Three Dimensional Geometry

$\operatorname{Let} \mathbf{b}_1=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{b}_2=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$

Normal vector to plane

$ \begin{aligned} & \mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 1 \\ 4 & 5 & -3 \end{array}\right| \\ = & \hat{\mathbf{i}}(-6-5)-\hat{\mathbf{j}}(-3-4)+\hat{\mathbf{k}}(5-8) \\ = & -11 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

∴ Direction cosines of normal to the plane

$ \begin{aligned} & =\frac{-11}{\sqrt{(-11)^2+7^2+(-3)^2}}, \\ & \frac{7}{\sqrt{(-11)^2+7^2+(-3)^2}}, \\ & \frac{-3}{\sqrt{(-11)^2+7^2+(-3)^2}} \\ & =\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}} \end{aligned} $

We know that, foot of perpendicular from a point $\left(x_1, y_1, z_1\right)$ on the plane $a x+b y+c z+d=0$ is $(x, y, z)$ where

$ \begin{aligned} \frac{x-x_1}{a} & =\frac{y-y_1}{b}=\frac{z-z_1}{c} \\ & =\frac{-\left(a x_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2} \end{aligned} $

Since, the foot of the perpendicular drawn from the point $(1,1,1)$ to the plane $\pi_1$ is $(1,3,5)$

$ \begin{aligned} & \therefore \frac{1-1}{a}=\frac{3-1}{b}=\frac{5-1}{c}=\frac{-(a+b+c+d)}{a^2+b^2+c^2} \\ & \Rightarrow \quad \frac{0}{a}=\frac{2}{b}=\frac{4}{c}=\frac{-(a+b+c+d)}{a^2+b^2+c^2} \\ & \Rightarrow \quad a=0, b=2 k, c=4 k . \end{aligned} $

Equation of plane $\pi_1$ through point $(1,3,5)$ and having directions of normal to plane are $a, b, c$ is

$ \begin{array}{rlr} & 0(x-1)+2 k(y-3)+4 k(z-5) =0 \\ \Rightarrow & (y-3)+2(z-5) =0 \\ \Rightarrow & y+2 z-13=0 & \ldots(\mathrm{i}) \tag{i} \end{array} $

And equation of plane $\pi_2$ is

$ \begin{aligned} & \quad\left|\begin{array}{ccc} x-2 & y-2 & z+1 \\ 3-2 & 4-2 & 2+1 \\ 3-2 & 3-2 & 0+1 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x-2 & y-2 & z+1 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{array}\right|=0 \\ & \Rightarrow(x-2)(2-3)-(y-2)(1-3) \\ & \quad+(z+1)(1-2)=0 \\ & \Rightarrow(x-2)+(y-2)(2)+(z+1)=0 \\ & \Rightarrow x-2 y+z-3=0 \end{aligned} $

∴ Angle between plane $\pi_1$ and $\pi_2$

$ \begin{aligned} & \theta=\cos ^{-1}\left(\left|\frac{0 \times 0+(1) \times(-2)+(2) \times(1)}{\sqrt{0+1+4} \sqrt{1+4+1}}\right|\right) \\ & \Rightarrow \quad \theta=\cos ^{-1}(0)=\frac{\pi}{2} \end{aligned} $

Given,

Position vector of a point $P$ is

$ (2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) $

Equation of plane,

$ (\mathbf{r}-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(\mathbf{a} \times \mathbf{b})=0 $

∴ Normal of plane $=\mathbf{a} \times \mathbf{b}$

Equation of line through

$\mathbf{P}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ and normal to $\mathbf{b}$ and lying on the plane is

$ \begin{aligned} & \mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\lambda((\mathbf{a} \times \mathbf{b}) \times \mathbf{b}) \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda((\mathbf{b} \cdot \mathbf{a}) \mathbf{b}-(\mathbf{b} . \mathbf{b}) \mathbf{a}) \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+ \\ & \lambda[(-5)(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-6(-\hat{\mathbf{i}}-2 \hat{\mathbf{k}})] \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \end{aligned} $

We have,

$ \begin{aligned} & L: \mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\ & P: \mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5 \end{aligned} $

$ \begin{aligned}Now, (\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}}) =1-5+4=0 \end{aligned} $

∴ Line is parallel to plane.

