3D Geometry
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 .
Let a line $L_1$ pass through the origin and be perpendicular to the lines
$\mathrm{L}_2: \overrightarrow{\mathrm{r}}=(3+\mathrm{t}) \hat{i}+(2 \mathrm{t}-1) \hat{j}+(2 \mathrm{t}+4) \hat{k}$ and
$\mathrm{L}_3: \overrightarrow{\mathrm{r}}=(3+2 \mathrm{~s}) \hat{i}+(3+2 \mathrm{~s}) \hat{j}+(2+\mathrm{s}) \hat{k}, \mathrm{t}, \mathrm{s} \in \mathbf{R}$.
If $(a, b, c), a \in \mathbf{Z}$, is the point on $\mathrm{L}_3$ at a distance of $\sqrt{17}$ from the point of intersection of $\mathrm{L}_1$ and $\mathrm{L}_2$, then $(\mathrm{a}+\mathrm{b}+\mathrm{c})^2$ is equal to $\_\_\_\_$ .
Explanation:
A pair of lines in 3D have the vector equations :
$ \begin{array}{|l|l|l|} \hline \boldsymbol{L}_2 & \boldsymbol{L}_3 & (\boldsymbol{t}, s) \\ \hline \overrightarrow{\boldsymbol{r}}=(3+\boldsymbol{t}) \hat{\boldsymbol{i}}+(2 \boldsymbol{t}-1) \hat{\boldsymbol{j}}+(2 \boldsymbol{t}+4) \hat{\boldsymbol{k}} & \overrightarrow{\boldsymbol{r}}=(3+2 \boldsymbol{s}) \hat{\boldsymbol{i}}+(3+2 \boldsymbol{s}) \hat{\boldsymbol{j}}+(2+\boldsymbol{s}) \hat{\boldsymbol{k}} & \in \mathbb{R} \\ \hline \end{array} $
A third line $\boldsymbol{L}_1$ passes through the origin and is ⟂ to both the lines $\boldsymbol{L}_2$ and $\boldsymbol{L}_3$ defined just above
The point of intersection of $\boldsymbol{L}_1$ and $\boldsymbol{L}_2$ is at a distance $\sqrt{17}$ units from another point with coordinates $(\boldsymbol{a}, \boldsymbol{b}, \boldsymbol{c})$ and lying on $\boldsymbol{L}_3$ where $a \in Z$
$\begin{aligned} & \mathrm{L}_2: \frac{\mathrm{x}-3}{1}=\frac{\mathrm{y}+1}{2}=\frac{\mathrm{z}-4}{2}=\mathrm{t} \\ & \mathrm{L}_5: \frac{\mathrm{x}-3}{2}=\frac{\mathrm{y}-3}{2}=\frac{\mathrm{z}-2}{1}=\mathrm{s} \\ & \text { vector } \perp \text { to } \mathrm{L}_2 \& \mathrm{~L}_3\end{aligned}$
$\begin{aligned} & \Rightarrow\left|\begin{array}{lll}i & j & k \\ 1 & 2 & 2 \\ 2 & 2 & 1\end{array}\right|=\hat{i}(-2)-\hat{j}(-3)+\hat{k}(-2) \\ & =-2 \hat{i}+3 \hat{j}-2 \hat{k} \\ & =-(2 \hat{i}-3 \hat{j}+2 \hat{k})\end{aligned}$
Now, $\mathrm{L}_1: \frac{\mathrm{x}}{2}=\frac{\mathrm{y}}{-3}=\frac{\mathrm{z}}{2}=\ell$
intersection of $\mathrm{L}_1$ \& $\mathrm{L}_2$
$ \Rightarrow \mathrm{t}+3=2 \ell, 2 \mathrm{t}-1=-3 \ell, 2 \mathrm{t}+4=2 \ell $
$\Rightarrow $ $t+3=2 t+4$
$ \begin{aligned} & t=-1 \\ & \Rightarrow P \equiv(2,-3,2) \end{aligned} $
A point on $\mathrm{L}_3$ is $\mathrm{Q}(2 \mathrm{~s}+3,2 \mathrm{~s}+3, \mathrm{~s}+2)$
$ \begin{aligned} & \Rightarrow P Q^2=(2 s+1)^2+(2 s+6)^2+s^2=17 \\ & \Rightarrow 9 s^2+28 s+20=0 \\ & \Rightarrow(9 s+10)(s+2)=0 \Rightarrow s=-2 \\ & \therefore Q \equiv(-1,-1,0) \equiv(a, b, c) \Rightarrow(a+b+c)^2=4 \end{aligned} $
Let the image of the point $\mathrm{P}(0,-5,0)$ in the line $\frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2}$ be the point R and the image of the point $\mathrm{Q}\left(0, \frac{-1}{2}, 0\right)$ in the line $\frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1}$ be the point S . Then the square of the area of the parallelogram PQRS is $\_\_\_\_$ .
Explanation:
Let us find the reflected points $R$ and $S$, and then use vectors to get the area of parallelogram $PQRS$.
1. Reflection of $P(0,-5,0)$ in the line
$ \frac{x-1}{2}=\frac{y}{1}=\frac{z+1}{-2} $
Write the line in parametric form:
$ x=1+2t,\quad y=t,\quad z=-1-2t $
So, a point on the line is
$ A(1,0,-1) $
and its direction vector is
$ \vec d_1=(2,1,-2) $
To reflect point $P$ in the line, first find the foot of perpendicular from $P$ to the line.
Let the foot be
$ M=A+t\vec d_1=(1+2t,\ t,\ -1-2t) $
Since $PM\perp \vec d_1$,
$ (\vec{PM})\cdot \vec d_1=0 $
Now,
$ \vec{PM}=M-P=(1+2t,\ t+5,\ -1-2t) $
So,
$ (1+2t,\,t+5,\,-1-2t)\cdot (2,1,-2)=0 $
$ 2(1+2t)+(t+5)+(-2)(-1-2t)=0 $
$ 2+4t+t+5+2+4t=0 $
$ 9+9t=0 $
$ t=-1 $
Hence,
$ M=(1+2(-1),\ -1,\ -1-2(-1))=(-1,-1,1) $
Since $M$ is the midpoint of $PR$,
$ R=2M-P $
$ R=2(-1,-1,1)-(0,-5,0)=(-2,3,2) $
So,
$ R=(-2,3,2) $
2. Reflection of $Q\left(0,-\frac12,0\right)$ in the line
$ \frac{x-1}{-1}=\frac{y+9}{4}=\frac{z+1}{1} $
Write the line in parametric form:
$ x=1-u,\quad y=-9+4u,\quad z=-1+u $
So, a point on the line is
$ B(1,-9,-1) $
and direction vector is
$ \vec d_2=(-1,4,1) $
Let the foot of perpendicular from $Q$ to this line be
$ N=B+u\vec d_2=(1-u,\ -9+4u,\ -1+u) $
Since $QN\perp \vec d_2$,
$ (\vec{QN})\cdot \vec d_2=0 $
Now,
$ \vec{QN}=N-Q=\left(1-u,\ -9+4u+\frac12,\ -1+u\right) =\left(1-u,\ -\frac{17}{2}+4u,\ -1+u\right) $
So,
$ \left(1-u,\ -\frac{17}{2}+4u,\ -1+u\right)\cdot (-1,4,1)=0 $
$ -(1-u)+4\left(-\frac{17}{2}+4u\right)+(-1+u)=0 $
$ -1+u-34+16u-1+u=0 $
$ 18u-36=0 $
$ u=2 $
Hence,
$ N=(1-2,\ -9+8,\ -1+2)=(-1,-1,1) $
Again, $N$ is the midpoint of $QS$, so
$ S=2N-Q $
$ S=2(-1,-1,1)-\left(0,-\frac12,0\right) =\left(-2,-2+\frac12,2\right) =\left(-2,-\frac32,2\right) $
So,
$ S=\left(-2,-\frac32,2\right) $
3. Check the parallelogram sides
Now,
$ P=(0,-5,0),\quad Q=\left(0,-\frac12,0\right),\quad R=(-2,3,2),\quad S=\left(-2,-\frac32,2\right) $
Take adjacent sides from $P$:
$ \vec{PQ}=Q-P=\left(0,\frac{9}{2},0\right) $
and
$ \vec{PR}=R-P=(-2,8,2) $
Also,
$ \vec{QS}=S-Q=(-2,-1,2) $
which is not equal to $\vec{PR}$, so let us check the correct order of parallelogram.
