3D Geometry
Let the foot of the perpendicular from the point (1, 2, 4) on the line ${{x + 2} \over 4} = {{y - 1} \over 2} = {{z + 1} \over 3}$ be P. Then the distance of P from the plane $3x + 4y + 12z + 23 = 0$ is :
The shortest distance between the lines
${{x - 3} \over 2} = {{y - 2} \over 3} = {{z - 1} \over { - 1}}$ and ${{x + 3} \over 2} = {{y - 6} \over 1} = {{z - 5} \over 3}$, is :
If two straight lines whose direction cosines are given by the relations $l + m - n = 0$, $3{l^2} + {m^2} + cnl = 0$ are parallel, then the positive value of c is :
If the plane $2x + y - 5z = 0$ is rotated about its line of intersection with the plane $3x - y + 4z - 7 = 0$ by an angle of ${\pi \over 2}$, then the plane after the rotation passes through the point :
If the lines $\overrightarrow r = \left( {\widehat i - \widehat j + \widehat k} \right) + \lambda \left( {3\widehat j - \widehat k} \right)$ and $\overrightarrow r = \left( {\alpha \widehat i - \widehat j} \right) + \mu \left( {2\widehat i - 3\widehat k} \right)$ are co-planar, then the distance of the plane containing these two lines from the point ($\alpha$, 0, 0) is :
Let $\overrightarrow a = \widehat i + \widehat j + 2\widehat k$, $\overrightarrow b = 2\widehat i - 3\widehat j + \widehat k$ and $\overrightarrow c = \widehat i - \widehat j + \widehat k$ be three given vectors. Let $\overrightarrow v $ be a vector in the plane of $\overrightarrow a $ and $\overrightarrow b $ whose projection on $\overrightarrow c $ is ${2 \over {\sqrt 3 }}$. If $\overrightarrow v \,.\,\widehat j = 7$, then $\overrightarrow v \,.\,\left( {\widehat i + \widehat k} \right)$ is equal to :
If the two lines ${l_1}:{{x - 2} \over 3} = {{y + 1} \over {-2}},\,z = 2$ and ${l_2}:{{x - 1} \over 1} = {{2y + 3} \over \alpha } = {{z + 5} \over 2}$ are perpendicular, then an angle between the lines l2 and ${l_3}:{{1 - x} \over 3} = {{2y - 1} \over { - 4}} = {z \over 4}$ is :
Let the plane 2x + 3y + z + 20 = 0 be rotated through a right angle about its line of intersection with the plane x $-$ 3y + 5z = 8. If the mirror image of the point $\left( {2, - {1 \over 2},2} \right)$ in the rotated plane is B(a, b, c), then :
Let p be the plane passing through the intersection of the planes $\overrightarrow r \,.\,\left( {\widehat i + 3\widehat j - \widehat k} \right) = 5$ and $\overrightarrow r \,.\,\left( {2\widehat i - \widehat j + \widehat k} \right) = 3$, and the point (2, 1, $-$2). Let the position vectors of the points X and Y be $\widehat i - 2\widehat j + 4\widehat k$ and $5\widehat i - \widehat j + 2\widehat k$ respectively. Then the points :
Let Q be the mirror image of the point P(1, 0, 1) with respect to the plane S : x + y + z = 5. If a line L passing through (1, $-$1, $-$1), parallel to the line PQ meets the plane S at R, then QR2 is equal to :
If the shortest distance between the lines ${{x - 1} \over 2} = {{y - 2} \over 3} = {{z - 3} \over \lambda }$ and ${{x - 2} \over 1} = {{y - 4} \over 4} = {{z - 5} \over 5}$ is ${1 \over {\sqrt 3 }}$, then the sum of all possible value of $\lambda$ is :
Let the points on the plane P be equidistant from the points ($-$4, 2, 1) and (2, $-$2, 3). Then the acute angle between the plane P and the plane 2x + y + 3z = 1 is :
Let a line with direction ratios $a,-4 a,-7$ be perpendicular to the lines with direction ratios $3,-1,2 b$ and $b, a,-2$. If the point of intersection of the line $\frac{x+1}{a^{2}+b^{2}}=\frac{y-2}{a^{2}-b^{2}}=\frac{z}{1}$ and the plane $x-y+z=0$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to _________.
