3D Geometry

Given $a\,.\,3 + ( - 4a)( - 1) + ( - 7)2b = 0$ ...... (1)

and $ab - 4{a^2} + 14 = 0$ ....... (2)

$ \Rightarrow {a^2} = 4$ and ${b^2} = 1$

$\therefore$ $L \equiv {{x + 1} \over 5} = {{y - 2} \over 3} = {z \over 1} = \lambda $ (say)

$\Rightarrow$ General point on line is $(5\lambda - 1,\,3\lambda + 2,\,\lambda )$ for finding point of intersection with $x - y + z = 0$ we get $(5\lambda - 1) - (3\lambda + 2) + (\lambda ) = 0$

$ \Rightarrow 3\lambda - 3 = 0 \Rightarrow \lambda = 1$

$\therefore$ Point at intersection $(4,\,5,\,1)$

$\therefore$ $\alpha + \beta + \gamma = 4 + 5 + 1 = 10$

d.r's of $RS = < \alpha , - 1,\beta > $

d.r's of $PQ = < {{90} \over {17}},{{60} \over {17}},{{94} \over {17}} > \, = \, < 45,30,47 > $

as PQ and RS are diagonals of rhombus

$\alpha (45) + 30( - 1) + 47(\beta ) = 0$

$ \Rightarrow 45\alpha + 47\beta = 30$

i.e., $\alpha = {{30 - 47\beta } \over {45}}$

for minimum integral value $\alpha = - 15$ and $\beta = 15$

$ \Rightarrow {\alpha ^2} + {\beta ^2} = 450$.

Equation of plane containing the line $4ax - y + 5z - 7a = 0 = 2x - 5y - z - 3$ can be written as

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

$(4a + 2\lambda )x - (1 + 5\lambda )y + (5 - \lambda )z - (7z + 3\lambda ) = 0$

Which is coplanar with the line

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

$4(4a + 2\lambda ) + (1 + 5\lambda ) - (7a + 3\lambda ) = 0$

$9a + 10\lambda + 1 = 0$ ..... (1)

$(4a + 2\lambda )1 + (1 + 5\lambda )2 + 5 - \lambda = 0$

$4a + 11\lambda + 7 = 0$ ...... (2)

$a = 1,\,\lambda = - 1$

Equation of plane is $x + 2y + 3z - 2 = 0$

Intersection with the line

${{x - 3} \over 7} = {{y - 2} \over { - 1}} = {{z - 3} \over { - 4}}$

$(7t + 3) + 2( - t + 2) + 3( - 4t + 3) - 2 = 0$

$ - 7t + 14 = 0$

$t = 2$

So, the required point is $(17,0, - 5)$

$\alpha + \beta + \gamma = 12$

Normal to plane $ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & a & { - 1} \cr { - 1} & 1 & { - a} \cr } } \right|$

$ = \widehat i(1 - {a^2}) - \widehat j( - a - 1) + \widehat k(1 + a)$

$ = (1 - a)\widehat i + \widehat j + \widehat k$

$\therefore$ Plane $(1 - a)(x - 1) + (y - 1) + z = 0$

Distance from (2, 1, 4) is $\sqrt 3 $ i.e.

$ \Rightarrow \left| {{{(1 - a) + 0 + 4} \over {\sqrt {{{(1 - a)}^2} + 1 + 1} }}} \right| = \sqrt 3 $

$ \Rightarrow 25 + {a^2} - 10a = 3{a^2} - 6a + 9$

$ \Rightarrow 2{a^2} + 4a - 16 = 0$

$ \Rightarrow {a^2} + 2a - 8 = 0$

$a = 2$ or $ - 4$

$\therefore$ ${a_{\max }} = 2$

$L:lx - y + 3(1 - l)z = 1$, $x + 2y - z = 2$ and plane containing the line $p:3x - 8y + 7z = 4$

Let $\overrightarrow n $ be the vector parallel to L.

then $\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr l & { - 1} & {3(1 - l)} \cr 1 & 2 & { - 1} \cr } } \right|$

$ = (6l - 5)\widehat i + (3 - 2l)\widehat j + (2l + 1)\widehat k$

$\because$ R containing L

$3(6l - 5) - 8(3 - 2l) + 7(2l + 1) = 0$

$18l + 16l + 14l - 15 - 24 + 7 = 0$

$\therefore$ $l = {{32} \over {48}} = {2 \over 3}$

Let $\theta$ be the acute angle between L and y-axis

$\therefore$ $\cos \theta = {{{5 \over 3}} \over {\sqrt {1 + {{25} \over 9} + {{49} \over 9}} }} = {5 \over {\sqrt {83} }}$

$\therefore$ $415{\cos ^2}\theta = 125$

$L:{{x + 1} \over 2} = {{y + 2} \over 3} = {{2 - 1} \over 2}$

Let $T(2t - 1,\,3t - 2,\,2t + 1)$

$\because$ $PT\,{ \bot ^r}\,QR$

$\therefore$ $2(2t - 5) + 3(3t - 4) + 2(2t - 6) = 0$

$17t = 34$

$\therefore$ $t = 2$

So $T(3,4,5)$

$\therefore$ $PT = \sqrt {1 + 4 + 4} = 3$

$\therefore$ $QT = \sqrt {26 - 9} = \sqrt {17} $

$\therefore$ Area of $\Delta PQR = {1 \over 2} \times 2\sqrt {17} \times 3 = 3\sqrt {17} $

$\therefore$ Square of $ar(\Delta PQR) = 153$.

