CUET PG 2022 — Mathematics PYQ

We are given that $\vec{a},\vec{b},$ and $\vec{c}$ are unit vectors, so:

Now, expand each squared magnitude term:

• $|\vec{a}-\vec{b}{|}^2=|\vec{a}{|}^2+|\vec{b}{|}^2-2(\vec{a}\cdot \vec{b})=1+1-2(\vec{a}\cdot \vec{b})=2-2(\vec{a}\cdot \vec{b})$

• $|\vec{b}-\vec{c}{|}^2=|\vec{b}{|}^2+|\vec{c}{|}^2-2(\vec{b}\cdot \vec{c})=1+1-2(\vec{b}\cdot \vec{c})=2-2(\vec{b}\cdot \vec{c})$

• $|\vec{c}-\vec{a}{|}^2=|\vec{c}{|}^2+|\vec{a}{|}^2-2(\vec{c}\cdot \vec{a})=1+1-2(\vec{c}\cdot \vec{a})=2-2(\vec{c}\cdot \vec{a})$

Adding these together, let the expression be $S$:

Step 2: Use the Square of the Sum Identity

Consider the magnitude of the sum of the three vectors:

Expanding this:

Substitute the values of the unit vectors:

Rearranging for the dot product sum:

Step 3: Find the Maximum Value of $S$

Go back to the expression for $S$:

To maximize $S$, we need to subtract the minimum possible value of the bracketed term. From Step 2, the minimum value of $2(\vec{a}\cdot \vec{b}+\dots )$ is $-3$.

Final Answer:

The expression $|\vec{a}-\vec{b}{|}^2+|\vec{b}-\vec{c}{|}^2+|\vec{c}-\vec{a}{|}^2$ does not exceed 9.