Observe:
$ \vec{PS}=S-P=\left(-2,\frac72,2\right) $
and
$ \vec{QR}=R-Q=\left(-2,\frac72,2\right) $
Thus, $PS\parallel QR$.
Also,
$ \vec{PQ}=\left(0,\frac92,0\right),\quad \vec{SR}=R-S=\left(0,\frac92,0\right) $
So the correct parallelogram order is $PQR S$, with adjacent sides
$ \vec{PQ}\quad \text{and} \quad \vec{PS} $
Hence area of parallelogram is
$ \left|\vec{PQ}\times \vec{PS}\right| $
4. Compute cross product
$ \vec{PQ}=\left(0,\frac92,0\right),\qquad \vec{PS}=\left(-2,\frac72,2\right) $
$ \vec{PQ}\times \vec{PS} = \begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & \frac92 & 0\\ -2 & \frac72 & 2 \end{vmatrix} $
$ = \hat i\left(\frac92\cdot 2-0\cdot \frac72\right) -\hat j(0\cdot 2-0\cdot (-2)) +\hat k\left(0\cdot \frac72-\frac92\cdot (-2)\right) $
$ =9\hat i+0\hat j+9\hat k $
So,
$ \left|\vec{PQ}\times \vec{PS}\right| =\sqrt{9^2+9^2} =\sqrt{162} =9\sqrt2 $
Therefore, the square of the area is
$ (9\sqrt2)^2=162 $
$ \boxed{162} $
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)
$8(8 k+1)+12(12 k+12)+16(16 k+5)=0$
$\begin{aligned} & 64 k+8+144 k+144+256 k+80=0 \\ & 464 k=-232 \\ & k=\frac{-232}{464}=\frac{-1}{2} \\ & Q(-4+2,6-5,-8+11) \\ & Q(-2,1,3) \\ & P Q=\sqrt{(10+2)^2+(-3)^2+(4)^2} \\ & P Q=\sqrt{12^2+3^2+4^2} \\ & P Q=\sqrt{169}=13 \end{aligned}$
Let the point $(-1, \alpha, \beta)$ lie on the line of the shortest distance between the lines $\frac{x+2}{-3}=\frac{y-2}{4}=\frac{z-5}{2}$ and $\frac{x+2}{-1}=\frac{y+6}{2}=\frac{z-1}{0}$. Then $(\alpha-\beta)^2$ is equal to _________.
Explanation:
$\overrightarrow{A B} \perp \vec{L}_1 \text { and } \overrightarrow{A B} \perp \vec{L}_2$
$\begin{aligned} & \overrightarrow{A B}=(-3 m-2+n+2,4 m+2-2 n+6,2 m+5-1) \\ & =(-3 m+n, 4 m-2 n+8,2 m+4) \\ & \overrightarrow{A B} \perp \vec{L}_1 \\ & \Rightarrow-3(-3 m+n)+4(4 m-2 n+8)+2(2 m+4)=0 \\ & (9 m+16 m+4 m)+(-3 n-8 n)+32+8=0 \\ & \Rightarrow 29 m-11 n+40=0 \quad \ldots(1) \\ & \overrightarrow{A B} \perp \vec{L}_2 \\ & \Rightarrow-1(-3 m+n)+2(4 m-2 n+8)+0(2 m+4)=0 \\ & \Rightarrow 3 m-n+8 m-4 n+16=0 \\ & \Rightarrow 11 m-5 n+16=0 \Rightarrow m=-1, n=1 \\ & \Rightarrow A \equiv(1,-2,3), \quad B \equiv(-3,-4,1) \\ & A B \text { line } \Rightarrow \frac{x-1}{2}=\frac{y+2}{1}=\frac{z-3}{1} \\ & \Rightarrow \alpha=-2, \quad \beta=3 \\ & \Rightarrow(\alpha-\beta)^2=25 \end{aligned}$
Consider a line $\mathrm{L}$ passing through the points $\mathrm{P}(1,2,1)$ and $\mathrm{Q}(2,1,-1)$. If the mirror image of the point $\mathrm{A}(2,2,2)$ in the line $\mathrm{L}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+6 \gamma$ is equal to __________.
Explanation:
$\begin{aligned} & A(2,2,2) \\ & P Q: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{-2} \end{aligned}$
General point,
$(k+1,-k+2,-2 k+1)$
$\begin{aligned} & \overrightarrow{O A}=(k+1-2) \hat{i}+(-k+2-2) \hat{j}+(-2 k+1-2) \hat{k} \\ & \overrightarrow{O A}=(k-1) \hat{i}-k \hat{j}+(-2 k-1) \hat{k} \\ & \overrightarrow{P Q}=1 \hat{i}-\hat{j}-2 \hat{k} \\ & \overrightarrow{O A} \cdot \overrightarrow{P Q}=0 \\ & (k-1)+k+2(2 k+1)=0 \\ & k-1+k+4 k+2=0 \\ & 6 k+1=0 \\ & k=\frac{-1}{6} \\ & 0\left(\frac{-1}{6}+1, \frac{+1}{6}+2,-2\left(\frac{-1}{6}\right)+1\right) \\ & 0\left(\frac{5}{6}, \frac{13}{6}, \frac{-8}{6}\right) \\ & 0\left(\frac{5}{6}, \frac{13}{6}, \frac{8}{6}\right)=\left(\frac{\alpha+2}{2}, \frac{\beta+2}{2}, \frac{\gamma+2}{2}\right) \end{aligned}$
$\begin{array}{l|l} \alpha+2=\frac{10}{6} & \beta+2=\frac{26}{6} \\ \alpha=\frac{10}{6}-2 & \beta=\frac{26-12}{6} \\ \alpha=\frac{-2}{6} & \beta=\frac{14}{6} \\ \alpha =-\frac{1}{3} & \beta=\frac{7}{3} \end{array}$
$\begin{aligned} & \gamma+2=\frac{16}{6} \\ & \gamma=\frac{16-12}{6} \\ & \gamma=\frac{4}{6} \\ & \Rightarrow \alpha+\beta+6 \gamma \\ & \Rightarrow \frac{-1}{3}+\frac{7}{3}+6 \times \frac{4}{6} \\ & \Rightarrow 2+4=6 \end{aligned}$
$ \begin{aligned} & \mathrm{L}_1: \overrightarrow{\mathrm{r}}=(\hat{i}+2 \hat{j}+3 \hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \text { and } \\\\ & \mathrm{L}_2: \overrightarrow{\mathrm{r}}=(4 \hat{i}+5 \hat{j}+6 \hat{k})+\mu(\hat{i}+\hat{j}-\hat{k}) \end{aligned} $
intersect $\mathrm{L}_1$ and $\mathrm{L}_2$ at $\mathrm{P}$ and $\mathrm{Q}$ respectively. If $(\alpha, \beta, \gamma)$ is the mid point of the line segment $\mathrm{PQ}$, then $2(\alpha+\beta+\gamma)$ is equal to ____________.
Explanation:
$P \equiv(\lambda+1,-\lambda+2, \lambda+3)$
$Q \equiv(\mu+4, \mu+5,-\mu+6)$
$\overrightarrow{P Q}=(\mu-\lambda+3, \mu+\lambda+3,-\mu-\lambda+3)$
$\overrightarrow{P Q} \cdot \vec{b}_1=0 \Rightarrow 3 \lambda+\mu=3$ ..........(i)
$\overrightarrow{P Q} \cdot \vec{b}_2=0 \Rightarrow 3 \mu+\lambda=3$ ..........(ii)
From (i) and (ii),
$ \begin{aligned} & P \equiv\left(\frac{5}{2}, \frac{1}{2}, \frac{9}{2}\right) \& Q \equiv\left(\frac{5}{2}, \frac{7}{2}, \frac{15}{2}\right) \\\\ & \alpha=\frac{5}{2}, \beta=\frac{4}{2}, \gamma=\frac{12}{2} \\\\ & 2(\alpha+\beta+\gamma)=21 \end{aligned} $
A line passes through $A(4,-6,-2)$ and $B(16,-2,4)$. The point $P(a, b, c)$, where $a, b, c$ are non-negative integers, on the line $A B$ lies at a distance of 21 units, from the point $A$. The distance between the points $P(a, b, c)$ and $Q(4,-12,3)$ is equal to __________.
