Question

The parallelopiped is determined by vectors \(\overset{̄}{a} = -2\hat{i}+5\hat{j}+3\hat{k}\), \(\overset{̄}{b} = \hat{i}+3\hat{j}-2\hat{k}\), \(\overset{̄}{c} = -3\hat{i}+\hat{j}+4\hat{k}\). The altitude of parallelopiped on the parallelogram base determined by vectors \(\overset{̄}{b}\) and \(\overset{̄}{c}\) is

• A. \(\frac{2\sqrt{3}}{5}\)
• B. \(\frac{2}{5}\)
• C. \(12\)
• D. \(\frac{5\sqrt{3}}{2}\)

Hint:

Altitude = volume divided by base area, using the triple product and the cross product.

Answer 1

The Correct Option is A

Step 1: Key idea:
Volume \(=|\vec a\cdot(\vec b\times\vec c)|\), base area \(=|\vec b\times\vec c|\), and altitude \(h=\dfrac{\text{volume}}{\text{base area}}\).

Step 2: Find b x c:
\(\vec b\times\vec c=\begin{vmatrix}\hat i&\hat j&\hat k\\1&3&-2\\-3&1&4\end{vmatrix}=\hat i(12+2)-\hat j(4-6)+\hat k(1+9)=14\hat i+2\hat j+10\hat k\).

Step 3: Volume and base area:
\(\vec a\cdot(\vec b\times\vec c)=-28+10+30=12\). \(|\vec b\times\vec c|=\sqrt{196+4+100}=\sqrt{300}=10\sqrt3\).

Step 4: Altitude:
\[ h=\dfrac{12}{10\sqrt3}=\dfrac{6}{5\sqrt3}=\dfrac{2\sqrt3}{5} \]
Option A.

Step 5: Why the other options are wrong.
12 is just the volume. \(2/5\) and \(5\sqrt3/2\) come from errors in the area of the base.

Final Answer:
The altitude is 2 sqrt 3 / 5. \[ \boxed{\text{(A) }\dfrac{2\sqrt3}{5}} \]