RD Sharma Class 12 Exercise 27 FBQ Straight line in space Solutions Maths

RD Sharma Class 12 Exercise 27 FBQ Straight line in space Solutions Maths - Download PDF Free Online

Edited By Lovekush kumar saini | Updated on Jan 25, 2022 09:48 AM IST

The class 12 students work hard on preparing for their public exams. They do not get much time in visiting tuitions for extra classes beyond the school timings. A subject like mathematics requires many tricks to solve the sums quickly during examinations. The Straight Line in Space is a challenging concept that is tricky for the students to solve. RD Sharma solutions Best solution guides like the RD Sharma Class 12th Chapter 27 FBQ lend a helping hand to the class 12 students to understand the concept in-depth.

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RD Sharma Class 12 Solutions Chapter27FBQ Straight line in space - Other Exercise

Straight Line in Space Excercise: 27 FBQ

Straight Line in Space Exercise Fill in the blanks Question 1

Answer : $\vec{r}=\lambda \hat{i}, \vec{r}=\mu \hat{\jmath}, \vec{r}=v \hat{k}$
Hint : Use vector equations
Given : The vector equations of OX,OY and OZ are
Solution:
$\text{Let}\text { OX,OY ad OZ be } \lambda, \mu \text { and } v$
$\therefore O \mathrm{X}=\lambda \hat{\imath} \quad, \mathrm{OY}=\mu \hat{\jmath}, \mathrm{OZ}=v \hat{k}$

Straight Line in Space Exercise Fill in the blanks Question 2

Answer : $(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(-3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k})$
Hint : Compare the equation with the formula
Given : $\frac{5-x}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
Solution : We have given line as
$\frac{x-5}{-3}=\frac{y+4}{7}=\frac{z-6}{2}$
By comparing with equation
$\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}$
The given line passes through the point (x1 , x2, x3)
i.e. (5 , -4 , 6) and direction ratios are (a ,b ,c) i.e. (-3 , 7 ,2)
Now , we can write vector equation of the line as
$\vec{A}=(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(-3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k}$

Straight Line in Space Exercise Fill in the blanks Question 3

Answer : $\vec{r}=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}+\lambda(-2 \hat{\imath}-5 \hat{\jmath}+13 \hat{k})$
Hint : Use the vector equation
Given : Point (3 , 4 , -7) and ( 1 , -1 , 6)
Solution : Given points are
$\begin{aligned} &\mathrm{A}(3,4,-7), \mathrm{B}(1,-1,6) \\ & \end{aligned}$
$\vec{A}=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k} \\$
$\vec{B}=\hat{i}-\hat{\jmath}+6 \hat{k}$
∴ vector equation
$\begin{aligned} \vec{r} &=(3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k})+\lambda[\hat{\imath}-\hat{\jmath}+6 \hat{k}-(3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k})] \\ & \end{aligned}$
$=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}+\lambda(-2 \hat{\imath}-5 \hat{\jmath}+13 \hat{k})$

Straight Line in Space Exercise Fill in the blanks question 4

Answer : $\pm \frac{1}{\sqrt{3}}$
Hint : $l^{2}+m^{2}+n^{2}=1$
Given : $l,l,l$ are direction cosine of a line ,
Solution : Since , direction cosine of a line are $l,l\; \text{and} \: l$
$\therefore l=l, \mathrm{~m}=l \text { and } \mathrm{n}=l$
We know that , $l^{2}+m^{2}+n^{2}=1$
$\begin{aligned} &=>l^{2}+l^{2}+l^{2}=1 \\ & \end{aligned}$
$\Rightarrow>l^{2}=\frac{1}{3} \\$
$\therefore l=\pm \frac{1}{\sqrt{3}}$

Straight Line in Space Exercise Fill in the blanks question 5

Answer : $l=\frac{1}{\sqrt{3}}, m=\frac{1}{\sqrt{3}}, n=\frac{-1}{\sqrt{3}}$
Hint : Use straight line equation
Given : $\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+1}{n}$ passes through the point ( -1 , 0 , 1)
$\frac{x-1}{l}=\frac{y-2}{m}=\frac{z+1}{n} \Rightarrow \frac{x-1}{1}=\frac{y-2}{1}=\frac{z+1}{-1}$

By using direction ratios of line
$\begin{gathered} l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}} \text { And } n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}} \\ \end{gathered}$
$\Rightarrow l=\frac{1}{\sqrt{3}}, m=\frac{1}{\sqrt{3}}, n=\frac{-1}{\sqrt{3}}$

Straight Line in Space Exercise Fill in the blanks question 6

Answer : $\frac{x+2}{1}=\frac{y-3}{1}=\frac{z-4}{1}$
Hint : Use equation of line
Given : (-2 , 3 , 4) and equally inclined with the coordinate axes
OX , OY and OZ are -----------
Solution : Equation of line passing through (x1 , y1 , z1)
And having direction cosines l,m,n is
$\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$
Also line is equally inclined to co-ordinate axes
Here $(x_{1}, y_{1}, z_{1})=(-2,3,4)$
∴ Equation of line is
$\Rightarrow \frac{x+2}{1}=\frac{y-3}{1}=\frac{z-4}{1}$