Explanation:
$\begin{aligned} & \frac{x-4}{12}=\frac{x+6}{4}=\frac{z+2}{6} \\ & \frac{x-4}{\frac{6}{7}}=\frac{y+6}{\frac{2}{7}}=\frac{z+2}{\frac{3}{7}}=21 \\ & \left(21 \times \frac{6}{7}+4, \frac{2}{7} \times 21-6, \frac{3}{7} \times 21-2\right) \\ & =(22,0,7)=(a, b, c) \\ & \therefore \sqrt{324+144+16}=22 \end{aligned}$
Let $\mathrm{Q}$ and $\mathrm{R}$ be the feet of perpendiculars from the point $\mathrm{P}(a, a, a)$ on the lines $x=y, z=1$ and $x=-y, z=-1$ respectively. If $\angle \mathrm{QPR}$ is a right angle, then $12 a^2$ is equal to _________.
Explanation:
$\begin{aligned} & \frac{x}{1}=\frac{y}{1}=\frac{z-1}{0}=r \rightarrow Q(r, r, 1) \\ & \frac{x}{1}=\frac{y}{-1}=\frac{z+1}{0}=k \rightarrow R(k,-k,-1) \\ & \overline{P Q}=(a-r) \hat{i}+(a-r) \hat{j}+(a-1) \hat{k} \\ & a=r+a-r=0 \\ & 2 a=2 r \rightarrow a=r \\ & \overline{P R}=(a-k) i+(a+k) \hat{j}+(a+1) \hat{k} \\ & a-k-a-k=0 \Rightarrow k=0 \\ & A s, P Q \perp P R \\ & (a-r)(a-k)+(a-r)(a+k)+(a-1)(a+1)=0 \\ & a=1 \text { or }-1 \\ & 12 a^2=12 \end{aligned}$
Let a line passing through the point $(-1,2,3)$ intersect the lines $L_1: \frac{x-1}{3}=\frac{y-2}{2}=\frac{z+1}{-2}$ at $M(\alpha, \beta, \gamma)$ and $L_2: \frac{x+2}{-3}=\frac{y-2}{-2}=\frac{z-1}{4}$ at $N(a, b, c)$. Then, the value of $\frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}$ equals __________.
Explanation:
$\begin{aligned} & \mathrm{M}(3 \lambda+1,2 \lambda+2,-2 \lambda-1) \quad \therefore \alpha+\beta+\gamma=3 \lambda+2 \\ & \mathrm{~N}(-3 \mu-2,-2 \mu+2,4 \mu+1) \quad \therefore \mathrm{a}+\mathrm{b}+\mathrm{c}=-\mu+1 \end{aligned}$
$\begin{aligned} & \frac{3 \lambda+2}{-3 \mu-1}=\frac{2 \lambda}{-2 \mu}=\frac{-2 \lambda-4}{4 \mu-2} \\ & 3 \lambda \mu+2 \mu=3 \lambda \mu+\lambda \\ & 2 \mu=\lambda \\ & 2 \lambda \mu-\lambda=\lambda \mu+2 \mu \\ & \lambda \mu=\lambda+2 \mu \\ &\Rightarrow \lambda \mu=2 \lambda \end{aligned}$
$\begin{aligned} \Rightarrow \quad & \mu=2 \quad(\lambda \neq 0) \\ \therefore \quad & \lambda=4 \\ & \alpha+\beta+\gamma=14 \\ & a+b+c=-1 \\ & \frac{(\alpha+\beta+\gamma)^2}{(a+b+c)^2}=196 \end{aligned}$
If $\mathrm{d}_1$ is the shortest distance between the lines $x+1=2 y=-12 z, x=y+2=6 z-6$ and $\mathrm{d}_2$ is the shortest distance between the lines $\frac{x-1}{2}=\frac{y+8}{-7}=\frac{z-4}{5}, \frac{x-1}{2}=\frac{y-2}{1}=\frac{z-6}{-3}$, then the value of $\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}$ is :
Explanation:
$\mathrm{L}_1: \frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}}{1 / 2}=\frac{\mathrm{z}}{-1 / 12}, \mathrm{~L}_2: \frac{\mathrm{x}}{1}=\frac{\mathrm{y}+2}{1}=\frac{\mathrm{z}-1}{\frac{1}{6}}$
$\mathrm{d}_1=$ shortest distance between $\mathrm{L}_1 ~\& \mathrm{~L}_2$
$\begin{aligned} & =\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\left(\vec{b}_1 \times \vec{b}_2\right)\right|}\right| \\ & d_1=2 \\ & \mathrm{~L}_3: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}+8}{-7}=\frac{\mathrm{z}-4}{5}, \mathrm{~L}_4: \frac{\mathrm{x}-1}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-6}{-3} \\ & \mathrm{~d}_2=\text { shortest distance between } \mathrm{L}_3 ~\& \mathrm{~L}_4 \\ & \mathrm{~d}_2=\frac{12}{\sqrt{3}} \text { Hence } \\ & =\frac{32 \sqrt{3} \mathrm{~d}_1}{\mathrm{~d}_2}=\frac{32 \sqrt{3} \times 2}{\frac{12}{\sqrt{3}}}=16 \end{aligned}$
Let O be the origin, and M and $\mathrm{N}$ be the points on the lines $\frac{x-5}{4}=\frac{y-4}{1}=\frac{z-5}{3}$ and $\frac{x+8}{12}=\frac{y+2}{5}=\frac{z+11}{9}$ respectively such that $\mathrm{MN}$ is the shortest distance between the given lines. Then $\overrightarrow{O M} \cdot \overrightarrow{O N}$ is equal to _________.
Explanation:
$\mathrm{L}_1: \frac{\mathrm{x}-5}{4}=\frac{\mathrm{y}-4}{1}=\frac{\mathrm{z}-5}{3}=\lambda \quad \operatorname{drs}(4,1,3)=\mathrm{b}_1$
$\begin{aligned} & \mathrm{M}(4 \lambda+5, \lambda+4,3 \lambda+5) \\ & \mathrm{L}_2: \frac{\mathrm{x}+8}{12}=\frac{\mathrm{y}+2}{5}=\frac{\mathrm{z}+11}{9}=\mu \\ & \mathrm{N}(12 \mu-8,5 \mu-2,9 \mu-11) \\ & \overrightarrow{\mathrm{MN}}=(4 \lambda-12 \mu+13, \lambda-5 \mu+6,3 \lambda-9 \mu+16) \quad \text{..... (1)} \end{aligned}$
Now
$\overrightarrow{\mathrm{b}}_1 \times \overrightarrow{\mathrm{b}}_2=\left|\begin{array}{ccc} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 4 & 1 & 3 \\ 12 & 5 & 9 \end{array}\right|=-6 \hat{\mathrm{i}}+8 \hat{\mathrm{k}} \quad \text{.... (2)}$
Equation (1) and (2)
$\therefore \frac{4 \lambda-12 \mu+13}{-6}=\frac{\lambda-5 \mu+6}{0}=\frac{3 \lambda-9 \mu+16}{8}$
I and II
$\lambda-5 \mu+6=0 \quad \text{.... (3)}$
I and III
$\lambda-3 \mu+4=0 \quad \text{.... (4)}$
Solve (3) and (4) we get
$\begin{aligned} \lambda= & -1, \mu=1 \\ \therefore \quad & \mathrm{M}(1,3,2) \\ & \mathrm{N}(4,3,-2) \\ \therefore \quad & \overrightarrow{\mathrm{OM}} \cdot \overrightarrow{\mathrm{ON}}=4+9-4=9 \end{aligned}$
A line with direction ratios $2,1,2$ meets the lines $x=y+2=z$ and $x+2=2 y=2 z$ respectively at the points $\mathrm{P}$ and $\mathrm{Q}$. If the length of the perpendicular from the point $(1,2,12)$ to the line $\mathrm{PQ}$ is $l$, then $l^2$ is __________.
