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

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 –$(3,-1,11);\frac{x}{z}=\frac{y-z}{3}=\frac{z-3}{4}$
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 $(2\gamma ,-3\gamma +2,4\gamma +3).$
Hence
Direction of PQ is
$=(2y-3),(-3y+2+1),(4y+3-11)$$=(2y-3),(-3y+3),(4y-8).$

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.

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

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

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

$\begin{array}{rl}& =\sqrt{(x_2-x_1)+{(y_2-y_1)}^2+{(z_2-z_1)}^2}\end{array}$

$\begin{array}{rl}& =\sqrt{{(\frac{94}{29}-3)}^2+{(\frac{-83}{29}+1)}^2+{(\frac{275}{29}-11)}^2}\end{array}$

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

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

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 $(2\gamma +1,-3\gamma -1,8\gamma -10)$
Hence,
Direction ratio of PQ is

$\begin{array}{rl}& =(2\gamma +1-1),(-3\gamma -1-0),(8\gamma -10-0)\\ & =(2\gamma ),(-3\gamma -1),(8\gamma -10)\end{array}$
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.

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

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

$\begin{array}{rl}& =\sqrt{(x_2-x_1)+{(y_2-y_1)}^2+{(z_2-z_1)}^2}\\ & =\sqrt{(1-3{)}^2+(0+4{)}^2+(-2-0{)}^2}\\ & =\sqrt{(-2{)}^2+(4{)}^2+(-2{)}^2}\\ & =\sqrt{4+16+4}\\ & =\sqrt{24}\\ & =2\sqrt{6}\text{ unit. }\end{array}$

Straight Line in Space Exercise 27.4 Question 3

Answer - The answer of this question is $D(\frac{5}{3},\frac{7}{3},\frac{17}{3})$
(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,
$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$
$\begin{array}{rl}& =\frac{x-4}{3-4}=\frac{y-7}{5-7}=\frac{z-1}{3-1}\\ & =\frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2}\end{array}$
Now,
$\begin{array}{rl}& =\frac{x-4}{-1}=\frac{y-7}{-2}=\frac{z-1}{2}=\gamma (let)\end{array}$$=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

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

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”.

$\begin{array}{rl}& a_1{a}_2+b_1{b}_2+c_1{c}_2=0\end{array}$

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

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

$\begin{array}{rl}=& \gamma =\frac{21}{9}\\ =& \gamma =\frac{7}{3}.\end{array}$

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

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

$\begin{array}{rl}& \bullet \mathrm{D}(-\gamma +4,-2\gamma +7,2\gamma +1)\\ & D(\frac{-7}{3}+4,-2\frac{7}{3}+7,2\frac{7}{3}+1)\\ & \text{ - }(\frac{5}{3},\frac{7}{3},\frac{17}{3})\end{array}$

Straight Line in Space Exercise 27.4 Question 4

Answer- The answer of the given question is $D(\frac{22}{9},-\frac{11}{9},\frac{5}{9})$
(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).$\begin{array}{rl}& =\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\\ \\ & =\frac{x-0}{2-0}=\frac{y+11}{-3+11}=\frac{z-3}{1-3}\\ \\ & =\frac{x}{2}=\frac{y+11}{8}=\frac{z-3}{-2}=\gamma (\text{ let })\\ \\ & =(x=2\gamma ,y=8\gamma -11,z=-2\gamma +3\end{array}$

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

Hence,

Direction ratios of AD

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

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”.

$\begin{array}{rl}& a_1{a}_2+b_1{b}_2+c_1{c}_2=0\end{array}$

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

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

$\begin{array}{r}=72\gamma -88=0\end{array}$

$\begin{array}{rl}=& \gamma =\frac{88}{72}\end{array}$

$\begin{array}{rl}=& \gamma =\frac{11}{9}\end{array}$

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

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

$\begin{array}{rl}& \bullet D(2\gamma ,8\gamma -11,-2\gamma +3\end{array}$

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

$\begin{array}{rlr}& \bullet & D(\frac{22}{9},\frac{-11}{9},\frac{5}{9})\end{array}$

Straight Line in Space Exercise 27.4 Question 5

Answer - The answer of this question is $\frac{3}{7}\sqrt{101}Units,Q(\frac{170}{49},\frac{78}{49},\frac{10}{49})$
(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,
$\begin{array}{rl}& \ast \frac{4-x}{2}=\frac{y}{6}=\frac{1-z}{3}=\gamma (\text{ let })\\ & \ast \mathrm{x}=4-2\gamma ,y=6\gamma ,z=1-3\gamma \end{array}$
:: Co-ordinates of $\begin{array}{rl}& \mathrm{Q}(-2\gamma ,6\gamma ,-3\gamma +1)\end{array}$

Hence

Direction of PQ is
$\begin{array}{rl}& =(-2\gamma +4-2),(6\gamma -3),(-3\gamma +1-4)\\ & =(-2\gamma +2),(6\gamma -3),(-3\gamma -3)\end{array}$
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.

