3D Geometry
The plane $2x-y+z=4$ intersects the line segment joining the points A ($a,-2,4)$ and B ($2,b,-3)$ at the point C in the ratio 2 : 1 and the distance of the point C from the origin is $\sqrt5$. If $ab < 0$ and P is the point $(a-b,b,2b-a)$ then CP$^2$ is equal to :
If the lines ${{x - 1} \over 1} = {{y - 2} \over 2} = {{z + 3} \over 1}$ and ${{x - a} \over 2} = {{y + 2} \over 3} = {{z - 3} \over 1}$ intersect at the point P, then the distance of the point P from the plane $z = a$ is :
The shortest distance between the lines ${{x - 1} \over 2} = {{y + 8} \over -7} = {{z - 4} \over 5}$ and ${{x - 1} \over 2} = {{y - 2} \over 1} = {{z - 6} \over { - 3}}$ is :
The foot of perpendicular of the point (2, 0, 5) on the line ${{x + 1} \over 2} = {{y - 1} \over 5} = {{z + 1} \over { - 1}}$ is ($\alpha,\beta,\gamma$). Then, which of the following is NOT correct?
The shortest distance between the lines $x+1=2y=-12z$ and $x=y+2=6z-6$ is :
The distance of the point P(4, 6, $-$2) from the line passing through the point ($-$3, 2, 3) and parallel to a line with direction ratios 3, 3, $-$1 is equal to :
Consider the lines $L_1$ and $L_2$ given by
${L_1}:{{x - 1} \over 2} = {{y - 3} \over 1} = {{z - 2} \over 2}$
${L_2}:{{x - 2} \over 1} = {{y - 2} \over 2} = {{z - 3} \over 3}$.
A line $L_3$ having direction ratios 1, $-$1, $-$2, intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively. Then the length of line segment $PQ$ is
If the foot of the perpendicular drawn from (1, 9, 7) to the line passing through the point (3, 2, 1) and parallel to the planes $x+2y+z=0$ and $3y-z=3$ is ($\alpha,\beta,\gamma$), then $\alpha+\beta+\gamma$ is equal to :
Let the plane containing the line of intersection of the planes
P1 : $x+(\lambda+4)y+z=1$ and
P2 : $2x+y+z=2$
pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of
the point (2$\lambda,\lambda,-\lambda$) from the plane P2 is :
The distance of the point (7, $-$3, $-$4) from the plane passing through the points (2, $-$3, 1), ($-$1, 1, $-$2) and (3, $-$4, 2) is :
The distance of the point ($-1,9,-16$) from the plane
$2x+3y-z=5$ measured parallel to the line
${{x + 4} \over 3} = {{2 - y} \over 4} = {{z - 3} \over {12}}$ is :
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$
Match each entry in List-I to the correct entries in List-II.
| List - I | List - II |
|---|---|
| (P) The value of $d\left(H_0\right)$ is | (1) $\sqrt{3}$ |
| (Q) The distance of the point $(0,1,2)$ from $H_0$ is | (2) $\frac{1}{\sqrt{3}}$ |
| (R) The distance of origin from $H_0$ is | (3) 0 |
| (S) The distance of origin from the point of intersection of planes $y=z, x=1$ and $H_0$ is | (4) $\sqrt{2}$ |
| (5) $\frac{1}{\sqrt{2}}$ |
The correct option is:
Let $Q$ be the foot of perpendicular drawn from the point $P(1,2,3)$ to the plane $x+2 y+z=14$. If $R$ is a point on the plane such that $\angle P R Q=60^{\circ}$, then the area of $\triangle P Q R$ is equal to :
If $(2,3,9),(5,2,1),(1, \lambda, 8)$ and $(\lambda, 2,3)$ are coplanar, then the product of all possible values of $\lambda$ is:
If the foot of the perpendicular from the point $\mathrm{A}(-1,4,3)$ on the plane $\mathrm{P}: 2 x+\mathrm{m} y+\mathrm{n} z=4$, is $\left(-2, \frac{7}{2}, \frac{3}{2}\right)$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios $3,-1,-4$, is equal to :
Let the lines
$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ and
$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$ be coplanar
and $\mathrm{P}$ be the plane containing these two lines.
