Animated Solution for Mathematics - Vector Algebra: If $\vec{P}=(a+1)\hat{i}+a\hat{j}+a\hat{k}$, $\vec{Q}=a\hat{i}+(a+1)\hat{j}+a\hat{k}$, $\vec{R}=a\hat{i}+a\hat{j}+(a+1)\hat{k}$ and $\vec{P},\vec{Q},\vec{R}$ are coplanar vectors and $3(\vec{P}\cdot \vec{Q}{)}^2-\lambda |\vec{R}\times \vec{Q}{|}^2=0$ then value of $\lambda$ is

A Vector Odyssey

Welcome, fellow traveler, to the fascinating world of vectors. Today, we are not just solving an equation; we are exploring the architecture of three-dimensional space.

We have three vectors, $\vec{P}$, $\vec{Q}$, and $\vec{R}$, defined by a parameter $a$. They are dancing in space, and we are told they are coplanar.

This means they are trapped on a single, flat sheet of paper, no matter how you rotate them. This is our starting point.

The Key to Coplanarity

When we say three vectors are coplanar, we are saying they do not span a volume. In the language of linear algebra, the scalar triple product, denoted as $[\vec{P}\vec{Q}\vec{R}]$, must be zero.

This product is calculated as the determinant of the matrix formed by the components of our vectors:

I know what you are thinking: 'Do I really have to expand this three-by-three determinant?' If you do it blindly, you might get lost in the algebra. But look closely at the symmetry; every row is a permutation of the others.

This is a gift! Let us use a column operation to simplify our lives. By adding column two and column three to column one ($C_1\to C_1+C_2+C_3$), we transform the first column into $(3a+1)$ for every row.

Now, we can factor out $(3a+1)$ from the determinant. The remaining determinant is simple, and it evaluates to $1$. Thus, we are left with $(3a+1)(1)=0$, which immediately gives us $a=-\frac{1}{3}$.

With $a=-\frac{1}{3}$ in our pocket, the vectors $\vec{P}$, $\vec{Q}$, and $\vec{R}$ are no longer abstract; they are concrete. Substituting this value, we get:

Now, the problem asks us to evaluate a specific expression involving a dot product and a cross product. First, the dot product $\vec{P}\cdot \vec{Q}$.

Multiplying the components, we get:

Next, the cross product $\vec{R}\times \vec{Q}$. This vector is perpendicular to the plane containing $\vec{R}$ and $\vec{Q}$.

Calculating the determinant, we find:

The squared magnitude of this vector is:

We have all the pieces of our puzzle. The given equation is $3(\vec{P}\cdot \vec{Q}{)}^2-\lambda |\vec{R}\times \vec{Q}{|}^2=0$.

Plugging in our values, we get:

This simplifies to:

The conclusion is inevitable: $\lambda =1$. You have navigated the geometry, mastered the determinant, and conquered the algebra. Well done!