- Exam
- JEE Advanced
- Year
- 2023
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Three Dimensional Geometry
JEE Advanced 2023 · Paper 1 · Mathematics · Three Dimensional Geometry
Let Q be the cube with the set of vertices \ ( x _ 1 , x _ 2 , x _...
Options
- A
- B
- C
- D
Correct answer
A.
Step-by-step solution
Plotting the diagram of cube we get,
Now equation of line will be,
\overset{\to}{r} = \overset{\to}{0} + \lambda\left(\overset{^}{i} + \overset{^}{j}\right)
And equation of diagonal will be,
\overset{\to}{r}1 = \overset{^}{j} + \mu\left(\overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right)
Now finding the shortest distance between line we get,
S . D = \left|\dfrac{\left(\overset{^}{j} - 0\right) \cdot \left[\left(\overset{^}{i} + \overset{^}{j}\right) \times \left(\overset{^}{i} - \overset{^}{j} + \overset{^}{k}\right)\right]}{\sqrt{{1}^{2} + {1}^{2} + {0}^{2}} \cdot \sqrt{{1}^{2} + {1}^{2} + {1}^{2}}}\right|
⇒ S . D = \left|\dfrac{\overset{^}{j} \cdot \left(\overset{^}{i} - \overset{^}{j} - 2\overset{^}{k}\right)}{\sqrt{6}}\right| = \dfrac{1}{\sqrt{6}}
Now in other case shortest distance will be zero.