Question

The acute angle between the line \(\overset{⃗}{r} = (\hat{i}+2\hat{j}+\hat{k})+λ(\hat{i}+\hat{j}+\hat{k})\) and the plane \(\overset{⃗}{r}\cdot (2\hat{i}+p\hat{j}+\hat{k}) = 8\) is \(sin^{-1}(\frac{\sqrt{2}}{3})\), then the value of \(p\) are...

• A. \(p = 1\) or \(p = 17\)
• B. \(p = -1\) or \(p = -17\)
• C. \(p = 6\) or \(p = 3\)
• D. \(p = -6\) or \(p = -3\)

Hint:

sin of the angle equals |d . n| / (|d||n|) with direction d and normal n.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The angle \(\phi\) between a line with direction \(\bar d\) and a plane with normal \(\bar n\) satisfies \(\sin\phi = \frac{|\bar d\cdot\bar n|}{|\bar d||\bar n|}\).

Step 2: Set up:
\(\bar d = (1,1,1)\) and \(\bar n = (2, p, 1)\). Then \(\bar d\cdot\bar n = 3 + p\), \(|\bar d| = \sqrt3\), \(|\bar n| = \sqrt{5 + p^2}\).
\[ \frac{|3 + p|}{\sqrt3\sqrt{5+p^2}} = \frac{\sqrt2}{3} \]

Step 3: Solve:
Square both sides: \(\frac{(3+p)^2}{3(5+p^2)} = \frac29\), so \(9(9 + 6p + p^2) = 6(5 + p^2)\).
\(81 + 54p + 9p^2 = 30 + 6p^2\), so \(3p^2 + 54p + 51 = 0\), that is \(p^2 + 18p + 17 = 0\).
\((p+1)(p+17) = 0\), so \(p = -1\) or \(p = -17\).

Step 4: Why the other options are wrong.
Option (A) has the positive values 1 and 17, which come from the sign of \(p\) being flipped. With \(p = 1\): \(\sin\phi = \frac{4}{\sqrt3\sqrt6} = \frac{4}{3\sqrt2}\), which is not \(\frac{\sqrt2}{3}\).

Final Answer:
\(p = -1\) or \(p = -17\), option (B). \[ \boxed{p=-1\text{ or }p=-17} \]