3D Geometry Syllabus Reduced

${{x - 1} \over 2} = {{y - 2} \over 3} = {{z + 1} \over 6} = \lambda $

$x = 2\lambda + 1,y = 3\lambda + 2,z = 6\lambda - 1$

for point of intersection of line & plane

$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$

$7\lambda = 7 \Rightarrow \lambda = 1$

point : (3, 5, 5)

(distance)² = ${(3 + 1)^2} + {(5 + 1)^2} + {(5 - 2)^2}$

$ = 16 + 36 + 9 = 61$

Since R(3, 5, $\gamma$) lies on the plane 2x $-$ y + z + 3 = 0.

Therefore, 6 $-$ 5 + $\gamma$ + 3 = 0

$\Rightarrow$ $\gamma$ = $-$4

Now,

dr's of line QS are 2, $-$1, 1

equation of line QS is

${{x - 1} \over 2} = {{y - 3} \over { - 1}} = {{z - 4} \over 1} = \lambda $ (say)

$ \Rightarrow F(2\lambda + 1, - \lambda + 3,\lambda + 4)$

F lies in the plane

$ \Rightarrow 2(2\lambda + 1) - ( - \lambda + 3) + (\lambda + 4) + 3$ = 0

$ \Rightarrow 4\lambda + 2 + \lambda - 3 + \lambda + 7 = 0$

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

$\Rightarrow$ F($-$1, 4, 3)

Since, F is mid-point of QS.

Therefore, coordinated of S are ($-$3, 5, 2).

So, SR = $\sqrt {36 + 0 + 36} = \sqrt {72} $

SR² = 72.

Containing the line $\left| {\matrix{ {x + 1} & {y - 1} & {z - 3} \cr 6 & 7 & 8 \cr 3 & 5 & 7 \cr } } \right| = 0$

$9(x + 1) - 18(y - 1) + 9(z - 3) = 0$

$x - 2y + z = 0$

$PQ = \left| {{{7 + 4 + 13} \over {\sqrt 6 }}} \right| = 4\sqrt 6 $

$P{Q^2} = 96$

To find the projection let's find the foot of perpendicular from $(1,3$, 4) to plane $x-2 y-z=3$


$ \begin{aligned} & \frac{x-1}{1}=\frac{y-3}{-2}=\frac{z-4}{-1}=\lambda_1 \\\\ & \left(\lambda_1+1\right)-2\left(-2 \lambda_1+3\right)-\left(-\lambda_1+4\right)=3 \\\\ & \Rightarrow 6 \lambda_1=12 \Rightarrow \lambda_1=2 \end{aligned} $

So, foot of perpendicular from $(1,3,4)$ to plane $x-2 y-z=3$ is $A$ $(3,-1,2)$.

Let us also find the intersection point of the plane and line

$ \begin{gathered} \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-4}{2}=\lambda_2 \\\\ \left(2 \lambda_2+1\right)-2\left(\lambda_2+3\right)-\left(2 \lambda_2+4\right)=3-2 \lambda_2=12 \Rightarrow \lambda_2=-6 \end{gathered} $

The intersection point of the plane and line is $B(-11,-3,-8)$ Line passing through $A$ and $B$ is

$ \begin{aligned} & \frac{x-3}{-14}=\frac{y+1}{-2}=\frac{z-2}{-10}=\mu \\\\ & \frac{x-3}{7}=\frac{y+1}{1}=\frac{z-2}{5}=\mu \end{aligned} $

Now, let's find the distance from $O(0,0,6)$ to this line $L$.

Let's say $C(7 \mu+3, \mu-1,5 \mu+2)$ is any point on $L$. Then,

$ \begin{aligned} & \{(7 \mu+3)-0\} \cdot 7+\{(\mu-1)-0\} \cdot 1+\{(5 \mu+2)-6\} \cdot 5=0 \\\\ & \Rightarrow 49 \mu+21+\mu-1+25 \mu-20=0 \Rightarrow \mu=0 \\\\ & \therefore C(3,-1,2) \\\\ & \text { Distance }=\sqrt{(3-0)^2+(-1-0)^2+(2-6)^2}=\sqrt{26} \\\\ & d^2=26 \end{aligned} $

$\overrightarrow {QR} : - {{x - 3} \over 1} = {{y + 4} \over { - 1}} = {{z + 5} \over { - 6}} = r$

$ \Rightarrow (x,y,z) \equiv (r + 3, - r - 4, - 6r - 5)$

Now, satisfying it in the given plane.

We get r = $-$2

so, required point of intersection is T(1, $-$2, 7).

Hence, PT = 7.