Explanation:
Let $\mathrm{P}(\mathrm{t}, \mathrm{t}-2, \mathrm{t})$ and $\mathrm{Q}(2 \mathrm{~s}-2, \mathrm{~s}, \mathrm{~s})$
D.R's of PQ are 2, 1, 2
$\begin{aligned} & \frac{2 \mathrm{~s}-2-\mathrm{t}}{2}=\frac{\mathrm{s}-\mathrm{t}+2}{1}=\frac{\mathrm{s}-\mathrm{t}}{2} \\ & \Rightarrow \mathrm{t}=6 \text { and } \mathrm{s}=2 \\ & \Rightarrow \mathrm{P}(6,4,6) \text { and } \mathrm{Q}(2,2,2) \\ & \mathrm{PQ}: \frac{\mathrm{x}-2}{2}=\frac{\mathrm{y}-2}{1}=\frac{\mathrm{z}-2}{2}=\lambda \end{aligned}$
Let $\mathrm{F}(2 \lambda+2, \lambda+2,2 \lambda+2)$
$\mathrm{A}(1,2,12)$
$\overrightarrow{\mathrm{AF}} \cdot \overrightarrow{\mathrm{PQ}}=0$
$\therefore \lambda=2$
So $F(6,4,6)$ and $A F=\sqrt{65}$
The lines $\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point $P$. If the distance of $\mathrm{P}$ from the line $\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}$ is $l$, then $14 l^2$ is equal to __________.
Explanation:
$\begin{aligned} & \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-7}{8}=\lambda \\ & \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} \\ & \Rightarrow \lambda+2=4 \mathrm{k}-3 \\ & -\lambda=3 \mathrm{k}-2 \\ & \Rightarrow \mathrm{k}=1, \lambda=-1 \\ & 8 \lambda+7=\mathrm{k}-2 \\ & \therefore \mathrm{P}=(1,1,-1) \end{aligned}$
Projection of $2 \hat{i}-2 \hat{k}$ on $2 \hat{i}+3 \hat{j}+\hat{k}$ is
$\begin{aligned} & =\frac{4-2}{\sqrt{4+9+1}}=\frac{2}{\sqrt{14}} \\ & \therefore l^2=8-\frac{4}{14}=\frac{108}{14} \\ & \Rightarrow 14 l^2=108 \end{aligned}$
parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point
$\mathrm{A}(8,-1,-19)$ from the plane $\mathrm{P}$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$
is equal to ______________.
Explanation:
$ \begin{aligned} & (2 x+y-z-3)+\lambda(5 x-3 y)+4 z+9)=0 \\\\ & (5 \lambda+2) x+(1-3 \lambda) y+(4 \lambda-1) z+9 \lambda-3=0 \\\\ & \overrightarrow{\mathrm{n}} \cdot \overrightarrow{\mathrm{b}}=0 \text { where } \overrightarrow{\mathrm{b}}(2,4,5) \\\\ & 2(5 \lambda+2)+4(1-3 \lambda)+5(4 \lambda-1)=0 \\\\ & \lambda=-\frac{1}{6} \end{aligned} $
Plane $7 x+9 y-10 z-27=0$
Equation of line $\mathrm{AB}$ is
$ \frac{\mathrm{x}-8}{-3}=\frac{\mathrm{y}+1}{4}=\frac{z+19}{12}=\lambda $
Let $B=(8-3 \lambda,-1+4 \lambda,-19+12 \lambda)$ lies on plane $P$
$ \begin{aligned} & \therefore 7(8-3 \lambda)+9(4 \lambda-1)-10(12 \lambda-19)=27 \\\\ & \lambda=2 \\\\ & \therefore \text { Point } B=(2,7,5) \\\\ & A B=\sqrt{6^2+8^2+24^2}=26 \end{aligned} $
Let the image of the point $\left(\frac{5}{3}, \frac{5}{3}, \frac{8}{3}\right)$ in the plane $x-2 y+z-2=0$ be P. If the distance of the point $Q(6,-2, \alpha), \alpha > 0$, from $\mathrm{P}$ is 13 , then $\alpha$ is equal to ___________.
Explanation:
$ \begin{aligned} & \frac{x-\frac{5}{3}}{1}=\frac{y-\frac{5}{3}}{-2}=\frac{\mathrm{z}-\frac{8}{3}}{1}=\frac{-2\left(1 \times \frac{5}{3}+(-2) \times \frac{8}{3}+1 \times \frac{8}{3}-2\right)}{1^2+2^2+1^2} =\frac{1}{3} \\\\ & \therefore x=2, y=1, \mathrm{z}=3 \\\\ &PQ^2 = 13^2=(6-2)^2+(-2-1)^2+(\alpha-3)^2 \\\\ & \Rightarrow(\alpha-3)^2=144 \Rightarrow \alpha=15(\because \alpha>0) \end{aligned} $
Let the plane $x+3 y-2 z+6=0$ meet the co-ordinate axes at the points A, B, C. If the orthocenter of the triangle $\mathrm{ABC}$ is $\left(\alpha, \beta, \frac{6}{7}\right)$, then $98(\alpha+\beta)^{2}$ is equal to ___________.
Explanation:
$ \begin{aligned} & \overrightarrow{\mathrm{AH}} \cdot \overrightarrow{\mathrm{BC}}=0 \\\\ & \left(\alpha+6, \beta, \frac{6}{7}\right) \cdot(0,2,3)=0 \\\\ & \beta=\frac{-9}{7} \\\\ & \overrightarrow{\mathrm{CH}} \cdot \overrightarrow{\mathrm{AB}}=0 \\\\ & \left(\alpha, \beta, \frac{-15}{7}\right) \cdot(6,-2,0)=0 \\\\ & 6 \alpha-2 \beta=0 \\\\ & \alpha=\frac{-3}{7} \end{aligned} $
$ 98(\alpha+\beta)^2=(98) \frac{(144)}{49}=288 $
Let the line $l: x=\frac{1-y}{-2}=\frac{z-3}{\lambda}, \lambda \in \mathbb{R}$ meet the plane $P: x+2 y+3 z=4$ at the point $(\alpha, \beta, \gamma)$. If the angle between the line $l$ and the plane $P$ is $\cos ^{-1}\left(\sqrt{\frac{5}{14}}\right)$, then $\alpha+2 \beta+6 \gamma$ is equal to ___________.
Explanation:
$ P: x+2 y+3 z=4$
Vector parallel to line : $\langle 1,2, \lambda\rangle=\bar{b}$
Normal vector to plane $P:<1,2,3\rangle=\bar{n}$
Angle between plane and line is $\theta$
Then, $\sin \theta=\frac{<1,2, \lambda>\cdot<1,2,3>}{\sqrt{1^2+2^2+\lambda^2} \cdot \sqrt{1^2+2^2+3^2}}$
$ \Rightarrow \frac{3}{\sqrt{14}}=\frac{1+4+3 \lambda}{\sqrt{\lambda^2+5} \sqrt{14}} \Rightarrow \lambda=\frac{2}{3} $
$ L_1 \equiv \frac{x-0}{3}=\frac{y-1}{6}=\frac{z-3}{2}=\mu $
Any point on line : $(3 \mu, 6 \mu+1,2 \mu+3)$
It lies on P
$ \begin{aligned} & \therefore 3 \mu+12 \mu+2+6 \mu+9=4 \\\\ & \Rightarrow \mu=\frac{-1}{3} \end{aligned} $
Hence, $\alpha=3 \mu=-1, \beta=6 \mu+1=-1, \gamma=2 \mu+3=\frac{7}{3}$
Now, $\alpha+2 \beta+6 \gamma=11$
Let a line $l$ pass through the origin and be perpendicular to the lines
$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$ and
$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$.
