If the line, $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 2}}{4}$$ meets the plane, $$x+2y+3z=15$$ at a point $$P,$$ then the distance of $$P$$ from the origin is :
A.
$$\frac{{\sqrt 5 }}{2}$$
B.
$$2\sqrt 5 $$
C.
$$\frac{9}{2}$$
D.
$$\frac{7}{2}$$
Answer :
$$\frac{9}{2}$$
Solution :
Let point on line be $$P\left( {2k + 1,\,3k - 1,\,4k + 2} \right)$$
Since, point $$P$$ lies on the plane $$x+2y+3z=15$$
$$2k+1+6k-2+12k+6=15$$
$$\eqalign{
& \Rightarrow k = \frac{1}{2} \cr
& \therefore P \equiv \left( {2,\,\frac{1}{2},\,4} \right) \cr} $$
Then the distance of the point $$P$$ from the origin is
$$OP = \sqrt {4 + \frac{1}{4} + 16} = \frac{9}{2}$$
Releted MCQ Question on Geometry >> Three Dimensional Geometry
Releted Question 1
The value of $$k$$ such that $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$ lies in the plane $$2x - 4y + z = 7,$$ is :
If the lines $$\frac{{x - 1}}{2} = \frac{{y + 1}}{3} = \frac{{z - 1}}{4}$$ and $$\frac{{x - 3}}{1} = \frac{{y - k}}{2} = \frac{z}{1}$$ intersect, then the value of $$k$$ is :
A plane which is perpendicular to two planes $$2x - 2y + z = 0$$ and $$x - y + 2z = 4,$$ passes through $$\left( {1,\, - 2,\,1} \right).$$ The distance of the plane from the point $$\left( {1,\,2,\,2} \right)$$ is :
Let $$P\left( {3,\,2,\,6} \right)$$ be a point in space and $$Q$$ be a point on the line $$\vec r = \left( {\hat i - \hat j + 2\hat k} \right) + \mu \left( { - 3\hat i + \hat j + 5\hat k} \right)$$
Then the value of $$\mu $$ for which the vector $$\overrightarrow {PQ} $$ is parallel to the plane $$x-4y+3z=1$$ is :