Let $$A\left( {\overrightarrow a } \right)$$ and $$B\left( {\overrightarrow b } \right)$$ be points on two skew line $$\overrightarrow r = \overrightarrow a + \overrightarrow \lambda $$ and $$\overrightarrow r = \overrightarrow b + u\overrightarrow q $$ and the shortest distance between the skew line is $$1$$, where $$\overrightarrow p $$ and $$\overrightarrow q $$ are unit vectors forming adjacent sides of a parallelogram enclosing an area of $$\frac{1}{2}$$ units. If an angle between $$AB$$ and the line of shortest distance is $${60^ \circ },$$ then $$AB = ?$$
A.
$$\frac{1}{2}$$
B.
$$2$$
C.
$$1$$
D.
$$\lambda \, \in \,R - \left\{ 0 \right\}$$
Answer :
$$2$$
Solution :
$$\eqalign{
& 1 = \left| {\left( {\overrightarrow b - \overrightarrow a } \right).\frac{{\left( {\overrightarrow p \times \overrightarrow q } \right)}}{{\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right| \cr
& \Rightarrow \left| {\overrightarrow a - \overrightarrow b } \right|\cos \,{60^ \circ } = 1 \cr
& \Rightarrow AB = 2 \cr} $$
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 :