Three Dimensional Geometry MCQ Questions & Answers in Geometry | Maths

It is obvious that the given line and plane are parallel.
Given point on the line is $$A\left( {2,\, - 2,\,3} \right).$$
$$B\left( {0,\,0,\,5} \right)$$   is a point on the plane. Therefore,
$$\overrightarrow {\,\,\,\,\,AB\,\,\,\,\,} = \left( {2 - 0} \right)\hat i + \left( { - 2 - 0} \right)\hat j + \left( {3 - 5} \right)\hat k$$
Then distance of $$B$$ from the plane $$=$$ projection of $$\overrightarrow {\,\,\,\,\,AB\,\,\,\,\,} $$   on vector $$\hat i + 5\hat j + \hat k$$
$$\eqalign{ & P = \left| {\frac{{\left( {2\hat i - 2\hat j - 2\hat k} \right).\left( {\hat i + 5\hat j + \hat k} \right)}}{{\sqrt {1 + 25 + 1} }}} \right| \cr & \Rightarrow P = \left| {\frac{{2 - 10 - 2}}{{\sqrt {27} }}} \right| \cr & \Rightarrow P = \frac{{10}}{{3\sqrt 3 }} \cr} $$

Given planes are
$$\eqalign{ & x - 2y + z = 1......\left( {\text{i}} \right) \cr & {\text{and }} - 3x + 6y - 3z = - 2 \cr & \equiv x - 2y + z = \frac{2}{3}......\left( {{\text{ii}}} \right) \cr} $$
Since both planes are parallel and $$a = 1,\,b = - 2,\,c = 1$$
$${\text{and }}{d_1} = - 1,\,{d_2} = \frac{{ - 2}}{3}$$
$$\therefore $$  Distance
$$\eqalign{ & = \left| {\frac{{{d_2} - {d_1}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right| \cr & = \left| {\frac{{1 - \frac{2}{3}}}{{\sqrt {1 + 4 + 1} }}} \right| \cr & = \frac{1}{{3\sqrt 6 }} \cr} $$

Equation of line $$PQ$$  is $$\frac{{x - 1}}{1} = \frac{{y + 2}}{4} = \frac{{z - 3}}{5}$$
$${\text{Let }}F\,{\text{be }}\left( {\lambda + 1,\,4\lambda - 2,\,5\lambda + 3} \right)$$

Since $$F$$ lies on the plane
$$\eqalign{ & \therefore 2\left( {\lambda + 1} \right) + 3\left( {4\lambda - 2} \right) - 4\left( {5\lambda + 3} \right) + 22 = 0 \cr & 2\lambda + 2 + 12\lambda - 6 - 20\lambda - 12 + 22 = 0 \cr & \Rightarrow - 6\lambda + 6 = 0 \cr & \Rightarrow \lambda = 1 \cr & \therefore F{\text{ is }}\left( {2,\,2,\,8} \right) \cr & PQ = 2PF = 2\sqrt {{1^2} + {4^2} + {5^2}} = 2\sqrt {42} \cr} $$

Line lies in the plane $$ \Rightarrow \left( {3,\, - 2,\, - 4} \right)$$   lie in the plane
$$ \Rightarrow 3l - 2m + 4 = 9{\text{ or }}3l - 2m = 5.....(1)$$
Also, $$l,\,m,\,-1$$   are dr's of line perpendicular to plane and $$2,\,-1,\,3$$   are dr's of line lying in the plane
$$ \Rightarrow 2l - m - 3 = 0{\text{ or 2}}l - m = 3.....(2)$$
Solving (1) and (2) we get $$l=1$$  and $$m=-1$$
$$ \Rightarrow {l^2} + {m^2} = 2$$

