NIMCET 2007 — Mathematics PYQ

Step 1: Use the first equality

We are given that:

$$\vec{a}\times \vec{b}=\vec{b}\times \vec{c}$$

Bring both terms to one side:

$$\vec{a}\times \vec{b}-\vec{b}\times \vec{c}=\vec{0}$$

Using the anti-commutative property of the vector cross product ($\vec{b}\times \vec{c}=-\vec{c}\times \vec{b}$), we can rewrite the equation as:

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

Now, factor out the common vector $\vec{b}$ from the right side using the distributive property:

$$(\vec{a}+\vec{c})\times \vec{b}=\vec{0}$$

Note: When the cross product of two vectors is zero, it implies that the vectors are parallel or collinear.

Therefore, $(\vec{a}+\vec{c})$ must be parallel to $\vec{b}$. We can express this relationship using a scalar multiplier $\lambda$:

$$\vec{a}+\vec{c}=\lambda \vec{b}\text{— (Equation 1)}$$

Step 2: Use the second equality

Next, we take the other part of the given condition:

$$\vec{b}\times \vec{c}=\vec{c}\times \vec{a}$$

Bring both terms to one side:

$$\vec{b}\times \vec{c}-\vec{c}\times \vec{a}=\vec{0}$$

Apply the anti-commutative property ($\vec{c}\times \vec{a}=-\vec{a}\times \vec{c}$):

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

Factor out the common vector $\vec{c}$ from the right side:

$$(\vec{b}+\vec{a})\times \vec{c}=\vec{0}$$

This implies that $(\vec{a}+\vec{b})$ is parallel to $\vec{c}$. We can express this using another scalar multiplier $\mu$:

$$\vec{a}+\vec{b}=\mu \vec{c}\text{— (Equation 2)}$$

Step 3: Analyze and Solve for Scalars

From Equation 1, add $\vec{b}$ to both sides to form the expression $\vec{a}+\vec{b}+\vec{c}$:

$$\vec{a}+\vec{b}+\vec{c}=\lambda \vec{b}+\vec{b}=(\lambda +1)\vec{b}$$

From Equation 2, add $\vec{c}$ to both sides to form the same expression:

$$\vec{a}+\vec{b}+\vec{c}=\mu \vec{c}+\vec{c}=(\mu +1)\vec{c}$$

Equating both expressions since they both equal $\vec{a}+\vec{b}+\vec{c}$:

$$(\lambda +1)\vec{b}=(\mu +1)\vec{c}$$

$$(\lambda +1)\vec{b}-(\mu +1)\vec{c}=\vec{0}$$

Since it is given in the problem statement that $\vec{a},\vec{b},\vec{c}$ are non-collinear vectors, $\vec{b}$ and $\vec{c}$ cannot be parallel to each other. For a linear combination of two non-collinear vectors to equal zero, their coefficients must individually equal zero:

$$\lambda +1=0⟹\lambda =-1$$

$$\mu +1=0⟹\mu =-1$$

Step 4: Final Calculation

Substitute $\lambda =-1$ back into the expression for $\vec{a}+\vec{b}+\vec{c}$:

$$\vec{a}+\vec{b}+\vec{c}=(-1+1)\vec{b}=0\cdot \vec{b}=\vec{0}$$

Thus, the value of $\vec{a}+\vec{b}+\vec{c}$ is $0$ (the zero vector).