RD Sharma Class 12 Exercise 27.4 Straight line in space Solutions Maths - Download PDF Free Online

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# RD Sharma Class 12 Exercise 27.4 Straight line in space Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 09:47 AM IST

RD Sharma books are one of the best materials available in the country for maths. They have a reputation for being comprehensive and detailed, so they are the most beneficial for exams. Many schools and teachers all over the country use this material for their exam paper preparation.

Also Read - RD Sharma Solutions For Class 9 to 12 Maths

The RD Sharma class 12th exercise 27.4 consists of 14 questions that are based on fundamentals. These questions are of Level-1 difficulty and can be solved easily with a bit of basic knowledge. Moreover, RD Sharma class 12th exercise 27.4 contains questions and answers in one place and can be a more accessible alternative for revision.

## Straight Line in Space Excercise: 27.4

Straight Line in Space Exercise 27.4 Question 1

Answer – The length of perpendicular is $\sqrt{\frac{4901}{841}}$ unit.
(Hints – Denominator terms of line equation).
Given $\left ( 3,-1,11 \right );\frac{x}{z}=\frac{y-z}{3}=\frac{z-3}{4}$
Solution – Let, PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.
Thus to find distance PQ we have to first find co-ordinator of Q .
$\frac{x}{z}=\frac{y-2}{3}=\frac{z-3}{4}=\gamma (let)$

Therefore ,Co-ordinates of Q $\left ( 2\gamma ,-3\gamma +2,4\gamma +3 \right ).$
Hence
Direction of PQ is
$=\left ( 2y-3 \right ),\left ( -3y+2+1 \right ),\left ( 4y+3-11 \right )$$=\left ( 2y-3 \right ),\left ( -3y+3 \right ),\left ( 4y-8 \right ).$

And by comparing with given line equation, direction ratios of the given line are

(Hint – denominator terms of line equation)

$=2,-3,4$
PQ is perpendicular to given line,
:: By “Condition of Perpendicularity”.
$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$
; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &\rightarrow 2(2 \gamma-3)+(-3)(-3 \gamma+3)+4(4 \gamma-8) \\ &\rightarrow 4 \gamma-6+9 \gamma-9+16 \gamma-32=0 \\ &\rightarrow 29 \gamma-47=0 \\ &\rightarrow \gamma=\frac{47}{29} \end{aligned}
::Co-ordinate of Q i.e foot of perpendicular , by putting the value of γ
In Q co-ordinate equation ,we got

$=Q\left \{ 2\left ( \frac{47}{29} \right ),-3\left ( \frac{47}{29} \right )+2,4\left ( \frac{47}{29} \right )+3 \right \}$

$=Q\left ( \frac{94}{29},\frac{-83}{29},\frac{275}{29} \right )$
Now
Distance between PQ,

\begin{aligned} &=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\left(\frac{94}{29}-3\right)^{2}+\left(\frac{-83}{29}+1\right)^{2}+\left(\frac{275}{29}-11\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\left(\frac{94-87}{29}\right)^{2}+\left(\frac{-83+29}{29}\right)^{2}+\left(\frac{275-319}{29}\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{49}{841}+\frac{2016}{841}+\frac{1936}{841}} \\ &=\sqrt{\frac{4901}{841}} \text { unit. } \end{aligned}

Straight Line in Space Exercise 27.4 Question 2

Answer – The answer of the given question is $2\sqrt{6}$.
(Hint – denominator terms of line equation).
Given – Point ( 1, 0 ,0 ) and equation of line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z-10}{8}$
Solution - Let PQ be the perpendicular drawn from the P to given line whose endpoint/foot is Q point.
$\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z-10}{8}=\gamma (let)$

:: Co-ordinates of Q $\left ( 2\gamma +1,-3\gamma -1,8\gamma -10 \right )$
Hence,
Direction ratio of PQ is

\begin{aligned} &=(2 \gamma+1-1),(-3 \gamma-1-0),(8 \gamma-10-0) \\ &=(2 \gamma),(-3 \gamma-1),(8 \gamma-10) \end{aligned}
and by comparing the given line equation, direction ratios of the given line are
( Hint – denominator terms of line equation).

= (2 , -3 , 8)
Since PQ is perpendicular at the given line.
Therefore, By “Conditions of Perpendicularity”.

