Question

If $\frac{1}{(x-11)(x-12)(x-13)}=\frac{A}{x-11}+\frac{B}{x-12}+\frac{C}{x-13}$, then the volume

of the parallelopiped whose adjacent sides are $A\hat{i}+B\hat{j}+2C\hat{k}$, $2A\hat{i}-B\hat{j}$ and $\hat{i}-3B\hat{j}+4C\hat{k}$ is

• A. $5$ cu. units
• B. $4$ cu. units
• C. $3$ cu. units
• D. $2$ cu. units

Answer 1

Answer: (A)

$5$ cu. units

Given that $\frac{1}{(x-11)(x-12)(x-13)}=\frac{A}{(x-11)}+\frac{B}{(x-12)}+\frac{C}{(x-13)}$

$A=\frac{1}{2}B=-1C=\frac{1}{2}$

Volume of parallelopiped $=\left[\overline{a}\overline{b}\overline{c}\right]$
Putting Values we get

$=\left|\begin{array}{ccc}A& B& 2C\\ 2A& -B& 0\\ 1& -3B& 4C\end{array}\right|$

$=-4ABC-B\left(8AC\right)+2C(-6AB+B)$

$=1+2+\frac{2}{2}(3-1)$

$=5$