Given, G is the centroid of $\Delta$ABC

$\therefore$ $G = \left( {{{0 + \alpha + \alpha } \over 3},\,{{\alpha + 0 + \alpha } \over 3},\,{{\alpha + \alpha + 0} \over 3}} \right)$

$ = \left( {{{2\alpha } \over 3},\,{{2\alpha } \over 3},\,{{2\alpha } \over 3}} \right)$

Also given, D is a point moving on the line $x + z - 3 = 0 = y$

Let $D = (h,\,0,\,k)$

$x + z - 3 = 0 \Rightarrow h + k - 3 = 0 \Rightarrow h = 3 - k$

$\therefore$ $D = (3 - k,\,0,\,k)$

Now, length of $GD = d$

$ = \sqrt {{{\left( {{{2\alpha } \over 3} - 3 + k} \right)}^2} + {{\left( {{{2\alpha } \over 3}} \right)}^2} + {{\left( {{{2\alpha } \over 3} - k} \right)}^2}} $

$ \Rightarrow {d^2} = {\left( {{{2\alpha } \over 3} - 3 + k} \right)^2} + {\left( {{{2\alpha } \over 3}} \right)^2} + {\left( {{{2\alpha } \over 3} - k} \right)^2}$

Differentiating both side with respect to k, we get

$2dd' = 2\left( {{{2\alpha } \over 3} - 3 + k} \right) + 0 + 2\left( {{{2\alpha } \over 3} - k} \right) \times ( - 1)$

For maximum or minimum value of k, $d' = 0$

$\therefore$ $0 = 2\left( {{{2\alpha } \over 3} - 3 + k} \right) - 2\left( {{{2\alpha } \over 3} - k} \right)$

$ \Rightarrow 2\left( {{{2\alpha } \over 3} - 3 + k} \right) = 2\left( {{{2\alpha } \over 3} - k} \right)$

$ \Rightarrow - 3 + k = - k$

$ \Rightarrow k = {3 \over 2}$

$\therefore$ $G{D^2} = {\left( {{{2\alpha } \over 3} - 3 + {3 \over 2}} \right)^2} + {\left( {{{2\alpha } \over 3}} \right)^2} + {\left( {{{2\alpha } \over 3} - {3 \over 2}} \right)^2} = {{57} \over 2}$

$ \Rightarrow {{{{(4\alpha - 9)}^2}} \over {36}} + {{16{\alpha ^2}} \over {36}} + {{{{(4\alpha - 9)}^2}} \over {36}} = {{57} \over 2}$

$ \Rightarrow {(4\alpha - 9)^2} + 16{\alpha ^2} + {(4\alpha - 9)^2} = 57 \times 18$

$ \Rightarrow 16{\alpha ^2} - 72\alpha + 81 + 16{\alpha ^2} + 16{\alpha ^2} - 72\alpha + 81 = 57 \times 18$

$ \Rightarrow 48{\alpha ^2} - 144\alpha + 162 = 1026$

$ \Rightarrow 24{\alpha ^2} - 72\alpha + 81 - 513 = 0$

$ \Rightarrow 24{\alpha ^2} - 72\alpha - 432 = 0$

$ \Rightarrow {\alpha ^2} - 3\alpha - 18 = 0$

$ \Rightarrow {\alpha ^2} - 6\alpha + 3\alpha - 18 = 0$

$ \Rightarrow \alpha (\alpha - 6) + 3(\alpha - 6) = 0$

$ \Rightarrow (\alpha - 6)(\alpha + 3) = 0$

$ \Rightarrow \alpha = 6,\, - 3$

Given, $\alpha > 0$

$\therefore$ Possible value of $\alpha = 6$.

Foot of perpendicular from P

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

$ \Rightarrow p' \equiv \left( {{2 \over 3},{7 \over 3},{{ - 2} \over 3}} \right)$

and foot of perpendicular from Q

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

$ \Rightarrow Q' \equiv \left( {{5 \over 3},{{ - 2} \over 3},{{10} \over 3}} \right)$

$P'Q' = \sqrt {{{(1)}^2} + {{(3)}^2} + {{(4)}^2}} = d = \sqrt {26} $

$ \Rightarrow {d^2} = 26$

Direction ratio of normal to ${P_1} \equiv < 2,1, - 3 > $

and that of ${P_2} \equiv \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & 1 & { - 5} \cr { - 1} & { - 2} & 5 \cr } } \right| = - 5\widehat i - \widehat j( - 5) + \widehat k(1)$

i.e. $ < - 5,5,1 > $

d.r's of line of intersection are along vector

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 3} \cr { - 5} & 5 & 1 \cr } } \right| = \widehat i(16) - \widehat j( - 13) + \widehat k(15)$

i.e. $ < 16,13,15 > $

$\therefore$ $\alpha + \beta = 13 + 15 = 28$

The point dividing PQ in the ratio 1 : 3 will be mid-point of P & foot of perpendicular from P on the line.

