Three Dimensional Geometry

Given equation of the line is

\( \frac{x-1}{2}=-y=\frac{2 z+1}{6} \)
\( \Rightarrow \frac{x-1}{2}=\frac{y-0}{-1}=\frac{2\left(z+\frac{1}{2}\right)}{6} \)
\( \Rightarrow \frac{x-1}{2}=\frac{y-0}{-1}=\frac{z+\frac{1}{2}}{3}\)

The direction ratios of the line are \((2,-1,3)\).

Let \(\frac{x-1}{2}=-y=\frac{2 z+1}{6}=\lambda\)

\(x=1+2 \lambda \Rightarrow \frac{d x}{d \lambda}=2\)

\(-y=\lambda \Rightarrow y=-\lambda\)

\(\Rightarrow \frac{d y}{d \lambda}=-1\)

\(\frac{2 z+1}{6}=\lambda \Rightarrow 2 z=6 \lambda-1\)

\(\Rightarrow z=3 \lambda-\frac{1}{2} \Rightarrow \frac{d z}{d \lambda}=3\)

Perpendicular to \(x, y, z \Rightarrow \alpha=\beta=\gamma=90^{\circ}\).

$\Rightarrow \cos \alpha =\cos \beta =\cos \gamma =0.$

But for any line: \(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\).
Here \(0+0+0 \neq 1 \Rightarrow\) impossible.
Thus Assertion is true.
Reason states the required identity and explains the impossibility.

Direction of line :

Position vector of point :

${\overrightarrow{r}}_0=\left(4,-3,7\right)$

Equation of line :

$\overrightarrow{r}={\overrightarrow{r}}_0+t\overrightarrow{d}$

Parametric form :

$\Rightarrow x=4+3t, y=-3+t, z=7+2t$

Distance from \(y\)-axis \(=\) shortest distance to line \((0, t, 0)\)

Choose point on \(y\)-axis \(Q=(0, b, 0)\)

\(\overrightarrow{Q P}=(a, 0, c) \)

Distance \(=|\overrightarrow{Q P}|=\sqrt{a^2+0^2+c^2}=\sqrt{a^2+c^2}\)

Let $P\left(a,b,c\right)$

Choose point on y-axis $Q=\left(0,b,0\right)$

$\overrightarrow{QP}=\left(a,0,c\right)$

Distance$=\overrightarrow{QP}=$ $\sqrt{a^2+0^2+c^2}=\sqrt{a^2+c^2}$

Vector equation for \(x\)-axis is \(\vec{r}=\lambda \hat{i}\) where \(x=\lambda\).

Let a point \(P(\lambda, 0,0)\) is on the \(x\)-axis, foot of perpendicular to the given point \(A(0,1,2)\).

then \(\overrightarrow{A P}\) is perpendicular to \(\lambda \hat{i}\).

\( \overrightarrow{A P} \cdot \lambda \hat{i}=0 \) where \( \overrightarrow{A P}=\lambda \hat{i}-\hat{j}-2 \hat{k} \)
\( (\lambda \hat{i}-\hat{j}-2 \hat{k}) \cdot(\lambda \hat{i}+0 \hat{j}+0 \hat{k})=0 \)
\(\Rightarrow \lambda^2=0\)

\(\Rightarrow \lambda=0\),

So the foot of the perpendicular drawn from the point \((0,1,2)\) is \((0,0,0)\).

Given, \(\alpha=30^{\circ}\), \(\beta=120^{\circ}\)

So, \(l=\cos \alpha=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)

\(m=\cos \beta=\cos 120^{\circ}=-\frac{1}{2}\)

Since, \(l^2+m^2+n^2=1\)

Therefore, \(\left(\frac{\sqrt{3}}{2}\right)^2+\left(-\frac{1}{2}\right)^2+n^2=1\)

\(\Rightarrow \frac{3}{4}+\frac{1}{4}+n^2=1 \Rightarrow n^2=0 \Rightarrow n=0\)

\(\Rightarrow \cos \gamma=n=0 \Rightarrow \gamma=90^{\circ}\)

Given line is $\frac{x}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{0}$.

Direction ratios of the given line are $1,-1,0$.

Direction ratios of the $y-$axis are $0,1,0$.

We have ${\overrightarrow{b}}_1=\hat{i}-\hat{j}$ and ${\overrightarrow{b}}_2=\hat{j}$.

Let angle between \(y-\) axis and given line is \(\theta\).

