$x+2y+3z=4$ is ${\cos ^{ - 1}}\left( {\sqrt {{5 \over {14}}} } \right),$ then $\lambda $ equals :
$B(1,6,3)$ in the line : ${x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$
Statement - 2 : The line ${x \over 1} = {{y - 1} \over 2} = {{z - 2} \over 3}$ bisects the line
segment joining $A(1,0,7)$ and $B(1, 6, 3)$
Statement-2 : The plane $x-y+z=5$ bisects the line segment joining $A(3, 1, 6)$ and $B(1, 3, 4).$
${x^2} + {y^2} + {z^2} + 6x - 8y - 2z = 13$ and
${x^2} + {y^2} + {z^2} - 10x + 4y - 2z = 8$ then a equals :
$\overrightarrow r = 2\widehat i - 2\widehat j + 3\widehat k + \lambda \left( {i - j + 4k} \right),$ and the plane
$\overrightarrow r .\left( {\widehat i + 5\widehat j + \widehat k} \right) = 5$ is
the plane $2x - y + \sqrt \lambda \,\,z + 4 = 0$ is such that $\sin \,\,\theta = {1 \over 3}$ then value of $\lambda $ is :
If the angle $\beta \,$, which it makes with y-axis, is such that $\,{\sin ^2}\beta = 3{\sin ^2}\theta ,$ then ${\cos ^2}\theta $ equals :
${x^2} + {y^2} + {z^2} + 7x - 2y - z = 13$ and
${x^2} + {y^2} + {z^2} - 3x + 3y + 4z = 8$
is the same as the intersection of one of the sphere and the plane
$\,2x + y + 2z = 8$ and $4x + 2y + 4z + 5 = 0$ is :
$x=1+s,y=-3$$ - \lambda s,$ $z = 1 + \lambda s$ and $x = {t \over 2},y = 1 + t,z = 2 - t,$ with parameters $s$ and $t$ respectively, are co-planar, then $\lambda $ equals :
${x^2} + {y^2} + {z^2} + 4x - 2y - 6z = 155$ is
${x^2} + {y^2} + {z^2} + 2x - 2y - 4z - 19 = 0$ is cut by the plane
$x+2y+2z+7=0$ is
${{x - 4} \over 1} = {{y - 7} \over 5} = {{z - 4} \over 4}$ is :
Let a line L passing through the point $\mathrm{P}(1,1,1)$ be perpendicular to the lines $\frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}$ and $\frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}$. Let the line L intersect the $y z-$ plane at the point Q . Another line parallel to L and passing through the point $\mathrm{S}(1,0,-1)$ intersects the $y z$-plane at the point R . Then the square of the area of the parallelogram PQRS is equal to $\_\_\_\_$ .
Explanation:
$ \begin{aligned} & \text { let } L_1: \frac{x-4}{4}=\frac{y-1}{1}=\frac{z-1}{1}=\lambda_1 \\ & L_2: \frac{x-17}{1}=\frac{y-71}{1}=\frac{z}{0}=\lambda_2 \end{aligned} $
given that $L$ is passing through $P(1,1,1)$
and perpendicular to $L_1$ and $L_2$.
dr's of $L=$ dr's of $L_1 \times$ dr's of $L_2$
$\begin{aligned} & =\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 4 & 1 & 1 \\ 1 & 1 & 0\end{array}\right| \\\\ = & \hat{i}(0-1)-\hat{j}(0-1)+\hat{k}(4-1) \\\\ = & -\hat{i}+\hat{j}+3 \hat{k}\end{aligned}$
Equation of line $L: \frac{x-1}{-1}=\frac{y-1}{1}=\frac{z-1}{3}=\lambda$
It is given that $L$ intersect $y z$ plane at $Q$ for $Q$ on $y-z$ plane $x=0 $
$\lambda=\frac{0-1}{-1}=1$
$\frac{y-1}{1}=\lambda \Rightarrow y-1=1 \Rightarrow y=2$
$\frac{z-1}{3}=\lambda \Rightarrow \frac{z-1}{3}=1 \Rightarrow z=4$
Point $Q$ is $(0,2,4)$
Another line parallel to $L$ let $L^{\prime}$ passing through $S(1,0,-1)$.
The area of parallelogram $P Q R S$
$ \begin{aligned} = & |\overrightarrow{P Q} \times \overrightarrow{P S}| \\ & \overrightarrow{P Q}=Q-P=(0,2,4)-(1,1,1)=(-1,1,3) \\ & \overrightarrow{P S}=S-P=(1,0,-1)-(1,1,1)=(0,-1,-2) \end{aligned} $
$=\hat{i}(-2+3)-\hat{j}(2-0)+\hat{k}(1-0)$
$=\hat{i}-2 \hat{j}+\hat{k}$
$ \text { Area }=|\overrightarrow{P Q} \times \overrightarrow{P S}|=\sqrt{1^2+(-2)^2+(1)^2}=\sqrt{6} $
Square of area $=6$
If the image of the point $\mathrm{P}(a, 2, a)$ in the line $\frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}$ is Q and the image
of Q in the line $\frac{x-2 b}{2}=\frac{y-a}{1}=\frac{z+2 b}{-5}$ is P , then $a+b$ is equal to $\_\_\_\_$ .
