Question

Let the three vectors $\vec{a}, \vec{b}$, and $\vec{c}$ be pairwise non-collinear vectors. If $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$ and if $\vec{b}+2\vec{c}$ is collinear with $\vec{a}$, then $\vec{a}+2\vec{b}+5\vec{c}$ is equal to

• A. $\vec{0}$
• B. $\vec{a}$
• C. $\vec{b}$
• D. $\vec{c}$
• E. $2\vec{a}$

Hint:

Logic Tip: A common property in such vector collinearity problems is that if $\vec{x} + \alpha\vec{y}$ is parallel to $\vec{z}$ and $\vec{y} + \beta\vec{z}$ is parallel to $\vec{x}$, their combined linear sum $(\vec{x} + \alpha\vec{y} + \alpha\beta\vec{z})$ often equals the zero vector $\vec{0}$. This instantly shows $\vec{a} + 2\vec{b} + 4\vec{c} = \vec{0}$, making $\vec{a} + 2\vec{b} + 5\vec{c} = \vec{c}$.

Answer 1

The Correct Option is D

Concept:
If two vectors $\vec{u}$ and $\vec{v}$ are collinear, then one can be expressed as a scalar multiple of the other: $\vec{u} = \lambda\vec{v}$ for some scalar $\lambda$.
Step 1: Express the collinearity conditions algebraically.
Since $\vec{a} + 2\vec{b}$ is collinear with $\vec{c}$, we can write: $$\vec{a} + 2\vec{b} = \lambda\vec{c} \quad \text{--- (Equation 1)}$$ Since $\vec{b} + 2\vec{c}$ is collinear with $\vec{a}$, we can write: $$\vec{b} + 2\vec{c} = \mu\vec{a} \quad \text{--- (Equation 2)}$$
Step 2: Solve for one of the vectors to find the scalars.
From Equation 1, isolate $\vec{a}$: $$\vec{a} = \lambda\vec{c} - 2\vec{b}$$ Substitute this expression for $\vec{a}$ into Equation 2: $$\vec{b} + 2\vec{c} = \mu(\lambda\vec{c} - 2\vec{b})$$ $$\vec{b} + 2\vec{c} = \mu\lambda\vec{c} - 2\mu\vec{b}$$
Step 3: Equate coefficients to find $\mu$ and $\lambda$.
Rearrange the terms to group $\vec{b}$ and $\vec{c}$: $$\vec{b} + 2\mu\vec{b} + 2\vec{c} - \mu\lambda\vec{c} = \vec{0}$$ $$(1 + 2\mu)\vec{b} + (2 - \mu\lambda)\vec{c} = \vec{0}$$ Because $\vec{b}$ and $\vec{c}$ are pairwise non-collinear, the only way a linear combination of them can equal the zero vector is if both coefficients are zero: $$1 + 2\mu = 0 \implies \mu = -\frac{1}{2}$$ $$2 - \mu\lambda = 0$$ Substitute $\mu = -\frac{1}{2}$ into the second coefficient equation: $$2 - \left(-\frac{1}{2}\right)\lambda = 0$$ $$2 + \frac{\lambda}{2} = 0 \implies \lambda = -4$$
Step 4: Evaluate the final expression.
Substitute $\lambda = -4$ back into Equation 1: $$\vec{a} + 2\vec{b} = -4\vec{c}$$ We need to find the value of $\vec{a} + 2\vec{b} + 5\vec{c}$. Substitute our known relationship: $$(\vec{a} + 2\vec{b}) + 5\vec{c} = (-4\vec{c}) + 5\vec{c} = \vec{c}$$