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$
; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=2(2 \gamma)+(-3)(-3 \gamma-1)+(8)(8 \gamma-10)=0 \\ &=4 \gamma-9 \gamma+3+64 \gamma+80=0 \\ &=77 \gamma+77=0 \\ &=\gamma=1 \end{aligned}
::Co-ordinate of Q i.e foot of perpendicular , by putting the value of γ in Q co-ordinate equation ,we get

\begin{aligned} &\mathrm{Q}\{2(1)+1,-3(1)-1,8(1)-10\} \\ &\mathrm{Q}\{3,-4,-2\} \\ \end{aligned}
NOW
Distance between PQ

\begin{aligned} &=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\ &=\sqrt{(1-3)^{2}+(0+4)^{2}+(-2-0)^{2}} \\ &=\sqrt{(-2)^{2}+(4)^{2}+(-2)^{2}} \\ &=\sqrt{4+16+4} \\ &=\sqrt{24} \\ &=2 \sqrt{6} \text { unit. } \end{aligned}

Straight Line in Space Exercise 27.4 Question 3

Answer - The answer of this question is $D\left ( \frac{5}{3},\frac{7}{3},\frac{17}{3} \right )$
(Hint – By using $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$)
Given – Perpendicular from A ( 1,0,3)drawn at line joining points B (4,7,1) and (3,5,3).
Solution – Let D be the foot of the perpendicular drawn from A (1,0,3) to line joining points B (4,7,1) and C(3,5,3).
Now,
$\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
\begin{aligned} &=\frac{x-4}{3-4}=\frac{y-7}{5-7}=\frac{z-1}{3-1} \\ &=\frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2} \end{aligned}
Now,
\begin{aligned} &=\frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2} =\gamma (let) \end{aligned}$=X=-\gamma +4,y=-2\gamma +7,z=2\gamma +1$

:: Co-ordinates of $D (-\gamma +4,-2y+7,2y+1)$

Hence

\begin{aligned} &=(-\gamma+4-1),(-2 \gamma+7-0),(2 \gamma-2) \\ &=(-\gamma+3),(-2 \gamma+7),(2 \gamma-2) \end{aligned}

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (-1,-2,2)

Since PQ is perpendicular at the given line.

Therefore, By “ Conditions of Perpendicularity”.

\begin{aligned} & a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\ \end{aligned}

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} =&-1(-\gamma+3)+(-2)(-2 \gamma+7)+2(2 \gamma-2)=0 \\ =& \gamma-3+4 \gamma-14+4 \gamma-4=0 \\ =& 9 \gamma-21=0 \\ \end{aligned}

\begin{aligned} =& \gamma=\frac{21}{9} \\ =& \gamma=\frac{7}{3} . \\ \end{aligned}

Co-ordinates of D i.e. foot of perpendicular

By putting value of γ in D co-ordinate equation. we get,

\begin{aligned} \quad & \bullet \operatorname{D}(-\gamma+4,-2 \gamma+7,2 \gamma+1) \\ \quad & \quad D\left(\frac{-7}{3}+4,-2 \frac{7}{3}+7,2 \frac{7}{3}+1\right) \\ \quad & \text { - }\left(\frac{5}{3}, \frac{7}{3}, \frac{17}{3}\right) \end{aligned}

Straight Line in Space Exercise 27.4 Question 4

Answer- The answer of the given question is $D\left ( \frac{22}{9},-\frac{11}{9},\frac{5}{9} \right )$
(Hint – By using the formula of a line joining the two points.)
Given- Perpendicular from A (1,0,4)drawn at line joining points B(0,-11,3) and C(2,
3,1).
Solution – D be the foot of the perpendicular drawn from A(1,0,4) to line joining
points B(0,-11,3) and C(2,-3,1).
Now, lets find the equation of the line which is formed by joining points B(0,-11,3) and C(2,-3,1).\begin{aligned} &=\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \\\\ &=\frac{x-0}{2-0}=\frac{y+11}{-3+11}=\frac{z-3}{1-3} \\\\ &=\frac{x}{2}=\frac{y+11}{8}=\frac{z-3}{-2}=\gamma(\text { let }) \\\\&=(x=2 \gamma, y=8 \gamma-11, z=-2 \gamma+3 \end{aligned}

::Co-ordinates of$D\left ( 2\gamma ,8\gamma -11,-2\gamma +3 \right )$.

Hence,

\begin{aligned} &=(2 \gamma-1),(8 \gamma-11-0),(-2 \gamma+3-4) \\\ &=(2 \gamma-1),(8 \gamma-11),(-2 \gamma-1) \end{aligned}

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (2,8,-2)

AD is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

\begin{aligned} & a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\ \end{aligned}

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} =2 &(2 \gamma-1)+(8)(8 \gamma-11)-2(-2 \gamma-1)=0 \\ \end{aligned}

\begin{aligned} =72 \gamma-88=0 \\ \end{aligned}

\begin{aligned} =& \gamma=\frac{88}{72} \\ \end{aligned}

\begin{aligned} =& \gamma=\frac{11}{9} \\ \end{aligned}

:: Co-ordinates of D i.e.foot of the perpendicular.

