The foot of the perpendicular drawn from the origin to a plane is the point $$\left( {1,\, - 3,\,1} \right).$$ What is the intercept cut on the $$x$$-axis by the plane ?
A.
$$1$$
B.
$$3$$
C.
$$\sqrt {11} $$
D.
$$11$$
Answer :
$$11$$
Solution :
Equation of plane passing through $$\left( {1,\, - 3,\,1} \right)$$ and whose normal $$\left( {1,\, - 3,\,1} \right)$$ is
$$\eqalign{
& 1\left( {x - 1} \right) - 3\left( {y + 3} \right) + 1\left( {z - 1} \right) = 0 \cr
& \Rightarrow x - 3y + z - 11 = 0 \cr
& \Rightarrow \frac{x}{{11}} - \frac{y}{{\frac{{11}}{3}}} + \frac{z}{{11}} = 0 \cr} $$
The above plane intercept the $$x$$-axis at $$11.$$
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 :