∴ Shortest distance between line and plane $=$ Distance of point $(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ from planer $\cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})-5=0$

$ \begin{aligned} & =\left|\frac{(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})-5}{|\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}}|}\right| \\ & =\left|\frac{2-10+3-5}{\sqrt{1+25+1}}\right|=\frac{10}{\sqrt{27}}=\frac{10}{3 \sqrt{3}} \end{aligned} $

We have, $A(-1,0,7), B(3,2, t)$ and $C(5, k,-2)$ are collinear.

$ \begin{aligned} & \therefore \quad \frac{3+1}{5-3}=\frac{2-0}{k-2}=\frac{t-7}{-2-t} \\ & \Rightarrow \quad 2=\frac{2}{k-2}=\frac{t-7}{-2-t} \Rightarrow k=3, \\ & t=1 \\ & \therefore \quad B=(3,2,1), C=(5,3,-2) \text { and } \\ & P(1,1,4) \end{aligned} $

Let $P$ divides $B C$ in the ratio $\lambda: 1$

$ \begin{array}{ll} \therefore & \frac{5 \lambda+3}{\lambda+1}=1 \\ \Rightarrow & 5 \lambda+3=\lambda+1 \Rightarrow 4 \lambda=-2 \end{array} $

$ \begin{aligned} &\Rightarrow \quad \lambda=-1 / 2\\ &\text { ∴ } \text { Required ratio }=-1: 2 \end{aligned} $

We have,

$ \begin{aligned} & 3 l+m+5 n=0 \text { and } 6 m n-2 n l+5 l m=0 \\ & \Rightarrow \quad 6 n(-3 l-5 n)-2 n l+5 l(-3 l-5 n)=0 \\ & \Rightarrow \quad-18 n l-30 n^2-2 n l-15 l^2-25 n l=0 \\ & \Rightarrow \quad-15 l^2-30 n^2-45 n l=0 \\ & \Rightarrow \quad l^2+3 n l+2 n^2=0 \\ & \Rightarrow \quad(l+n)(l+2 n)=0 \\ & \Rightarrow \quad l=-n,-2 n \end{aligned} $

When $l=-n$

$ -3 n+m+5 n=0 \Rightarrow m=-2 n $

So, d'r are

$ -n,-2 n, n \text { or } 1,2,-1 $

When $l=-2 n$

$ -6 n+m+5 n=0 \Rightarrow m=n $

So, d'r are

$ \begin{aligned} & -2 n, n, n \\ & -2,1,1 \\ \therefore \cos \theta & =\frac{1 \times(-2)+2 \times 1+(-1) \times 1}{\sqrt{1+4+1} \sqrt{4+1+1}}=\frac{-1}{6} \\ \therefore|\cos \theta| & =\frac{1}{6} \end{aligned} $

Equation of plane $O A B$ is given by,

$ \begin{aligned} &\left|\begin{array}{ccc} x-0 & y-0 & z-0 \\ 1-0 & 2-0 & 1-0 \\ 2-0 & 1-0 & 3-0 \end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{lll} x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right|=0 \\ & x(6-1)-y(3-2)+z(1-4)=0 \\ & 5 x-y-3 z=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

Equation of plane $A B C$ is given by,

$ \begin{aligned} \left|\begin{array}{ccc} x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1 \end{array}\right| & =0 \\ \left|\begin{array}{ccc} x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{array}\right| & =0 \end{aligned} $

$ \begin{gathered} (x-1)(-1+2)-(y-2)(1+4) \\ +(z-1)(-1-2)=0 \\ x-1-5 y+10-3 z+3=0 \\ x-5 y-3 z+12=0 \quad \ldots(\mathrm{ii}) \end{gathered} $

Then angle between planes represented by Eqs. (i) and (ii) is given by,

$ \begin{aligned} \cos \theta & =\frac{(5)(1)+(-1)(-5)+(-3)(-3)}{\sqrt{25+1+9} \sqrt{1+25+9}} \\ & =\frac{19}{35} \end{aligned} $