Question

Shortest distance between lines \(\frac{(x-5)}{4}\)=\(\frac{(y-3)}{6}\)=\(\frac{(z-2)}{4}\) and \(\frac{(x-3)}{7}=\frac{(y-2)}{5}=\frac{(z-9)}{6}\) is ?

• A. \(\frac{190}{37}\)
• B. \(\frac{190}{\sqrt{756}}\)
• C. \(\frac{37}{190}\)
• D. \(\frac{756}{\sqrt{190}}\)

Hint:

To calculate shortest distance, find the vector perpendicular to both direction vectors and project the joining vector onto it.

Answer 1

The Correct Option is B

To find the shortest distance between two skew lines, we can use the vector projection method. Here's how:
Let the first line be L₁ and the second line be L₂.
L₁: \(\frac{(x - 5)}{ 4 }= \frac{(y - 3) }{6} = \frac{(z - 2)}{4}\)
L₂: \(\frac{(x - 3)}{ 7 }= \frac{(y - 2) }{5} = \frac{(z - 9)}{6}\)
Let A and B be any two points on L₁, and let C and D be any two points on L₂. Then, the vector connecting A to B is parallel to L₁, and the vector connecting C to D is parallel to L₂.
Let's first find the direction vectors of L₁ and L₂:
L₁ direction vector: <4, 6, 4>
L₂ direction vector: <7, 5, 6>
Next, let's choose two points on L₁ and L₂, respectively. For simplicity, we can choose the points where all three parameters are 0.
A = (5, 3, 2)
B = (9, 9, 6)
C = (3, 2, 9)
D = (10, 7, 15)
Now, let's find the vector connecting A to B and the vector connecting C to D.
AB = <9 - 5, 9 - 3, 6 - 2> = <4, 6, 4>
CD = <10 - 3, 7 - 2, 15 - 9> = <7, 5, 6>
To find the shortest distance between L₁ and L₂, we need to find the component of AB that is perpendicular to CD. We can do this using vector projection:
\(proj_{CD}(AB)\) = (AB . CD_unit) * \(CD_{unit}\)
where CD_unit is the unit vector in the direction of CD, given by:
\(CD_{unit}\) = \(\frac{CD}{ ||CD||}\)
First, let's find ||CD||:
||CD|| = \(\sqrt{(7^2 + 5^2 + 6^2)} = \sqrt{110}\)
Next, let's find CD_unit:
\(CD_{unit}\) = \(\frac{CD}{ ||CD||}\) = \(7\sqrt{1.10}, 5\sqrt{110} , 6\sqrt{110}\)
Now, let's find the dot product AB . CD_unit:
AB . CD_unit = \(47\sqrt{110}+ 65\sqrt{110} +4*6\sqrt{110}=58\sqrt{110}\)
Finally, let's find the vector projection proj_CD(AB):
proj_CD(AB) = (AB . CD_unit) * CD_unit = \(58\sqrt{110}\)* <\(7\sqrt{110},(5\sqrt{110}),(6\sqrt{110})\)> = \(<\frac{406}{110},\frac{290}{110},\frac{348}{110}>\)
The length of proj_CD(AB) gives us the shortest distance between L₁ and L₂:
||proj_CD(AB)|| = \(\sqrt{(\frac{406}{110})^2+(\frac{290}{110})^2+(\frac{348}{110})^2}\)) = \(\frac{190}{√756}\) = \(10\sqrt{\frac{190}{9}}\)
Therefore, the answer is \(\frac{190}{√756}\).
Answer. B