$\overrightarrow {BA} = (\widehat i + 4\widehat j - 5\widehat k)$

$\overrightarrow {BA} \times \overrightarrow l = \overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 3} & 2 & 1 \cr 1 & 4 & { - 5} \cr } } \right|$

$a\widehat i + b\widehat j + c\widehat k = - 14\widehat i - \widehat j(14) + \widehat k( - 14)$

a = 1, b = 1, c = 1

Plane is (x $-$ 2) + (y $-$ 3) + (z + 2) = 0

$ \Rightarrow $ x + y + z $-$ 3 = 0

$ \therefore $ d = $\sqrt 3 $ $\Rightarrow$ d² = 3

$\left| {\matrix{ {k + 1} & 4 & 6 \cr 1 & 2 & 3 \cr 3 & 2 & 1 \cr } } \right| = 0$

$ \Rightarrow $ $(k + 1)[2 - 6] - 4[1 - 9] + 6[2 - 6] = 0$

$ \Rightarrow $ $k = 1$

Equation of plane :

$\left| {\matrix{ {x - 1} & {y - 0} & {z - 1} \cr {1 - 1} & 2 & {1 - 1} \cr {1 - 0} & {0 - 1} & {1 + 2} \cr } } \right| = 0$

$ \Rightarrow 3x - z - 2 = 0$

$\overrightarrow a = \alpha \widehat i + \beta \widehat j + \gamma \widehat k$ || to 3x $-$ z $-$ 2 = 0

$ \Rightarrow 3\alpha - 8 = 0$ ..... (1)

$\overrightarrow a \bot \widehat i + \widehat j + 3\widehat k$

$ \Rightarrow \alpha + 2\beta + 3\gamma = 0$ ...... (2)

$\overrightarrow a .(\widehat i + \widehat j + 2\widehat k) = 0$

$\Rightarrow$ $\alpha$ + $\beta$ + 2$\gamma$ = 2 ........ (3)

On solving 1, 2 & 3

$\alpha$ = 1, $\beta$ = $-$5, $\gamma$ = 3

So, ($\alpha$ $-$ $\beta$ + $\gamma$)² = 81

Given equation of plane in vector form is $\overrightarrow r \,.\,(2\widehat i - \widehat j + \widehat k) - b = 0$

Its Cartesian form will be

$2x - y + z = b$ ...... (i)

$\because$ R is the mid-point of PQ.

$\therefore$ $R \equiv {{P + Q} \over 2} \Rightarrow R \equiv \left( { - 1,4,{{a + 2} \over 2}} \right)$

$\because$ R lies on the plane (i).

$\therefore$ $ - 2 - 4 + {{a + 2} \over 2} = b \Rightarrow a + 2 = 2b + 12$

$ \Rightarrow a = 2b + 10$ ....... (ii)

$\because$ Direction ratio's of QP is $(1 - ( - 3),3 - 5,a - 2)$

i.e. $(4, - 2,a - 2)$

and direction ratios of normal to the given plane are (2, $-$1, 1)

$\because$ n and QP are parallel.

$\therefore$ ${2 \over 4} = {{ - 1} \over { - 2}} = {1 \over {a - 2}}$

$\therefore$ $a - 2 = 2 \Rightarrow a = 4$

From Eq. (ii), b = $-$3

$\therefore$ $|a + b| = |4 - 3| = |1| = 1$

Equation of required plane is $\left| {\matrix{ {x - 1} & {y + 6} & {z + 5} \cr 3 & 4 & 2 \cr 4 & { - 3} & 7 \cr } } \right| = 0$

Since, (1, $-$1, $\alpha$) lies on it,

So, replace x by 1, y by ($-$1) and z and $\alpha$.

$\left| {\matrix{ 0 & 5 & {\alpha + 5} \cr 3 & 4 & 2 \cr 4 & { - 3} & 7 \cr } } \right| = 0$

$ \Rightarrow 5\alpha + 38 = 0 \Rightarrow 5\alpha = - 38$

$\therefore$ $\left| {5\alpha } \right| = \left| { - 38} \right| = 38$

Normal of plane = $\overrightarrow {PQ} = - 2\widehat i + 6\widehat j - 6\widehat k$

a = $-$2, b = 6, c = $-$6

& equation of plane is

$-$2x + 6y $-$ 6z + d = 0

$ M(3,0, - 2)$ is the midpoint of the line which present on the plane
which satisfy the plane

