\frac{9}{9}=-\frac{7}{9}$$

• Magnitudes ($|\mathbf{a}|$ and $|\mathbf{b}|$):

  $$|\mathbf{a}|=\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2+1^2}=\sqrt{\frac{1}{9}+\frac{1}{9}+1}=\sqrt{\frac{11}{9}}=\frac{\sqrt{11}}{3}$$
  $$|\mathbf{b}|=\sqrt{{\left(\frac{1}{3}\right)}^2+{\left(\frac{1}{3}\right)}^2+(-1{)}^2}=\sqrt{\frac{11}{9}}=\frac{\sqrt{11}}{3}$$

• Final Calculation:

  $$\cos \theta =\frac{-7/9}{(\sqrt{11}/3)\cdot (\sqrt{11}/3)}=\frac{-7/9}{11/9}=-\frac{7}{11}$$

Since the magnitude of the angle in these problems is often sought as an absolute value or based on the specific orientation provided in choices: