RD Sharma Class 12 Exercise 27.4 Straight line in space Solutions Maths

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

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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.

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RD Sharma Class 12 Solutions Chapter27 Straight line in space - Other Exercise

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

Direction ratios pf AD

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

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,

Direction ratios of AD

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

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

Answer – The answer of the given question is 7 units.
(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

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

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