$ \therefore $ d = $-$6



Now equation of plane is

$-$2x + 6y $-$ 6z $-$ 6 = 0

x $-$ 3y + 3z + 3 = 0

$ \Rightarrow $ (a² + b² + c² + d²)_min = 1² + 9 + 9 + 9 = 28

The equation of the planes parallel to the plane x $-$ 2y + 2z $-$ 3 = 0

$x - 2y + 2z + \lambda = 0$

Now given

$d = {{\left| {1 - 4 + 6 + \lambda } \right|} \over {\sqrt 9 }} = 1$

$\left| {\lambda + 3} \right| = 3$

$\lambda + 3 = \pm 3 \Rightarrow \lambda = 0, - 6$

So planes are : $x - 2y + 2z - 6 = 0$

and $x - 2y + 2z = 0$

$b - d = - 2 + 6 = 4$

$c - a = 2 - 1 = 1$

$ \therefore $ $ {{b - d} \over {c - a}} = k$

$ \Rightarrow k = 4$

Let point P is ($\alpha$, $\beta$, $\gamma$)

${\left( {{{\alpha + \beta + \gamma } \over {\sqrt 3 }}} \right)^2} + {\left( {{{l\alpha - n\gamma } \over {\sqrt {{l^2} + {n^2}} }}} \right)^2} + {\left( {{{\alpha - 2\beta + \gamma } \over {\sqrt 6 }}} \right)^2} = 9$

Locus is ${{{{(x + y + z)}^2}} \over 3} + {{{{(\ln - nz)}^2}} \over {{l^2} + {n^2}}} + {{{{(x - 2y + z)}^2}} \over 6} = 9$

${x^2}\left( {{1 \over 2} + {{{l^2}} \over {{l^2} + {n^2}}}} \right) + {y^2} + {z^2}\left( {{1 \over 2} + {{{n^2}} \over {{l^2} + {n^2}}}} \right) + 2zx\left( {{1 \over 2} - {{\ln } \over {{l^2} + {n^2}}}} \right) - 9 = 0$

Since its given that ${x^2} + {y^2} + {z^2} = 9$

After solving l = n,

then, l $-$ n = 0

Equation of plane can be written using family of planes : P₁ + $\lambda$P₂ = 0

(2x $-$ 7y + 4z $-$ 3) + $\lambda$ (3x $-$ 5y + 4z + 11) = 0

It passes through ($-$2, 1, 3)

$ \therefore $ ($-$4 + 7 + 12 $-$ 3) + $\lambda$ ($-$6 $-$ 5 + 12 + 11) = 0

$-$2 + $\lambda$ (12) = 0

$\lambda$ = ${1 \over 6}$.

$ \therefore $ 12x $-$ 42y + 24z $-$ 18 + 3x $-$ 5y + 4z + 11 = 0

15x $-$ 47y + 28z $-$ 7 = 0

$ \therefore $ a = 15, b = $-$47, c = 28

$ \therefore $ 2a + b + c $-$ 7 = 30 $-$ 47 + 28 $-$ 7 = 4

Given line L,

${{x - 1} \over 3} = {{2 - y} \over m} = {{z + 3} \over 1}$

$ \Rightarrow $ ${{x - 1} \over 3} = {{y - 2} \over -m} = {{z + 3} \over 1}$

$ \therefore $ D.R of line = <3, -m, 1>

D.R of parallel line PQ will also be same.

$ \therefore $ Equation of line PQ,

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

Pt. $Q(3\lambda + 1, - m\lambda - 2,\lambda + 3)$ lie on plane

$(3\lambda + 1) + 2( - m\lambda - 2) - 3(\lambda + 3) + 10 = 0$

$ \Rightarrow 3\lambda - 2m\lambda - 3\lambda + 1 - 4 - 9 + 10 = 0$

$ \Rightarrow - 2m\lambda = 2$

$m\lambda = - 1 \Rightarrow \lambda = - {1 \over m}$

$Q\left[ { - {3 \over m} + 1, - 1, - {1 \over m} + 3} \right]$

Given, $PQ = \sqrt {{7 \over 2}} $

$ \Rightarrow $ $\sqrt {{{\left( { - {3 \over m}} \right)}^2} + 1 + {{\left( { - {1 \over m}} \right)}^2}} = \sqrt {{7 \over 2}} $

$ \Rightarrow {{10 + {m^2}} \over {{m^2}}} = {7 \over 2}$

$ \Rightarrow 20 + 2{m^2} = 7{m^2}$

$ \Rightarrow $ ${m^2} = 4 \Rightarrow |m| = 2$

$\overrightarrow {AB} \bot \overrightarrow {PQ} $

$\left[ {(4 - \lambda )\widehat i - 4\widehat j + \widehat k} \right].\left[ { + \widehat i + 5\widehat j - 6\widehat k} \right] = 0$

