3D Geometry

The two lines intersect if shortest distance between them is zero $i.e.$

${{\left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}} = 0$

$ \Rightarrow \left( {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}} \right).{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$

where ${\overrightarrow a _1} = \widehat i + 2\widehat j + 3\widehat k,{\overrightarrow b _1} = k\widehat i + 2\widehat j + 3\widehat k$

$\,\,\,\,\,\,\,{\overrightarrow a _2} = 2\widehat i + 3\widehat j + \widehat k,\,\,{\widehat b_2} = 3\widehat i + k\widehat j + 2\widehat k$

$ \Rightarrow \left| {\matrix{ 1 & 1 & { - 2} \cr k & 2 & 3 \cr 3 & k & 2 \cr } } \right| = 0$

$ \Rightarrow 1\left( {4 - 3k} \right) - 1\left( {2k - 9} \right) - 2\left( {{k^2} - 6} \right) = 0$

$ \Rightarrow - 2{k^2} - 5k + 25 = 0 \Rightarrow k = - 5$ $\,\,\,\,$ or $\,\,\,\,$ ${5 \over 2}$

As $k$ is an integer, therefore $k=-5$

For given sphere center is $\left( {3,6,1} \right)$

Coordinates of one end of diameter of the sphere are $\left( {2,3,5} \right).$

Let the coordinates of the other end of diameter are $\left( {\alpha ,\beta ,\gamma } \right)$

$\therefore$ ${{\alpha + 2} \over 2} = 3,\,\,{{\beta + 3} \over 2} = 6,\,\,{{\gamma + 5} \over 2} = 1$

$ \Rightarrow \alpha = 4,\,\,\beta = 9\,\,$ and $\,\,\gamma = - 3$

$\therefore$ Coordinate of other end of diameter are $\left( {4,9, - 3} \right)$

Let the direction cosines of line $L$ be $l,m,n,$

then $2l+3m+n=0$ $\,\,\,\,\,\,\,....\left( i \right)$

and $l + 3m + 2n = 0\,\,\,\,\,\,\,\,\,\,....\left( {ii} \right)$

on solving equation $(i)$ and $(ii),$ we get

${l \over {6 - 3}} = {m \over {1 - 4}} = {n \over {6 - 3}}$

$\,\,\,\,\,\,\,\,\,\, \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3}$

Now $ \Rightarrow {l \over 3} = {m \over { - 3}} = {n \over 3} = {{\sqrt {{l^2} + {m^2} + {n^2}} } \over {\sqrt {{3^2} + {{\left( { - 3} \right)}^2} + {3^2}} }}$

As ${l^2} + {m^2} + {n^2} = 1$

$\therefore$ ${l \over 3} = {m \over { - 3}} = {n \over 3} = {1 \over {\sqrt {27} }}$

$ \Rightarrow l = {3 \over {\sqrt {27} }} = {1 \over {\sqrt 3 }},\,\,m = - {1 \over {\sqrt 3 }},n = {1 \over {\sqrt 3 }}$

Line $L,$ makes an angle $\alpha $ with $+ve$ $x$-axis

$\therefore$ $l = \cos \,\alpha \,\,\,\, \Rightarrow \,\,\,\cos \alpha \,\, = {1 \over {\sqrt 3 }}$

Let the angle of line makes with the positive direction of $z$-axis is $\alpha $ direction cosines of line with the $+ve$ directions of $x$-axis, $y$-axis, and $z$-axis is $l,$ $m,$ $n$ respectively.