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

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

$=\sqrt{(x_2-x_1)+{(y_2-y_1)}^2+{(z_2-z_1)}^2}$

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

$\begin{array}{rl}& =\sqrt{\frac{5184}{2401}+\frac{4761}{2401}+\frac{34596}{2401}}\end{array}$

$\begin{array}{rl}& =\sqrt{\frac{44541}{2401}}\end{array}$

$\begin{array}{rl}& =\sqrt{\frac{909}{49}}\end{array}$

$\begin{array}{rl}& =\frac{3}{7}\sqrt{101}\text{ units. }\end{array}$

Straight Line in Space Exercise 27.4 Question 6

Answer – The answer of the given question is $(-4,1,-3)\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z-1}{-2}$
(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.
$\begin{array}{rl}& \ast \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}=\gamma (\text{ let })\\ & \ast \mathrm{x}=\gamma -5,y=4\gamma -3,z=-9\gamma +6\end{array}$
:: Co-ordinates of $\begin{array}{rl}& \mathrm{Q}(\gamma -5,4\gamma -3,-9\gamma +6)\end{array}$

Hence,

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

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.

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

:: Co-ordinates of Q.,

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

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

Now,

So, equation of perpendicular PQ is

$\begin{array}{rl}& =\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}\\ & =\frac{x-2}{-4-2}=\frac{y-4}{1-4}=\frac{z+1}{-3+1}\\ & =\frac{x-2}{-6}=\frac{y-4}{-3}=\frac{z+1}{-2}.\end{array}$

Straight Line in Space Exercise 27.4 Question 7

Answer - The answer of the given question is $\sqrt{\frac{2109}{110}}Units,Q(\frac{188}{110},\frac{351}{110},\frac{195}{110})$
Hint – By using two point formula).
Given – Point (5,4,-1) and equation of the line $\hat{r}=\hat{i}+\gamma (2\hat{i}+9\hat{j}+5\hat{x})$
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
$\Rightarrow \hat{r}=\hat{i}+\gamma (z\hat{i}+9\hat{j}+5\hat{x}$
$\begin{array}{c}\Rightarrow x\hat{ı}+y\hat{ȷ}+z\hat{x}\\ \Rightarrow \hat{i}+\gamma (z\hat{i}+9\hat{j}+5\hat{x}\\ \Rightarrow x\hat{ı}+y\hat{ȷ}+z\hat{x}\\ \Rightarrow (1+2\gamma )\hat{i}+(9\gamma )\hat{j}+(5\gamma )\hat{x}\end{array}$

On comparing both side we get,

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

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

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

Hence

Direction ratio of PQ is

$\begin{array}{rl}& =(2\gamma +1-5),(9\gamma -4),(5\gamma +1)\\ & =(2\gamma -4),(9\gamma -4),(5\gamma +1)\end{array}$
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.

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

:: Co-ordinates of Q.,

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

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

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

By putting the value of γ in Q

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

Now,

Distance between PQ

$\begin{array}{rl}& =\sqrt{(x_2-x_1)+{(y_2-y_1)}^2+{(z_2-z_1)}^2}\end{array}$

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

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

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 $\hat{i}+(6\hat{j}+3\hat{x})$and equation line $\hat{r}=\hat{j}+2\hat{x}+\gamma (\hat{i}+2\hat{j}+3\hat{x}).$
Solution – Let PQ be the perpendicular drawn from $P(\hat{i}+6\hat{j}+3\hat{x})$ to given line whose endpoint/foot is Q point.
Q is on line.
$=\hat{r}=\hat{ȷ}+2\hat{x}+\gamma (\hat{ı}+2\hat{ȷ}+3\hat{x})$
$=\gamma \hat{i}+(2\gamma +1)\hat{j}+(3\gamma +2)\hat{x}isthepositionvectorofQ$
Hence,
$=\overrightarrow{PQ}=PositionvectorofQ-Positionvectorof$
$=\overrightarrow{PQ}=\gamma \hat{ı}+(2\gamma +1)\hat{ȷ}+(3\gamma +2)\hat{x}-(\hat{ı}+6\hat{ȷ}+3\hat{x})$
$=\overrightarrow{PQ}=(\gamma -1)\hat{ı}+(2\gamma +1)\hat{ȷ}+(3\gamma +2)\hat{x}-\hat{ı}-6\hat{ȷ}-3\hat{x}$$=\overrightarrow{PQ}=(\gamma -1)\hat{ı}+(2\gamma -5)\hat{ȷ}+(3\gamma -1)\hat{x}$

:: PQ is perpendicular on line

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

:: Their dot product is zero.

