Question

The height of parallelopiped whose adjacent edges are the vectors $\overline{a},\overline{b},\overline{c}$ is

• A. $6{\left(\frac{2}{3}\right)}^{1/2}$
• B. $4{\left(\frac{3}{2}\right)}^{1/2}$
• C. $4{\left(\frac{2}{3}\right)}^{1/2}$
• D. $8{\left(\frac{2}{3}\right)}^{1/2}$

Answer 1

Answer: (C)

$4{\left(\frac{2}{3}\right)}^{1/2}$

Given $\left|\overline{a}\right|=\left|\overline{b}\right|=\left|\overline{c}\right|=4or{\left|\overline{a}\right|}^2={\left|\overline{b}\right|}^2={\left|\overline{c}\right|}^2=16$
and $\overline{a}\cdot \overline{b}=\overline{b}\cdot \overline{c}=\overline{c}\cdot \overline{a}=8$
$\therefore \cos {\theta }_1=\cos {\theta }_2=\cos {\theta }_3=\frac{1}{2}$
$\Rightarrow {\theta }_1={\theta }_2={\theta }_3=\frac{\pi }{3}$
and $\overline{a}\cdot \overline{a}=\overline{b}\cdot \overline{b}=\overline{c}\cdot \overline{c}=16,{\overline{a}}^2={\overline{b}}^2={\overline{c}}^2=16$
Also $\left[\overline{a}\overline{b}\overline{c}\right]\left[\overline{x}\overline{y}\overline{z}\right]=\left|\begin{array}{ccc}\overline{a}\cdot \overline{x}& \overline{b}\cdot \overline{x}& \overline{c}\cdot \overline{x}\\ \overline{a}\cdot \overline{y}& \overline{b}\cdot \overline{y}& \overline{c}\cdot \overline{y}\\ \overline{a}\cdot \overline{z}& \overline{b}\cdot \overline{z}& \overline{c}\cdot \overline{z}\end{array}\right|$
${\left[\overline{a}\overline{b}\overline{c}\right]}^2=\left[\overline{a}\overline{b}\overline{c}\right]\left[\overline{a}\overline{b}\overline{c}\right]=\left|\begin{array}{ccc}\overline{a}\cdot \overline{a}& \overline{b}\cdot \overline{a}& \overline{c}\cdot \overline{a}\\ \overline{a}\cdot \overline{b}& \overline{b}\cdot \overline{b}& \overline{c}\cdot \overline{b}\\ \overline{a}\cdot \overline{c}& \overline{b}\cdot \overline{c}& \overline{c}\cdot \overline{c}\end{array}\right|$
$\Rightarrow {\left[\overline{a}\overline{b}\overline{c}\right]}^2=\left|\begin{array}{ccc}16& 8& 8\\ 8& 16& 8\\ 8& 8& 16\end{array}\right|=512\left|\begin{array}{ccc}2& 1& 1\\ 1& 2& 1\\ 1& 1& 2\end{array}\right|=1024\times 2$
$\left|\left[\overline{a}\overline{b}\overline{c}\right]\right|=32\sqrt{2}$ ......( * )
Volume of parallelopiped $=$ (Area of base) $\times$ height
$\Rightarrow 32\sqrt{2}=2$ (Are. of equilateral triangle) $\times$ height
$\Rightarrow 32\sqrt{2}=2\frac{\sqrt{3}}{4}(side{)}^2\times height$
$\Rightarrow 32\sqrt{2}=2\times 4\sqrt{3}\times height$
$\Rightarrow 4\sqrt{\frac{2}{3}}=height$
$\therefore$ Choice (c) is correct answer