Straight Line in Space Exercise Fill in the blanks Question 7

Answer : $0^{\circ}$
Hint : Use vector dot product
Given : $a, b, c \text { and } \frac{1}{b c}, \frac{1}{c a}, \frac{1}{a b}$
Solution : Let $\overrightarrow{m 1}$ and $\overrightarrow{m 2}$ be vectors parallel to the two given lines.
$\therefore \overrightarrow{m 1}=a \hat{i}+b \hat{j}+c \hat{k}$
$And\; \overrightarrow{m 2}=\frac{1}{b c} i+\frac{1}{c a} j+\frac{1}{a b} \hat{k}$
Let θ be angle between given lines
$\therefore \theta \text { is angle between } \overrightarrow{m 1} \text { and } \overrightarrow{m 2}$
$\therefore \cos \theta=\frac{\overrightarrow{m 1} \cdot \overrightarrow m 2}{|\overrightarrow{m 1}||\overrightarrow{m 2}|}$ $=\frac{a\left(\frac{1}{b c}\right)+b\left(\frac{1}{c a}\right)+c\left(\frac{1}{a b}\right)}{\sqrt{a^{2}+b^{2}+c^{2}} \sqrt{\left(\frac{1}{b c}\right)^{2}+\left(\frac{1}{c a}\right)^{2}+\left(\frac{1}{a b}\right)^{2}}}=1$
$\therefore \theta=0^{\circ}$

Straight Line in Space Exercise Fill in the blanks Question 8

Answer : $\frac{x-a}{b}=\frac{y-b}{c}=\frac{z-c}{a}$
Hint : Use equation for straight line
Given : $(a , b , c)\; and \; (a-b , b-c , c-a )$
Solution :

Equation of line passing through $\left ( x,y,z \right )$ and having drs ;l,m,n is
$\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$
Hence $\left ( x,y,z \right )= \left ( a,b,c \right )$
So, Equation of line $AB= \frac{x-a}{b}=\frac{y-b}{c}=\frac{z-c}{a}$

Straight Line in Space Exercise Fill in the blanks Question 9

Answer : $\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$
Hint : Use direction equation of $\text { cosine }\left\{\frac{x_{2}-x_{1}}{P Q}, \frac{y_{2}-y_{1}}{P Q}, \frac{z_{2}-z_{1}}{P Q}\right\}$
Given : ( a , b, c) and parallel to z- axis
Solution : The direction cosine of line parallel to Z-axis are ( 0 , 0 , 1 )
Since, the line passes through ( a, b , c)
Thus , its equation is $\frac{x-a}{0}=\frac{y-b}{0}=\frac{z-c}{1}$

Straight Line of Space Excercise Fill in the blanks Question 10

Answer : $-\frac{10}{7}$
Hint : Use vector dot product and vector equation of line.
Given : $\begin{aligned} &\frac{x-1}{-3}=\frac{y-2}{2 k}=\frac{z-3}{2} \text { and } \frac{x-1}{3 k}=\frac{5-y}{-1}=\frac{6-z}{5} \\ & \end{aligned}$
$\text { let } \overrightarrow{\mathrm{b} 1}=-3 \hat{\imath}+2 \mathrm{k} \hat{\mathrm{j}}+2 \hat{k} \text { and } \overrightarrow{\mathrm{b} 2}=3 \mathrm{k} \hat{\imath}+\hat{\jmath}-5 \hat{k}$
$\overrightarrow{\mathrm{b} 1} \cdot \overrightarrow{\mathrm{b} 2}=0$ as the lines are at right angle
$\begin{aligned} &-9 k+2 k-10=0 \\ & \end{aligned}$
$-7 k=10 \\$
$k=-\frac{10}{7}$

Straight Line of Space Excercise Fill in the blanks Question 11

Answer : $\frac{x-2}{1}=\frac{y-1}{1}=\frac{z+1}{\sqrt{2}}$
Hint : Compare the given equation
Given : ( 2 , 1 , -1 ) and making angles $\frac{\pi}{3}, \frac{\pi}{3} \text { and } \frac{\pi}{4}$
Solution : So comparing with general Cartesian from of Eq.
a = 2 , b =1 , c = -1
another line passing through P ( 1 , 1 , 2)
we can assume it as position vector and parallel to given line so normal vector will be same .
Eq. of line is given by $\frac{x-x_{1}}{l}=\frac{y-y_{1}}{m}=\frac{z-z_{1}}{n}$
Here $\begin{gathered} l ,m, n(1,1, \sqrt{2}) \\ \end{gathered}$
$\frac{x-2}{1}=\frac{y-1}{1}=\frac{z+1}{\sqrt{2}}$

Straight Line of Space Excercise Fill in the blanks Question 12

Answer : y = 0 , z = 0
Hint : x will not be 0
Given : The equation of x- axis in unsymmetrical form
Solution : y = 0 & z = 0
Point ( x , 0 , 0) points on z axis.

