3D Geometry

The D.C. of the line are ${1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }},{1 \over {\sqrt 3 }}$.

We find that any point on the line at a distance $t$ from $P(2, - 1,2)$ is

$\left( {2 + {t \over {\sqrt 3 }}, - 1 + {t \over {\sqrt 3 }},2 + {t \over {\sqrt 3 }}} \right)$ which lies on $2x + y + z = 9 $

$\Rightarrow t = \sqrt 3 $.

We see that any point on the line is given as

$Q \equiv \{ (1 - 3\mu ),(\mu - 1),(5\mu + 2)\} $

$\overrightarrow {PQ} \equiv \{ - 3\mu - 2,\mu - 3,5\mu - 4\} $

Now, we have

$1( - 3\mu - 2) - 4(\mu - 3) + 3(5\mu - 4) = 0$

$ \Rightarrow - 3\mu - 2 - 4\mu + 12 + 15\mu - 12 = 0$

That is, $8\mu = 2 \Rightarrow \mu = {1 \over 4}$

The equation of the plane passing through the point ($-1,-2,-1$) and whose normal is perpendicular to both the given lines L$_1$ and L$_2$ written as

$(x + 1) + 7(y + 2) - 5(z + 1) = 0$

i.e., $x + 7y - 5z + 10 = 0$

The distance of the point (1, 1, 1) from the plane

$ = \left| {{{1 + 7 - 5 + 10} \over {\sqrt {1 + 49 + 25} }}} \right|$

$ = {{13} \over {\sqrt {75} }}$ units

We have,

${P_1}:x - y + z = - 1$

${P_2}:x + y - z = - 1$

${P_3}:x - 3y + 3z = 2$

Let dr's of the lines of L$_1$, L$_2$ and L$_3$ are ${a_1},{b_1},{c_1}:{a_2},{b_2},{c_2}$ and ${a_3},{b_3},{c_3}$ respectively.

Therefore,

${a_1} + {b_1} - {c_1} = 0$

${a_1} - 3{b_1} + 3{c_1} = 0$

$ \Rightarrow {{{a_1}} \over 0} = {{{b_1}} \over { - 4}} = {{{c_1}} \over { - 4}}$

${a_1},{b_1},{c_1} = 0,1,1$

again ${a_2} - {b_2} + {c_2} = 0$

${a_2} - 3{b_2} + 3{c_2} = 0$

${{{a_2}} \over 0} = {{{b_2}} \over { - 2}} = {{{c_2}} \over { - 2}}$

${a_2},{b_2},{c_2} = 0,1,1$

Again ${a_3} - {b_3} + {c_3} = 0$

${a_3} + {b_3} - {c_3} = 0$

$ \Rightarrow {{{a_3}} \over 0} = {{{b_3}} \over 2} = {{{c_3}} \over 2} \Rightarrow {a_3},{b_3},{c_3} = 0,1,1$

L$_1$, L$_2$ and L$_3$ are parallel.

The line of intersection of given plane is

$3 x-6 y-2 z-15=0$

$2 x+y-2 z=5 \Rightarrow 2 x+y-2 z-5=0$

For $z=0$, we obtain

$x=3$ and $y=-1$

$\therefore$ line passes through $(3,-1,0)$

Let $a, b, c$ be the d'rs of line of intersection, then $3 a-6 b-2 c=0$ and $2 a+b-2 c=0$

Solving the above equation using cross product method

$a: b: c=14: 2: 15$

$\therefore$ Equation of line is

$\frac{x-3}{14}=\frac{y+1}{2}=\frac{z}{15}=t$

Whose parametric form is $x=3+14 t, y=-1+2 t, z=15 t$

$\therefore$ Statement $\mathrm{I}$ is false

Since $\mathrm{dr}$'s of line intersection of given planes are $14,2,15$

$\therefore 14 \hat{i}+2 \hat{j}+15 \hat{k}$ is parallel to this line.