Explanation:
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 $(4,\,5,\,1)$
$\therefore$ $\alpha + \beta + \gamma = 4 + 5 + 1 = 10$
Let $\mathrm{P}(-2,-1,1)$ and $\mathrm{Q}\left(\frac{56}{17}, \frac{43}{17}, \frac{111}{17}\right)$ be the vertices of the rhombus PRQS. If the direction ratios of the diagonal RS are $\alpha,-1, \beta$, where both $\alpha$ and $\beta$ are integers of minimum absolute values, then $\alpha^{2}+\beta^{2}$ is equal to ____________.
Explanation:
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$.
Let the line $\frac{x-3}{7}=\frac{y-2}{-1}=\frac{z-3}{-4}$ intersect the plane containing the lines $\frac{x-4}{1}=\frac{y+1}{-2}=\frac{z}{1}$ and $4 a x-y+5 z-7 a=0=2 x-5 y-z-3, a \in \mathbb{R}$ at the point $P(\alpha, \beta, \gamma)$. Then the value of $\alpha+\beta+\gamma$ equals _____________.
Explanation:
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 $(17,0, - 5)$
$\alpha + \beta + \gamma = 12$
The largest value of $a$, for which the perpendicular distance of the plane containing the lines $ \vec{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+a \hat{j}-\hat{k})$ and $\vec{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-a \hat{k})$ from the point $(2,1,4)$ is $\sqrt{3}$, is _________.
Explanation:
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$
$\therefore$ ${a_{\max }} = 2$
The plane passing through the line $L: l x-y+3(1-l) z=1, x+2 y-z=2$ and perpendicular to the plane $3 x+2 y+z=6$ is $3 x-8 y+7 z=4$. If $\theta$ is the acute angle between the line $L$ and the $y$-axis, then $415 \cos ^{2} \theta$ is equal to _____________.
Explanation:
$L:lx - y + 3(1 - l)z = 1$, $x + 2y - z = 2$ 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
$\therefore$ $\cos \theta = {{{5 \over 3}} \over {\sqrt {1 + {{25} \over 9} + {{49} \over 9}} }} = {5 \over {\sqrt {83} }}$
$\therefore$ $415{\cos ^2}\theta = 125$
Let $\mathrm{Q}$ and $\mathrm{R}$ be two points on the line $\frac{x+1}{2}=\frac{y+2}{3}=\frac{z-1}{2}$ at a distance $\sqrt{26}$ from the point $P(4,2,7)$. Then the square of the area of the triangle $P Q R$ is ___________.
Explanation:
$L:{{x + 1} \over 2} = {{y + 2} \over 3} = {{2 - 1} \over 2}$
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 $ar(\Delta PQR) = 153$.
The line of shortest distance between the lines $\frac{x-2}{0}=\frac{y-1}{1}=\frac{z}{1}$ and $\frac{x-3}{2}=\frac{y-5}{2}=\frac{z-1}{1}$ makes an angle of $\cos ^{-1}\left(\sqrt{\frac{2}{27}}\right)$ with the plane $\mathrm{P}: \mathrm{a} x-y-z=0$, $(a>0)$. If the image of the point $(1,1,-5)$ in the plane $P$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta-\gamma$ is equal to _________________.