$4 - \lambda - 20 - 6 = 0$

$ \Rightarrow $ $\lambda $ = -22

Now, ${\lambda \over {11}} = - 2$

$ \Rightarrow {\left( {{\lambda \over {11}}} \right)^2} - {{4\lambda } \over {11}} - 4$

$ \Rightarrow 4 + 8 - 4 = 8$

${l_1}:\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$

${l_1}:{{x - 3} \over 1} = {{y + 1} \over 2} = {{z - 4} \over 2} \Rightarrow $ D.R. of ${l_1} = 1,2,2$

${l_2}:\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$

${l_2}:{{x - 3} \over 2} = {{y - 3} \over 2} = {{z - 2} \over 1} \Rightarrow $ D.R. of ${l_2} = 2,2,1$

D.R. of l is $ \bot $ to l₁ & k₂

$ \therefore $ D.R. of $l\,||\,({l_1} \times {l_2}) \Rightarrow ( - 2,3 - 2)$

$ \therefore $ Equation of $l:{x \over 2} = {y \over { - 3}} = {z \over 2}$

Solving l & l₁

$(2\lambda , - 3\lambda ,2\lambda ) = (\mu + 3,2\mu - 1,2\mu + \mu )$

$ \Rightarrow 2\lambda = \mu + 3$

$ - 3\lambda = 2\mu - 1$

$2\lambda = 2\mu + 4$

$ \Rightarrow \mu + 3 = 2\mu + 4$

$\mu = - 1$

$\lambda = 1$

$P(2, - 3,2)$ {intersection point}

Let, $Q(2v + 3,2v + 3,v + 2)$ be point on l₂

Now, $PQ = \sqrt {{{(2v + 3 - 2)}^2} + {{(2v + 3 + 3)}^2} + {{(v + 2 - 2)}^2}} = \sqrt {17} $

$ \Rightarrow {(2v + 1)^2} + {(2v + 6)^2} + {(v)^2} = 17$

$ \Rightarrow 9{v^2} + 28v + 36 + 1 - 17 = 0$

$ \Rightarrow 9{v^2} + 28v + 20 = 0$

$ \Rightarrow 9{v^2} + 18v + 10v + 20 = 0$

$ \Rightarrow (9v + 10)(v + 2) = 0$

$ \Rightarrow v = - 2$ (rejected), $ - {{10} \over 9}$ (accepted)

$Q\left( {3 - {{20} \over 9},3 - {{20} \over 9},2 - {{10} \over 9}} \right)$

$\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$

$ \therefore $ $18(a + b + c)$

$ = 18\left( {{7 \over 9},{7 \over 9},{8 \over 9}} \right)$

$ = 44$

${{x - \lambda } \over 1} = {{y - {1 \over 2}} \over {{1 \over 2}}} = {z \over { - {1 \over 2}}}$

${{x - \lambda } \over 2} = {{y - {1 \over 2}} \over 1} = {2 \over { - 1}}$ ....... (1)

Point on line = $\left( {\lambda ,{1 \over 2},0} \right)$

${x \over 1} = {{y + 2\lambda } \over 1} = {{z - \lambda } \over 1}$ ....... (2)

Point on line = $(0, - 2\lambda ,\lambda )$

Distance between skew lines $ = {{\left[ {{{\overrightarrow a }_2} - {{\overrightarrow a }_1}{{\overrightarrow b }_1}{{\overrightarrow b }_2}} \right]} \over {\left| {{{\overrightarrow b }_1} \times {{\overrightarrow b }_2}} \right|}}$

$\left| {\matrix{ \lambda & {{1 \over 2} + 2\lambda } & { - \lambda } \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|$

$\overline {\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 1 & 1 \cr } } \right|} $

$ = {{\left| { - 5\lambda - {3 \over 2}} \right|} \over {\sqrt {14} }} = {{\sqrt 7 } \over {2\sqrt 2 }}$

$ = |10\lambda + 3| = 7 \Rightarrow \lambda = - 1$

$ \Rightarrow |\lambda | = 1$

Equation of plane P is

$(x + 4y - z + 7) + \lambda (3x + y + 5z - 8) = 0$

$ \Rightarrow x(1 + 3\lambda ) + y(4 + \lambda ) + z( - 1 + 5\lambda ) + (7 - 8\lambda ) = 0$

${{1 + 3\lambda } \over a} = {{4 + \lambda } \over b} = {{5\lambda - 1} \over 6} = {{7 - 8\lambda } \over { - 15}}$

