Let `r=a+lambdal and r=b+mum` br be two lines in space, where `a= 5hat(i)+hat(j)+2hat(k), b=-hat(i)+7hat(j)+8hat(k), l=-4hat(i)+hat(j)-hat(k), and m=2hat(i)-5hat(j)-7hat(k)`, then the position vector of a point which lies on both of these lines, is

Options

A. `hat(i)+2hat(j)+hat(k)`
B. `2hat(i)+hat(j)+hat(k)`
C. `hat(i)+hat(j)+2hat(k)`
D. None of these

Answer

To find the position vector of a point that lies on both lines given by \( r = a + \lambda l \) and \( r = b + \mu m \), we will follow these steps:

### Step 1: Write the equations of the lines
The first line is given by:
\[ r_1 = a + \lambda l \]
where \( a = 5\hat{i} + \hat{j} + 2\hat{k} \) and \( l = -4\hat{i} + \hat{j} - \hat{k} \).

Substituting the values, we get:
\[ r_1 = (5 - 4\lambda)\hat{i} + (1 + \lambda)\hat{j} + (2 - \lambda)\hat{k} \]

The second line is given by:
\[ r_2 = b + \mu m \]
where \( b = -\hat{i} + 7\hat{j} + 8\hat{k} \) and \( m = 2\hat{i} - 5\hat{j} - 7\hat{k} \).

Substituting the values, we get:
\[ r_2 = (-1 + 2\mu)\hat{i} + (7 - 5\mu)\hat{j} + (8 - 7\mu)\hat{k} \]

### Step 2: Equate the position vectors
To find the intersection point, we set \( r_1 = r_2 \):
\[ (5 - 4\lambda)\hat{i} + (1 + \lambda)\hat{j} + (2 - \lambda)\hat{k} = (-1 + 2\mu)\hat{i} + (7 - 5\mu)\hat{j} + (8 - 7\mu)\hat{k} \]

This gives us three equations:
1. \( 5 - 4\lambda = -1 + 2\mu \) (Equation 1)
2. \( 1 + \lambda = 7 - 5\mu \) (Equation 2)
3. \( 2 - \lambda = 8 - 7\mu \) (Equation 3)

### Step 3: Solve the equations
From Equation 1:
\[ 5 + 1 = 4\lambda + 2\mu \implies 6 = 4\lambda + 2\mu \implies 2\lambda + \mu = 3 \quad \text{(Equation 4)} \]

From Equation 2:
\[ \lambda + 5\mu = 6 \quad \text{(Equation 5)} \]

Now we have two equations (Equation 4 and Equation 5):
1. \( 2\lambda + \mu = 3 \)
2. \( \lambda + 5\mu = 6 \)

### Step 4: Solve for \( \lambda \) and \( \mu \)
From Equation 4, we can express \( \mu \) in terms of \( \lambda \):
\[ \mu = 3 - 2\lambda \]

Substituting this into Equation 5:
\[ \lambda + 5(3 - 2\lambda) = 6 \\ \lambda + 15 - 10\lambda = 6 \\ -9\lambda = -9 \\ \lambda = 1 \]

Now substituting \( \lambda = 1 \) back into Equation 4 to find \( \mu \):
\[ 2(1) + \mu = 3 \\ \mu = 3 - 2 = 1 \]

### Step 5: Verify the values in Equation 3
Now we check if these values satisfy Equation 3:
\[ 2 - \lambda = 8 - 7\mu \\ 2 - 1 = 8 - 7(1) \\ 1 = 1 \quad \text{(True)} \]

### Step 6: Find the position vector
Now we substitute \( \lambda = 1 \) into either line equation to find the intersection point. Using \( r_1 \):
\[ r_1 = 5 - 4(1)\hat{i} + 1 + 1\hat{j} + 2 - 1\hat{k} \\ = (5 - 4)\hat{i} + (1 + 1)\hat{j} + (2 - 1)\hat{k} \\ = 1\hat{i} + 2\hat{j} + 1\hat{k} \]

Thus, the position vector of the intersection point is:
\[ \boxed{1\hat{i} + 2\hat{j} + 1\hat{k}} \]