Explanation:
$ \left|\begin{array}{lll} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & 1 & 1 \\ 2 & 2 & 1 \end{array}\right|=-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}} $
angle between line and plane is $\cos ^{-1} \sqrt{\frac{2}{27}}=\alpha$
$ \cos \alpha=\sqrt{\frac{2}{27}}, \sin \alpha=\frac{5}{3 \sqrt{3}} $
DR's normal to plane $(1,-1,-1)$
$ \sin \alpha=\left|\frac{-a-2+2}{\sqrt{4+4+1} \sqrt{a^2+1+1}}\right|=\frac{5}{3 \sqrt{3}} $
$\sqrt{3}|a|=5 \sqrt{a^2+2}$
$ 3 a^2=25 a^2+50 $
No value of (a)
Consider a triangle ABC whose vertices are A(0, $\alpha$, $\alpha$), B($\alpha$, 0, $\alpha$) and C($\alpha$, $\alpha$, 0), $\alpha$ > 0. Let D be a point moving on the line x + z $-$ 3 = 0 = y and G be the centroid of $\Delta$ABC. If the minimum length of GD is $\sqrt {{{57} \over 2}} $, then $\alpha$ is equal to ____________.
Explanation:
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 $\alpha = 6$.
Let d be the distance between the foot of perpendiculars of the points P(1, 2, $-$1) and Q(2, $-$1, 3) on the plane $-$x + y + z = 1. Then d2 is equal to ___________.
Explanation:
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$
Let ${P_1}:\overrightarrow r \,.\,\left( {2\widehat i + \widehat j - 3\widehat k} \right) = 4$ be a plane. Let P2 be another plane which passes through the points (2, $-$3, 2), (2, $-$2, $-$3) and (1, $-$4, 2). If the direction ratios of the line of intersection of P1 and P2 be 16, $\alpha$, $\beta$, then the value of $\alpha$ + $\beta$ is equal to ________________.
Explanation:
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 > $
$\therefore$ $\alpha + \beta = 13 + 15 = 28$
Let the image of the point P(1, 2, 3) in the line $L:{{x - 6} \over 3} = {{y - 1} \over 2} = {{z - 2} \over 3}$ be Q. Let R ($\alpha$, $\beta$, $\gamma$) be a point that divides internally the line segment PQ in the ratio 1 : 3. Then the value of 22 ($\alpha$ + $\beta$ + $\gamma$) is equal to __________.
Explanation:
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$
Let the mirror image of the point (a, b, c) with respect to the plane 3x $-$ 4y + 12z + 19 = 0 be (a $-$ 6, $\beta$, $\gamma$). If a + b + c = 5, then 7$\beta$ $-$ 9$\gamma$ is equal to ______________.
Explanation:
${{x - a} \over 3} = {{y - b} \over { - 4}} = {{z - c} \over {12}} = {{ - 2(3a - 4b + 12c + 19)} \over {{3^2} + {{( - 4)}^2} + {{12}^2}}}$
${{x - a} \over 3} = {{y - b} \over { - 4}} = {{z - c} \over {12}} = {{ - 6a + 8b - 24c - 38} \over {169}}$
$(x,y,z) \equiv (a - 6,\,\beta ,\gamma )$
${{(a - 6) - a} \over 3} = {{\beta - b} \over { - 4}} = {{\gamma - c} \over {12}} = {{ - 6a + 8b - 24c - 38} \over {169}}$
${{\beta - b} \over { - 4}} = - 2 \Rightarrow \beta = 8 + b$
${{\gamma - c} \over {12}} = - 2 \Rightarrow \gamma = - 24 + c$
${{ - 6a + 8b - 24c - 38} \over {169}} = - 2$
$ \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$
Let l1 be the line in xy-plane with x and y intercepts ${1 \over 8}$ and ${1 \over {4\sqrt 2 }}$ respectively, and l2 be the line in zx-plane with x and z intercepts $ - {1 \over 8}$ and $ - {1 \over {6\sqrt 3 }}$ respectively. If d is the shortest distance between the line l1 and l2, then d$-$2 is equal to _______________.