If $\mathrm{P}$ is the point of intersection of $l$ and $l_{1}$, and $\mathrm{Q}(\propto, \beta, \gamma)$ is the foot of perpendicular from P on $l_{2}$, then $9(\alpha+\beta+\gamma)$ is equal to _____________.
Explanation:
$l_{1}: \vec{r}=(\hat{\imath}-11 \hat{\jmath}-7 \hat{k})+\lambda(\hat{i}+2 \hat{\jmath}+3 \hat{k}), \lambda \in \mathbb{R}$ and
$l_{2}: \vec{r}=(-\hat{\imath}+\hat{\mathrm{k}})+\mu(2 \hat{\imath}+2 \hat{\jmath}+\hat{\mathrm{k}}), \mu \in \mathbb{R}$.
Let direction ratio of line $l$ be $a, b$ and $c$ Equation of line $l$$ \begin{aligned} \vec{r} & =(0 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}+0 \hat{\mathbf{k}})+\delta(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \\\\ & =\delta(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \end{aligned} $
As, line $l$ is perpendicular to $l_1$ and $l_2$,
$ a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 3 \\ 2 & 2 & 1 \end{array}\right|=-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} $
$\therefore$ Equation of line $l: \vec{r}=\delta(-4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})$
As, $P$ is the intersecting point of $l$ and $l_1$
$ -4 \delta=1+\lambda, 5 \delta=-11+2 \lambda,-2 \delta=-7+3 \lambda $
After solving the above three equation, we get
$ \delta=-1 \text { and } \lambda=3 $
$\therefore$ Co-ordinate of point $P$ is $(4,-5,2)$.
$Q$ is a point on line $l_2$
Let co-ordinate of $Q$ be $(-1+2 \mu, 2 \mu, 1+\mu)$
$ \begin{gathered} \overrightarrow {PQ}=(-5+2 \mu) \hat{\mathbf{i}}+(2 \mu+5) \hat{\mathbf{j}}+(\mu-1) \hat{\mathbf{k}} \\\\ \overrightarrow {PQ} \cdot(2 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})=0 \left[\because \overrightarrow {PQ} \perp l_2\right]\\\\ 2(-5+2 \mu)+2(2 \mu+5)+\mu-1=0 \\\\ 9 \mu-1=0 \Rightarrow \mu=\frac{1}{9} \end{gathered} $
$ \begin{aligned} & \therefore \alpha=-1+\frac{2}{9}=\frac{-7}{9}, \beta=2 \times \frac{1}{9}=\frac{2}{9}, r=1+\frac{1}{9}=\frac{10}{9} \\\\ & \text { Hence, } 9(\alpha+\beta+\gamma)=9\left(-\frac{7}{9}+\frac{2}{9}+\frac{10}{9}\right)=5 \end{aligned} $
Let the foot of perpendicular from the point $\mathrm{A}(4,3,1)$ on the plane $\mathrm{P}: x-y+2 z+3=0$ be N. If B$(5, \alpha, \beta), \alpha, \beta \in \mathbb{Z}$ is a point on plane P such that the area of the triangle ABN is $3 \sqrt{2}$, then $\alpha^{2}+\beta^{2}+\alpha \beta$ is equal to ___________.
Explanation:
We have, equation of plane $P$ is $x-y+2 z+3=0$
Perpendicular distance from $A$ to a plane $P$ i.e. N
$=\frac{|4-3+2(1)+3|}{\sqrt{1^2+(-1)^2+2^2}}$
$B(5, \alpha, \beta)$ lies on the plane $P$, so
$5-\alpha+2 \beta+3=0 \Rightarrow \alpha=2 \beta+8$ .........(i)
Direction ratio of $A N$ is $<1,-1,2>$
Equation of $A N$ is $\frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}$
Co-ordinate of $N: \frac{x-4}{1}=\frac{y-3}{-1}=\frac{z-1}{2}=\frac{-(4-3+2+3)}{1+1+4}$ $=-1$
$\Rightarrow x=3, y=4, z=-1$
$\therefore$ Co-ordinate of $N$ is $(3,4,-1)$
$ \begin{aligned} & \text { Also, } \operatorname{Area of}(\triangle \mathrm{ABN})=3 \sqrt{2} \\\\ & \Rightarrow \frac{1}{2} \times \mathrm{AN} \times \mathrm{BN}=3 \sqrt{2} \\\\ & \Rightarrow \frac{1}{2} \times \sqrt{6} \times \mathrm{BN}=3 \sqrt{2} \Rightarrow \mathrm{BN}=2 \sqrt{3} \\\\ & \text { or } \mathrm{BN}^2=12 \end{aligned} $
$ \begin{aligned} & \Rightarrow(5-3)^2+(\alpha-4)^2+(\beta+1)^2=12 \\\\ & 4+(2 \beta+4)^2+(\beta+1)^2=12(\text { using eqn (i)) } \\\\ & \Rightarrow 4 \beta^2+16+16 \beta+\beta^2+1+2 \beta=8 \\\\ & \Rightarrow 5 \beta^2+18 \beta+9=0 \end{aligned} $
$ \begin{aligned} & \Rightarrow(5 \beta+3)(\beta+3)=0 \\\\ & \Rightarrow \beta=-3 \left[\because \beta \in z \text {, so rejecting } \beta=\frac{-3}{5}\right]\\\\ &\text { and } \alpha=2 \beta+8=2(-3)+8=2\\\\ & \text { also } \alpha^2+\beta^2+\alpha \beta=9+4-6=7 \end{aligned} $
Let $\mathrm{P}_{1}$ be the plane $3 x-y-7 z=11$ and $\mathrm{P}_{2}$ be the plane passing through the points $(2,-1,0),(2,0,-1)$, and $(5,1,1)$. If the foot of the perpendicular drawn from the point $(7,4,-1)$ on the line of intersection of the planes $P_{1}$ and $P_{2}$ is $(\alpha, \beta, \gamma)$, then $\alpha+\beta+\gamma$ is equal to ___________.
Explanation:
$ \begin{aligned} & \left|\begin{array}{ccc} x-5 & y-1 & z-1 \\ 3 & 2 & 1 \\ 3 & 1 & 2 \end{array}\right|=0 \\\\ & \Rightarrow(x-5)(4-1)-(y-1)(6-3)+(z-1)(3-6)=0 \\\\ & \Rightarrow 3 x-15-3 y+3-3 z+3=0 \\\\ & \Rightarrow 3 x-3 y-3 z-9=0 \\\\ & \Rightarrow x-y-z=3 .......(i) \end{aligned} $
Now, direction ratios of line of intersection of $P_1$ and $\mathrm{P}_2$ is
$ \begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -1 \\ 3 & -1 & -7 \end{array}\right| \\\\ & =\hat{i}(7-1)-\hat{j}(-7+3)+\hat{k}(-1+3) \\\\ & =6 \hat{i}+4 \hat{j}+2 \hat{k} \end{aligned} $
At $z=0, x-y=3$ [from (i)]
$ 3 x-y=11 $
On solving, we get
$ x=4 \text { and } y=1 $
So, equation of line is
$ \frac{x-4}{6}=\frac{y-1}{4}=\frac{z-2}{6}=k $
$ \begin{aligned} & \therefore(\alpha, \beta, \gamma)=(6 k+4,4 k+1,2 k) \\\\ & \Rightarrow(6)(\alpha-7)+4(\beta-4)+2(\gamma+1)=0 \\\\ & \Rightarrow 6(6 k+4-7)+4(4 k+1-4)+2(2 k+1)=0 \\\\ & \Rightarrow 36 k-18+16 k-12+4 k+4=0 \\\\ & \Rightarrow 56 k=26 \Rightarrow k=\frac{1}{2} \\\\ & \text { So, } \alpha=7, \beta=3 \text { and } \gamma=1 \\\\ & \therefore \alpha+\beta+\gamma=7+3+1=11 \end{aligned} $
Let $\lambda_{1}, \lambda_{2}$ be the values of $\lambda$ for which the points $\left(\frac{5}{2}, 1, \lambda\right)$ and $(-2,0,1)$ are at equal distance from the plane $2 x+3 y-6 z+7=0$. If $\lambda_{1} > \lambda_{2}$, then the distance of the point $\left(\lambda_{1}-\lambda_{2}, \lambda_{2}, \lambda_{1}\right)$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$ is ____________.
