Question

Suppose direction cosines of two lines are given by $ul+vm+wn=0$ and $a{l}^2+b{m}^2+c{n}^2=0$ where $u,v,w,a,b,c$ are arbitrary constants and $1,m,n$ are direction cosines of the line. The given lines will be perpendicular, if

• A. $\sum u^2(a-b)=0$
• B. $\sum u^2(b-c)=0$
• C. $\sum u^2(a+b)=0$
• D. $\sum u^2(b+c)=0$

Answer 1

The correct option is D $\sum u^2(b+c)=0$

Direction cosines of the two lines are given by $ul+vm+wn=0$ ------$\left(1\right)$

and $a{l}^2+b{m}^2+c{n}^2=0$ ------$\left(2\right)$

Eliminating $n$ from $\left(1\right)$ and $\left(2\right)$ gives

$a{l}^2+b{m}^2+c{\left(\frac{ul+vm}{-w}\right)}^2=0$

$\Rightarrow w^{2}a{l}^2+w^{2}b{m}^2+c(ul+vm{)}^2=0$

$\Rightarrow (a{w}^2+c{u}^2){\left(\frac{l}{m}\right)}^2+2uvc\left(\frac{l}{m}\right)+(b{w}^2+c{v}^2)=0$

$\frac{l_1}{m_1}$ and $\frac{l_2}{m_2}$ are roots of above equation.

Product of roots $\frac{l_1{l}_2}{m_1{m}_2}=\frac{b{w}^2+c{v}^2}{a{w}^2+c{u}^2}$ -----$\left(3\right)$

Similarly, eliminating $l$ from $\left(1\right)$ and $\left(2\right)$ gives,

$\frac{m_1{m}_2}{n_1{n}_2}=\frac{a{w}^2+c{u}^2}{a{v}^2+b{u}^2}$ -----$\left(4\right)$

From $\left(3\right)$ and $\left(4\right)$, we get

$\frac{l_1{l}_2}{b{w}^2+c{v}^2}=\frac{m_1{m}_2}{a{w}^2+c{u}^2}=\frac{n_1{n}_2}{a{v}^2+b{u}^2}$

Lines are perpendicular if $l_1{l}_2+m_1{m}_2+n_1{n}_2=0$

$\therefore u^2(b+c)+v^2(c+a)+w^2(a+b)=0$

Hence, option D.