If $p$ is the shortest distance between the lines $\frac{x+1}{7} = \frac{y+1}{-6} = z+1$ and $\overset{⃗}{r} = (3\hat{i}+5\hat{j}+7\hat{k})+μ(\hat{i}-2\hat{j}+\hat{k})$ then $[p]$ is... ,(where $[\,.\,]$ denotes the greatest integer function.)

Question

If the shortest distance between the lines $\frac{x - \lambda}{-2} = \frac{y - 2}{1} = \frac{z - 1}{1}$ and $\frac{x - \sqrt{3}}{1} = \frac{y - 1}{-2} = \frac{z - 2}{1}$ is 1, then the sum of all possible values of $ \lambda $ is:

Answer 1

Step 1: Plane Method:
The shortest distance is the distance from $P_1$ to the plane through line 2 parallel to line 1. The normal to that plane is $\vec n=\vec d_1\times\vec d_2\propto(2,3,4)$.

Step 2: Equation:
Plane through $P_2(3,5,7)$: $2(x-3)+3(y-5)+4(z-7)=0$, i.e. $2x+3y+4z=49$.

Step 3: Distance:
From $P_1(-1,-1,-1)$: $\dfrac{|2(-1)+3(-1)+4(-1)-49|}{\sqrt{4+9+16}}=\dfrac{58}{\sqrt{29}}=2\sqrt{29}=\sqrt{116}\approx10.77$. The greatest integer is 10. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } 10} \]

If \(p\) is the shortest distance between the lines \(\frac{x+1}{7} = \frac{y+1}{-6} = z+1\) and \(\overset{⃗}{r} = (3\hat{i}+5\hat{j}+7\hat{k})+μ(\hat{i}-2\hat{j}+\hat{k})\) then \([p]\) is... ,(where \([\,.\,]\) denotes the greatest integer function.)

Show Hint

The Correct Option is C

Solution and Explanation

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