A line with positive direction cosines passes through the point $\mathrm{P}(2,1,2)$ and makes equal angles…
A line with positive direction cosines passes through the point $\mathrm{P}(2,1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $2 x+y+\mathrm{z}=9$ at point Q . The length of the line segment PQ equals $\qquad$ units.
$\frac{5}{\sqrt{3}}$
$2 \sqrt{3}$
$\frac{4}{\sqrt{3}}$
$4 \sqrt{3}$
Solution
1. Direction Cosines: If a line makes equal angles with coordinate axes, its direction ratios are proportional to $(1,1,1)$.
2. Equation of the Line: Parametric form is:
$\frac{x-2}{1}=\frac{y-1}{1}=\frac{z-2}{1}=t$
or,
$x=2+t, y=1+t, z=2+t$
3. Point of Intersection with Plane: Substituting $x=2+t, y=1+t, z=2+t$ into $2 x+$ $y+z=9$ :
$\begin{gathered}
2(2+t)+(1+t)+(2+t)=9 \\
4+2 t+1+t+2+t=9 \quad \Rightarrow \quad 4 t+7=9 \\
t=\frac{2}{4}=\frac{1}{2}
\end{gathered}$ Substituting $t=\frac{1}{2}$ back, the coordinates of Q are:
$Q=\left(2+\frac{1}{2}, 1+\frac{1}{2}, 2+\frac{1}{2}\right)=\left(\frac{5}{2}, \frac{3}{2}, \frac{5}{2}\right)$
4. Length of PQ: Using the distance formula:
$\begin{gathered}
P Q=\sqrt{\left(\frac{5}{2}-2\right)^2+\left(\frac{3}{2}-1\right)^2+\left(\frac{5}{2}-2\right)^2} \\
P Q=\sqrt{\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^2}=\sqrt{3 \cdot\left(\frac{1}{2}\right)^2}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}
\end{gathered}$ Answer: $2 \sqrt{3}$, Option 2.