Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

As the point $$\left( {3,\,2,\,0} \right)$$  lies on the given line $$\frac{{x - 4}}{1} = \frac{{y - 7}}{5} = \frac{{z - 4}}{4}$$
$$\therefore $$ There can be infinite many planes passing through this line. But here out of the four options only first option is satisfied by the coordinates of both the points $$\left( {3,\,2,\,0} \right)$$  and $$\left( {4,\,7,\,4} \right)$$
$$\therefore \,\,x - y + z = 1$$    is the required plane.

$$\eqalign{ & {\text{Let the coordinates of point}}\,P\,{\text{be}}\left( {x,\,y,\,z} \right) \cr & {\text{Here, }}P{A^2} = {\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} + {\left( {z - 5} \right)^2} \cr & P{B^2} = {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z + 7} \right)^2} \cr & {\text{By the given condition}}\,P{A^2} + P{B^2} = 2{K^2} \cr & {\text{We have}} \cr & {\left( {x - 3} \right)^2} + {\left( {y - 4} \right)^2} + {\left( {z - 5} \right)^2} + {\left( {x + 1} \right)^2} + {\left( {y - 3} \right)^2} + {\left( {z + 7} \right)^2} = 2{K^2} \cr & {\text{i}}{\text{.e}}{\text{., }}2{x^2} + 2{y^2} + 2{z^2} - 4x - 14y + 4z = 2{K^2} - 109 \cr} $$

As the line $$\frac{{x - 4}}{1} = \frac{{y - 2}}{1} = \frac{{z - k}}{2}$$     lies in the plane $$2x-4y+z=7,$$    the point $$\left( {4,\,2,\,k} \right)$$   through which line passes must also lie on the given plane and hence $$2 \times 4 - 4 \times 2 + k = 7\,\,\, \Rightarrow k = 7$$

General point on given line $$ \equiv P\left( {3r + 2,\,4r - 1,\,12r + 2} \right)$$
Point $$P$$ must satisfy equation of plane
$$\eqalign{ & \left( {3r + 2} \right) - \left( {4r - 1} \right) + \left( {12r + 2} \right) = 16 \cr & \Rightarrow 11r + 5 = 16 \cr & \Rightarrow r = 1 \cr & P\left( {3 \times 1 + 2,\,4 \times 1 - 1,\,12 \times 1 + 2} \right)\,\,\,\,\, = P\left( {5,\,3,\,14} \right) \cr} $$
distance between $$P$$ and (1, 0, 2)
$$D = \sqrt {{{\left( {5 - 1} \right)}^2} + {3^2} + {{\left( {14 - 2} \right)}^2}} = 13$$

Suppose $$P$$ divides $$QR$$  in ratio $$\lambda :1$$
Then, co-ordinates of $$P$$ are $$\left( {\frac{{5\lambda + 2}}{{\lambda + 1}},\,\frac{{\lambda + 2}}{{\lambda + 1}},\,\frac{{ - 2\lambda + 1}}{{\lambda + 1}}} \right)$$
It is given that the $$x$$-coordinate of $$P$$ is $$4.$$
i.e., $$\frac{{5\lambda + 2}}{{\lambda + 1}} = 4 \Rightarrow \lambda = 2$$
So, $$z$$-coordinate of $$P$$ is $$\frac{{ - 2\lambda + 1}}{{\lambda + 1}} = \frac{{ - 4 + 1}}{{2 + 1}} = - 1$$

$$\eqalign{ & {L_1} \equiv \overrightarrow {{r_1}} = \left( {2\overrightarrow i + \overrightarrow j + 2\overrightarrow k } \right) + t\left( { - 3\overrightarrow i + 2\overrightarrow j + 6\overrightarrow k } \right) \cr & {L_1} \equiv \overrightarrow {{r_2}} = \left( {\overrightarrow i + 2\overrightarrow j - \overrightarrow k } \right) + s\left( {4\overrightarrow i - \overrightarrow j + 8\overrightarrow k } \right) \cr & \therefore \,{\text{angle between}}\,{L_1}{\text{ and}}\,{L_2}{\text{ is given by}} \cr & \Rightarrow \cos \,\theta = \frac{{\left( { - 3\overrightarrow i + 2\overrightarrow j + 6\overrightarrow k } \right).\left( {4\overrightarrow i - \overrightarrow j + 8\overrightarrow k } \right)}}{{\sqrt {9 + 4 + 36} \,\sqrt {16 + 1 + 64} }} \cr & \Rightarrow \cos \,\theta = \frac{{ - 12 - 2 + 48}}{{\sqrt {49} \,\sqrt {81} }} \cr & \Rightarrow \cos \,\theta = \frac{{34}}{{7 \times 9}} \cr & \Rightarrow \cos \,\theta = \frac{{34}}{{63}} \cr & \Rightarrow \theta = {\cos ^{ - 1}}\left( {\frac{{34}}{{63}}} \right) \cr} $$