Compare given line equation with

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

$\begin{array}{rl}& =\overrightarrow{PQ}\cdot \hat{b}=0\end{array}$

$\begin{array}{rl}& =\{(\gamma -1)\hat{ı}+(2\gamma -5)\hat{ȷ}+(3\gamma -1)\hat{x}\}\cdot \{(\hat{ı}+2\hat{ȷ}+3\hat{x})\}=0\end{array}$

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

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

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

$\begin{array}{rl}& \gamma \hat{i}+(2\gamma +1)\hat{j}+(3\gamma +2)\hat{x}\end{array}$

$\begin{array}{rl}& =\hat{ı}+(2+1)\hat{ȷ}+(3+2)\hat{x}\\ & =\hat{ı}+3\hat{ȷ}+5\hat{x};\text{ foot of perpendicular }\end{array}$

Putting the value of γ in PQ.

$\begin{array}{rl}& =\overrightarrow{PQ}=(\gamma -1)\hat{ı}+(2\gamma -5)\hat{ȷ}+(3\gamma -1)\hat{x}\\ & =\overrightarrow{PQ}=(1-1)\hat{ı}+(2-5)\hat{ȷ}+(3-1)\hat{x}\\ & =\overrightarrow{PQ}=-3\hat{ȷ}+2\hat{x}\end{array}$

Now, Magnitude of PQ , we know that

$\begin{array}{rl}& |AB|=\sqrt{x^2+y^2+z^2}\end{array}$

Hence

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

Straight Line in Space Exercise 27.4 Question 9

Answer- The answer of the given question is $(\frac{-4}{7},\frac{12}{7},\frac{15}{7})\cdot \hat{r}=(-\hat{ı}+3\hat{ȷ}+2\hat{k})+\gamma \{(\frac{-4}{7}\hat{ı}+\frac{12}{7}\hat{ȷ}+\frac{15}{7}\hat{k})-(\hat{ı}+3\hat{ȷ}+2\hat{k})\}$
(Hint- By comparing both the equation).
Given – Point P(-1,3,2) and equation of line $\hat{r}=(2\hat{j}+3\hat{k})+\gamma (2\hat{i}+\hat{j}+3\hat{k}).$
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$\begin{array}{rl}& =\hat{r}=(2\hat{ȷ}+3\hat{k})+\gamma (2\hat{ı}+\hat{ȷ}+3\hat{k})\\ & =x\hat{ı}+y\hat{ȷ}+z\hat{k}=(2\hat{ȷ}+3\hat{k})+\gamma (2\hat{ı}+\hat{ȷ}+3\hat{k})\\ & =x\hat{ı}+y\hat{ȷ}+z\hat{k}=(2\gamma )\hat{ı}+(\gamma +2)\hat{ȷ}+(3\gamma +3)\hat{k}\end{array}$

On comparing both sides we get,

$=\frac{x}{2}=\frac{y-2}{1}=\frac{z-3}{3}=\gamma ;equationofline$

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

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

$\begin{array}{rl}& \mathrm{Q}(\gamma ,\gamma +2,3\gamma +3)\end{array}$

Hence

Direction of PQ is

$\begin{array}{rl}& =(2\gamma +1),(\gamma +2-1),(3\gamma +3-2)\\ & =(2\gamma +1),(\gamma -1),(3\gamma +1)\end{array}$
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”.
$\begin{array}{rl}& a_1{a}_2+b_1{b}_2+c_1{c}_2=0\end{array}$

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

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

:: Co-ordinates of Q,

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

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

Position vector of Q

$\begin{array}{rl}& =\frac{-4}{7}\hat{ı}+\frac{12}{7}\hat{ȷ}+\frac{15}{7}\hat{k}\end{array}$

Now, Equation of line passing through two points with position vector $\hat{a}and\hat{b}isgivenby$

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

$=Here,$

$\begin{array}{c}=\hat{a}=-\hat{ı}+3\hat{ȷ}+2\hat{k}\end{array}$

$\begin{array}{c}=\text{ and }\hat{b}=\frac{-4}{7}\hat{ı}+\frac{12}{7}\hat{ȷ}+\frac{15}{7}\hat{k}\end{array}$

$\begin{array}{c}=\hat{r}=(-\hat{ı}+3\hat{ȷ}+2\hat{k})+\gamma [(\frac{-4}{7}\hat{ı}+\frac{12}{7}\hat{ȷ}+\frac{15}{7}\hat{k})-(-\hat{ı}+3\hat{ȷ}+2\hat{k})].\end{array}$

Straight Line in Space Exercise 27.4 Question 10

Answer – The answer of the question is $Q(\frac{-3}{2},\frac{-1}{2},4)$
(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.$\begin{array}{rl}& \ast \frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}=\gamma (\text{ let })\\ & \ast \mathrm{x}=-\gamma -2,y=3\gamma +1,z=-2\gamma +3\\ & ::\text{ Co-ordinates of }\mathrm{Q}(-\gamma -2,3\gamma +1,-2\gamma +3)\end{array}$

Hence,

Direction of PQ is

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

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.

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

:: Co-ordinates of Q ,

By putting the value of $\gamma$ in Q

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