Explanation:
For line 1 :
$ \frac{x}{2}=\frac{y+a}{1}=\frac{z}{1}=\lambda $
general point $M$ on line is $(2 \lambda, \lambda-a, \lambda)$
$ \begin{aligned} & P \text { is }(a, 2, a) \\ & \overrightarrow{P M}=(2 \lambda-a, \lambda-a-2, \lambda-a) \end{aligned} $
since $P M$ is perpendicular to line 1 , their dot product is 0
$ \begin{aligned} & \overrightarrow{P M} \cdot(2,1,1)=0 \\ & 2(2 \lambda-a)+1(\lambda-a-2)+1(\lambda-a)=0 \\ & 6 \lambda-4 a-2=0 \Rightarrow \lambda=\frac{2 a+1}{3} \\ & \text { midpoint } M=\left(\frac{4 a+2}{3}, \frac{1-a}{3}, \frac{2 a+1}{3}\right) \end{aligned} $
for line 2:
$ \frac{x-2 b}{2}=\frac{y-a}{1}=\frac{z+2 b}{-5} $
since $P$ is the image of $Q$, midpoint $M$ lies on line 2
$ \begin{aligned} & \frac{\frac{4 a+2}{3}-2 b}{2}=\frac{\frac{1-a}{3}-a}{1}=\frac{\frac{2 a+1}{3}+2 b}{-5} \\ & \frac{4 a+2-6 b}{6}=\frac{1-4 a}{3}=\frac{2 a+1+6 b}{-15} \end{aligned} $
from first two parts:
$ \begin{aligned} & 4 a+2-6 b=2(1-4 a) \\ & 4 a+2-6 b=2-8 a \Rightarrow 12 a=6 b \Rightarrow b=2 a \end{aligned} $
from last two parts:
$ \begin{aligned} & -5(1-4 a)=2 a+1+6 b \\ & -5+20 a=2 a+1+6(2 a) \\ & 20 a-5=14 a+1 \\ & 6 a=6 \Rightarrow a=1 \\ & b=2(1)=2 \\ & a+b=1+2=3 \end{aligned} $
the value of $a+b$ is 3 .
Explanation:
$\begin{aligned} & L_1: x+2=y-1=z=\ell \\ & L_2: \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}=m \\ & L_3: \frac{x}{-3}=\frac{y-3}{5}=\frac{z-2}{1}=n \end{aligned}$
$\begin{aligned} &\text { Point of intersection of } L_1 \text { and } L_2\\ &\left.\begin{array}{r} \ell-2=5 \mathrm{~m}+3 \\ \ell+1=-\mathrm{m} \\ \ell=\mathrm{m}+1 \end{array}\right\} \ell=0, \mathrm{~m}=-1 \quad \mathrm{~A}(-2,1,0) \end{aligned}$
Point of intersection of $L_2$ and $L_3$
$\left.\begin{array}{l} 5 m+3=-3 n \\ -m=3 n+3 \\ m+1=n+2 \end{array}\right\} m=0, n=-1, B(3,0,1)$
Point of intersection $L_3$ and $L_4$
$\left.\begin{array}{r} -3 \mathrm{n}=\ell-2 \\ 3 \mathrm{n}+3=\ell+1 \\ \mathrm{n}+2=\ell \end{array}\right\} \ell=2, \mathrm{n}=0, \mathrm{C}(0,3,2)$
$\begin{aligned} & \operatorname{Ar}(\triangle \mathrm{ABC})=\left|\frac{1}{2}\right| \begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ -5 & 1 & -1 \\ -3 & 3 & 1 \end{array}| | \\ & \mathrm{A}=\frac{1}{2}|\hat{\mathrm{i}}(4)-\hat{\mathrm{j}}(-8)+\hat{\mathrm{k}}(-12)| \\ & \mathrm{A}=\frac{1}{2} \sqrt{16+64+144}=\sqrt{56} \\ & \mathrm{~A}^2=56 \end{aligned}$
Let P be the image of the point $\mathrm{Q}(7,-2,5)$ in the line $\mathrm{L}: \frac{x-1}{2}=\frac{y+1}{3}=\frac{z}{4}$ and $\mathrm{R}(5, \mathrm{p}, \mathrm{q})$ be a point on $L$. Then the square of the area of $\triangle P Q R$ is _________.
