Question

Let $a,b$ and $c$ be distinct positive numbers. If the vectors $a\hat{i}+a\hat{j}+c\hat{k},\hat{i}+\hat{k}$ and $c\hat{i}+c\hat{j}+b\hat{k}$ are co-planar, then $c$ is equal to

• A. $\frac{2}{\frac{1}{a}+\frac{1}{b}}$
• B. $\sqrt{ab}$
• C. $\frac{a+b}{2}$
• D. $\frac{1}{a}+\frac{1}{b}$

Answer 1

The correct option is B $\sqrt{ab}$

$\because a\hat{i}+a\hat{j}+c\hat{k},\hat{i}+\hat{k}$ and $c\hat{i}+c\hat{j}+b\hat{k}$ are co-planar.
$\therefore \left|\begin{array}{ccc}a& a& c\\ 1& 0& 1\\ c& c& b\end{array}\right|=0$
$C_1\to C_1-C_2$
$\left|\begin{array}{ccc}0& a& c\\ 1& 0& 1\\ 0& c& b\end{array}\right|=0$
$\Rightarrow -a(b-0)+c\left(c\right)=0$
$\Rightarrow c^2=ab$
$\therefore c=\sqrt{ab}$