Animated Solution for Mathematics - Vector Algebra: $A(2,6,2),B(-4,0,\lambda ),C(2,3,-1)$ and $D(4,5,0),|\lambda |\le 5$ are the vertices of a quadrilateral ABCD. If its area is 18 square units, then $5-6\lambda$ is equal to

Welcome, future engineer! Today, we are not just solving a coordinate geometry problem; we are navigating the elegant landscape of 3D space. Imagine you are standing in a room, and four points are floating in the air, forming a quadrilateral $ABCD$.

One of these points, $B$, is elusive—it contains a variable $\lambda$. Our goal is to pin down this variable and uncover the hidden geometry.

When dealing with a quadrilateral in 3D space, many students immediately reach for the triangle area formula. While that works, it is often tedious. Instead, let us embrace the power of vectors.

The area of any quadrilateral in 3D can be elegantly expressed using its diagonals, $\overrightarrow{d_1}$ and $\overrightarrow{d_2}$. The formula is:

Here, our diagonals are $\overrightarrow{AC}$ and $\overrightarrow{BD}$. This formula is a masterpiece of efficiency, as it bypasses the need to calculate the area of individual triangles and goes straight to the heart of the shape's orientation in space.

Let us define our diagonals. To find the vector $\overrightarrow{AC}$, we subtract the position vector of $A(2,6,2)$ from $C(2,3,-1)$:

Next, we tackle $\overrightarrow{BD}$. With $B(-4,0,\lambda )$ and $D(4,5,0)$, we perform the subtraction $D-B$:

We now have our two vectors. They are the building blocks of our area calculation.

Now, we invoke the cross product. This is where the magic happens. We set up the determinant:

Expanding this determinant is a test of precision. Let us take it step by step:

- For $\hat{i}$: $(-3)(-\lambda )-(-3)(5)=3\lambda +15$ - For $\hat{j}$: $-(0(-\lambda )-(-3)(8))=-(0+24)=-24$ - For $\hat{k}$: $(0)(5)-(-3)(8)=0+24=24$

So, our cross product vector is $(3\lambda +15)\hat{i}-24\hat{j}+24\hat{k}$.

We know the area is $18$. Plugging this into our formula, $\frac{1}{2}|\overrightarrow{AC}\times \overrightarrow{BD}|=18$, we get $|\overrightarrow{AC}\times \overrightarrow{BD}|=36$. Squaring both sides to eliminate the square root, we get:

Taking the square root, $3\lambda +15=\pm 12$. This gives us two paths: $3\lambda =-3$ (so $\lambda =-1$) or $3\lambda =-27$ (so $\lambda =-9$).

But wait! The problem gave us a crucial constraint: $|\lambda |\le 5$. If we choose $\lambda =-9$, the absolute value is $9$, which is greater than $5$. Thus, we must reject it. Our only valid solution is $\lambda =-1$.

We are asked to find the value of $5-6\lambda$. Substituting our valid $\lambda =-1$:

And there you have it! Through careful vector construction and rigorous constraint checking, we have arrived at the answer. Keep practicing these steps—the beauty of JEE physics and math lies in this very process of logical deduction.