Explanation:
${{x - {1 \over 8}} \over {{1 \over 8}}} = {y \over { - {1 \over {4\sqrt 2 }}}} = {z \over 0}\,\,\,\,\,\,\,\,\,\,\,\,\,$ ______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} }}$
${d^{ - 2}} = 51$
Let the lines
${L_1}:\overrightarrow r = \lambda \left( {\widehat i + 2\widehat j + 3\widehat k} \right),\,\lambda \in R$
${L_2}:\overrightarrow r = \left( {\widehat i + 3\widehat j + \widehat k} \right) + \mu \left( {\widehat i + \widehat j + 5\widehat k} \right);\,\mu \in R$,
intersect at the point S. If a plane ax + by $-$ z + d = 0 passes through S and is parallel to both the lines L1 and L2, then the value of a + b + d is equal to ____________.
Explanation:
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$
$a + b + d = 7 - 2 + 0 = 5$
Let a line having direction ratios, 1, $-$4, 2 intersect the lines ${{x - 7} \over 3} = {{y - 1} \over { - 1}} = {{z + 2} \over 1}$ and ${x \over 2} = {{y - 7} \over 3} = {z \over 1}$ at the points A and B. Then (AB)2 is equal to ___________.
Explanation:
So, DR's of $A B \propto 3 \lambda-2 \mu+7,-(\lambda+3 \mu+6), \lambda-\mu$ $-2$
Clearly $\frac{3 \lambda-2 \mu+7}{1}=\frac{\lambda+3 \mu+6}{4}=\frac{\lambda-\mu-2}{2}$
$\Rightarrow 5 \lambda-3 \mu=-16$
And $\lambda-5 \mu=10$
From (i) and (ii) we get $\lambda=-5, \mu=-3$
So, $A$ is $(-8,6,-7)$ and $B$ is $(-6,-2,-3)$
$ A B=\sqrt{4+64+16} \Rightarrow(A B)^{2}=84 $
If the shortest distance between the lines
$\overrightarrow r = \left( { - \widehat i + 3\widehat k} \right) + \lambda \left( {\widehat i - a\widehat j} \right)$
and $\overrightarrow r = \left( { - \widehat j + 2\widehat k} \right) + \mu \left( {\widehat i - \widehat j + \widehat k} \right)$ is $\sqrt {{2 \over 3}} $, then the integral value of a is equal to ___________.
Explanation:
$\vec{a}_{1}-\vec{a}_{2}=-\hat{i}+\hat{j}+\hat{k}$
Shortest distance $=\left|\frac{\left(\vec{a}_{1}-\vec{a}_{2}\right) \cdot\left(\vec{b}_{1} \times \vec{b}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|$
$ \begin{aligned} &\Rightarrow \quad \sqrt{\frac{2}{3}}=\frac{2(a-1)}{\sqrt{a^{2}+1+(a-1)^{2}}} \\\\ &\Rightarrow 6\left(a^{2}-2 a+1\right)=2 a^{2}-2 a+2 \\\\ &\Rightarrow \quad(a-2)(2 a-1)=0 \Rightarrow a=2 \text { because } a \in z . \end{aligned} $
planes 3x $-$ 2y + 4z $-$ 7 = 0
and x + 5y $-$ 2z + 9 = 0, be
$\alpha$x + $\beta$y + $\gamma$z + 3 = 0, then $\alpha$ + $\beta$ + $\gamma$ is equal to :
${{x - \alpha } \over 1} = {{y - 1} \over 2} = {{z - 1} \over 3}$ and ${{x - 4} \over \beta } = {{y - 6} \over 3} = {{z - 7} \over 3}$, lies on the plane x + 2y $-$ z = 8, then $\alpha$ $-$ $\beta$ is equal to :
${{x - 3} \over 2} = {{y - 1} \over 1} = {{z - 2} \over 1}$.
Let Q be the mirror image of the point (2, 3, $-$1) with respect to L. Let a plane P be such that it passes through Q, and the line L is perpendicular to P. Then which of the following points is on the plane P?
$3 + {{x - 11} \over {{{(y - 19)}^2}{{(z - 12)}^2}}} + {{y - 19} \over {{{(x - 11)}^2}{{(z - 12)}^2}}} + {{z - 12} \over {{{(x - 11)}^2}{{(y - 19)}^2}}} - {{x + y + z} \over {14(x - 11)(y - 19)(z - 12)}}$ is equal to :