Then, $\begin{array}{rcl}\cos \theta & =& \frac{{\overrightarrow{b}}_1·{\overrightarrow{b}}_2}{\left|{\overrightarrow{b}}_1\right|·\left|{\overrightarrow{b}}_2\right|}\end{array}$

$\begin{array}{rcl}& =& \frac{(\hat{i}-\hat{j})·\left(\hat{j}\right)}{\sqrt{1^2+{\left(-1\right)}^2}·\sqrt{1^2}}\end{array}$$\begin{array}{rcl}& =& -\frac{1}{\sqrt{2}}\end{array}$

$\begin{array}{rcl}& \Rightarrow & \cos \theta =\cos \left(\frac{3\pi }{4}\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & \theta =\frac{3\pi }{4}\end{array}$

For a point to lie on the line:
\(y=-5+\mu\) and \(z=-6-3 \mu\) with the same \(\mu\).
Testing \((1,-5,-6)\) :
From \(x\) : \(1=1+5 \mu \Rightarrow \mu=0\)
Then \(y=-5+0=-5, z=-6-3 \cdot 0=-6\)

$\overrightarrow{d}=(3,1,2)$

${\overrightarrow{r}}_0=(4,-3,7)$

$\overrightarrow{r}={\overrightarrow{r}}_0+t\overrightarrow{d}$

$\Rightarrow x=4+3t, y=-3+1·t, z=7+2t$

\(l=\cos \alpha=\cos 30^{\circ}=\frac{\sqrt{3}}{2}\)

\(m=\cos \beta=\cos 120^{\circ}=-\frac{1}{2}\)

\(l^2+m^2+n^2=1\)

\(\left(\frac{\sqrt{3}}{2}\right)^2+\left(-\frac{1}{2}\right)^2+n^2=1\)

\(\frac{3}{4}+\frac{1}{4}+n^2=1 \Rightarrow n^2=0 \Rightarrow n=0\)

\(\cos \gamma=n=0 \Rightarrow \gamma=90^{\circ}\)

The line passes through the point $\left(x_1,y_1,z_1\right)=(1,-3,2)$.

The line is parallel to the line with the vector equation, $\overrightarrow{\mathrm{r}}=\left(2+\lambda \right)\hat{i}+\lambda \hat{j}+\left(2\lambda -1\right)\hat{k}$.

$\Rightarrow \overrightarrow{\mathrm{r}}=\left(2\hat{i}-\hat{k}\right)+\lambda \left(\hat{i}+\hat{j}+2\hat{k}\right)$

The Cartesian equation of a line passing through a point $\left(x_1,y_1,z_1\right)$ with a direction ratios $⟨a,b,c⟩$ is given by the formula:

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

The Cartesian equation of the new line is:

$$\begin{gathered} \frac{x-1}{1}=\frac{y-(-3)}{1}=\frac{z-2}{2} \\ \Rightarrow \frac{x-1}{1}=\frac{y+3}{1}=\frac{z-2}{2} \end{gathered}$$

The line passes through the point $\left(x_1,y_1,z_1\right)=(1,-3,2)$.

The line is parallel to the line with the vector equation, $\mathrm{r}=(2+\lambda )i+\lambda j+(2\lambda -1)k$.

$\Rightarrow \mathrm{r}=(2i-k)+\lambda (i+j+2k)$

The Cartesian equation of a line passing through a point $\left(x_1,y_1,z_1\right)$ with a direction ratios $⟨a,b,c⟩$ is given by the formula:

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}$

The Cartesian equation of the new line is:

$$\begin{gathered} \frac{x-1}{1}=\frac{y-(-3)}{1}=\frac{z-2}{2} \\ \frac{x-1}{1}=\frac{y+3}{1}=\frac{z-2}{2} \end{gathered}$$

Given line is $\frac{x}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}}{0}$.

Direction ratios of the given line are $1,-1,0$.

Direction ratios of the $y-$axis are $0,1,0$.

We have ${\overrightarrow{b}}_1=\hat{i}-\hat{j}$ and ${\overrightarrow{b}}_2=\hat{j}$.

