Q.Find $\lambda$ and $\mu$ if $(2\hat{i}+6\hat{j}+27\hat{k})\times(\hat{i}+\lambda\hat{j}+\mu\hat{k})=\vec{0}$.

Concept: For two non-zero vectors, their cross product equals the zero vector if and only if the vectors are parallel (collinear). This means one is a scalar multiple of the other.

Let $\vec{a} = 2\hat{i}+6\hat{j}+27\hat{k}$ and $\vec{b} = \hat{i}+\lambda\hat{j}+\mu\hat{k}$.

Since $\vec{a} \times \vec{b} = \vec{0}$, the vectors are parallel. Therefore, the ratios of corresponding components must be equal:

$\frac{2}{1} = \frac{6}{\lambda} = \frac{27}{\mu}$ …

The cross product of two vectors is zero if and only if the vectors are parallel (or one of them is the zero vector). Geometrically, the cross product measures the area of the parallelogram they span — when they point in the same or exactly opposite direction, that area collapses to zero.

So the problem reduces to: find $\lambda$ and $\mu$ such that $(2\hat{i}+6\hat{j}+27\hat{k})$ is parallel to $(\hat{i}+\lambda\hat{j}+\mu\hat{k})$.

1. Set up the proportionality condition. If two vectors $\vec{a}$ and $\vec{b}$ are parallel, there exists some scalar $k$ such that $\vec{a} = k \vec{b}$. Here:

2. Equate the coefficients of $\hat{i}$.

   From the $\hat{i}$ components: $2 = k \cdot 1$, so $k = 2$.

3. Use $k$ to find $\lambda$.

   Equate the $\hat{j}$ components: $6 = k \cdot \lambda = 2\lambda$, hence $\lambda = 3$.

4. Use $k$ to find $\mu$.

   Equate the $\hat{k}$ components: $27 = k \cdot \mu = 2\mu$, hence $\mu = \frac{27}{2}$. …

Method: Finding Unknowns from a Vanishing Cross Product

A zero cross product means the two vectors are parallel — use proportional components to solve for unknowns.

Steps

Step 1: Convert $\times=\vec{0}$ into parallelism.

(for non-zero vectors).

Step 2: Fix $k$ from a fully-known component pair. …

Common Mistakes

Mistake 1: Equating components without a common scalar.

Why it's wrong: setting $2=\lambda$ or $27=\mu$ ignores the proportionality constant $k$. Correct approach: write $\vec{a}=k\vec{b}$, find $k=2$ from the $\hat{i}$ terms, then use it everywhere.

Mistake 2: Thinking $\vec{a}\times\vec{b}=\vec{0}$ forces a zero vector. …