$ \therefore $ 15 - 75$\lambda $ = 42 - 48$\lambda $

$ \Rightarrow $ -27 = 27$\lambda $

$ \Rightarrow $ $\lambda $ = -1

$ \therefore $ Plane is $(x + 4y - z + 7) - 1 (3x + y + 5z - 8) = 0$

$ \Rightarrow $ $2x - 3y + 6z - 15 = 0$

Distance of (3, 2, -1) from the plane P

= ${{\left| {6 - 6 - 6 - 15} \right|} \over 7} = {{21} \over 7} = 3$

Given lines,

$\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ parallel to $(\widehat i + \widehat j)$

Let, $\overrightarrow {{n_1}} = (\widehat i + \widehat j)$

and $\overrightarrow r = - \widehat j + \mu (\widehat j - \widehat k)$ parallel to $(\widehat j - \widehat k)$

Let, $\overrightarrow {{n_2}} = (\widehat j - \widehat k)$

$ \therefore $ Normal of plane, $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $

$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & 0 \cr 0 & 1 & { - 1} \cr } } \right|$

$ = - \widehat i + \widehat j + \widehat k$

Line $\overrightarrow r = \widehat i + \lambda (\widehat i + \widehat j)$ is on the plane so, point on the line (1, 0, 0) will be also on the plane.

$ \therefore $ Equation of the plane,

$ - 1(x - 1) + 1(y - 0) + 1(z - 0) = 0$

$ \Rightarrow x - y - z - 1 = 0$

Foot of perpendicular from (x₁, y₁, z₁) on the plane,

${{x - {x_1}} \over a} = {{y - {y_1}} \over b} = {{z - {z_1}} \over c} = - {{(a{x_1} + b{y_1} + c{z_1} + d)} \over {{a^2} + {b^2} + {c^2}}}$

Here foot of perpendicular is drawn from M(1, 0, 1),

$ \therefore $ ${{x - 1} \over 1} = {{y - 0} \over { - 1}} = {{z - 1} \over { - 1}} = - {{(1 - 0 - 1 - 1)} \over 3}$

$ \therefore $ $x - 1 = {1 \over 3} \Rightarrow x = {4 \over 3}$

${y \over { - 1}} = {1 \over 3} \Rightarrow y = - {1 \over 3}$

${{z - 1} \over { - 1}} = {1 \over 3} \Rightarrow z = {2 \over 3}$

According to the question,

$x = \alpha $, $y = \beta $, $z = \gamma $

$ \therefore $ $\alpha = {4 \over 3}$, $\beta = - {1 \over 3}$, $\gamma = {2 \over 3}$

$ \therefore $ $3(\alpha + \beta + \gamma ) = 3\left( {{4 \over 3} - {1 \over 3} + {2 \over 3}} \right) = 5$

Required distance = perpendicular distance of plane 23x – 10y – 2z + 48 = 0 either from (–1, 3, –1) or (–3, –2, 1)

$ \Rightarrow $ $\left| {{{ - 23 - 30 + 2 + 48} \over {\sqrt {{{\left( {23} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|$ = ${k \over {\sqrt {633} }}$

$ \Rightarrow $ $\left| {{3 \over {\sqrt {633} }}} \right|$ = ${k \over {\sqrt {633} }}$

$ \Rightarrow $ k = 3

Let A (1, – 1, 3), B(2, – 4, 11), C (–1, 2, 3) & D (3, –2, 10)

$ \therefore $ $\overrightarrow {AB} = \widehat i - 3\widehat j + 8\widehat k$

$ \Rightarrow $ $\overrightarrow {CD} = 4\widehat i - 4\widehat j + 7\widehat k$

Projection of $\overrightarrow {AB} $ on $\overrightarrow {CD} $ = ${{\overrightarrow {AB} .\overrightarrow {CD} } \over {\left| {\overrightarrow {CD} } \right|}}$

= ${{4 + 12 + 56} \over {\sqrt {16 + 16 + 49} }}$

= ${{72} \over 9}$

= 8

Direction Ratio of PQ are

= (${5 \over 3} - 1$, ${7 \over 3} - 0$, ${{17} \over 3} - 3$)

= (2, 7, 8)

Direction ratio of line QA are

= ($\alpha - {5 \over 3}$, $7 - {7 \over 3}$, 1 - ${{17} \over 3}$)

= (3$\alpha $ – 5, 14, –14)

PQ is perpendicular to line QA

$ \therefore $ $\overrightarrow {PQ} .\overrightarrow {QA} $ = 0

$ \Rightarrow $ 2(3$\alpha $ – 5) + 7.14 + (–14).8 = 0

$ \Rightarrow $ $\alpha $ = 4