# Find the distance of the point (2, 12, 5) from the point of intersection of the line $\overrightarrow r = 2\hat i - 4\hat j + 2 \hat k + \lambda ( 3\hat i + 4 \hat j + 2 \hat k )$ and the plane $\overrightarrow r. ( \hat i - 2 \hat j + \hat k ) =0.$

$(a)\;13\;units\qquad(b)\;1\;units\qquad(c)\;12\;units\qquad(d)\;14\;units$

Equation of the given line is
$\overrightarrow r = 2\hat i - 4\hat j + 2 \hat k + \lambda ( 3\hat i + 4 \hat j + 2 \hat k )$ -----------(1)
Given equation of the plane is
$\overrightarrow r = ( \hat i - 2 \hat j+ \hat k ) = 0$--------(2)
Substituting the value of $\overrightarrow r$ in eqn 2 we get
$( 2\hat i - 4\hat j + 2 \hat k) + \lambda ( 3\hat i + 4 \hat j + 2 \hat k ) = ( \hat i - 2 \hat j + \hat k ) = 0$
$\Rightarrow [ (2+3 \lambda) \hat i + (4 \lambda - 4 ) \hat j + (2+2 \lambda) \hat k ].[ \hat i - 2 \hat j + \hat k ] = 0$
$\Rightarrow (2+ 3 \lambda)(1)+(4 \lambda-4)(-2)+(2+2 \lambda)(1)=0$
$2+ 3 \lambda-8\lambda+8+2+2 \lambda = 0$ Hence the equation of the line is
$\overrightarrow r = (2\hat i - 4\hat j + 2 \hat k) + 4 ( 3\hat i + 4 \hat j + 2 \hat k )$
(i.e.,) $\overrightarrow r = 2\hat i - 4\hat j + 2 \hat k +12 \hat i + 16 \hat j + 8 \hat k$
$= 14 \hat i + 18 \hat k + 10 \hat k$
The point of intersection of the given line and the plane is given by the co-ordinate
(14, 12, 10), the given point is ( 2, 12, 5)
hence the distance between the point is
$d = \sqrt{ (14-2)^2 + (12-12)^2+ (10-5)^2}$
=$\sqrt{ 12^2+0+5^2}$
=$\sqrt{144+25}$
=$\sqrt{144+25}$
=$\sqrt{169}$
$= 13$
hence the distance is 13 units
answered Mar 31, 2014
edited Apr 1, 2014