Statement 2 is true.

(i) We need to find the value of $\tan t$ where $t = \sum\limits_{i=1}^\infty \tan^{-1}\left(\frac{1}{2i^2}\right)$.

We rewrite the sum: $\tan^{-1}\left(\frac{1}{2i^2}\right)$ can be written as $\tan^{-1}\left(\frac{2}{(2i+1)-(2i-1)}\right)$.

Using the formula: $\tan^{-1} A - \tan^{-1} B = \tan^{-1}\left(\frac{A-B}{1+AB}\right)$, we turn the sum into a telescoping series: $\sum\limits_{i=1}^\infty [\tan^{-1}(2i+1) - \tan^{-1}(2i-1)]$.

When you write out the terms, most of them cancel out. What is left is: $t = \lim_{n \to \infty} [\tan^{-1}(2n+1) - \tan^{-1}(1)]$.

As $n$ becomes very large, $\tan^{-1}(2n+1)$ approaches $\frac{\pi}{2}$, so: $t = \frac{\pi}{2} - \tan^{-1}(1)$.

We know that $\tan^{-1}(1) = \frac{\pi}{4}$, so: $t = \frac{\pi}{2} - \frac{\pi}{4} = \frac{\pi}{4}$.

So, $\tan t = \tan \left(\frac{\pi}{4}\right) = 1$.

(ii) The sides $a, b, c$ of a triangle are in arithmetic progression (AP). We have: $\cos \theta_1 = \frac{a}{b+c}, \cos \theta_2 = \frac{b}{a+c}, \cos \theta_3 = \frac{c}{a+b}$.

We know: $\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}$.

So, $\cos \theta_1 = \frac{1-\tan^2\left(\frac{\theta_1}{2}\right)}{1+\tan^2\left(\frac{\theta_1}{2}\right)} = \frac{a}{b+c}$.

Using componendo and dividendo, we get: $\tan^2\left(\frac{\theta_1}{2}\right) = \frac{b+c-a}{a+b+c}$.

Similarly, $\tan^2\left(\frac{\theta_3}{2}\right) = \frac{a+b-c}{a+b+c}$.

So if we add them: $\tan^2\left(\frac{\theta_1}{2}\right) + \tan^2\left(\frac{\theta_3}{2}\right) = \frac{b+c-a + a+b-c}{a+b+c} = \frac{2b}{a+b+c}$.

If $a,b,c$ are in AP, then $a + c = 2b$, so $a + b + c = 3b$.

Therefore, $\tan^2\left(\frac{\theta_1}{2}\right) + \tan^2\left(\frac{\theta_3}{2}\right) = \frac{2b}{3b} = \frac{2}{3}$.

(iii) The line passes through (0, 1, 0) and is perpendicular to the plane $x + 2y + 2z = 0$.

The normal vector to the plane is (1, 2, 2). So, the direction vector for the line is (1, 2, 2).

The equation of the line is: $\frac{x-0}{1} = \frac{y-1}{2} = \frac{z-0}{2} = r$.

Let $P(r, 2r+1, 2r)$ be a point on the line. To find the value of $r$ where the vector from the origin (0, 0, 0) to the point on the line is perpendicular to the direction vector (1, 2, 2), set up:

$(r, 2r+1, 2r) \cdot (1, 2, 2) = 0$

That is: $r \times 1 + (2r+1)\times 2 + 2r \times 2 = 0$

So: $r + 4r + 2 + 4r = 0$ $9r + 2 = 0$ $r = -\frac{2}{9}$

So the point is: $\left(-\frac{2}{9}, \frac{5}{9}, -\frac{4}{9}\right)$

The distance from the origin to this point is: $\sqrt{\left(-\frac{2}{9}\right)^2 + \left(\frac{5}{9}\right)^2 + \left(-\frac{4}{9}\right)^2} = \sqrt{\frac{4+25+16}{81}} = \sqrt{\frac{45}{81}} = \frac{\sqrt{5}}{3}$

