Question

Find the point of intersection of the lines $\vec{r} = \hat{i} - \hat{j} + 6\hat{k} + \lambda(3\hat{i} - \hat{k})$, and $\vec{r} = -3\hat{j} + 3\hat{k} + \mu(\hat{i} + 2\hat{j} - \hat{k})$. Also, find the vector equation of the line passing through the point of intersection of the given lines and perpendicular to both the lines.

Answer 1

The lines do not intersect.

Answer 2

To find the point of intersection of the two lines, we first write their parametric equations by equating the components of the position vector $\vec{r}$.

Given lines are:

1. $\vec{r}_1 = \hat{i} - \hat{j} + 6\hat{k} + \lambda(3\hat{i} - \hat{k})$
2. $\vec{r}_2 = -3\hat{j} + 3\hat{k} + \mu(\hat{i} + 2\hat{j} - \hat{k})$

Part 1: Finding the point of intersection

Let a general point on the first line be $P_1(x_1, y_1, z_1)$ and on the second line be $P_2(x_2, y_2, z_2)$.

From Line 1: $x_1 = 1 + 3\lambda$ $y_1 = -1$ $z_1 = 6 - \lambda$

From Line 2: $x_2 = \mu$ $y_2 = -3 + 2\mu$ $z_2 = 3 - \mu$

For the lines to intersect, there must exist values of $\lambda$ and $\mu$ such that $P_1 = P_2$. Equating the corresponding components:

1. $1 + 3\lambda = \mu$ (Equation from x-components)
2. $-1 = -3 + 2\mu$ (Equation from y-components)
3. $6 - \lambda = 3 - \mu$ (Equation from z-components)

Let's solve these equations:

From Equation (2): $-1 = -3 + 2\mu$ $2\mu = -1 + 3$ $2\mu = 2$ $\mu = 1$

Now, substitute the value of $\mu = 1$ into Equation (1): $1 + 3\lambda = 1$ $3\lambda = 0$ $\lambda = 0$

Finally, we must check if these values of $\lambda = 0$ and $\mu = 1$ satisfy Equation (3).

Substitute $\lambda = 0$ and $\mu = 1$ into Equation (3): LHS: $6 - \lambda = 6 - 0 = 6$ RHS: $3 - \mu = 3 - 1 = 2$

Since LHS $\neq$ RHS ($6 \neq 2$), the values of $\lambda$ and $\mu$ obtained from the first two equations do not satisfy the third equation. This means that the system of equations is inconsistent, and there are no common values of $\lambda$ and $\mu$ for which the lines intersect.

Therefore, the given lines do not intersect. They are skew lines.

Part 2: Vector equation of the line passing through the point of intersection and perpendicular to both lines

Since the lines do not intersect, there is no "point of intersection". Consequently, the second part of the question, which asks for a line passing through this non-existent point, cannot be answered as stated.

The problem as stated has no solution for the point of intersection.

Explanation of the solution:

1. Express general points on each line in parametric form using parameters $\lambda$ and $\mu$.
2. Equate corresponding coordinates (x, y, z) to form a system of three linear equations in $\lambda$ and $\mu$.
3. Solve two of these equations to find unique values for $\lambda$ and $\mu$.
4. Substitute these values into the third equation. If the third equation is satisfied, the lines intersect, and the point can be found by substituting $\lambda$ or $\mu$ back into their respective line equations.
5. If the third equation is not satisfied (as in this case, $6 \neq 2$), the lines do not intersect; they are skew lines.
6. Since there is no point of intersection, the subsequent part of the question asking for a line passing through this point cannot be answered.