By putting the value of $\gamma$ in D co-ordinate equation, we get

\begin{aligned} & \bullet \quad D(2 \gamma, 8 \gamma-11,-2 \gamma+3\\ \end{aligned}

\begin{aligned} & \bullet & \quad D\left(2 \frac{11}{9}, 8\left(\frac{11}{9}\right)-11,-2\left(\frac{11}{9}\right)+3\right) \\ \end{aligned}

\begin{aligned} & \bullet & \quad D\left(\frac{22}{9}, \frac{-11}{9}, \frac{5}{9}\right) \end{aligned}

Straight Line in Space Exercise 27.4 Question 5

Answer - The answer of this question is $\frac{3}{7}\sqrt{101}Units ,Q\left ( \frac{170}{49},\frac{78}{49},\frac{10}{49} \right )$
(Hint – By using the distance between two point formula).
Given – Point (2,3,4) and equation of the line $\frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}$
Solution – Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.
Now,
\begin{aligned} &* \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}=\gamma(\text { let }) \\ &* \mathrm{x}=4-2 \gamma, y=6 \gamma, z=1-3 \gamma \\ \end{aligned}
:: Co-ordinates of \begin{aligned} &\mathrm{Q}(-2 \gamma, 6 \gamma,-3 \gamma+1) \\ \end{aligned}

Hence

Direction of PQ is
\begin{aligned} &=(-2 \gamma+4-2),(6 \gamma-3),(-3 \gamma+1-4) \\ &=(-2 \gamma+2),(6 \gamma-3),(-3 \gamma-3) \end{aligned}
And by comparing with given line equation, direction ratios of the given line are
( Hint – denominator terms of line equation).
= (-2,6,-3)
PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.
$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$
; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=-2(-2 \gamma+2)+(6)(6 \gamma-3)-3(-3 \gamma-3)=0 \\ &=4 \gamma-4+36 \gamma-18+9 \gamma+9=0 \\ &=49 \gamma-13=0 \\ &=\gamma=\frac{13}{49} \end{aligned}
:: Co-ordinates of Q.,
By putting the value of γ in Q co-ordinate equation, we get

\begin{aligned} &=\mathrm{Q}\left\{-2 \frac{13}{49}+4,6 \frac{13}{49},-3 \frac{13}{49}+1\right\} \\ &=\mathrm{Q} \gamma\left\{\frac{170}{49}, \frac{78}{49}, \frac{10}{49}\right\} \end{aligned}
Now,
Distance between PQ

$=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

\begin{aligned} &=\sqrt{\left(\frac{170}{49}-2\right)^{2}+\left(\frac{78}{49}-3\right)^{2}+\left(\frac{10}{49}-4\right)^{2}} \\ &=\sqrt{\left(\frac{72}{49}\right)^{2}+\left(\frac{69}{49}\right)^{2}+\left(\frac{168}{49}\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{5184}{2401}+\frac{4761}{2401}+\frac{34596}{2401}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{44541}{2401}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{909}{49}} \\ \end{aligned}

\begin{aligned} &=\frac{3}{7} \sqrt{101} \text { units. } \end{aligned}

Straight Line in Space Exercise 27.4 Question 6

Answer – The answer of the given question is $\left ( -4,1,-3 \right )\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z-1}{-2}$
(Hint – By using two point formula).
Given – Point (2,4,-1) and equation of the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$
Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.
\begin{aligned} &* \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\gamma(\text { let }) \\ &* \quad \mathrm{x}=\gamma-5, y=4 \gamma-3, z=-9 \gamma+6 \\ \end{aligned}
:: Co-ordinates of \begin{aligned} &\mathrm{Q}(\gamma-5,4 \gamma-3,-9 \gamma+6) \\ \end{aligned}

Hence,

Direction of PQ is
\begin{aligned} &=(\gamma-5-2),(4 \gamma-3-4),(-9 \gamma+6+1) \\ &=(\gamma+7),(4 \gamma-7),(-9 \gamma+7) \end{aligned}

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (1,4,9)

PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=1(\gamma-7)+(4)(4 \gamma-7)-9(-9 \gamma+7)=0 \\ &=\gamma-7+16 \gamma-28+81 \gamma-63=0 \\ &=98 \gamma-98=0 \\ &=\gamma=1 \\ \end{aligned}

:: Co-ordinates of Q.,

By putting the value of γ in Q co-ordinate equation, we get

\begin{aligned} &=Q\{(1)-5,4(1)-3,-9(1)+6\} \\ &=Q \gamma\{-4,1,-3\} \\ \end{aligned}

Now,

So, equation of perpendicular PQ is

\begin{aligned} &=\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \\ &=\frac{x-2}{-4-2}=\frac{y-4}{1-4}=\frac{z+1}{-3+1} \\ &=\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z+1}{-2} . \end{aligned}

