Q74 / 160 · TS EAMCET 2022 (20 Jul Shift 2) · Mathematics

Three non-coplanar vectors $\bar{a}, \bar{b}, \bar{c}$ are the coterminous edges of a parallelepiped. If $\bar{a}$ and $\bar{b}$ determine the base of the parallelepiped then its height is

  • A
    $\frac{|[\bar{a} \bar{b} \bar{c}]|}{|\bar{b} \times \bar{c}|}$
  • Correct answer
    B
    $\frac{|[\bar{a} \bar{b} \bar{c}]|}{|\bar{a} \times \bar{b}|}$
  • C
    $\frac{|[\bar{a} \bar{b} \bar{c}]|}{|\bar{a} \times \bar{c}|}$
  • D
    $\frac{|[\bar{a} \bar{b} \bar{c}]|}{|\bar{b}+\bar{c}|}$

Solution

We know that volume of parallelopiped is $\left[\begin{array}{lll}\overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}}\end{array}\right]$. $\therefore$ Height is parallel to the normal vector to the base. $\Rightarrow \overline{\mathrm{x}}=\lambda(\overline{\mathrm{a}} \times \overline{\mathrm{b}})$ $\Rightarrow|\overline{\mathrm{x}}|=|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|$ Hence height of parallelepiped is $\frac{\left|\left[\begin{array}{lll} \overline{\mathrm{a}} & \overline{\mathrm{b}} & \overline{\mathrm{c}} \end{array}\right]\right|}{|\overline{\mathrm{a}} \times \overline{\mathrm{b}}|}$

Hence, the correct answer is option: 'B'.

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