Q.A line passing through the points $A(1, 2, 3)$ and $B(5, 8, 11)$ intersects the line $\vec{r} = 4\hat{i} + \hat{j} + \lambda(5\hat{i} + 2\hat{j} + \hat{k})$. Find the co-ordinates of the point of intersection. Hence, write the equation of a line passing through the point of intersection and perpendicular to both the lines.

The key idea is that the point of intersection of two lines satisfies both their equations, and a line perpendicular to two given lines has a direction vector found by the cross product of their individual direction vectors.

1. The line passing through $A(1, 2, 3)$ and $B(5, 8, 11)$ has a direction vector $\vec{d_1} = \vec{AB} = (5-1)\hat{i} + (8-2)\hat{j} + (11-3)\hat{k} = 4\hat{i} + 6\hat{j} + 8\hat{k}$, which can be simplified to $2\hat{i} + 3\hat{j} + 4\hat{k}$. Its equation is $\vec{r_1} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 4\hat{k})$. The second line is given by $\vec{r_2} = (4\hat{i} + \hat{j}) + \lambda(5\hat{i} + 2\hat{j} + \hat{k})$.
2. For the point of intersection, equate the components: $1 + 2\mu = 4 + 5\lambda$ $2 + 3\mu = 1 + 2\lambda$ $3 + 4\mu = \lambda$ Solving these equations (e.g., substitute $\lambda$ from the third equation into the second) yields $\mu = -1$ and $\lambda = -1$.
3. Substitute $\mu = -1$ into $\vec{r_1}$ to find the point of intersection: $P = (\hat{i} + 2\hat{j} + 3\hat{k}) + (-1)(2\hat{i} + 3\hat{j} + 4\hat{k}) = (1-2)\hat{i} + (2-3)\hat{j} + (3-4)\hat{k} = -\hat{i} - \hat{j} - \hat{k}$. The coordinates of the point of intersection are $(-1, -1, -1)$.
4. The direction vector of a line perpendicular to both given lines is the cross product of their direction vectors: …

To solve this problem, we need to perform two main tasks: first, find the point where the two given lines intersect, and second, determine the equation of a new line that passes through this intersection point and is simultaneously perpendicular to both original lines.

The core idea for finding the intersection point is that if two lines intersect, there must be a common point that lies on both lines. In vector form, this means their position vectors must be equal for specific values of their respective parameters. This equality will lead to a system of linear equations that we can solve.

For the second part, a line perpendicular to two other lines must have a direction vector that is perpendicular to the direction vectors of both original lines. The cross product of two vectors yields a vector that is perpendicular to both of them. Thus, taking the cross product of the direction vectors of the two given lines will give us the direction vector for our new line. With a point (the intersection point) and a direction vector, we can then write the equation of the required line.

1. Write the vector equation of the line passing through $A(1, 2, 3)$ and $B(5, 8, 11)$.

A line passing through two points $A(\vec{a})$ and $B(\vec{b})$ can be represented by the vector equation $\vec{r} = \vec{a} + \mu(\vec{b} - \vec{a})$, where $\mu$ is a scalar parameter.

Given points are $A(1, 2, 3)$ and $B(5, 8, 11)$.

The position vector of point A is $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$.

The position vector of point B is $\vec{b} = 5\hat{i} + 8\hat{j} + 11\hat{k}$.

The direction vector of the line passing through A and B is $\vec{d_1} = \vec{b} - \vec{a}$:

$\vec{d_1} = (5\hat{i} + 8\hat{j} + 11\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k})$

$\vec{d_1} = (5-1)\hat{i} + (8-2)\hat{j} + (11-3)\hat{k}$

$\vec{d_1} = 4\hat{i} + 6\hat{j} + 8\hat{k}$

We can simplify this direction vector by taking out a common factor of 2:

$\vec{d_1} = 2(2\hat{i} + 3\hat{j} + 4\hat{k})$.

Let's use the simplified direction vector $\vec{d_1}' = 2\hat{i} + 3\hat{j} + 4\hat{k}$ for the line equation, using a new parameter $\mu'$.

The equation of the line passing through A and B (Line 1) is:

The second line (Line 2) is given as:

Its direction vector is $\vec{d_2} = 5\hat{i} + 2\hat{j} + \hat{k}$.

2. Find the coordinates of the point of intersection.

For the lines to intersect, their position vectors must be equal for some values of $\mu'$ and $\lambda$.

Equating the two vector equations:

$(\hat{i} + 2\hat{j} + 3\hat{k}) + \mu'(2\hat{i} + 3\hat{j} + 4\hat{k}) = 4\hat{i} + \hat{j} + \lambda(5\hat{i} + 2\hat{j} + \hat{k})$

Rearranging terms to group the $\hat{i}$, $\hat{j}$, and $\hat{k}$ components:

$(1 + 2\mu')\hat{i} + (2 + 3\mu')\hat{j} + (3 + 4\mu')\hat{k} = (4 + 5\lambda)\hat{i} + (1 + 2\lambda)\hat{j} + \lambda\hat{k}$

Equating the coefficients of $\hat{i}$, $\hat{j}$, and $\hat{k}$ gives us a system of three linear equations:

1. $1 + 2\mu' = 4 + 5\lambda \implies 2\mu' - 5\lambda = 3$
2. $2 + 3\mu' = 1 + 2\lambda \implies 3\mu' - 2\lambda = -1$
3. $3 + 4\mu' = \lambda$

We can solve this system. Substitute Equation 3 into Equation 1:

$2\mu' - 5(3 + 4\mu') = 3$

$2\mu' - 15 - 20\mu' = 3$

$-18\mu' = 18$

$\mu' = -1$

Now, substitute the value of $\mu'$ back into Equation 3 to find $\lambda$:

$\lambda = 3 + 4(-1)$

$\lambda = 3 - 4$

$\lambda = -1$

To find the coordinates of the point of intersection, substitute $\lambda = -1$ into the equation for Line 2 (or $\mu' = -1$ into Line 1):

$\vec{r} = 4\hat{i} + \hat{j} + (-1)(5\hat{i} + 2\hat{j} + \hat{k})$ …