Question

If \(d\) is the distance of point (2, 5, 10) from the plane containing the lines \(\overset{̄}{r} = (4\hat{j}-\hat{k})+λ(\hat{i}+2\hat{j}-2\hat{k})\) and \(\overset{̄}{r} = (2\hat{i}+\hat{j})+μ(\hat{i}+2\hat{j}-2\hat{k})\), then \(d^2 =\)

• A. \(145\)
• B. \(13\)
• C. \(90\)
• D. \(79\)

Hint:

The two lines are parallel, so the normal is the cross product of their common direction and the vector joining one point on each line.

Answer 1

The Correct Option is C

Step 1: Data
Both lines have direction \(\vec d = \hat i+2\hat j-2\hat k\). Points on them: \(P_1(0,4,-1)\) and \(P_2(2,1,0)\).

Step 2: Normal
\(\overrightarrow{P_1P_2} = (2,-3,1)\). \(\vec n = \vec d\times\overrightarrow{P_1P_2} = \begin{vmatrix}\hat i&\hat j&\hat k\\1&2&-2\\2&-3&1\end{vmatrix} = (2-6)\hat i-(1+4)\hat j+(-3-4)\hat k = (-4,-5,-7)\).

Step 3: Plane
Use \((4,5,7)\) and the point \((0,4,-1)\): \(4x+5(y-4)+7(z+1) = 0\), that is \(4x+5y+7z-13 = 0\).

Step 4: Distance
For \((2,5,10)\): \(8+25+70-13 = 90\). The norm is \(\sqrt{16+25+49} = \sqrt{90}\). So \(d = \frac{90}{\sqrt{90}} = \sqrt{90}\) and \(d^2 = 90\). Option (C).

Final Answer:
d squared is 90. \[ \boxed{\text{(C)}\ 90} \]