So,

$$ \text{Altitude}=\frac{\text{Volume of parallelepiped}}{\text{Area of base}} $$

The volume is given by the scalar triple product:

$$ \text{Volume}=|[\vec a \ \vec b \ \vec c]|=|(\vec a \times \vec b)\cdot \vec c| $$

First find $\vec a \times \vec b$.

Given,

$$ \vec a=\hat i+\hat j+\hat k $$

$$ \vec b=2\hat i+4\hat j-\hat k $$

So,

$$ \vec a \times \vec b= \begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 1 & 1\\ 2 & 4 & -1 \end{vmatrix} $$

$$ \vec a \times \vec b = \hat i(-1-4)-\hat j(-1-2)+\hat k(4-2) $$

$$ \vec a \times \vec b=-5\hat i+3\hat j+2\hat k $$

Now,

$$ |\vec a \times \vec b| =\sqrt{(-5)^2+3^2+2^2} $$

$$ =\sqrt{25+9+4} $$

$$ =\sqrt{38} $$

So, area of the base is $\sqrt{38}$ square units.

Now find the volume:

$$ (\vec a \times \vec b)\cdot \vec c =(-5\hat i+3\hat j+2\hat k)\cdot(\hat i+\hat j+3\hat k) $$

$$ =-5+3+6 $$

$$ =4 $$

So,

$$ \text{Volume}=4 $$

Hence,

$$ \text{Altitude}=\frac{4}{\sqrt{38}} $$

Since,

$$ \sqrt{38}=\sqrt{2\times 19}=\sqrt2\sqrt{19} $$

we get

$$ \frac{4}{\sqrt{38}} =\frac{4}{\sqrt2\sqrt{19}} =\frac{2\sqrt2}{\sqrt{19}} $$

So, the altitude is

$$ \boxed{\frac{2\sqrt2}{\sqrt{19}}\text{ units}} $$

Correct option: Option A.

Vector Algebra

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