Question

The altitude of the parallelopiped, whose coterminous edges are the vectors \(\overset{⃗}{a} = \hat{i}+\hat{j}+\hat{k}\), \(\overset{⃗}{b} = 2\hat{i}+4\hat{j}-\hat{k}\), \(\overset{⃗}{c} = \hat{i}+\hat{j}+3\hat{k}\), where \(\overset{⃗}{a}\), \(\overset{⃗}{b}\) are the sides of the base of parallelopiped, is

• A. \(\frac{2\sqrt{2}}{\sqrt{19}}\) units
• B. \(\frac{\sqrt{2}}{\sqrt{19}}\) units
• C. \(\frac{\sqrt{19}}{\sqrt{2}}\) units
• D. \(\frac{\sqrt{19}}{2\sqrt{2}}\) units

Hint:

Altitude = volume divided by base area = |a.(b x c)| / |a x b|.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept
Volume of the parallelepiped = base area x altitude. The base is formed by \(\vec a, \vec b\), so the base area is \(|\vec a \times \vec b|\) and the volume is \(|(\vec a \times \vec b)\cdot\vec c|\).

Step 2: Base vector
\[ \vec a \times \vec b = \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 1 & 1 \\ 2 & 4 & -1 \end{vmatrix} = (-1 - 4)\hat i - (-1 - 2)\hat j + (4 - 2)\hat k = -5\hat i + 3\hat j + 2\hat k \]
\[ |\vec a \times \vec b| = \sqrt{25 + 9 + 4} = \sqrt{38} \]

Step 3: Volume and height
\[ (\vec a \times \vec b)\cdot\vec c = -5 + 3 + 6 = 4 \]
\[ h = \frac{4}{\sqrt{38}} = \frac{4}{\sqrt2\sqrt{19}} = \frac{2\sqrt2}{\sqrt{19}} \]
Options (B), (C) and (D) do not equal \(4/\sqrt{38}\).

Final Answer:
The altitude is \(\frac{2\sqrt2}{\sqrt{19}}\) units, option (A). \[ \boxed{\frac{2\sqrt{2}}{\sqrt{19}}} \]