Straight Line in Space Exercise 27.4 Question 7

Answer - The answer of the given question is $\sqrt{\frac{2109}{110}}Units ,Q\left ( \frac{188}{110},\frac{351}{110},\frac{195}{110} \right )$
Hint – By using two point formula).
Given – Point (5,4,-1) and equation of the line $\hat{r}=\hat{i}+\gamma \left ( 2\hat{i}+9\hat{j}+5\hat{x} \right )$
Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.
As we know positive vector is given by
$\hat{r}=x\hat{i}+y\hat{j}+z\hat{x}$
:: position vector of point P is
$5\hat{i}+4\hat{j}-1\hat{x}$
Ad from a given line we get
$\Rightarrow \hat{r}=\hat{i}+\gamma ( z\hat{i}+9\hat{j}+5\hat{x}$
$\begin{gathered} \Rightarrow x \hat{\imath}+y \hat{\jmath}+z \hat{x} \\ \Rightarrow \hat{i}+\gamma(z \hat{i}+9 \hat{j}+5 \hat{x} \\ \Rightarrow x \hat{\imath}+y \hat{\jmath}+z \hat{x} \\ \Rightarrow(1+2 \gamma) \hat{i}+(9 \gamma) \hat{j}+(5 \gamma) \hat{x} \end{gathered}$

On comparing both side we get,

\begin{aligned} &=\mathrm{x}=(1+2 \gamma), y=(9 \gamma), z=(5 \gamma)\\ &=\frac{x-1}{2}=\frac{y}{9}=\frac{z}{5}=\gamma ; \text { equation of line }\\ \end{aligned}

Thus, Co-ordinates of Q i.e. General point of the given line.

\begin{aligned} &\mathrm{Q}\{(1+2 \gamma),(9 \gamma),(5 \gamma)\} \end{aligned}

Hence

Direction ratio of PQ is

\begin{aligned} &=(2 \gamma+1-5),(9 \gamma-4),(5 \gamma+1) \\ &=(2 \gamma-4),(9 \gamma-4),(5 \gamma+1) \end{aligned}
And by comparing with given line equation, direction ratios of the given line are
( Hint – denominator terms of line equation).
= (2,9,5).
PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=2(2 \gamma-4)+(9)(9 \gamma-4)+5(5 \gamma+1)=0 \\\\ &=4 \gamma-8+81 \gamma-36+25 \gamma+5=0 \\\\ &=110 \gamma-39=0 \\\\ &=\gamma=\frac{39}{110} \\ \end{aligned}

:: Co-ordinates of Q.,

By putting the value of γ in Q co-ordinate equation, we get

\begin{aligned} &=Q\{(1)-5,4(1)-3,-9(1)+6\} \\\\ &=Q \gamma\{-4,1,-3\} \\ \end{aligned}

:: Co-ordinates of Q i.e. foot of perpendicular

By putting the value of γ in Q

\begin{aligned} &=Q\left\{2\left(\frac{39}{110}\right)+1,9\left(\frac{39}{110}\right), 5\left(\frac{39}{110}\right)\right\} \\\\ &=Q\left\{\left(\frac{188}{110}, \frac{351}{110}, \frac{195}{110}\right)\right\} \\ \end{aligned}

Now,

Distance between PQ

\begin{aligned} &=\sqrt{\left(x_{2}-x_{1}\right)+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \end{aligned}

\begin{aligned} &=\sqrt{\left(\frac{188}{110}-5\right)^{2}+\left(\frac{351}{110}-4\right)^{2}+\left(\frac{195}{110}+1\right)^{2}} \\ &=\sqrt{\left(\frac{-362}{110}\right)^{2}+\left(\frac{-89}{110}\right)^{2}+\left(\frac{305}{110}\right)^{2}} \\ \end{aligned}

\begin{aligned} &=\sqrt{\frac{131044}{12100}+\frac{7921}{12100}+\frac{93025}{12100}} \\ &=\sqrt{\frac{231990}{12100}} \\ &=\sqrt{\frac{2109}{110} \text { units. }} \end{aligned}

Straight Line in Space Exercise 27.4 Question 8

Answer – The answer of the given question $\sqrt{13}Units.$
(Hint - $\mid AB\mid =\sqrt{x^{2}+y^{2}+z^{2}}.$
Given – Point with positive vector $\hat{i}+\left ( 6\hat{j}+3\hat{x} \right )$and equation line $\hat{r}=\hat{j}+2\hat{x}+\gamma \left ( \hat{i}+2\hat{j}+3\hat{x} \right ).$
Solution – Let PQ be the perpendicular drawn from $P\left ( \hat{i} +6\hat{j}+3\hat{x} \right )$ to given line whose endpoint/foot is Q point.
Q is on line.
$=\hat{r}=\hat{\jmath}+2 \hat{x}+\gamma(\hat{\imath}+2 \hat{\jmath}+3 \hat{x})$
$=\gamma \hat{i}+(2 \gamma+1) \hat{j}+(3 \gamma+2) \hat{x} \; is \; the position \; vector \; of \; Q$
Hence,
$=\overrightarrow{P Q}= Position vector of Q- Position vector of$
$=\overrightarrow{P Q}=\gamma \hat{\imath}+(2 \gamma+1) \hat{\jmath}+(3 \gamma+2) \hat{x}-(\hat{\imath}+6 \hat{\jmath}+3 \hat{x})$
$=\overrightarrow{P Q}=(\gamma-1) \hat{\imath}+(2 \gamma+1) \hat{\jmath}+(3 \gamma+2) \hat{x}-\hat{\imath}-6 \hat{\jmath}-3 \hat{x}$$=\overrightarrow{P Q}=(\gamma-1) \hat{\imath}+(2 \gamma-5) \hat{\jmath}+(3 \gamma-1) \hat{x}$

