Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The direction cosines of two lines are connected by the relations $l+m+n=0$ and $mn=2l(m+n)$. Which of the following can be TRUE?

Given the relations:

1. $l+m+n=0$

2. $mn=2l(m+n)$

From equation (1), we can write $m+n=-l$.

Substitute this into equation (2):

$mn=2l(-l)$

$mn=-2l^2$

We also know that for direction cosines, $l^2+m^2+n^2=1$.

From $m+n=-l$, square both sides:

$(m+n{)}^2=(-l{)}^2$

$m^2+n^2+2mn=l^2$

Substitute $mn=-2l^2$:

$m^2+n^2+2(-2l^2)=l^2$

$m^2+n^2-4l^2=l^2$

$m^2+n^2=5l^2$

Now substitute $m^2+n^2=5l^2$ into $l^2+m^2+n^2=1$:

$l^2+5l^2=1$

$6l^2=1$

$l^2=\frac{1}{6}$

$l=\pm \frac{1}{\sqrt{6}}$

Since $l^2=\frac{1}{6}$, we have $mn=-2l^2=-2\left(\frac{1}{6}\right)=-\frac{1}{3}$.

We have $m+n=-l$ and $mn=-\frac{1}{3}$.

Consider a quadratic equation $x^2-(m+n)x+mn=0$, whose roots are $m$ and $n$.

$x^2-(-l)x-\frac{1}{3}=0$

$x^2+lx-\frac{1}{3}=0$

For $m$ and $n$ to be real, the discriminant must be non-negative:

$D=l^2-4(1)\left(-\frac{1}{3}\right)\ge 0$

$l^2+\frac{4}{3}\ge 0$

Since $l^2=\frac{1}{6}$, this condition is satisfied: $\frac{1}{6}+\frac{4}{3}=\frac{1+8}{6}=\frac{9}{6}=\frac{3}{2}\ge 0$.

The direction cosines are real.

The possible values for $l$ are $\pm \frac{1}{\sqrt{6}}$.

The possible values for $mn$ are $-\frac{1}{3}$.

The possible values for $m+n$ are $\mp \frac{1}{\sqrt{6}}$.

Thus, the relations are consistent with the properties of direction cosines.