The line, $$\frac{{x - 2}}{3} = \frac{{y + 1}}{2} = \frac{{z - 1}}{{ - 1}}$$ intersects the curve $$xy = {c^2},\,z = 0$$ if $$c$$ is equal to :
A.
$$ \pm 1$$
B.
$$ \pm \frac{1}{3}$$
C.
$$ \pm \sqrt 5 $$
D.
None
Answer :
$$ \pm \sqrt 5 $$
Solution :
We have, $$z = 0$$ for the point where the line intersects the curve.
Therefore,
$$\eqalign{
& \frac{{x - 2}}{3} = \frac{{y + 1}}{2} = \frac{{0 - 1}}{{ - 1}} \cr
& \Rightarrow \frac{{x - 2}}{3} = 1{\text{ and }}\frac{{y + 1}}{2} = 1 \cr
& \Rightarrow x = 5{\text{ and }}y = 1 \cr} $$
Put these value in $$xy = {c^2},$$ we get,
$$5 = {c^2} \Rightarrow c = \pm 5$$
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 :