:: PQ is perpendicular on line

$\hat{i}=\hat{j}+2\hat{x}+\gamma \left ( \hat{i}+2\hat{j}+3\hat{x} \right )$

:: Their dot product is zero.

Compare given line equation with

$\hat{r}=\hat{a}+\gamma \hat{b}$

\begin{aligned} &=\overrightarrow{P Q} \cdot \hat{b}=0 \\ \end{aligned}

\begin{aligned} &=\{(\gamma-1) \hat{\imath}+(2 \gamma-5) \hat{\jmath}+(3 \gamma-1) \hat{x}\} \cdot\{(\hat{\imath}+2 \hat{\jmath}+3 \hat{x})\}=0 \\ \end{aligned}

\begin{aligned} &=(\gamma-1)(1)+(2 \gamma-5)(2)+(3 \gamma-1)(3)=0 \\ \end{aligned}

\begin{aligned} &=\gamma-1+4 \gamma-10+9 \gamma-3=0 \\\\ &=14 \gamma-140=0 \\\\ &=\gamma=1 \end{aligned}

Hence, Position vector of Q ,By putting the value of $\gamma$

\begin{aligned} &\qquad \gamma \hat{i}+(2 \gamma+1) \hat{j}+(3 \gamma+2) \hat{x} \\ \end{aligned}

\begin{aligned} &=\hat{\imath}+(2+1) \hat{\jmath}+(3+2) \hat{x} \\ &=\hat{\imath}+3 \hat{\jmath}+5 \hat{x} ; \text { foot of perpendicular } \\ \end{aligned}

Putting the value of γ in PQ.

\begin{aligned} &=\overrightarrow{P Q}=(\gamma-1) \hat{\imath}+(2 \gamma-5) \hat{\jmath}+(3 \gamma-1) \hat{x} \\ &=\overrightarrow{P Q}=(1-1) \hat{\imath}+(2-5) \hat{\jmath}+(3-1) \hat{x} \\ &=\overrightarrow{P Q}=-3 \hat{\jmath}+2 \hat{x} \\ \end{aligned}

Now, Magnitude of PQ , we know that

\begin{aligned} &|A B|=\sqrt{x^{2}+y^{2}+z^{2}} \\ \end{aligned}

Hence

\begin{aligned} &=|\mathrm{PQ}|=\sqrt{0^{2}+(-3)^{2}+2^{2}} \\\\ &=|\mathrm{PQ}|=\sqrt{13} \text { units. } \end{aligned}

Straight Line in Space Exercise 27.4 Question 9

Answer- The answer of the given question is $\left(\frac{-4}{7}, \frac{12}{7}, \frac{15}{7}\right) \cdot \hat{r}=(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})+\gamma\left\{\left(\frac{-4}{7} \hat{\imath}+\frac{12}{7} \hat{\jmath}+\frac{15}{7} \hat{k}\right)-(\hat{\imath}+3 \hat{\jmath}+2 \hat{k})\right\}$
(Hint- By comparing both the equation).
Given – Point P(-1,3,2) and equation of line $\hat{r}=\left ( 2\hat{j}+3\hat{k} \right )+\gamma \left ( 2\hat{i}+\hat{j}+3\hat{k} \right ).$
Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.
As we know position vector is given by
$\hat{r}=x\hat{i}+y\hat{j}+z\hat{k}$
:: Position vector is given by
$=-\hat{i}=3\hat{j}+2\hat{k}$
And, from a given line, we get\begin{aligned} &=\hat{r}=(2 \hat{\jmath}+3 \hat{k})+\gamma(2 \hat{\imath}+\hat{\jmath}+3 \hat{k}) \\ &=x \hat{\imath}+y \hat{\jmath}+z \hat{k}=(2 \hat{\jmath}+3 \hat{k})+\gamma(2 \hat{\imath}+\hat{\jmath}+3 \hat{k}) \\ &=x \hat{\imath}+y \hat{\jmath}+z \hat{k}=(2 \gamma) \hat{\imath}+(\gamma+2) \hat{\jmath}+(3 \gamma+3) \hat{k} \end{aligned}

