NIMCET 2023 — Mathematics PYQ

1. Simplify the Vector Expression Algebraically

Before substituting the actual vector values of $\vec{a}$ and $\vec{b}$, we can simplify the inner vector triple product term using the expansion formula:

$$\vec{X}\times (\vec{Y}\times \vec{Z})=(\vec{X}\cdot \vec{Z})\vec{Y}-(\vec{X}\cdot \vec{Y})\vec{Z}$$

Let's expand the core expression inside the square brackets:

$$[(\vec{a}\times \vec{b})\times (\vec{a}-2\vec{b})]$$

Distribute the cross product across the terms inside the parentheses:

$$=[(\vec{a}\times \vec{b})\times \vec{a}]-[(\vec{a}\times \vec{b})\times 2\vec{b}]$$

$$=[(\vec{a}\times \vec{b})\times \vec{a}]-2[(\vec{a}\times \vec{b})\times \vec{b}]$$

Using the identity $(\vec{X}\times \vec{Y})\times \vec{Z}=(\vec{Z}\cdot \vec{X})\vec{Y}-(\vec{Z}\cdot \vec{Y})\vec{X}$:

1. For the first term: $[(\vec{a}\times \vec{b})\times \vec{a}]=(\vec{a}\cdot \vec{a})\vec{b}-(\vec{a}\cdot \vec{b})\vec{a}=|\vec{a}{|}^2\vec{b}-(\vec{a}\cdot \vec{b})\vec{a}$

2. For the second term: $[(\vec{a}\times \vec{b})\times \vec{b}]=(\vec{b}\cdot \vec{a})\vec{b}-(\vec{b}\cdot \vec{b})\vec{a}=(\vec{a}\cdot \vec{b})\vec{b}-|\vec{b}{|}^2\vec{a}$

Substitute these expansions back:

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

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

2. Alternative Property Check (Scalar Triple Product)

Notice that the final scalar expression is a dot product of $(2\vec{a}+\vec{b})$ with a vector resulting from a cross product with $(\vec{a}\times \vec{b})$.

Any vector of the form $(\vec{a}\times \vec{b})\times \vec{V}$ lies entirely within the plane containing both vectors $\vec{a}$ and $\vec{b}$.

Let's compute the components directly to evaluate the value faster:

• Magnitude of $\vec{a}$:

  $$|\vec{a}|=\sqrt{{\left(\frac{1}{\sqrt{5}}\right)}^2+{\left(\frac{-2}{\sqrt{5}}\right)}^2}=\sqrt{\frac{1}{5}+\frac{4}{5}}=1⟹|\vec{a}{|}^2=1$$

• Magnitude of $\vec{b}$:

  $$|\vec{b}|=\sqrt{{\left(\frac{2}{\sqrt{14}}\right)}^2+{\left(\frac{1}{\sqrt{14}}\right)}^2+{\left(\frac{3}{\sqrt{14}}\right)}^2}=\sqrt{\frac{4}{14}+\frac{1}{14}+\frac{9}{14}}=\sqrt{\frac{14}{14}}=1⟹|\vec{b}{|}^2=1$$

• Dot Product $\vec{a}\cdot \vec{b}$:

  $$\vec{a}\cdot \vec{b}=\frac{1}{\sqrt{5}\sqrt{14}}\cdot [(1)(2)+(-2)(1)+(0)(3)]=\frac{1}{\sqrt{70}}\cdot [2-2+0]=0$$

Since $\vec{a}\cdot \vec{b}=0$, the two vectors are orthogonal.

3. Substitute Values into Simplified Expression

Since $\vec{a}\cdot \vec{b}=0$, $|\vec{a}{|}^2=1$, and $|\vec{b}{|}^2=1$, the expression simplifies significantly:

$$[(\vec{a}\times \vec{b})\times (\vec{a}-2\vec{b})]=\left(1-2(0)\right)\vec{b}+\left(2(1)-0\right)\vec{a}=\vec{b}+2\vec{a}$$

Now, take the dot product with the front term $(2\vec{a}+\vec{b})$:

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

$$=|2\vec{a}+\vec{b}{|}^2$$

$$=4|\vec{a}{|}^2+|\vec{b}{|}^2+4(\vec{a}\cdot \vec{b})$$

Substitute our values:

$$=4(1)+1+4(0)=4+1=5$$