Question

Given $\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}$ for a $\Delta ABC$ with usual notation. If $\frac{\cos A}{\alpha }=\frac{\cos B}{\beta }=\frac{\cos C}{\gamma }$, then d.r.'s of the line perpendicular to the lines $\frac{x}{\alpha }=\frac{y}{\beta }=\frac{z}{\gamma }$ and $\frac{x-1}{\beta }=\frac{y}{\alpha }=\frac{z+1}{\gamma }$ are

• A. $(25,-25,-26)$
• B. $(25,25,-26)$
• C. $(-25,25,-26)$
• D. $(25,25,26)$

Answer 1

The correct option is B $(25,25,-26)$

$\frac{b+c}{11}=\frac{c+a}{12}=\frac{a+b}{13}=\lambda$

$\therefore$ $b+c=11\lambda \cdots \left(1\right)c+a=12\lambda \cdots \left(2\right)a+b=13\lambda \cdots \left(3\right)$

$\Rightarrow a+b+c=18\lambda \cdots \left(4\right)$
Now subtract $\left(1\right)$ from $\left(4\right)$, we get
$a=7\lambda \Rightarrow b=6\lambda ,c=5\lambda$

By cosine rule,
$\cos A=\frac{36{\lambda }^2+25{\lambda }^2-49{\lambda }^2}{2\times 6\lambda \times 5\lambda }=\frac{1}{5}$

$\cos B=\frac{25{\lambda }^2+49{\lambda }^2-36{\lambda }^2}{2\times 5\lambda \times 7\lambda }=\frac{19}{35}$

$\cos C=\frac{49{\lambda }^2+36{\lambda }^2-25{\lambda }^2}{2\times 7\lambda \times 6\lambda }=\frac{5}{7}$

$\because \frac{\cos A}{\alpha }=\frac{\cos B}{\beta }=\frac{\cos C}{\gamma }$
$\Rightarrow \frac{1}{5\alpha }=\frac{19}{35\beta }=\frac{5}{7\gamma }=k\therefore \alpha =\frac{1}{5k}=\frac{7}{35k},\beta =\frac{19}{35k},$
and $\gamma =\frac{5}{7k}=\frac{25}{35k}$

So, $(\alpha ,\beta ,\gamma )=(7,19,25)$
$\therefore$ d.r.'s of line perpendicular to the lines $\frac{x}{\alpha }=\frac{y}{\beta }=\frac{z}{\gamma }$ and $\frac{x-1}{\beta }=\frac{y}{\alpha }=\frac{z+1}{\gamma }$ are
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 7& 19& 25\\ 19& 7& 25\end{array}\right|=12(25\hat{i}+25\hat{j}-26\hat{k})$
i.e., $(25,25,-26)$