On comparing both sides we get,

$=\frac{x}{2}=\frac{y-2}{1}=\frac{z-3}{3}= \gamma ;equation of line$

$=x=2 \gamma, y=\gamma+2, z=3 \gamma+3$

Thus, co-ordinate of Q i.e. general point on the given line.

\begin{aligned} &\mathrm{Q}(\gamma, \gamma+2,3 \gamma+3) \\ \end{aligned}

Hence

Direction of PQ is

\begin{aligned} &=(2 \gamma+1),(\gamma+2-1),(3 \gamma+3-2) \\ &=(2 \gamma+1),(\gamma-1),(3 \gamma+1) \\ \end{aligned}
And by comparing with given line equation, direction ratios of the given line are
( Hint – denominator terms of line equation).
= (2,1,3).
PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.
\begin{aligned} &\qquad a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \end{aligned}

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=2(2 \gamma+1)+(\gamma-1)+3(3 \gamma+1)=0 \\ &=14 \gamma+4=0 \\ &=\gamma=\frac{-4}{14} \\ &=\gamma=\frac{-2}{7} \\ \end{aligned}

:: Co-ordinates of Q,

By putting the value of γ in Q co-ordinate equation, we get

\begin{aligned} &=2 \gamma, \gamma+2,3 \gamma+3 \\ &=Q\left\{2\left(\frac{-2}{7}\right),\left(\frac{-2}{7}\right)+2,3\left(\frac{-2}{7}\right)+3\right\} \\ &=Q\left(\frac{-4}{7}, \frac{12}{7}, \frac{15}{7}\right) \\ \end{aligned}

Position vector of Q

\begin{aligned} &=\frac{-4}{7} \hat{\imath}+\frac{12}{7} \hat{\jmath}+\frac{15}{7} \hat{k} \end{aligned}

Now, Equation of line passing through two points with position vector $\hat{a}\; and\; \hat{b} \: is \: given \: by$

$\begin{gathered} \quad \hat{r}=\hat{a}+\gamma(\hat{b}-\hat{a}) \\ \end{gathered}$

$= Here,$

$\begin{gathered} =\hat{a}=-\hat{\imath}+3 \hat{\jmath}+2 \hat{k} \\ \end{gathered}$

$\begin{gathered} =\text { and } \hat{b}=\frac{-4}{7} \hat{\imath}+\frac{12}{7} \hat{\jmath}+\frac{15}{7} \hat{k} \\ \end{gathered}$

$\begin{gathered} =\hat{r}=(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})+\gamma\left[\left(\frac{-4}{7} \hat{\imath}+\frac{12}{7} \hat{\jmath}+\frac{15}{7} \hat{k}\right)-(-\hat{\imath}+3 \hat{\jmath}+2 \hat{k})\right] . \end{gathered}$

Straight Line in Space Exercise 27.4 Question 10

Answer – The answer of the question is $Q\left ( \frac{-3}{2},\frac{-1}{2},4 \right )$
(Hint – By putting the value of $\gamma$ ).
Given – Point (0,2,7) and equation of the line $\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}.$
Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.
Thus to find Distance PQ we have to first find co-ordinates of Q.\begin{aligned} &* \frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}=\gamma(\text { let }) \\ &* \quad \mathrm{x}=-\gamma-2, y=3 \gamma+1, z=-2 \gamma+3 \\ &:: \text { Co-ordinates of } \mathrm{Q}(-\gamma-2,3 \gamma+1,-2 \gamma+3) \end{aligned}

Hence,

Direction of PQ is

\begin{aligned} &=(-\gamma-2-0),(3 \gamma+1-2),(-2 \gamma+3-7) \\\\ &=(-\gamma-2),(3 \gamma-1),(-2 \gamma-4) \end{aligned}

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (-1,3,-2)

:: PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=-1(-\gamma-2)+(3)(3 \gamma-1)-2(-2 \gamma-4)=0 \\ &=\gamma+2+9 \gamma-3+4 \gamma+8=0 \\ &=\gamma=\frac{-1}{2} . \end{aligned}

:: Co-ordinates of Q ,

By putting the value of $\gamma$ in Q

\begin{aligned} &=Q\left\{-\left(\frac{-1}{2}\right)-3\left(\frac{-1}{2}\right)+1,-2\left(\frac{-1}{2}\right)+3\right\} \\ &=Q\left\{\frac{-3}{2}, \frac{-1}{2}, 4\right\} \end{aligned}