Explanation:
from plane $2 x+3 y-6 z+7=0$
$ \begin{aligned} & \therefore\left|\frac{2\left(\frac{5}{2}\right)+3(1)-6(\lambda)+7}{\sqrt{2^2+3^2+6^2}}\right|=\left|\frac{2(-2)+3(0)-6(1)+7}{\sqrt{2^2+3^2+6^2}}\right| \\\\ & \Rightarrow|5+3-6 \lambda+7|=|-4-6+7| \\\\ & \Rightarrow|15-6 \lambda|=|-3| \\\\ & \Rightarrow 15-6 \lambda= \pm 3 \\\\ & \Rightarrow 15-6 \lambda=3 \text { or } 15-6 \lambda=-3 \\\\ & \Rightarrow 6 \lambda=12 \text { or } 6 \lambda=18 \\\\ & \Rightarrow \lambda=2 \text { or } \lambda=3 \end{aligned} $
$ \begin{aligned} & \because \lambda_1>\lambda_2 \\\\ & \therefore \lambda_1=3 \text { and } \lambda_2=2 \end{aligned} $
So, point will be $(1,2,3)$
Let $\mathrm{M}_0=(1,2,3)$
$M_1$ is the point through which line passes i.e, $(5,1,-7)$
and $\vec{s}=\hat{i}+2 \hat{j}+2 \hat{k}$
$ \therefore \overrightarrow{\mathrm{M}_0 \mathrm{M}_1}=4 \hat{i}-\hat{j}-10 \hat{k} $
Now, required distance $=\left|\frac{\overrightarrow{\mathrm{M}_0 \mathrm{M}_1} \times \vec{s}}{|\vec{s}|}\right|$
$ \begin{aligned} & =\frac{|(4 \hat{i}-\hat{j}-10 \hat{k}) \times(\hat{i}+2 \hat{j}+2 \hat{k})|}{\sqrt{1+4+4}} \\\\ & =\frac{|18 \hat{i}-18 \hat{j}+9 \hat{k}|}{3}=9 \end{aligned} $
If the lines $\frac{x-1}{2}=\frac{2-y}{-3}=\frac{z-3}{\alpha}$ and $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}$ intersect, then the magnitude of the minimum value of $8 \alpha \beta$ is _____________.
Explanation:
$ \begin{aligned} & \frac{x-1}{2} =\frac{2-y}{-3}=\frac{z-3}{\alpha} \\\\ &\Rightarrow \frac{x-1}{2} =\frac{y-2}{3}=\frac{z-3}{\alpha}=\lambda .........(i) \end{aligned} $
Any point on the line (i)
$ x=2 \lambda+1, y=3 \lambda+2, z=\alpha \lambda+3 $
and line $\frac{x-4}{5}=\frac{y-1}{2}=\frac{z}{\beta}=\mu$ ............(ii)
Any point on line (ii)
$ \Rightarrow x=5 \mu+4, y=2 \mu+1, z=\beta \mu $
Since, given lines intersects
$ \begin{aligned} & \therefore 2 \lambda+1=5 \mu+4 ..........(iii)\\\\ & 3 \lambda+2=2 \mu+1 ............(iv)\\\\ & \text { and } \alpha \lambda+3=\beta \mu ..........(iv) \end{aligned} $
On solving (iii) and (iv), we get
$ \lambda=-1, \mu=-1 $
On putting value of $\lambda$ and $\mu$ in (v), we get
$ \begin{array}{cc} & \alpha(-1)+3=-\beta \\\\ &\Rightarrow \alpha=\beta+3 \end{array} $
Now,
$ \begin{aligned} & 8 \alpha \beta=8 (\beta+3)(\beta) \\\\ &= 8\left(\beta^2+3 \beta\right) \\\\ & =8\left(\beta^2+3 \beta+\frac{9}{4}-\frac{9}{4}\right) \\\\ & =8\left\{\left(\beta+\frac{3}{2}\right)^2-\frac{9}{4}\right\} \end{aligned} $
$=8\left(\beta+\frac{3}{2}\right)^2-18$
Here, minimum value $=-18$
$\therefore$ Magnitude of the minimum value of $8 \alpha \beta$ is 18 .
Let the image of the point $\mathrm{P}(1,2,3)$ in the plane $2 x-y+z=9$ be $\mathrm{Q}$. If the coordinates of the point $\mathrm{R}$ are $(6,10,7)$, then the square of the area of the triangle $\mathrm{PQR}$ is _____________.
Explanation:
$ \begin{aligned} & 2 x-y+z=9 \\\\ & \therefore \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1} =\frac{-2(2 \times 1+(-1)(2)+(1)(3)(-9)}{(2)^2+(-1)^2+(1)^2} \end{aligned} $
$ \begin{aligned} & \Rightarrow \frac{x-1}{2}=\frac{y-2}{-1}=\frac{z-3}{1}=\frac{-2(-6)}{6}=2 \\\\ & \Rightarrow x=5, y=0, z=5 \end{aligned} $
Now, $\overrightarrow {PQ} =4 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and
$\overrightarrow{P R}=5 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}$
$\therefore$ Area of the $\triangle P Q R=\frac{1}{2}|\overrightarrow{P Q} \times \overrightarrow{P R}|$
Now, $\overrightarrow{P Q} \times \overrightarrow{P R}$
$ \begin{aligned} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 4 & -2 & 2 \\ 5 & 8 & 4 \end{array}\right| \\\\ & =\hat{\mathbf{i}}(-8-16)-\hat{\mathbf{j}}(16-10)+\hat{\mathbf{k}}(32+10) \\\\ & =-24 \hat{\mathbf{i}}-6 \hat{\mathbf{j}}+42 \hat{\mathbf{k}} \end{aligned} $
$ \begin{aligned} \therefore \frac{1}{2} \mid(-24 \hat{\mathbf{i}} & -6 \hat{\mathbf{j}}+42 \hat{\mathbf{k}})|=|-12 \hat{\mathbf{i}}-3 \hat{\mathbf{j}}+21 \hat{\mathbf{k}} \mid \\\\ & =\sqrt{(-12)^2+(-3)^2+(21)^2} \\\\ & =\sqrt{144+9+441}=\sqrt{594} \end{aligned} $
$\therefore$ The square of the area of the $\triangle P Q R=594$
The point of intersection $\mathrm{C}$ of the plane $8 x+y+2 z=0$ and the line joining the points $\mathrm{A}(-3,-6,1)$ and $\mathrm{B}(2,4,-3)$ divides the line segment $\mathrm{AB}$ internally in the ratio $\mathrm{k}: 1$. If $\mathrm{a}, \mathrm{b}, \mathrm{c}(|\mathrm{a}|,|\mathrm{b}|,|\mathrm{c}|$ are coprime) are the direction ratios of the perpendicular from the point $\mathrm{C}$ on the line $\frac{1-x}{1}=\frac{y+4}{2}=\frac{z+2}{3}$, then $|\mathrm{a}+\mathrm{b}+\mathrm{c}|$ is equal to ___________.
