Animated Solution for Mathematics - Vector Algebra: If $\vec{a},\vec{b},\vec{c}$ are non-coplanar vectors and $\lambda$ is a real number, then the vectors $\vec{a}+2\vec{b}+3\vec{c}$, $\lambda \vec{b}+4\vec{c}$ and $(2\lambda -1)\vec{c}$ are non coplanar for

Imagine you are standing in the center of a room. You have three vectors, $\vec{a}$, $\vec{b}$, and $\vec{c}$, pointing in different directions. The problem states they are non-coplanar.

This is a powerful statement! It means these three vectors do not lie flat on a single sheet of paper; they span the entire three-dimensional space.

Think of them as the edges of a parallelepiped. If they are non-coplanar, the volume of this parallelepiped is non-zero, which mathematically means their scalar triple product, denoted as $[\vec{a},\vec{b},\vec{c}]$, is not equal to zero. This is our foundation.

Now, we are introduced to three new vectors: ${\vec{v}}_1=\vec{a}+2\vec{b}+3\vec{c}$ ${\vec{v}}_2=\lambda \vec{b}+4\vec{c}$ ${\vec{v}}_3=(2\lambda -1)\vec{c}$

These are linear combinations of our original basis vectors. The question asks us to find the condition on $\lambda$ such that these new vectors are also non-coplanar.

If they are non-coplanar, their scalar triple product must also be non-zero: $[\vec{v}_1, \vec{v}_2, \vec{v}_3] eq 0$. Here is the magic: we can express this scalar triple product as the determinant of the coefficient matrix multiplied by the scalar triple product of the basis vectors.

Since $[\vec{a}, \vec{b}, \vec{c}] eq 0$, the condition for non-coplanarity simplifies beautifully to the requirement that the determinant of the coefficient matrix must be non-zero.

Let's construct this matrix $D$ by extracting the coefficients of $\vec{a}$, $\vec{b}$, and $\vec{c}$ for each vector. For ${\vec{v}}_1$, the coefficients are $1,2,3$. For ${\vec{v}}_2$, they are $0,\lambda ,4$. For ${\vec{v}}_3$, they are $0,0,2\lambda -1$.

Our matrix looks like this:

Look closely at this structure. All the elements below the main diagonal are zero. This is an upper triangular matrix!

In the world of JEE, this is a massive shortcut. You do not need to perform a long, tedious expansion. The determinant of an upper triangular matrix is simply the product of its diagonal elements.

So, $D=1\cdot \lambda \cdot (2\lambda -1)=\lambda (2\lambda -1)$.

We have established that for the vectors to be non-coplanar, $D eq 0$. Therefore, $\lambda(2\lambda - 1) eq 0$.

This implies that neither $\lambda$ nor $(2\lambda -1)$ can be zero. Solving these simple equations, we find $\lambda eq 0$ and $\lambda eq \frac{1}{2}$.

This means that $\lambda$ can be any real number except for these two specific values. The vectors are non-coplanar for all values of $\lambda$ except $\lambda \in \{0,\frac{1}{2}\}$.

This is the elegance of linear algebra—what seemed like a complex 3D problem collapses into a simple algebraic condition once you recognize the underlying structure. Keep this in mind: whenever you see linear combinations of basis vectors, think of the determinant of the coefficients. It is your most powerful tool.