Explanation:
$\begin{aligned} &\text { Let } \mathrm{R}(2 \lambda+1,3 \lambda-1,4 \lambda)\\ &\begin{aligned} & 2 \lambda+1=5 \\ & \lambda=2 \\ & \mathrm{R}(5,5,8) \\ & \text { let } \mathrm{T}(2 \lambda+1,3 \lambda-1,4 \lambda) \\ & \overrightarrow{\mathrm{QT}}=(2 \lambda-6) \hat{\mathrm{i}}+(3 \lambda+1) \hat{\mathrm{j}}+(4 \lambda-5) \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{~b}}=2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{QT}} \cdot \overrightarrow{\mathrm{~b}}=0 \end{aligned}\\ &4 \lambda-12+9 \lambda+3+16 \lambda-20=0 \end{aligned}$
$\begin{aligned} & \lambda=1 \\ & \mathrm{~T}(3,2,4) \\ & \mathrm{QT}=\sqrt{33} \quad \mathrm{RT}=\sqrt{29} \\ & (\text { area of } \Delta \mathrm{PQR})^2=\left(\frac{1}{2} \sqrt{29} \cdot 2 \sqrt{33}\right)^2 \\ & =957 \end{aligned}$
Let $\mathrm{L}_1: \frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}$ and $\mathrm{L}_2: \frac{x-2}{2}=\frac{y}{0}=\frac{z+4}{\alpha}, \alpha \in \mathbf{R}$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A(1,1,-1)$ on $L_2$, then the value of $26 \alpha(\mathrm{~PB})^2$ is _________ .
Explanation:
Explanation
To find the value of $26 \alpha(\mathrm{PB})^2$, we proceed as follows:
Intersection Point $ B $ of $\mathrm{L}_1$ and $\mathrm{L}_2$
We are given the equations of the lines:
$ \mathrm{L}_1 : \frac{x-1}{3} = \frac{y-1}{-1} = \frac{z+1}{0} $
$ \mathrm{L}_2 : \frac{x-2}{2} = \frac{y}{0} = \frac{z+4}{\alpha} $
For $\mathrm{L}_1$, $ z+1 = 0 $, which implies $ z = -1 $.
For $\mathrm{L}_2$, since $ y = 0 $, the direction ratios can be matched using the parameter $\mu$:
$ \begin{aligned} & x = 3\lambda + 1, \quad y = -\lambda + 1, \quad z = -1, \\ & x = 2\mu + 2, \quad y = 0, \quad z = \alpha \mu - 4. \end{aligned} $
Setting the coordinates equal for intersection:
$ \begin{aligned} & 3\lambda + 1 = 2\mu + 2, \\ & -\lambda + 1 = 0, \\ & -1 = \alpha \mu - 4. \end{aligned} $
From $-\lambda + 1 = 0$, we find $\lambda = 1$.
Substituting $\lambda = 1$ into $3\lambda + 1 = 2\mu + 2$ gives:
$ 3(1) + 1 = 2\mu + 2 \implies \mu = 1. $
Also, substituting $\mu = 1$ into $-1 = \alpha \mu - 4$ gives:
$ -1 = \alpha(1) - 4 \implies \alpha = 3. $
So, the intersection point $B$ is:
$ B(4, 0, -1). $
Foot of Perpendicular from $A$ to $\mathrm{L}_2$
The point $P$ on $\mathrm{L}_2$ is given by:
$ P(2\delta + 2, 0, 3\delta - 4). $
For vector $\overrightarrow{AP} = (2\delta + 1, -1, 3\delta - 3)$, since $AP$ is perpendicular to $\mathrm{L}_2$, the dot product should be zero:
$ \begin{aligned} (2\delta + 1)\cdot 2 + (-1)\cdot 0 + (3\delta - 3)\cdot 3 &= 0. \end{aligned} $
Simplifying gives:
$ 4\delta + 2 + 9\delta - 9 = 0 \implies 13\delta = 7 \implies \delta = \frac{7}{13}. $
So, the coordinates of point $P$ are:
$ P\left(\frac{40}{13}, 0, \frac{-31}{13}\right). $
Distance $\mathrm{PB}$ and Calculation
The vector $\overrightarrow{PB}$ is:
$ \overrightarrow{PB} = \left(4 - \frac{40}{13}, 0 - 0, -1 - \left(\frac{-31}{13}\right)\right). $
Calculating the components:
$ \overrightarrow{PB} = \left(\frac{12}{13}, 0, \frac{18}{13}\right). $
The square of the distance $(PB)^2$ is:
$ \left(\frac{12}{13}\right)^2 + \left(0\right)^2 + \left(\frac{18}{13}\right)^2 = \frac{144}{169} + \frac{324}{169} = \frac{468}{169}. $
Thus, $26 \alpha (PB)^2$ is:
$ 26 \times 3 \times \frac{468}{169} = 216. $
So, the final value is 216.