Explanation:
Given line $\mathrm{AB}: \frac{\mathrm{x}-2}{5}=\frac{\mathrm{y}-4}{10}=\frac{\mathrm{z}+3}{-4}=\lambda$
Any point on line $(5 \lambda+2,10 \lambda+4,-4 \lambda-3)$
Point of intersection of line and plane
$ \begin{aligned} & 8(5 \lambda+2)+10 \lambda+4-8 \lambda-6=0 \\\\ & \lambda=-\frac{1}{3} \end{aligned} $
$\mathrm{C}\left(\frac{1}{3}, \frac{2}{3},-\frac{5}{3}\right)$
$\mathrm{L}: \frac{\mathrm{x}-1}{-1}=\frac{\mathrm{y}+4}{2}=\frac{\mathrm{z}+2}{3}=\mu$
$\begin{aligned} & \overrightarrow{\mathrm{CD}}=\left(-\mu+\frac{2}{3}\right) \hat{\mathrm{i}}+\left(2 \mu-\frac{14}{3}\right) \hat{\mathrm{j}}+\left(3 \mu-\frac{1}{3}\right) \hat{\mathrm{k}} \\\\ & \left(-\mu+\frac{2}{3}\right)(-1)+\left(2 \mu-\frac{14}{3}\right) 2+\left(3 \mu-\frac{1}{3}\right) 3=0 \\\\ & \Rightarrow \mu=\frac{11}{14} \\\\ & \overrightarrow{\mathrm{CD}}=\frac{-5}{42}, \frac{-130}{42}, \frac{85}{42} \\\\ & \text { Direction ratios } \rightarrow(-1,-26,17) \\\\ & |\mathrm{a}+\mathrm{b}+\mathrm{c}|=10\end{aligned}$
Let $\alpha x+\beta y+\gamma z=1$ be the equation of a plane passing through the point $(3,-2,5)$ and perpendicular to the line joining the points $(1,2,3)$ and $(-2,3,5)$. Then the value of $\alpha \beta y$ is equal to _____________.
Explanation:
$ a(x-3)+b(y+2)+c(z-5)=0 $
Dr's of plane : $3 \hat{i}-\hat{j}-2 \hat{k}$
$ \begin{aligned} & <3,-1,-2> \\\\ & P: 3(x-3)-1(y+2)-2(z-5)=0 \\\\ & 3 x-9-y-2-2 z+10=0 \\\\ & 3 x-y-2 z=1 \\\\ & \therefore \alpha=3, \beta=-1, \gamma=-2 \\\\ & \Rightarrow \alpha \beta \gamma=6 \end{aligned} $
Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$. Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$. If $\alpha$ is the area of triangle $P Q R$, then $\alpha^{2}$ is equal to __________.
Explanation:
Let $P \equiv\left(2 r_{1}+1,-r_{1}, r_{1}+3\right)$
$P$ lies on $2 x+y+3 z=16$
$\therefore 2\left(2 r_{1}+1\right)+\left(-r_{1}-1\right)+3\left(r_{1}+3\right)=16$
$r_{1}=1$
$P \equiv(3,-2,4)$
$R \equiv(1,-1,-3)$
Let $Q \equiv\left(2 r_{2}+1,-r_{2}-1, r_{2}+3\right)$
$D R$ s of $Q R \equiv\left(2 r_{2}-r_{2} r_{2}+6\right)$
DRs of $L \equiv(2,-1,1)$
$Q R \perp L \Rightarrow 4 r_{2}+r_{2}+r_{2}+6=0$
$r_{2}=-1$
$Q \equiv(-1,0,2)$
$\overrightarrow{Q P} \times \overrightarrow{R P}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 4 & -2 & 2 \\ 2 & -1 & 7\end{array}\right|=-12 \hat{i}-24 \hat{j}+0 \hat{k}$
$\alpha=[P Q R]=\frac{1}{2}|\overrightarrow{Q P} \times \overrightarrow{R P}|=\frac{1}{2} \times 12 \sqrt{5}$
$=6 \sqrt{5}$
$\alpha^{2}=180$
Let $\theta$ be the angle between the planes $P_{1}: \vec{r} \cdot(\hat{i}+\hat{j}+2 \hat{k})=9$ and $P_{2}: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15$. Let $\mathrm{L}$ be the line that meets $P_{2}$ at the point $(4,-2,5)$ and makes an angle $\theta$ with the normal of $P_{2}$. If $\alpha$ is the angle between $\mathrm{L}$ and $P_{2}$, then $\left(\tan ^{2} \theta\right)\left(\cot ^{2} \alpha\right)$ is equal to ____________.
Explanation:
$ \begin{aligned} & P_{2}: \vec{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})=15 \\\\ & \text { then } \cos \theta=\frac{3}{\sqrt{6} \cdot \sqrt{6}}=\frac{1}{2} \\\\ & \begin{aligned} & \therefore \theta=\frac{\pi}{3}, \text { Now } \alpha=\frac{\pi}{2}-\theta \\\\ & \therefore \tan ^{2} \theta \cdot \cot ^{2} \alpha=\tan ^{4} \theta \\\\ &=(\sqrt{3})^{4}=9 \end{aligned} \end{aligned} $
Explanation:
$L:{{x - 2} \over 1} = {{y - 3} \over { - 1}} = {{z - 1} \over { - 1}} = \lambda $
Any point on L can be taken as
$B(\lambda + 2, - \lambda + 3, - \lambda + 1)$
Let $A(5,3,8)$
So, $AB\,.\,(\widehat i - \widehat j - \widehat k) = 0$
$[(\lambda - 3)\widehat i - \lambda \widehat j - (\lambda + 7)\widehat k]\,.\,[\widehat i - \widehat j - \widehat k] = 0$
$\lambda - 3 + \lambda + \lambda + 7 = 0$
$\therefore$ $\lambda = {{ - 4} \over 3}$
$\overrightarrow {AB} = {{13} \over 3}\widehat i + {4 \over 3}\widehat i - {{17} \over 3}\widehat k$
$|\overrightarrow {AB} | = \sqrt {{{169} \over 9} + {{16} \over 9} + {{289} \over 9}} $
$ = {{\sqrt {474} } \over 3} = \alpha $
$3{\alpha ^2} = {{474} \over 9} \times 3 = 158$
If the equation of the plane passing through the point $(1,1,2)$ and perpendicular to the line $x-3 y+ 2 z-1=0=4 x-y+z$ is $\mathrm{A} x+\mathrm{B} y+\mathrm{C} z=1$, then $140(\mathrm{C}-\mathrm{B}+\mathrm{A})$ is equal to ___________.
Explanation:
Line of intersection of the planes $x - 3y + 2z - 1 = 0$ and $4x - y + z = 0$ is normal $(\overrightarrow n )$ to the required plane.
$\overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 3} & 2 \cr 4 & { - 1} & 1 \cr } } \right| = - \widehat i + 7\widehat j + 11\widehat k$
Equation of plane is
$ - x + 7y + 11z = \lambda $
It passes through (1, 1, 2)
$\therefore$ $\lambda = 28$
So, the plane is
$ - x + 7y + 11z = 28$
$ \Rightarrow {{ - 1} \over {28}}x + {7 \over {28}}y + {{11} \over {28}}z = 1$
$A = {{ - 1} \over {28}},B = {7 \over {28}},C = {{11} \over {28}}$
$140(C - B + A) = 15$
If $\lambda_{1} < \lambda_{2}$ are two values of $\lambda$ such that the angle between the planes $P_{1}: \vec{r}(3 \hat{i}-5 \hat{j}+\hat{k})=7$ and $P_{2}: \vec{r} \cdot(\lambda \hat{i}+\hat{j}-3 \hat{k})=9$ is $\sin ^{-1}\left(\frac{2 \sqrt{6}}{5}\right)$, then the square of the length of perpendicular from the point $\left(38 \lambda_{1}, 10 \lambda_{2}, 2\right)$ to the plane $P_{1}$ is ______________.
