Question

A parallelopiped is determined by vectors \(\vec{a}=\hat{i}+2\hat{k},\vec{b}=\hat{i}+\hat{j}-\hat{k}\) and \(\vec{c}=2\hat{i}+2\hat{j}+3\hat{k}\). The length (in units) of its altitude drawn to the base determined by vectors \(\vec{b}\) and \(\vec{c}\) is

• A. \(\sqrt{2}\)
• B. \(5\sqrt{2}\)
• C. \(\frac{1}{\sqrt{2}}\)
• D. \(\frac{1}{2}\)

Hint:

The altitude equals the volume (scalar triple product) divided by the area of the base, which is the magnitude of b cross c.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept
Height \(=\frac{\text{Volume}}{\text{Base area}}=\frac{|\vec{a}\cdot(\vec{b}\times\vec{c})|}{|\vec{b}\times\vec{c}|}\).

Step 2: Cross product
\(\vec{b}\times\vec{c}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\1&1&-1\\2&2&3\end{vmatrix}=(3+2)\hat{i}-(3+2)\hat{j}+(2-2)\hat{k}=5\hat{i}-5\hat{j}\).

Step 3: Volume and base area
\(\vec{a}\cdot(\vec{b}\times\vec{c})=(1)(5)+0+0=5\) and \(|\vec{b}\times\vec{c}|=5\sqrt{2}\).

Step 4: Altitude
\(h=\frac{5}{5\sqrt{2}}=\frac{1}{\sqrt{2}}\).

Final Answer:
The altitude is 1/√2 units. \[ \boxed{\text{(C) }\dfrac{1}{\sqrt{2}}} \]