35. (a) Represent the equations of lines $l_1$ and $l_2$ in vector form and check whether they are intersecting or not. $l_1: \frac{x+3}{-3} = \frac{y-1}{1} = \frac{z-5}{5}$ $l_2: \frac{x+1}{-1} = \frac{2-y}{-2} = \frac{z-5}{5}$ (vector form, lines, intersection, parametric equations, direction vectors)

Step 1: Convert Line 1 to Vector Form

A line in Cartesian form $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$ passes through the point $(x_1, y_1, z_1)$ and has direction ratios $(a, b, c)$. The vector form of this line is $\vec{r} = (x_1 \hat{i} + y_1 \hat{j} + z_1 \hat{k}) + \lambda (a \hat{i} + b \hat{j} + c \hat{k})$. For $l_1$, comparing it with the standard form, we identify the point and direction vector.

Step 2: Identify Point and Direction Vector for $l_1$

From the given equation for line $l_1: \frac{x - (-3)}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5}$, we can see that it passes through the point $(-3, 1, 5)$ and has direction ratios $(-3, 1, 5)$. So, the position vector $\vec{a}_{1}$ is $-3\hat{i} + \hat{j} + 5\hat{k}$, and the direction vector $\vec{b}_{1}$ is $-3\hat{i} + \hat{j} + 5\hat{k}$.

Step 3: Convert Line 2 to Vector Form

Similarly, we convert the Cartesian equation of line $l_2$ into its vector form. First, we need to adjust the $y$ term in the equation to match the standard form $\frac{y-y_1}{b}$.

Step 4: Adjust $l_2$ Cartesian Form

The given equation for line $l_2$ is $\frac{x+1}{-1} = \frac{2-y}{-2} = \frac{z-5}{5}$. To get it into the standard form $(y-y_1)/b$, we rewrite $\frac{2-y}{-2}$ as $\frac{-(y-2)}{-2} = \frac{y-2}{2}$. Now the equation is in the correct format.

Step 5: Identify Point and Direction Vector for $l_2$

From the adjusted equation for line $l_2: \frac{x - (-1)}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5}$, we identify that it passes through the point $(-1, 2, 5)$ and has direction ratios $(-1, 2, 5)$. So, the position vector $\vec{a}_{2}$ is $-1\hat{i} + 2\hat{j} + 5\hat{k}$, and the direction vector $\vec{b}_{2}$ is $-1\hat{i} + 2\hat{j} + 5\hat{k}$.

Step 6: Check for Intersection: Condition for Intersecting Lines

Two lines intersect if the shortest distance between them is zero. This condition is met if the scalar triple product of the difference in position vectors and the cross product of their direction vectors is zero. We calculate $\vec{a}_{2} - \vec{a}_{1}$, $\vec{b}_{1} \times \vec{b}_{2}$, and then their dot product.

Step 7: Calculate $\vec{a}_{2} - \vec{a}_{1}$

We subtract the position vector of the first line from the position vector of the second line to get the vector connecting a point on $l_1$ to a point on $l_2$. This results in a vector $2\hat{i} + \hat{j}$.

Step 8: Calculate $\vec{b}_{1} \times \vec{b}_{2}$

Next, we compute the cross product of the direction vectors $\vec{b}_{1}$ and $\vec{b}_{2}$. This vector is perpendicular to both $\vec{b}_{1}$ and $\vec{b}_{2}$, and its magnitude is related to the area of the parallelogram formed by the vectors.

Step 9: Calculate $(\vec{a}_{2} - \vec{a}_{1}) \cdot (\vec{b}_{1} \times \vec{b}_{2})$

Finally, we compute the dot product of the vector $(\vec{a}_{2} - \vec{a}_{1})$ and the cross product $(\vec{b}_{1} \times \vec{b}_{2})$. Since the result is 0, the lines intersect.

Step 10: Conclusion about Intersection

The scalar triple product being zero indicates that the vectors $\vec{a}_{2} - \vec{a}_{1}$, $\vec{b}_{1}$, and $\vec{b}_{2}$ are coplanar. When this condition is met, and the lines are not parallel (which they are not, as $\vec{b}_{1}$ is not a scalar multiple of $\vec{b}_{2}$), the lines intersect.

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