Let the direction cosines of line $$L$$ be $$l,\,m,\,n,$$   then
$$\eqalign{ & 2l + 3m + n = 0......\left( {\text{i}} \right) \cr & {\text{and }}l + 3m + 2n = 0......\left( {{\text{ii}}} \right) \cr} $$
on solving equations $$\left( {\text{i}} \right)$$ and $$\left( {\text{ii}} \right),$$  we get
$$\eqalign{ & \frac{l}{{6 - 3}} = \frac{m}{{1 - 4}} = \frac{n}{{6 - 3}} \Rightarrow \frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3} \cr & {\text{Now, }}\frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3} = \frac{{\sqrt {{l^2} + {m^2} + {n^2}} }}{{\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }} \cr & \because \,{l^2} + {m^2} + {n^2} = 1 \cr & \therefore \,\frac{l}{3} = \frac{m}{{ - 3}} = \frac{n}{3} = \frac{1}{{\sqrt {27} }} \cr & \Rightarrow l = \frac{3}{{\sqrt {27} }} = \frac{1}{{\sqrt 3 }},\,m = - \frac{1}{{\sqrt 3 }},\,n = \frac{1}{{\sqrt 3 }} \cr} $$
Line $$L$$, makes an angle $$\alpha $$ with +ve $$x$$-axis
$$\therefore \,l = \cos \,\alpha \Rightarrow \cos \,\alpha = \frac{1}{{\sqrt 3 }}$$

The two lines intersect if shortest distance between them is zero i.e.
$$\frac{{\left( {{{\vec a}_2} - {{\vec a}_1}} \right).{{\vec b}_1} \times {{\vec b}_2}}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}} = 0\,\,\,\,\,\, \Rightarrow \left( {{{\vec a}_2} - {{\vec a}_1}} \right).{{\vec b}_1} \times {{\vec b}_2} = 0$$
Where,
$$\eqalign{ & {{\vec a}_1} = \hat i + 2\hat j + 3\hat k,\,\,\,\,{{\vec b}_1} = k\hat i + 2\hat j + 3\hat k \cr & {{\vec a}_2} = 2\hat i + 3\hat j + \hat k,\,\,\,\,{{\vec b}_2} = 3\hat i + k\hat j + 2\hat k \cr} $$
\[ \Rightarrow \left| \begin{array}{l} 1\,\,\,\,\,1\,\,\,\, - 2\\ k\,\,\,\,\,2\,\,\,\,\,\,\,\,\,3\\ 3\,\,\,\,\,k\,\,\,\,\,\,\,\,\,2 \end{array} \right| = 0\]
$$\eqalign{ & \Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0 \cr & \Rightarrow - 2{k^2} - 5k + 25 = 0 \cr & \Rightarrow k = - 5{\text{ or }}\frac{5}{2} \cr} $$
$$\because \,\,k$$  is an integer, therefore $$k=-5$$

Let $$PQ$$  be the shortest distance vector between $${l_1}$$ and $${l_2}.$$
Now, $${l_1}$$ passes through $${A_1}\left( {\overrightarrow {{a_1}} } \right)$$  and is parallel to $${\overrightarrow {{b_1}} }$$ and $${l_2}$$ passes through $${A_2}\left( {\overrightarrow {{a_2}} } \right)$$  and is parallel to $${\overrightarrow {{b_2}} }.$$
Since, $$PQ$$  is perpendicular to both $${l_1}$$ and $${l_2}$$ it is parallel to $$\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} .$$

Let $${\hat n}$$ be the unit vector along $$PQ.$$
Then, $$\hat n = \frac{{\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} }}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}$$
Let $$d$$ be the shortest distance between the given lines $${l_1}$$ and $${l_2}.$$
$$\overrightarrow {\left| {PQ} \right|} = d$$   and $$\overrightarrow {\left| {PQ} \right|} = d\,\hat n$$
Next $$PQ$$  being the line of shortest distance between $${l_1}$$ and $${l_2}$$ is the projection of the line joining the points $${A_1}\left( {\overrightarrow {{a_1}} } \right)$$  and $${A_2}\left( {\overrightarrow {{a_2}} } \right)$$  on $${\hat n}$$ ;
$$\left| {\overrightarrow {PQ} } \right| = \left| {\overrightarrow {{A_1}} \overrightarrow {{A_2}} .\hat n} \right| \Rightarrow d = \frac{{\left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right).\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}$$

Let $$\ell ,\,m,\,n$$  are the direction cosines of a line that is inclined equally at $$\alpha $$ to the +ve direction of axes.
Now, $$\ell = \cos \,\alpha ,\,m = \cos \,\alpha ,\,n = \cos \,\alpha $$
$$\eqalign{ & {\text{Also, }}{\ell ^2} + {m^2} + {n^2} = 1 \cr & \Rightarrow 3\,{\cos ^2}\alpha = 1 \cr & \Rightarrow \cos \,\alpha = \frac{1}{{\sqrt 3 }} \cr} $$
$$\therefore $$  dc’s of the line are : $$\left\langle {\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }},\,\frac{1}{{\sqrt 3 }}} \right\rangle $$