Question

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.)

• A. \(5\)
• B. \(20\)
• C. \(10\)
• D. \(8\)

Hint:

Use the formula |(a2 - a1) . (d1 x d2)| / |d1 x d2|.

Answer 1

The Correct Option is C

Step 1: Write the Lines:
Line 1: \(\dfrac{x+1}7=\dfrac{y+1}{-6}=\dfrac{z+1}1\), point \(P_1(-1,-1,-1)\), direction \(\vec d_1=(7,-6,1)\).
Line 2: point \(P_2(3,5,7)\), direction \(\vec d_2=(1,-2,1)\).

Step 2: Cross Product:
\[ \vec d_1\times\vec d_2=\big((-6)(1)-(1)(-2),\ (1)(1)-(7)(1),\ (7)(-2)-(-6)(1)\big)=(-4,-6,-8) \]
\(|\vec d_1\times\vec d_2|=\sqrt{16+36+64}=\sqrt{116}\).

Step 3: Distance:
\(\overrightarrow{P_1P_2}=(4,6,8)\). Then \((4,6,8)\cdot(-4,-6,-8)=-116\).
\[ p=\frac{116}{\sqrt{116}}=\sqrt{116}\approx10.77 \]

Step 4: Greatest Integer:
Since \(10^2=100<116<121=11^2\), \(10<\sqrt{116}<11\), so \([p]=10\). Option (C).

Final Answer:
\([p]=10\), option (C). \[ \boxed{\text{(C) } 10} \]