Question
A line with positive direction cosines passes through the point $$P\left( {2,\, - 1,\,2} \right)$$ and makes equal angles with the coordinate axes. The line meets the plane $$2x+y+z=9$$ at point $$Q.$$ The length of the line segment $$PQ$$ equals -
A.
$$1$$
B.
$$\sqrt 2 $$
C.
$$\sqrt 3 $$
D.
$$2$$
Answer :
$$\sqrt 3 $$
Solution :
The line has +ve and equal direction cosines, these are $$\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}$$ or direction ratios are 1, 1, 1. Also the lines passes through $$P\left( {2,\, - 1,\,2} \right).$$
$$\therefore $$ Equation of line is $$\frac{{x - 2}}{1} = \frac{{y + 1}}{1} = \frac{{z - 2}}{1} = \lambda \,\,\left( {{\text{say}}} \right)$$
Let $$Q\left( {\lambda + 2,\,\lambda - 1,\,\lambda + 2} \right)$$ be a point on this line where it meets the plane $$2x+y+z=9$$
Then $$Q$$ must satisfy the equation of plane
$${\text{i}}{\text{.e}}{\text{.,}}\,\,\,2\left( {\lambda + 2} \right) + \lambda - 1 + \lambda + 2 = 9\,\, \Rightarrow \lambda = 1$$
$$\therefore \,\,Q$$ has coordintes $$\left( {3,\,0,\,3} \right)$$
Hence the length of line segments $$PQ$$
$$ = \sqrt {{{\left( {2 - 3} \right)}^2} + {{\left( { - 1 - 0} \right)}^2} + {{\left( {2 - 3} \right)}^2}} = \sqrt 3 $$