$\begin{array}{rcl}\cos \theta & =& \frac{{\overrightarrow{b}}_1·{\overrightarrow{b}}_2}{\left|{\overrightarrow{b}}_1\right|·\left|{\overrightarrow{b}}_2\right|}\end{array}$

$\begin{array}{rcl}& =& \frac{(\hat{i}-\hat{j})·\left(\hat{j}\right)}{\sqrt{1^2+{\left(-1\right)}^2}·\sqrt{1^2}}\end{array}$

$\begin{array}{rcl}& =& -\frac{1}{\sqrt{2}}\end{array}$

$\begin{array}{rcl}\cos \theta & =& \cos \left(\frac{3\pi }{4}\right)\end{array}$

$\begin{array}{rcl}& \Rightarrow & \theta =\frac{3\pi }{4}\end{array}$

Given: Direction cosines \(l=m=n=\sqrt{3} k\).

We know that, $l^2+m^2+n^2=1$

$\Rightarrow (\sqrt{3}k{)}^2+(\sqrt{3}k{)}^2+(\sqrt{3}k{)}^2=1$

$\Rightarrow 3k^2+3k^2+3k^2=1$

$\Rightarrow 9k^2=1$

$\Rightarrow k=\pm \frac{1}{3}$

Let direction cosines \((l, m, n)=(\cos \alpha, \cos \beta, \cos \gamma)\)

\(\alpha=\frac{3 \pi}{4}, \beta=\frac{\pi}{3}, \gamma=\theta\)

\(l^2+m^2+n^2=1\)

\(\left(\cos \frac{3 \pi}{4}\right)^2+\left(\cos \frac{\pi}{3}\right)^2+(\cos \theta)^2=1\)

\(\left(-\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2+(\cos \theta)^2=1\)

\(\frac{1}{2}+\frac{1}{4}+(\cos \theta)^2=1\)

\((\cos \theta)^2=\frac{1}{4} \Rightarrow \cos \theta= \pm \frac{1}{2}\)

\(\theta=\frac{\pi}{3}\) or \(\frac{2 \pi}{3}\)

So, option B is correct.

Direction cosines \(l=m=n=\sqrt{3} k\).

$l^2+m^2+n^2=1$

$(\sqrt{3}k{)}^2+(\sqrt{3}k{)}^2+(\sqrt{3}k{)}^2=1$

$3k^2+3k^2+3k^2=1$

$9k^2=1$

$\Rightarrow k=\pm \frac{1}{3}$

x -axis: \((t, 0,0)\)

Direction vector of \(x\)-axis: \(\vec{d}=\langle 1,0,0\rangle\)

Foot \(F(t, 0,0)\) such that \(\overrightarrow{F P} \perp \vec{d}, P(0,1,2)\)

\(\overrightarrow{F P}=\langle 0-t, 1-0,2-0\rangle=\langle-t, 1,2\rangle\)

\(\overrightarrow{F P} \cdot \vec{d}=0 \Rightarrow(-t) \cdot 1+1 \cdot 0+2 \cdot 0=0\)

\(\Rightarrow t=0\)

$x=1+5\mu \Rightarrow \mu =\frac{x-1}{5}$

For a point to lie on the line:
\(y=-5+\mu\) and \(z=-6-3 \mu\) with the same \(\mu\).
Test \((1,-5,-6)\) :
From \(x\) : \(1=1+5 \mu \Rightarrow \mu=0\)
Then \(y=-5+0=-5, z=-6-3 \cdot 0=-6\)

Let direction cosines \((l, m, n)=(\cos \alpha, \cos \beta, \cos \gamma)\)

\(\alpha=\frac{3 \pi}{4}, \beta=\frac{\pi}{3}, \gamma=\theta\)

We know, \(l^2+m^2+n^2=1\)

\(\left(\cos \frac{3 \pi}{4}\right)^2+\left(\cos \frac{\pi}{3}\right)^2+(\cos \theta)^2=1\)

\(\left(-\frac{1}{\sqrt{2}}\right)^2+\left(\frac{1}{2}\right)^2+(\cos \theta)^2=1\)

\(\frac{1}{2}+\frac{1}{4}+(\cos \theta)^2=1\)

\((\cos \theta)^2=\frac{1}{4} \Rightarrow \cos \theta= \pm \frac{1}{2}\)

\(\theta=\frac{\pi}{3}\) or \(\frac{2 \pi}{3}\)

but \(\theta\) must be acute.

So, option B is correct.

Perpendicular to \(x, y, z \Rightarrow \alpha=\beta=\gamma=90^{\circ}\).

$\Rightarrow \cos \alpha =\cos \beta =\cos \gamma =0.$

But for any line: \(\cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1\).
Here \(0+0+0 \neq 1 \Rightarrow\) impossible.
Thus Assertion is true.
Reason states the required identity and explains the impossibility.