Then which of the following points does NOT lie on P?
A plane P is parallel to two lines whose direction ratios are $-2,1,-3$ and $-1,2,-2$ and it contains the point $(2,2,-2)$. Let P intersect the co-ordinate axes at the points $\mathrm{A}, \mathrm{B}, \mathrm{C}$ making the intercepts $\alpha, \beta, \gamma$. If $\mathrm{V}$ is the volume of the tetrahedron $\mathrm{OABC}$, where $\mathrm{O}$ is the origin, and $\mathrm{p}=\alpha+\beta+\gamma$, then the ordered pair $(\mathrm{V}, \mathrm{p})$ is equal to :
The foot of the perpendicular from a point on the circle $x^{2}+y^{2}=1, z=0$ to the plane $2 x+3 y+z=6$ lies on which one of the following curves?
If the length of the perpendicular drawn from the point $P(a, 4,2)$, a $>0$ on the line $\frac{x+1}{2}=\frac{y-3}{3}=\frac{z-1}{-1}$ is $2 \sqrt{6}$ units and $Q\left(\alpha_{1}, \alpha_{2}, \alpha_{3}\right)$ is the image of the point P in this line, then $\mathrm{a}+\sum\limits_{i=1}^{3} \alpha_{i}$ is equal to :
If the line of intersection of the planes $a x+b y=3$ and $a x+b y+c z=0$, a $>0$ makes an angle $30^{\circ}$ with the plane $y-z+2=0$, then the direction cosines of the line are :
If the plane $P$ passes through the intersection of two mutually perpendicular planes $2 x+k y-5 z=1$ and $3 k x-k y+z=5, k<3$ and intercepts a unit length on positive $x$-axis, then the intercept made by the plane $P$ on the $y$-axis is :
A vector $\vec{a}$ is parallel to the line of intersection of the plane determined by the vectors $\hat{i}, \hat{i}+\hat{j}$ and the plane determined by the vectors $\hat{i}-\hat{j}, \hat{i}+\hat{k}$. The obtuse angle between $\vec{a}$ and the vector $\vec{b}=\hat{i}-2 \hat{j}+2 \hat{k}$ is :
The length of the perpendicular from the point $(1,-2,5)$ on the line passing through $(1,2,4)$ and parallel to the line $x+y-z=0=x-2 y+3 z-5$ is :
A plane $E$ is perpendicular to the two planes $2 x-2 y+z=0$ and $x-y+2 z=4$, and passes through the point $P(1,-1,1)$. If the distance of the plane $E$ from the point $Q(a, a, 2)$ is $3 \sqrt{2}$, then $(P Q)^{2}$ is equal to :
The shortest distance between the lines $\frac{x+7}{-6}=\frac{y-6}{7}=z$ and $\frac{7-x}{2}=y-2=z-6$ is :
Let $\mathrm{P}$ be the plane containing the straight line $\frac{x-3}{9}=\frac{y+4}{-1}=\frac{z-7}{-5}$ and perpendicular to the plane containing the straight lines $\frac{x}{2}=\frac{y}{3}=\frac{z}{5}$ and $\frac{x}{3}=\frac{y}{7}=\frac{z}{8}$. If $\mathrm{d}$ is the distance of $\mathrm{P}$ from the point $(2,-5,11)$, then $\mathrm{d}^{2}$ is equal to :
The distance of the point (3, 2, $-$1) from the plane $3x - y + 4z + 1 = 0$ along the line ${{2 - x} \over 2} = {{y - 3} \over 2} = {{z + 1} \over 1}$ is equal to :