Straight Line in Space Exercise 27.4 Question 11

Answer - The answer of the question is $Q\left ( 1,1,-1 \right )$
(Hint – By comparing with line equation).
Given – Point $(1,2,-3)$ and equation of the line $\frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}$
Solution - Let PQ be the perpendicular drawn from P to given line whose endpoint/foot is Q point.
Thus to find Distance PQ we have to first find co-ordinates of Q.\begin{aligned} &\text { * } \frac{x+1}{2}=\frac{y-3}{-2}=\frac{z}{-1}=\gamma(\text { let }) \\ &\text { * }\mathrm{x}=2 \gamma-1, y=-2 \gamma+3, z=-\gamma \\ &:: \text { Co-ordinates of } \mathrm{Q}(2 \gamma-1,-2 \gamma+3,-\gamma) \end{aligned}

Hence,

Direction ratio of PQ is

\begin{aligned} &=(2 \gamma-1-1),(-2 \gamma+3-2),(-\gamma+3) \\\\ &=(2 \gamma-2),(-2 \gamma+1),(-\gamma+3) \end{aligned}

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

= (2,-2,-1)

:: PQ is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

$a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0$

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=2(2 \gamma-2)+(-2)(-2 \gamma+1)-1(-\gamma+3)=0 \\ &=4 \gamma-4+4 \gamma-2+\gamma-3=0 \\ &=9 \gamma-9=0 \\ &=\gamma=1 \\ \end{aligned}
:: Co-ordinates of Q ,
By putting the value of $\gamma$ in Q

\begin{aligned} &=Q\{2(1)-1,-2(1)+3,-1\} \\\\ &=Q\{1,1,-1\} \end{aligned}

Straight Line in Space Exercise 27.4 Question 12

Answer - The answer of the given question is $\frac{x-7}{4}=\frac{y-4}{1}=\frac{z+1}{2}$
(Hint – By using two point formula).
Given – Line passing through the point $A(0,6,-9) and B(-3,-6,3) and C (7,4,-1).$
Solution – $\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}$
Hence equation of line AB\begin{aligned} &=\frac{x-0}{-3-0}=\frac{y-6}{-6-6}=\frac{z+9}{3+9} \\\\ &=\frac{x}{-3}=\frac{y-6}{-12}=\frac{z+9}{12}=\gamma(l e t) \\\\ &=x=-3 \gamma, y=-12 \gamma+6, z=12 \gamma-9 \\\\ &:: \text { Co-ordinates of point } D(-3 \gamma,(-12 \gamma+6),(12 \gamma-9) \end{aligned}

Hence,

Direction ratio of CD is

\begin{aligned} &=(-3 \gamma-7),(-12 \gamma+6-4),(12 \gamma-9+1) \\ &=(-3 \gamma-7),(-12 \gamma+2),(12 \gamma-8) \\ \end{aligned}

And by comparing with given line equation, direction ratios of the given line are

( Hint – denominator terms of line equation).

$= (-3,-12,12)$

CD is perpendicular at the given line. Therefore, By “ Conditions of Perpendicularity”.

\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \\ \end{aligned}

; whereas a terms and b terms are direction ratio of lines which are perpendicular to each other.

\begin{aligned} &=-3(-3 \gamma-7)+(-12)(-12 \gamma+2)+12(12 \gamma-8)=0 \\ &=9 \gamma+21+144 \gamma-24+144 \gamma-96=0 \\ &=297 \gamma-99=0 \\ &=\gamma=\frac{1}{3} \\ \end{aligned}

:: Co-ordinates of D,

By putting the value of $\gamma$ in D , we get

\begin{aligned} &(-3 \gamma),(-12 \gamma+6),(12 \gamma-9) \\ &=D\left[-3\left(\frac{1}{3}\right),\left(-12\left(\frac{1}{3}\right)+6\right),\left(12\left(\frac{1}{3}\right)-9\right)\right. \\ &=D\{-1,2,-5\} \end{aligned}

Hence, Equation of line

\begin{aligned} &=\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}} \\ &=\frac{x-7}{-1-7}=\frac{y-4}{2-4}=\frac{z+1}{-5+1} \\ &=\frac{x-7}{-8}=\frac{y-4}{-2}=\frac{z+1}{-4} \\ &=\frac{x-7}{4}=\frac{y-4}{1}=\frac{z+1}{2} \end{aligned}

Straight Line in Space Exercise 27.4 Question 13

(Hint – By using cross product).
Given – Point $(2,4,-1)$ and equation of line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$
Solution – Let Q be a point through which line passes. Thus for given equation of line co-ordinates of Q is $Q(-5,-3,6).$
Hence,
Line parallel to $\vec{b}=\hat{i}+4\hat{j}-9\hat{k}$
Now,\begin{aligned} &=\overrightarrow{P Q}=(-5 \hat{\imath}-3 \hat{\jmath}+6 \hat{k})-(2 \hat{\imath}+4 \hat{\jmath}-\hat{k}) \\ &=\overrightarrow{P Q}=(-7 \hat{\imath}-7 \hat{\jmath}+7 \hat{k}) \\ \end{aligned}

Now, let’s find cross product of the two vectors.

