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Question: The point of intersection of the line passing through \[2\overline i + 3\overline j - \overline k ,3...

The point of intersection of the line passing through 2i‾+3j‾−k‾,3i‾+4j‾−2k‾2\overline i + 3\overline j - \overline k ,3\overline i + 4\overline j - 2\overline k and the line passing through i‾−2j‾+3k‾,i‾−6j‾+6k‾\overline i - 2\overline j + 3\overline k ,\overline i - 6\overline j + 6\overline k is
A) i‾+j‾+k‾\overline i + \overline j + \overline k
B) i‾+2j‾\overline i + 2\overline j
C) j‾+2k‾\overline j + 2\overline k
D) 2i‾+j‾2\overline i + \overline j

Explanation

Solution

Here we have to find the intersection of the two lines. For that, we will first find the vector equation of the first line using the two given position vectors. Then we will find the vector equation of the second line using the two given position vectors of two points. We will then obtain the vector equation of two lines. Further we will solve these two equations to find their intersection.

Complete step by step solution:
We know the vector equation of a line passing through two given points is r→=a→+λ(a→−b→)\overrightarrow r = \overrightarrow a + \lambda \left( {\overrightarrow a - \overrightarrow b } \right).
Here a→\overrightarrow a and b→\overrightarrow b are the position vectors of the two points,λ\lambda is some real number and r→\overrightarrow r is the position vector of any arbitrary point.
Given positions vectors are 2i‾+3j‾−k‾2\overline i + 3\overline j - \overline k and 3i‾+4j‾−2k‾3\overline i + 4\overline j - 2\overline k .
Therefore we will assume:
a→=2i^+3j^−k^b→=3i^+4j^−2k^\begin{array}{l}\overrightarrow a = 2\widehat i + 3\widehat j - \widehat k\\\\\overrightarrow b = 3\widehat i + 4\widehat j - 2\widehat k\end{array}
Now, the vector equation of the line is given as
r1→=2i^+3j^−k^+λ(2i^+3j^−k^−(3i^+4j^−2k^))\overrightarrow {{r_1}} = 2\widehat i + 3\widehat j - \widehat k + \lambda \left( {2\widehat i + 3\widehat j - \widehat k - \left( {3\widehat i + 4\widehat j - 2\widehat k} \right)} \right)
On further simplification, we get
r1→=2i^+3j^−k^+λ(−i^−j^+k^)\overrightarrow {{r_1}} = 2\widehat i + 3\widehat j - \widehat k + \lambda \left( { - \widehat i - \widehat j + \widehat k} \right)…………….. (1)\left( 1 \right)
This is the vector equation of the first line.
We will now find the vector equation of the second line.
Given positions vectors are i‾−2j‾+3k‾\overline i - 2\overline j + 3\overline k and i‾−6j‾+2k‾\overline i - 6\overline j + 2\overline k .
Therefore we can assume:
a→=i^−2j^+3k^b→=i^−6j^+6k^\begin{array}{l}\overrightarrow a = \widehat i - 2\widehat j + 3\widehat k\\\\\overrightarrow b = \widehat i - 6\widehat j + 6\widehat k\end{array}
Now, the vector equation of the line is:-
r2→=i^−2j^+3k^+α(i^−2j^+3k^−(i^−6j^+6k^))\overrightarrow {{r_2}} = \widehat i - 2\widehat j + 3\widehat k + \alpha \left( {\widehat i - 2\widehat j + 3\widehat k - \left( {\widehat i - 6\widehat j + 6\widehat k} \right)} \right)
On further simplification, we get
\Rightarrow $$$$\overrightarrow {{r_2}} = \widehat i - 2\widehat j + 3\widehat k + \alpha \left( {4\widehat j - 3\widehat k} \right)…………….. (2)\left( 2 \right)
This is the vector equation of the second line.
The intersection of the two lines will be:-
r1→=r2→\overrightarrow {{r_1}} = \overrightarrow {{r_2}}
Now substituting the values of r1→\overrightarrow {{r_1}} and r2→\overrightarrow {{r_2}} , we get
2i^+3j^−k^+λ(−i^+−j^+k^)=i^−2j^+3k^+α(4j^−3k^)2\widehat i + 3\widehat j - \widehat k + \lambda \left( { - \widehat i + - \widehat j + \widehat k} \right) = \widehat i - 2\widehat j + 3\widehat k + \alpha \left( {4\widehat j - 3\widehat k} \right)
We will simplify the equation further by combining the like terms.
\Rightarrow $$$$\left( {2 - \lambda } \right)\widehat i + \left( {3 - \lambda } \right)\widehat j + \left( { - 1 + \lambda } \right)\widehat k = \widehat i + \left( { - 2 + 4\alpha } \right)\widehat j + \left( {3 - 3\alpha } \right)\widehat k
On comparing the coefficients ofi^\widehat i,j^\widehat jand k^\widehat k, we get
2−λ=12 - \lambda = 1 ……..(3)\left( 3 \right)
3−λ=−2+4α3 - \lambda = - 2 + 4\alpha ………….(4)\left( 4 \right)
−1+λ=3−3α- 1 + \lambda = 3 - 3\alpha…………..(5)\left( 5 \right)
On further simplification of equation (3)\left( 3 \right) , we get
2−λ=1 ⇒λ=1\begin{array}{l}2 - \lambda = 1\\\ \Rightarrow \lambda = 1\end{array}
Putting value ofλ\lambda in equation (4)\left( 4 \right), we get
3−1=−2+4α ⇒α=1\begin{array}{l}3 - 1 = - 2 + 4\alpha \\\ \Rightarrow \alpha = 1\end{array}
We will put the value of λ\lambda in equation (1)\left( 1 \right) to get the required point.
Point of intersection =2i^+3j^−k^+1×(−i^−j^+k^)=i^+2j^ = 2\widehat i + 3\widehat j - \widehat k + 1 \times \left( { - \widehat i - \widehat j + \widehat k} \right) = \widehat i + 2\widehat j
Therefore, the point of intersection is i‾+2j‾\overline i + 2\overline j .

Thus, the correct option is B.

Note:
The point of intersection of two lines is defined as a common point shared by both the lines and intersecting lines are defined as two or more lines in a plane that cross each other and share a common point.