$\therefore$ $l = \cos {\pi \over 4},m = \cos {\pi \over 4},\,\,n = cos\,\alpha $

as we know that, ${l^2} + {m^2} + {n^2} = 1$

$\therefore$ ${\cos ^2}{\pi \over 4} + {\cos ^2}{\pi \over 4} + {\cos ^2}\alpha = 1$

$ \Rightarrow {1 \over 2} + {1 \over 2} + {\cos ^2}\alpha = 1$

$ \Rightarrow {\cos ^2}\alpha = 0 \Rightarrow \alpha = {\pi \over 2}$

Hence, angle with positive direction of the $z$-axis is ${\pi \over 2}$

Equation of lines

${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$

${{x - b'} \over {a'}} = {y \over 1} = {{z - d'} \over {c'}}$

Line are perpendicular

$ \Rightarrow aa' + 1 + cc' = 0$

$E{q^n}\,\,\,\,$ of $\,\,\,\,PN: $-

${{x + 1} \over 1} = {{y - 3} \over { - 2}} = {{z - 4} \over 0} = \lambda $

$N\left( {\lambda - 1, - 2\lambda + 3 - 4} \right)$

It lies on $x-2y=0$

$ \Rightarrow \lambda - 1 + 4\lambda - 6 = 0$

$ \Rightarrow \lambda = 7/5$

$N\left( {{2 \over 5},{1 \over 5},4} \right)$

$N$ is mid point of $PP'$

$\therefore$ $\alpha - 1 = {4 \over 5},\beta + 3 = {2 \over 5},r + 4 = 8$

$ \Rightarrow \alpha = {9 \over 5},\beta = {{ - 13} \over 5},r = 4$

$\therefore$ Image is $\left( {{9 \over 5},{{ - 13} \over 5},4} \right)$

Perpendicular distance of center $\left( {{1 \over 2},0, - {1 \over 2}} \right)$

from $x+2y-2=4$ is given by

${{\left| {{1 \over 2} + {1 \over 2} - 4} \right| = \sqrt {{3 \over 2}} } \over {\sqrt 6 }}$

radius of sphere

$ = \sqrt {{1 \over 4} + {1 \over 4} + 2} = \sqrt {{5 \over 2}} $

$\therefore$ radius of circle

$ = \sqrt {{5 \over 2} - {3 \over 2}} = 1.$

The given lines are $2x = 3y = - z$

or $\,\,\,\,\,\,\,\,\,\,{x \over 3} = {y \over 2} = {z \over { - 6}}$ $\,\,\,\left[ {} \right.$ Dividing by $6$ $\left. {} \right]$

and $6x = - y = - 4z$

or $\,\,\,\,\,\,\,\,\,{x \over 2} = {y \over { - 12}} = {z \over { - 3}}$ $\,\,\,\,\left[ {} \right.$ Dividing by $12$ $\left. {} \right]$

$\therefore$ Angle between two lines is

$\cos \theta = {{3.2 + 2.\left( { - 12} \right) + \left( { - 6} \right).\left( { - 3} \right)} \over {\sqrt {{3^2} + {2^2} + {{\left( { - 6} \right)}^2}} \sqrt {{2^2} + {{\left( { - 12} \right)}^2} + {{\left( { - 3} \right)}^2}} }}$

$ = {{6 - 24 + 18} \over {\sqrt {49} \sqrt {157} }} = 0 \Rightarrow \theta = {90^ \circ }$

Centers of given spheres are $\left( { - 3,4,1} \right)$ and $\left( {5, - 2,1} \right).$

Mid point of centers is $\left( {1,1,1} \right).$

Satisfying this in the equation of plane, we get

$2a - 3a + 4a + 6 = 0$

$ \Rightarrow a = - 2$

A point on lines is $\left( {2, - 2,3} \right)$ its perpendicular distance

from the plane $x + 5y + z - 5 = 0$ is

$ = \left| {{{2 - 10 + 3 - 5} \over {\sqrt {1 + 25 + 1} }}} \right| = {{10} \over {3\sqrt 3 }}$

If $\theta $ is the angle between line and plane then $\left( {{\pi \over 2} - 0} \right)$

is the angle between line and normal to plane given by

$\cos \left( {{\pi \over 2} - 0} \right) = {{\left( {\widehat i + 2\widehat j + 2\widehat k} \right).\left( {2\widehat i - \widehat j + \sqrt \lambda \widehat k} \right)} \over {3\sqrt {4 + 1 + \lambda } }}$