Explanation:
${P_1}:\overrightarrow r \,.\,(3\widehat i - 5\widehat j + \widehat k) = 7$
${P_2}:\overrightarrow r \,.\,(\lambda \widehat i + \widehat j - 3\widehat k) = 9$
Let angle between P1 and P2 is $\theta$
Then $\cos \theta = {{3\lambda - 5 - 3} \over {\sqrt {35} \sqrt {{\lambda ^2} + 10} }}$
But $\sin \theta = {{2\sqrt 6 } \over 5}$
$\therefore$ ${{{{(3\lambda - 8)}^2}} \over {35({\lambda ^2} + 10)}} = 1 - {{24} \over {25}}$
$ \Rightarrow 5(9{\lambda ^2} + 64 - 48\lambda ) = 7{\lambda ^2} + 70$
$ \Rightarrow 38{\lambda ^2} - 240\lambda + 250 = 0$
$ \Rightarrow 19{\lambda ^2} - 120\lambda + 125 = 0$
$ \Rightarrow (19\lambda - 25)(\lambda - 5) = 0$
$\therefore$ ${\lambda _1} = {{25} \over {19}},{\lambda _2} = 5$
So, point (50, 50, 2)
$\therefore$ $d = {{|150 - 250 + 2 - 7|} \over {\sqrt {35} }} = 315$
Let the equation of the plane P containing the line $x+10=\frac{8-y}{2}=z$ be $ax+by+3z=2(a+b)$ and the distance of the plane $P$ from the point (1, 27, 7) be $c$. Then $a^2+b^2+c^2$ is equal to __________.
Explanation:
$\because$ the plane $a x+b y+3 z=2(a+b)$
$\Rightarrow \mathrm{b}=2 \mathrm{a}$
$\&$ dot product of d.r.'s is zero
$\therefore \mathrm{a}-2 \mathrm{~b}+3=0$ $\therefore \mathrm{a}=1 \& \mathrm{~b}=2$
Distance from $(1,27,7)$ is
$c=\frac{1+54+21-6}{\sqrt{14}}=\frac{70}{\sqrt{14}}=5 \sqrt{14}$
$\therefore \mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}=1+4+350$
$=355$
Let the co-ordinates of one vertex of $\Delta ABC$ be $A(0,2,\alpha)$ and the other two vertices lie on the line ${{x + \alpha } \over 5} = {{y - 1} \over 2} = {{z + 4} \over 3}$. For $\alpha \in \mathbb{Z}$, if the area of $\Delta ABC$ is 21 sq. units and the line segment $BC$ has length $2\sqrt{21}$ units, then $\alpha^2$ is equal to ___________.
Explanation:
A. $\left(\mathrm{O}_{1} 2, \alpha\right)$
$\sqrt{(2 \alpha+5)^{2}+(2 \alpha+20)^{2}+(2 \alpha-5)^{2}}=\sqrt{21} \sqrt{38}$
$\Rightarrow 12 \alpha^{2}+80 \alpha+450=798$
$\Rightarrow 12 \alpha^{2}+80 \alpha-348=0$
$\Rightarrow \alpha=3 \Rightarrow \alpha^{2}=9$
If the shortest distance between the line joining the points (1, 2, 3) and (2, 3, 4), and the line ${{x - 1} \over 2} = {{y + 1} \over { - 1}} = {{z - 2} \over 0}$ is $\alpha$, then 28$\alpha^2$ is equal to ____________.
Explanation:
$L_{1}: \frac{(x-1)}{1}=\frac{(y-2)}{1}=\frac{(2-3)}{1}$
$L_{2}: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-2}{0}$
$\vec{b}_{1}=\hat{i}+\hat{j}+\hat{k}$
$\vec{b}_{2}=2 \hat{i}-\hat{j}+0 \hat{k}$
$ \begin{aligned} & \overrightarrow{a_{1}}-\overrightarrow{a_{2}}=0 \hat{i}-3 \hat{j}-\hat{k} \\\\ & d=\left|\frac{\left(\bar{a}_{1}-\bar{a}_{2}\right) \cdot\left(n_{1} \times n_{2}\right)}{\left|n_{1} \times n_{2}\right|}\right| \\\\ &=\left|\frac{6-3}{\sqrt{9+1+4}}\right|=\frac{3}{\sqrt{14}}=\alpha \\\\ & 28 \alpha^{2}=\frac{28 \times 9}{14}=18 \end{aligned} $
Let the equation of the plane passing through the line $x - 2y - z - 5 = 0 = x + y + 3z - 5$ and parallel to the line $x + y + 2z - 7 = 0 = 2x + 3y + z - 2$ be $ax + by + cz = 65$. Then the distance of the point (a, b, c) from the plane $2x + 2y - z + 16 = 0$ is ____________.
Explanation:
$(x-2 y-z-5)+\lambda(x+y+3 z-5)=0$
$\because \quad$ it's parallel to the line
$x+y+2 z-7=0=2 x+3 y+z-2$
So, vector along the line $\left|\begin{array}{lll}\hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 2 \\ 2 & 3 & 1\end{array}\right|$
$ =-5 \hat{i}+3 \hat{j}+\hat{k} $
$\because $ Plane is parallel to line
$\therefore -5(1+\lambda)+3(-2+\lambda)+1(-1+3 \lambda)=0$
$ \lambda=12 $
So, by (i)
$13 x+10 y+35 z=65$
$\therefore a=13, b=10, c=35$ and $d=\frac{26+20-35+16}{\sqrt{9}}=9$
If the shortest between the lines ${{x + \sqrt 6 } \over 2} = {{y - \sqrt 6 } \over 3} = {{z - \sqrt 6 } \over 4}$ and ${{x - \lambda } \over 3} = {{y - 2\sqrt 6 } \over 4} = {{z + 2\sqrt 6 } \over 5}$ is 6, then the square of sum of all possible values of $\lambda$ is :
Explanation:
$\frac{x+\sqrt{6}}{2}=\frac{y-\sqrt{6}}{3}=\frac{z-\sqrt{6}}{4}$
$\frac{x-\lambda}{3}=\frac{y-2 \sqrt{6}}{4}=\frac{2+2 \sqrt{6}}{5}$ is 6
Vector along line of shortest distance
$=\left|\begin{array}{lll}\mathrm{i} & \mathrm{j} & \mathrm{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|, \Rightarrow-\hat{\mathrm{i}}+2 \hat{\mathrm{j}}-\mathrm{k}$ (its magnitude is $\sqrt{6}$ )
Now $\frac{1}{\sqrt{6}}\left|\begin{array}{ccc}\sqrt{6}+\lambda & \sqrt{6} & -3 \sqrt{6} \\ 2 & 3 & 4 \\ 3 & 4 & 5\end{array}\right|=\pm 6$
$\Rightarrow \lambda=-2 \sqrt{6}, 10 \sqrt{6}$
So, square of sum of these values
$ =(10 \sqrt{6}-2 \sqrt{6})^2=(8 \sqrt{6})^2=384 $
The shortest distance between the lines ${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$ and ${{x - 6} \over 3} = {{1 - y} \over 2} = {{z + 8} \over 0}$ is equal to ________
Explanation:
For ${L_1}:$${{x - 2} \over 3} = {{y + 1} \over 2} = {{z - 6} \over 2}$
$\therefore$ Point on line is $A(2,-1,6)$
Parallel vector to this line is,
$\overrightarrow p = 3\widehat i + 2\widehat j + 2\widehat k$
For ${L_2}:$${{x - 6} \over 3} = {{y - 1} \over -2} = {{z + 8} \over 0}$
$\therefore$ Point on line is $B(6,1,-8)$
Parallel vector to this line is,
$\overrightarrow q = 3\widehat i - 2\widehat j + 0\widehat k$
Now shortest distance between this two lines is,
$ = \left| {{{\overrightarrow {AB} \,.\,\left( {\overrightarrow p \times \overrightarrow q } \right)} \over {\left| {\overrightarrow p \times \overrightarrow q } \right|}}} \right|$
Now, $\overrightarrow {AB} = 4 \widehat i + 2\widehat j -14\widehat k$
$\overrightarrow p \times \overrightarrow q = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 2 & 2 \cr 3 & -2 & 0 \cr } } \right|$
$ = 4\widehat i +6\widehat j -12\widehat k$
$\therefore$ $|\vec{p} \times \vec{q}|=\sqrt{16+36+144}=\sqrt{196}=14$
and $\overrightarrow {AB} \,.\,\left( {\overrightarrow p \times \overrightarrow q } \right) = + 16 + 12 + 168 = +196$
$\therefore$ Shortest distance $ = \left| {{{ 196} \over {14 }}} \right| = 14$
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 $ar(\Delta PQR) = 153$.