A parallelepiped is formed by planes drawn through the points $$\left( {2,\,4,\,5} \right)$$ and $$\left( {5,\,9,\,7} \right)$$ parallel to the coordinate planes. The length of the diagonal of the parallelepiped is :
A.
$$8{\text{ units}}$$
B.
$$4{\text{ units}}$$
C.
$$7{\text{ units}}$$
D.
$$11{\text{ units}}$$
Answer :
$$7{\text{ units}}$$
Solution :
$$\eqalign{
& {\text{The length of the edges are given by}} \cr
& a = 5 - 2 = 3, \cr
& b = 9 - 3 = 6, \cr
& c = 7 - 5 = 2 \cr
& {\text{So length of the diagonal is}} \cr
& \sqrt {{a^2} + {b^2} + {c^2}} = \sqrt {9 + 36 + 4} = 7{\text{ units}} \cr} $$
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 :