Question

The point of intersection of the line passing through \[2\hat i + 3\hat j - \hat k\] , $3\hat i + 4\hat j - 2\hat k$ and the line passing through $\hat i - 2\hat j + 3\hat k$ , $\hat i - 6\hat j + 6\hat k$ is
A. $\hat i + \hat j + \hat k$
B. $\hat i + 2\hat j$
C. $\hat j + 2\hat k$
D. $2\hat i + \hat j$

Answer 1

Hint: Firstly, we can find the vector equation of lines passing through the given points using the equation \[r = \vec a + \lambda \left( {\vec b - \vec a} \right)\] where $\vec a$ and $\vec b$ are vector equation of the points through which the line passes. Then we can find the point of intersection by equating the equations of the two lines, and on substitution of the variable in any of the two lines.

Complete step-by-step answer:
Equation of a line in vector form is given by \[r = \vec a + \lambda \vec b\] where $\vec a$ is any point on the line and $\vec b$ is a vector parallel to the line. Vector from point $\vec a$ to point $\vec b$ is given by $\vec p = \vec b - \vec a$ . If $\vec a$ and $\vec b$ are points on the line, vector $\vec p = \vec b - \vec a$ is parallel vector
For the line passing through $\vec a = 2\hat i + 3\hat j - \hat k$ , $\vec b = 3\hat i + 4\hat j - 2\hat k$
Parallel vector is given by
  $
  {{\vec p}_1} = \vec b - \vec a \\
   = \left( {3\hat i + 4\hat j - 2\hat k} \right) - \left( {2\hat i + 3\hat j - \hat k} \right) \\
   = \left( {3 - 2} \right)\hat i + \left( {4 - 3} \right)\hat j + \left( { - 2 + 1} \right)\hat k \\
   = \hat i + \hat j - \hat k \\
 $
Then equation for the line is given by,
 \[{r_1} = \vec a + t{\vec p_1}\]
 $
   \Rightarrow {r_1} = \left( {2\hat i + 3\hat j - \hat k} \right) + t\left( {\hat i + \hat j - \hat k} \right) \\
   \Rightarrow {r_1} = \left( {2 + t} \right)\hat i + \left( {3 + t} \right)\hat j - \left( {1 + t} \right)\hat k \\
 $
For the line passing through $\vec c = \hat i - 2\hat j + 3\hat k$ , $\vec d = \hat i - 6\hat j + 6\hat k$ ,
Parallel vector is given by,
 \[
  {{\vec p}_2} = \vec d - \vec c \\
   = \left( {\hat i - 6\hat j + 6\hat k} \right) - \left( {\hat i - 2\hat j + 3\hat k} \right) \\
   = \left( {1 - 1} \right)\hat i + \left( { - 6 + 2} \right)\hat j + \left( {6 - 3} \right)\hat k \\
   = - 04\hat j + 3\hat k \\
 \]
Then the equation for the line is given by,
 \[{r_2} = \vec c + s{\vec p_2}\]
 $
   \Rightarrow {r_2} = \left( {\hat i - 2\hat j + 3\hat k} \right) + s\left( { - 4\hat j + 3\hat k} \right) \\
   \Rightarrow {r_2} = \hat i - \left( {2 + 4s} \right)\hat j + \left( {3 + 3s} \right)\hat k \\
 $
Now we have vector equations of two lines. As the lines are intersecting, we can equate the vector component to find the value of t and s.
 $2 + t = 1$ … (1)
 $3 + t = - \left( {2 + 4s} \right)$ … (2)
 $ - \left( {1 + t} \right) = 3 + 3s$ … (3)
From (1), we get,
 $t = 1 - 2 = - 1$
By putting $t = - 1$ in (2), we get
 $
  3 - 1 = - 2 - 4s \\
   \Rightarrow 4s = - 2 - 2 = - 4 \\
   \Rightarrow s = - 1 \\
 $
Using the value of s and t in any of the line’s equations will give the point of intersection.
Putting $t = - 1$ in ${r_1}$ , we get
 $
  P = \left( {2 - 1} \right)\hat i + \left( {3 - 1} \right)\hat j - \left( {1 - 1} \right)\hat k \\
   = \hat i + 2\hat j \\
 $
Therefore, the point of intersection of the two lines is $\hat i + 2\hat j$ .
So, the correct answer is option B.

Note: The concepts of vectors are used in this problem. We must know to form equations of lines in vector form and to find the vector connecting two points. When we have 3 equations and 2 unknowns, use the third equation to validate our solution. If we interchange the order of the points while taking the parallel vector, the vector will be in the opposite direction. Then the values of the variable may change, but the point of intersection will be the same. If we change the point in the equation of the line, then also the point of intersection will be the same. We must take care about the sign of the vectors while doing operations.