Animated Solution for Mathematics - Three Dimensional Geometry: The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=z$ and $\frac{7-x}{2}=y-2=z-6$ is

Before we begin, we must address the most dangerous trap in this problem: the form of the line equations. You are given the line:

Many students will look at the denominator $2$ and immediately assume the direction vector component is $2$. However, the standard form requires the variable to have a coefficient of $+1$.

We must rewrite $7-x$ as $-(x-7)$. This transforms the equation into:

Now, we can safely extract our direction vector $\overrightarrow{d_2}=-2\hat{i}+\hat{j}+\hat{k}$ and point $B(7,2,6)$. For the first line, $\frac{x+7}{-6}=\frac{y-6}{7}=z$, the work is already done for us. We identify point $A(-7,6,0)$ and direction vector $\overrightarrow{d_1}=-6\hat{i}+7\hat{j}+\hat{k}$.

To find the shortest distance, we need a vector that is perpendicular to both lines. This is the common normal, which we find using the cross product: $\vec{n}=\overrightarrow{d_1}\times \overrightarrow{d_2}$.

We set up the determinant with $\hat{i},\hat{j},\hat{k}$ in the first row and the components of our direction vectors in the subsequent rows:

Expanding this, we get $\vec{n}=\hat{i}(7-1)-\hat{j}(-6+2)+\hat{k}(-6+14)$. This simplifies beautifully to $\vec{n}=6\hat{i}+4\hat{j}+8\hat{k}$.

Now, we connect the two lines with a vector $\overrightarrow{AB}$ joining point $A(-7,6,0)$ to point $B(7,2,6)$. This gives us:

The shortest distance ($SD$) is the projection of this vector $\overrightarrow{AB}$ onto our normal vector $\vec{n}$, defined by the formula:

Calculating the dot product, we have:

The magnitude of our normal vector is:

Finally, we compute the distance:

Simplifying this, we get $\sqrt{4\times 29}=2\sqrt{29}$. You have successfully navigated the 3D void and found the shortest path.