$$\eqalign{ & {\text{Line }}1\,:\frac{x}{1} = \frac{y}{1} = \frac{{z - 1}}{0} = r,\,Q\left( {r,\,r,\,1} \right) \cr & {\text{Line 2}}\,:\frac{x}{1} = \frac{y}{{ - 1}} = \frac{{z + 1}}{0} = k,\,R\left( {k,\, - k,\, - 1} \right) \cr & \overrightarrow {PQ} = \left( {\lambda - r} \right)\hat i + \left( {\lambda - r} \right)\hat j + \left( {\lambda - 1} \right)\hat k \cr & \overrightarrow {PQ} \,\,{\text{is perpendicular to Line }}1 \cr & \Rightarrow \lambda - r + \lambda - r \Rightarrow \lambda = r \cr & \overrightarrow {PR} = \left( {\lambda - k} \right)\hat i + \left( {\lambda + k} \right)\hat j + \left( {\lambda + 1} \right)\hat k \cr & \overrightarrow {PR} \,\,{\text{is perpendicular to Line }}2 \cr & \Rightarrow \lambda - k - \lambda - k = 0 \Rightarrow k = 0 \cr & {\text{Now, }}\overrightarrow {PQ} \bot \overrightarrow {PR} \cr & \Rightarrow \left( {\lambda - r} \right)\left( {\lambda - k} \right) + \left( {\lambda - r} \right)\left( {\lambda + k} \right) + \left( {\lambda - 1} \right)\left( {\lambda + 1} \right) = 0 \cr & \Rightarrow \lambda = \pm 1 \cr & {\text{For }}\lambda = 1,{\text{ points }}P{\text{ and }}Q{\text{ coincide}}{\text{.}} \cr & \therefore \,\lambda = - 1 \cr} $$

$$\eqalign{ & {\text{We have,}} \cr & {\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1......\left( {\text{i}} \right) \cr & \Rightarrow 2\,{\cos ^2}\alpha + 2\,{\cos ^2}\beta + 2\,{\cos ^2}\gamma = 2 \cr & \Rightarrow 2\,{\cos ^2}\alpha - 1 + 2\,{\cos ^2}\beta - 1 + 2\,{\cos ^2}\gamma - 1 = 2 - 3 \cr & \Rightarrow \cos \,2\alpha + \cos \,2\beta + \cos \,2\gamma = - 1 \cr & {\text{Hence statement }}{\bf{I}}{\text{ is correct}}{\text{.}} \cr & {\text{and now from }}\left( {\text{i}} \right), \cr & 1 - {\sin ^2}\alpha + 1 - {\sin ^2}\beta + 1 - {\sin ^2}\gamma = 1 \cr & \Rightarrow {\sin ^2}\alpha + {\sin ^2}\beta + {\sin ^2}\gamma = 2 \cr & {\text{Hence, only statement }}{\bf{I}}{\text{ is correct}}{\text{.}} \cr} $$

Since $$L$$ is the foot of perpendicular from $$P$$ on the $$xy$$ -plane, $$z$$-coordinate is zero in the $$xy$$ -plane. Hence, coordinates of $$L$$ are $$\left( {6,\,7,\,0} \right).$$

The required plane is
$$\left( {2 + 3\lambda } \right)x + \left( { - 8 - 5\lambda } \right)y + \left( {4 + 4\lambda } \right)z + P - 10\lambda = 0$$
Compare the coefficients with the plan. We get,
$$\eqalign{ & 4 + 4\lambda = 0 \Rightarrow \lambda = - 1 \cr & x + 3y + 0z + 13 = 0 \cr} $$
Then we get $$p = 3.$$

$$\eqalign{ & \because \,P{A^2} - P{B^2} = k \cr & \therefore \,\left[ {{{\left( {x - 2} \right)}^2} + {{\left( {y - 3} \right)}^2} + {{\left( {z - 4} \right)}^2}} \right] - \left[ {{{\left( {x + 2} \right)}^2} + {{\left( {y - 5} \right)}^2} + {{\left( {z + 4} \right)}^2}} \right] = k \cr & {\text{or, }} - 8x + 4y - 16z - 16 = k \cr} $$

Equation of line is $$\frac{x}{1} = \frac{{y - 1}}{2} = \frac{{z - 2}}{3}$$
Any point on this line is $$\left( {K,\,2K + 1,\,3K + 2} \right)$$
If this is the foot of perpendicular from $$\left( {1,\,6,\,3} \right)$$   then $$d.r$$  of this perpendicular are $$\left\langle {K - 1,\,2K - 5,\,3K - 1} \right\rangle $$
Now, using Condition of perpendicularity we have
$$\eqalign{ & \left( {K - 1} \right)1 + \left( {2K - 5} \right)2 + \left( {3K - 1} \right)3 = 0 \cr & \Rightarrow K - 1 + 4K - 10 + 9K - 3 = 0 \cr & \Rightarrow K = 1 \cr} $$
Hence, Required foot of perpendicular is $$\left( {1,\,3,\,5} \right)$$