The square of the distance of the image of the point $(6,1,5)$ in the line $\frac{x-1}{3}=\frac{y}{2}=\frac{z-2}{4}$, from the origin is __________.
Explanation:
$\begin{aligned} & \overrightarrow{P A}=(3 \lambda-5) \hat{i}+(2 \lambda-1) \hat{j}+(4 \lambda-3) \hat{k} \\ & (3 \lambda-5) 3+(2 \lambda-1) 2+(4 \lambda-3) 4=0 \\ & \Rightarrow 9 \lambda-15+4 \lambda-2+16 \lambda-12=0 \\ & \Rightarrow 29 \lambda=29 \\ & \therefore \lambda=1 \\ & \therefore \quad A(4,2,6) \\ & \therefore \quad P^{\prime}: \text { mirror image of } P \\ & \Rightarrow P^{\prime}(2,3,7) \\ & \left(O P^{\prime}\right)^2=4+9+49 \\ & =62 \\ & \end{aligned}$
Let $\mathrm{P}(\alpha, \beta, \gamma)$ be the image of the point $\mathrm{Q}(1,6,4)$ in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$. Then $2 \alpha+\beta+\gamma$ is equal to ________
Explanation:
$\begin{aligned} & \overrightarrow{Q R} \cdot(\hat{i}+2 \hat{j}+3 \hat{k})=0 \\ & (t-1)+(2 t-5) \times 2+(3 t-2) \times 3=0 \Rightarrow t=\frac{17}{14} \\ & \Rightarrow R \equiv\left(\frac{17}{14}, \frac{48}{14}, \frac{79}{14}\right) \\ & \Rightarrow \frac{\alpha+1}{2}=\frac{17}{14}, \frac{\beta+6}{2}=\frac{48}{14}, \frac{\gamma+4}{2}=\frac{79}{14} \\ & 2 \alpha+\beta+\gamma=\frac{68}{14}-2+\frac{96}{14}-6+\frac{158}{14}-4=11 \end{aligned}$
If the shortest distance between the lines $\frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1}$ and $\frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4}$ is $\frac{44}{\sqrt{30}}$, then the largest possible value of $|\lambda|$ is equal to _________.
Explanation:
$\begin{aligned} & L_1: \frac{x-\lambda}{3}=\frac{y-2}{-1}=\frac{z-1}{1} \\ & L_2: \frac{x+2}{-3}=\frac{y+5}{2}=\frac{z-4}{4} \end{aligned}$
$\begin{aligned} & n_1 \times n_2=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 1 \\ -3 & 2 & 4 \end{array}\right| \\ & =-6 \hat{i}-15 \hat{j}+3 \hat{k} \\ & d=\left|\frac{[(\lambda+2) \hat{i}+7 \hat{j}-3 \hat{k}][-6 \hat{i}-15 \hat{j}+3 \hat{k}]}{|-6 \hat{i}-15 \hat{j}+3 \hat{k}|}\right|=\frac{44}{\sqrt{30}} \\ & \left|\frac{-6 \lambda-12-105-9}{\sqrt{270}}\right|=\frac{44}{\sqrt{30}} \\ & |6 \lambda+126|=132 \\ & |\lambda+21|=22 \\ & \lambda+21= \pm 22 \\ & |\lambda|_{\max }=43 \end{aligned}$
Let $P$ be the point $(10,-2,-1)$ and $Q$ be the foot of the perpendicular drawn from the point $R(1,7,6)$ on the line passing through the points $(2,-5,11)$ and $(-6,7,-5)$. Then the length of the line segment $P Q$ is equal to _________.
Explanation:
$\begin{aligned} & P(10,-2,-1) \\ & M N: \frac{x-2}{8}=\frac{y+5}{-12}=\frac{z-11}{16} \end{aligned}$
General point
$(8k + 2, - 12k - 5,16k + 11)$
$\overrightarrow {RQ} = (8k + 2 - 1)\widehat i + ( - 12k - 5 - 7)\widehat j + (16k + 11 - 6)\widehat k$
$\overrightarrow {RQ} = (8k + 1)\widehat i - (12k + 12)\widehat j + (16k + 5)\widehat k$
$\overrightarrow {RQ} \,.\,\overrightarrow {MN} = 0$ (as both are perpendicular)
$8(8 k+1)+12(12 k+12)+16(16 k+5)=0$
$\begin{aligned} & 64 k+8+144 k+144+256 k+80=0 \\ & 464 k=-232 \\ & k=\frac{-232}{464}=\frac{-1}{2} \\ & Q(-4+2,6-5,-8+11) \\ & Q(-2,1,3) \\ & P Q=\sqrt{(10+2)^2+(-3)^2+(4)^2} \\ & P Q=\sqrt{12^2+3^2+4^2} \\ & P Q=\sqrt{169}=13 \end{aligned}$