$=\vec{b} \times \overrightarrow{P Q}=\begin{vmatrix} \hat{i} & \hat{\hat{j}} & \hat{k}\\ 1 & 4 &-9 \\ -7 &-7 &7 \end{vmatrix}$

\begin{aligned} &=\vec{b} \times \overrightarrow{P Q}=-35 \hat{\imath}+56 \hat{\jmath}+21 \hat{k} \\\\ &=\text { The magnitude of this cross product } \\\\ &=|\vec{b} \times \overrightarrow{P Q}|=\sqrt{1225+3136+441} \\\\ &=|\vec{b} \times \overrightarrow{P Q}|=\sqrt{4802} \end{aligned}

And magnitude of $\vec{b}$

\begin{aligned} &=|\vec{b}|=\sqrt{1+16+81} \\ &=|\vec{b}|=\sqrt{98} \\ \end{aligned}

Thus, distance of Point from line is

\begin{aligned} &=\mathrm{d}=\frac{|\vec{b} \times \overrightarrow{P Q}|}{|\vec{b}|} \\ &=\mathrm{d}=\frac{\sqrt{4802}}{\sqrt{98}} \\ &=\mathrm{d}=7 \text { units. } \end{aligned}

The benefits of RD Sharma class 12th exercise 27.4 material are:

• Students can refer to RD Sharma class 12 chapter 27 exercise 27.4 to clear their doubts and use it as a guide for their RD Sharma math textbook. The answers are prepared by subject experts who are easy to understand. RD Sharma class 12 solutions chapter 27 exercise 27.4 is a paperless and convenient alternative to textbooks as it covers all topics and can be accessed through any device with a browser.

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RD Sharma Chapter-wise Solutions

1. How can I benefit from this material?

As it contains questions and answers in the same place, students can use this material as a guide to prepare well for their exams and score good marks.

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##### Copy Writer

In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.

5 Jobs Available
##### Journalist

Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.

3 Jobs Available
##### Publisher

For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.

3 Jobs Available
##### Vlogger

In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.

3 Jobs Available
##### Editor

Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.

3 Jobs Available

Advertising managers consult with the financial department to plan a marketing strategy schedule and cost estimates. We often see advertisements that attract us a lot, not every advertisement is just to promote a business but some of them provide a social message as well. There was an advertisement for a washing machine brand that implies a story that even a man can do household activities. And of course, how could we even forget those jingles which we often sing while working?

2 Jobs Available
##### Photographer

Photography is considered both a science and an art, an artistic means of expression in which the camera replaces the pen. In a career as a photographer, an individual is hired to capture the moments of public and private events, such as press conferences or weddings, or may also work inside a studio, where people go to get their picture clicked. Photography is divided into many streams each generating numerous career opportunities in photography. With the boom in advertising, media, and the fashion industry, photography has emerged as a lucrative and thrilling career option for many Indian youths.

2 Jobs Available
##### Social Media Manager

A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.

2 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### Quality Controller

A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.

A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.

3 Jobs Available
##### Production Manager

Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.

3 Jobs Available

A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.

2 Jobs Available
##### Quality Systems Manager

A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.

2 Jobs Available
##### Merchandiser

A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.

2 Jobs Available
##### Procurement Manager

The procurement Manager is also known as  Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.

2 Jobs Available
##### Production Planner

Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.

2 Jobs Available
##### Information Security Manager

Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack

3 Jobs Available
##### Computer Programmer

Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.

3 Jobs Available
##### Product Manager

A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.

3 Jobs Available
##### ITSM Manager

ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.

3 Jobs Available
##### .NET Developer

.NET Developer Job Description: A .NET Developer is a professional responsible for producing code using .NET languages. He or she is a software developer who uses the .NET technologies platform to create various applications. Dot NET Developer job comes with the responsibility of  creating, designing and developing applications using .NET languages such as VB and C#.

2 Jobs Available
##### Corporate Executive

Are you searching for a Corporate Executive job description? A Corporate Executive role comes with administrative duties. He or she provides support to the leadership of the organisation. A Corporate Executive fulfils the business purpose and ensures its financial stability. In this article, we are going to discuss how to become corporate executive.

2 Jobs Available
##### DevOps Architect

A DevOps Architect is responsible for defining a systematic solution that fits the best across technical, operational and and management standards. He or she generates an organised solution by examining a large system environment and selects appropriate application frameworks in order to deal with the system’s difficulties.

2 Jobs Available
##### Cloud Solution Architect

Individuals who are interested in working as a Cloud Administration should have the necessary technical skills to handle various tasks related to computing. These include the design and implementation of cloud computing services, as well as the maintenance of their own. Aside from being able to program multiple programming languages, such as Ruby, Python, and Java, individuals also need a degree in computer science.

2 Jobs Available