$\cos \left( {{\pi \over 2} - \theta } \right) = {{2 - 2 + 2\sqrt \lambda } \over {3 \times \sqrt 5 + \lambda }}$

$ \Rightarrow \sin \theta = {{2\sqrt \lambda } \over {3\sqrt 5 + \lambda }} = {1 \over 3}$

$ \Rightarrow 4\lambda = 5 + \lambda \Rightarrow \lambda = {5 \over 3}$

Concept : If a line makes the angle $\alpha ,\beta ,\gamma $ with x, y, z axis respectively then $${\cos ^2}\alpha + {\cos ^2}\beta + {\cos ^2}\gamma = 1$$

In this question given that the line makes angle θ with x and z-axis and β with y−axis.

$\therefore\: cos^2\theta+cos^2\beta+cos^2\theta=1$

$\Rightarrow\:2cos^2\theta=1-cos^2\beta$

$ \Rightarrow 2{\cos ^2}\theta = {\sin ^2}\beta $

But given that $sin^2\beta=3sin^2\theta$

$\therefore$ $2{\cos ^2}\theta = 3{\sin ^2}\theta $

$ \Rightarrow 2{\cos ^2}\theta = 3\left( {1 - {{\cos }^2}\theta } \right)$

$ \Rightarrow 2{\cos ^2}\theta = 3 - 3{\cos ^2}\theta $

$ \Rightarrow 5{\cos ^2}\theta = 3$

$ \Rightarrow {\cos ^2}\theta = {3 \over 5}$

The equation of spheres are

${S_1}:{x^2} + {y^2} + {z^2} + 7x - 2y - z - 13 = 0$ and

${S_2}:{x^2} + {y^2} + {z^2} - 3x + 3y + 4z - 8 = 0$

Their plane of intersection is ${S_1} - {S_2} = 0$

$ \Rightarrow 10x - 5y - 5z - 5 = 0$

$ \Rightarrow 2x - y - z = 1$

The planes are $2x + y + 2x - 8 = 0\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

and $4x + 2y + 4z + 5 = 0$

or $2x + y + 2z + {5 \over 2} = 0\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

$\therefore$ Distance between $\left( 1 \right)$ and $\,\left( 2 \right)$

$ = \left| {{{{5 \over 2} + 8} \over {\sqrt {{2^2} + {1^2} + {2^2}} }}} \right|$

$ = \left| {{{21} \over {2\sqrt 9 }}} \right|$

$ = {7 \over 2}$

Let a point on the line

$x = y + a = z$ is $\left( {\lambda ,\lambda - a,\lambda } \right)$

and a point on the line

$x + a = 2y = 2z$ is $\left( {\mu - a,{\mu \over 2},{\mu \over 2}} \right),$

then direction ratio of the line joining these points are

$\lambda - \mu + a,\,\,\lambda - a - {\mu \over 2},\,\,\lambda - {\mu \over 2}$

If it represents the required line, then

${{\lambda - \mu + a} \over 2}$

$ = {{\lambda - a - {\mu \over 2}} \over 1}$

$ = {{\lambda - {\mu \over 2}} \over 2}$

on solving we get $\lambda = 3a,\,\mu = 2a$

$\therefore$ The required points of intersection are

$\left( {3a,3a - a,3a} \right)$ and $\left( {2a - a,{{2a} \over 2},{{2a} \over 2}} \right)$

or $\left( {3a,2a,3a} \right)$ and $\left( {a,a,a} \right)$

The given lines are

$x - 1 = {{y + 3} \over { - \lambda }} = {{z - 1} \over \lambda } = s.........\left( 1 \right)$

and $2x = y - 1 = {{z - 2} \over { - 1}} = t..........\left( 2 \right)$

The lines are coplanar, if

$\left| {\matrix{ {0 - \left( { - 1} \right)} & { - 1 - 3} & { - 2 - \left( { - 1} \right)} \cr 1 & { - \lambda } & \lambda \cr {{1 \over 2}} & 1 & { - 1} \cr } } \right| = 0$