We know that equation of plane passes through ( $1,-2,1$ )

$ a(x-1)+b(y+2)+c(z+1)=0 $

above plane is $\perp^r$ on $2 x-2 y+z=0$

$ \therefore \quad 2 a-2 b+c=0 $

Similarly   $ a-b+2 c=0 $

(as it is $\perp^r$ on $x-y+2 z=0$ )

$ \frac{a}{1}=\frac{b}{1}=\frac{c}{0} $

$\therefore \quad$ equation of plane is $x+y+1=0$

Distance of the plane from the point ( $1,2,2$ )

$ \begin{aligned} & =\frac{1+2+1}{\sqrt{1^2+1^2}} \\ & =2 \sqrt{2} \end{aligned} $

Let vector AO be parallel to line of intersection of planes $P_1$ and $P_2$ through origin.

$\therefore$ Normal to plane $\mathrm{P}_1$ is

$ \vec{a}_1=(2 \hat{j}+3 \hat{k}) \times(4 \hat{j}-3 \hat{k})=-18 \hat{i} $

Normal to plane $P_2$ is

$ \begin{aligned} \vec{a}_2 & =(\hat{j}-\hat{k}) \times(3 \hat{i}+3 \hat{j}) \\ & =3 \hat{i}-3 \hat{j}-3 \hat{k} \end{aligned} $

$\therefore \quad \overrightarrow{\mathrm{OA}}$ is parallel to $\pm\left(\overrightarrow{n_1} \times \overrightarrow{n_2}\right)$

$ =54 \hat{j}-54 \hat{k} $

Angle between $54(\hat{j}-\hat{k})$ and $(2 \hat{i}+\hat{j}-2 \hat{k})$

is $\cos \theta= \pm\left(\frac{54+108}{3 \times 54-\sqrt{2}}\right)= \pm \frac{1}{\sqrt{2}}$

$ \theta=\frac{\pi}{4}, \frac{3 \pi}{4} $

Required equation of plane is

$62 x+29 y+19 z-105=0$

The required line is the intersection of planes

$2 x-y+z-3=0$ ..... (i)

and $3 x+y+z-5=0$ .... (ii)

Thus, the plane containing required line is

$(2 x-y+z-3)+\lambda(3 x+y+z-5)=0$

$\Rightarrow(2+3 \lambda) x+(\lambda-1) y+(\lambda+1) z+(-5 \lambda-3)=0$

As its distance from point $(2,1,-1)$ is $\frac{1}{\sqrt{6}}$

$\Rightarrow\left|\frac{(3 \lambda+2) 2+(\lambda-1) 1-(\lambda+1)+(-5 \lambda-3)}{\sqrt{11 \lambda^{2}+12 \lambda+6}}\right|=\frac{1}{\sqrt{6}}$

$\Rightarrow \frac{6 \lambda+4+\lambda-1-\lambda-1-5 \lambda-3}{\sqrt{11 \lambda^{2}+12 \lambda+6}}=\frac{1}{\sqrt{6}}$

$\Rightarrow 6(\lambda-1)^{2}=11 \lambda^{2}+12 \lambda+6$ (Squaring both side)

$\Rightarrow 6\left(\lambda^{2}+1-2 \lambda\right)=11 \lambda^{2}+12 \lambda+6$

$\Rightarrow \quad 5 \lambda^{2}+24 \lambda=0$

$\Rightarrow \quad \lambda(5 \lambda+24)=0$

$\Rightarrow \quad \lambda=0, \frac{-24}{5}$

Required planes are

$\begin{aligned} & \begin{array}{l} 2 x-y+z-3=0 \text { and } \\ 62 x+29 y+19 z-105=0 \\ \text { Or } \\ 3 x+y+z-5=0 \\ 62 x+29 y+19 z-105=0 \end{array} \\ \end{aligned}$