Question

Let \(\overset{⃗}{a} = λ\hat{i}+\hat{j}+\hat{k},\overset{⃗}{b} = 2\hat{i}+4\hat{j}+4\hat{k},\overset{⃗}{c} = \hat{i}+μ\hat{j}+\hat{k}\)
If \(\overset{⃗}{a}\) is parallel to \(\overset{⃗}{b}\) and \(\overset{⃗}{b}\) is perpendicular to \(\overset{⃗}{c}\) then \(λ-μ = \ldots\)

• A. \(-2\)
• B. \(-1\)
• C. \(2\)
• D. \(1\)

Hint:

Find |c| from the first two conditions, then use the angle between a x b and c.

Answer 1

The Correct Option is C

Step 1: Find |c|:
\(|\vec a| = \sqrt{1 + 4 + 4} = 3\). From \(|\vec c - \vec a|^2 = 8\): \(|\vec c|^2 - 2\vec a\cdot\vec c + 9 = 8\). Using \(\vec a\cdot\vec c = |\vec c|\): \(|\vec c|^2 - 2|\vec c| + 1 = 0\), so \(|\vec c| = 1\).

Step 2: Find |a x b|:
\(\vec a\times\vec b = \begin{vmatrix}\hat i & \hat j & \hat k\\1 & 2 & -2\\1 & -1 & 1\end{vmatrix} = (2 - 2)\hat i - (1 + 2)\hat j + (-1 - 2)\hat k = -3\hat j - 3\hat k\).
So \(|\vec a\times\vec b| = 3\sqrt2\).

Step 3: Magnitude of the double product:
\[ |(\vec a\times\vec b)\times\vec c| = |\vec a\times\vec b||\vec c|\sin60^\circ = 3\sqrt2\cdot1\cdot\frac{\sqrt3}{2} = \frac{3\sqrt6}{2} = 3\sqrt{\frac32} \]

Final Answer:
The magnitude is \(3\sqrt{\frac32}\), option (C). \[ \boxed{3\sqrt{\frac{3}{2}}} \]