${c_2} \to {c_2} + {c_3};\left| {\matrix{ 1 & { - 5} & 1 \cr 1 & 0 & \lambda \cr {{1 \over 2}} & 0 & { - 1} \cr } } \right| = 0$

$ \Rightarrow 5\left( { - 1 - {\lambda \over 2}} \right) = 0 \Rightarrow \lambda = - 2$

Shortest distance $=$ perpendicular distance between the plane and sphere $=$ distance of plane from center of sphere $-$ radius

$ = \left| {{{ - 2 \times 12 + 4 \times 1 + 3 \times 3 - 327} \over {\sqrt {144 + 9 + 16} }}} \right| - \sqrt {4 + 1 + 9 + 155} $

$ = 26 - 13 = 13$

Equation of planes be ${x \over a} + {y \over b} + {z \over c} = 1\,\,\& \,\,{x \over {a'}} + {y \over {b'}} + {z \over {c'}} = 1$

So the distance from (0, 0, 0) to both the plane is same.

$ \therefore $ $\left| {{{ - 1} \over {\sqrt {{1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}}} }}} \right| = \left| {{{ - 1} \over {\sqrt {{1 \over {a{'^2}}} + {1 \over {b{'^2}}} + {1 \over {c{'^2}}}} }}} \right|$

$ \Rightarrow $ ${1 \over {{a^2}}} + {1 \over {{b^2}}} + {1 \over {{c^2}}} - {1 \over {{{a'}^{2}}}} - {1 \over {{{b'}^{2}}}} - {1 \over {{{c'}^{2}}}} = 0$

Center of sphere $ = \left( { - 1,1,2} \right)$

Radius of sphere $\sqrt {1 + 1 + 4 + 19} = 5$

Perpendicular distance from center to the plane

$OC = d = \left| {{{ - 1 + 2 + 4 + 7} \over {\sqrt {1 + 4 + 4} }}} \right| = {{12} \over 3} = 4.$

$A{C^2} = A{O^2} - O{C^2} = {5^2} - {4^2} = 9$

$ \Rightarrow AC = 3$

Coplanar if

$\left| {\matrix{ {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{l_1}} & {{m_1}} & {{n_1}} \cr {{l_2}} & {{m_2}} & {{n_2}} \cr } } \right| = 0$

$\therefore$ $\left| {\matrix{ 1 & { - 1} & { - 1} \cr 1 & 1 & { - k} \cr k & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow \left| {\matrix{ 0 & 0 & { - 1} \cr 2 & {1 + k} & { - k} \cr {k + 2} & 1 & 1 \cr } } \right| = 0$

${k^2} + 3k = 0 \Rightarrow k\left( {k + 3} \right) = 0$

or $k = 0$ or $-3$

${{x - b} \over a} = {y \over 1} = {{z - d} \over c};$

${{x - b'} \over {a'}}$ $ = {y \over 1} = {{z - d'} \over c'}$

For perpenedicularity of lines $aa' + 1 + cc' = 0$

As the point $\left( {3,2,0} \right)$ lies on the given line

${{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4}$

$\therefore$ 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.

Equation of plane through $\left( {1,0,0} \right)$ is

$a\left( {x - 1} \right) + by + cz = 0\,\,\,\,\,\,\,\,\,\,...\left( i \right)$

$(i)$ passes through $\left( {0,1,0} \right).$

$ - a + b = 0 \Rightarrow b = a;$

Also, $\cos {45^ \circ }$

$ = {{a + a} \over {\sqrt {2\left( {2{a^2} + {c^2}} \right)} }} \Rightarrow 2a = \sqrt {2{a^2} + {c^2}} $

$ \Rightarrow 2{a^2} = {c^2}$

$ \Rightarrow c = \sqrt 2 a.$

So $d.r$ of normal area $a,$ $a\sqrt {2a} $ i.e. $1,$ $1,\sqrt 2 .$