$\therefore$ Let a point on line be $\lambda$

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

$ \Rightarrow P'(3\lambda + 6,\,2\lambda + 1,\,3\lambda + 2)$

as P' is foot of perpendicular

$(3\lambda + 5)3 + (2\lambda - 1)2 + (3\lambda - 1)3 = 0$

$ \Rightarrow 22\lambda + 15 - 2 - 3 = 0$

$ \Rightarrow \lambda = {{ - 5} \over {11}}$

$\therefore$ $P'\left( {{{51} \over {11}},{1 \over {11}},{7 \over {11}}} \right)$

Mid-point of $PP' \equiv \left( {{{{{51} \over {11}} + 1} \over 2},{{{1 \over {11}} + 2} \over 2},{{{7 \over {11}} + 3} \over 2}} \right)$

$ \equiv \left( {{{62} \over {22}},{{23} \over {22}},{{40} \over {22}}} \right) \equiv (\alpha ,\beta ,\gamma )$

$ \Rightarrow 22(\alpha ,\beta ,\gamma ) = 62 + 23 + 40 = 125$

${{x - a} \over 3} = {{y - b} \over { - 4}} = {{z - c} \over {12}} = {{ - 2(3a - 4b + 12c + 19)} \over {{3^2} + {{( - 4)}^2} + {{12}^2}}}$

${{x - a} \over 3} = {{y - b} \over { - 4}} = {{z - c} \over {12}} = {{ - 6a + 8b - 24c - 38} \over {169}}$

$(x,y,z) \equiv (a - 6,\,\beta ,\gamma )$

${{(a - 6) - a} \over 3} = {{\beta - b} \over { - 4}} = {{\gamma - c} \over {12}} = {{ - 6a + 8b - 24c - 38} \over {169}}$

${{\beta - b} \over { - 4}} = - 2 \Rightarrow \beta = 8 + b$

${{\gamma - c} \over {12}} = - 2 \Rightarrow \gamma = - 24 + c$

${{ - 6a + 8b - 24c - 38} \over {169}} = - 2$

$ \Rightarrow 3a - 4b + 12c = 150$ ..... (1)

$a + b + c = 5$

$3a + 3b + 3c = 15$ ...... (2)

Applying (1) - (2)

$ - 7b + 9c = 135$

$7b - 9c = - 135$

$7\beta - 9\gamma = 7(8 + b) - 9( - 24 + c)$

$ = 56 + 216 + 7b - 9c$

$ = 56 + 216 - 135 = 137$

${{x - {1 \over 8}} \over {{1 \over 8}}} = {y \over { - {1 \over {4\sqrt 2 }}}} = {z \over 0}\,\,\,\,\,\,\,\,\,\,\,\,\,$ ______L₁

or ${{x - {1 \over 8}} \over 1} = {y \over { - \sqrt 2 }} = {z \over 0}$ ..... (i)

Equation of L₂

${{x + {1 \over 8}} \over { - 6\sqrt 3 }} = {y \over 0} = {z \over 8}$ ...... (ii)

$d = \left| {{{(\overrightarrow c - \overrightarrow a )\,.\,(\overrightarrow b \times \overrightarrow d )} \over {\left| {\overrightarrow b \times \overrightarrow d } \right|}}} \right|$

$ = {{\left( {{1 \over 4}\widehat i} \right)\,.\,\left( {4\sqrt 2 \widehat i + 4\widehat j + 3\sqrt 6 \widehat k} \right)} \over {\sqrt {{{\left( {4\sqrt 2 } \right)}^2} + {4^2} + {{\left( {3\sqrt 6 } \right)}^2}} }}$

$ = {{\sqrt 2 } \over {\sqrt {32 + 16 + 54} }} = {1 \over {\sqrt {51} }}$

${d^{ - 2}} = 51$

As plane is parallel to both the lines we have d.r's of normal to the plane as <7, $-$2, $-$1>

$\left( {from\,\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 3 \cr 1 & 1 & 5 \cr } } \right| = 7\widehat i - \widehat j(2) + \widehat k( - 1)} \right)$

Also point of intersection of lines is $2\widehat i + 4\widehat j + 6\widehat k$

$\therefore$ Equation of plane is

$7(x - 2) - 2(y - 4) - 1(z - 6) = 0$

$ \Rightarrow 7x - 2y - z = 0$

$a + b + d = 7 - 2 + 0 = 5$