Straight Line in Space Exercise Fill in the blanks Question 13

Answer : $\frac{x}{0}=\frac{y}{1}=\frac{z}{0}$
Hint : y-axis lies on xy plane and yz.
Given : The equation of y-axis in symmetrical form.
Solution : As y-axis lies on xy plane and yz
So , its distance from xy and yz plane is O .
∴ By basic definition of 3-D coordinate we can say that
x-coordinate and z-coordinate are 0.
∴ any point on y-axis has x and z coordinate = 0
∴ x = 0 and z = 0 can be considered as the equation of y-axis.
Hence , the correct answer is $\frac{x}{0}=\frac{y}{1}=\frac{z}{0}$

Straight Line in Space Exercise Fill in the blanks Question 14

Answer : The required answer is $2$
Hint : The condition of coplanar is
$(\overrightarrow{a 2}-\overrightarrow{a 1}) \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0$ -----1
Given : $\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a} 1}+\lambda \overrightarrow{\mathrm{b} 1} \text { and } \overrightarrow{\mathrm{r}}=2 \overrightarrow{\mathrm{a} 2}+\lambda \overrightarrow{\mathrm{b} 2}$ ------2
Solution : From Eq.1 and Eq.2
$\begin{aligned} &(\overrightarrow{a 2}-\overrightarrow{a 1}) \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0 \\ & \end{aligned}$
$(2 \overrightarrow{a 2}-\overrightarrow{a 1}) \cdot\lfloor{b 2} \times \overrightarrow{b 1})=0$
$\begin{gathered} 2(\overrightarrow{a 2} \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1}))-\overrightarrow{a 1} \cdot(\vec{b} 2 \times \overrightarrow{b 1})=0 \\ \end{gathered}$
$2 \overrightarrow{a 2} \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})-\overrightarrow{a 1} \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0$ -------3
Now $\frac {[\overrightarrow{a 1}\overrightarrow{b 1}\overrightarrow{b2} ]}{[\overrightarrow{a 2}\overrightarrow {b1}\overrightarrow{b2} ]}$ $= \frac{\overrightarrow{a1}.\left ( \overrightarrow{b1}\times \overrightarrow{b2}\right )}{\overrightarrow{a2}.\left ( \overrightarrow{b1}\times \overrightarrow{b2}\right )}$ [Using scalar triple formula]
$= \frac{2\overrightarrow {a2}\cdot \left ( \overrightarrow{b1}\times\overrightarrow{b2} \right )}{\overrightarrow {a2}\cdot \left ( \overrightarrow{b1}\times\overrightarrow{b2} \right )}$ Substituting the values in Numerator
$= 2$

Straight Line in Space Exercise Fill in the blanks Question 15

Answer : The value of k is $2$
Hint : The condition of coplanar is
$(\overrightarrow{a 2}-\overrightarrow{a 1}) \cdot(\overrightarrow{b 2} \times\overrightarrow{b 1})=0$
Given : The lines are ,
$\overrightarrow{\mathrm{r}}=2 \overrightarrow{\mathrm{a} 1}-3 \overrightarrow{\mathrm{a} 2}+\lambda \overrightarrow{\mathrm{b} 1}$ -------1
$\overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a} 2}+\mu \overrightarrow{\mathrm{b} 2}$ ------2
Solution : From the condition of co planarity
$\begin{aligned} &(\overrightarrow{a 2}-\overrightarrow{a 1}) \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0 \\ \end{aligned}$
$\Rightarrow (\overrightarrow{a 2}-2 \vec{a} 1+3 \overrightarrow{a 2}) \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0 \\$
$\Rightarrow (4 \vec{a} 2-2 \vec{a}) \cdot(\overrightarrow{b 2} \times \overrightarrow{b 1})=0$
$\Rightarrow 2(2 \overrightarrow{\mathrm{a} 2}-\overrightarrow{\mathrm{a} 1}) \cdot \overrightarrow{\mathrm{b} 2} \times \overrightarrow{\mathrm{b} 1})=0 \\$
$\Rightarrow(2 \overrightarrow{\mathrm{a} 2}-\overrightarrow{\mathrm{a} 1}) \cdot(\overrightarrow{\mathrm{b} 2} \times \overrightarrow{\mathrm{b} 1})=0 \\$
$\Rightarrow 2 \overrightarrow{\mathrm{a} 2} \cdot(\overrightarrow{\mathrm{b} 2} x \overrightarrow{\mathrm{b} 1})-\overrightarrow{\mathrm{a} 1} \cdot(\overrightarrow{\mathrm{b} 2} x \overrightarrow{\mathrm{b} 1})=0 \\$
$\Rightarrow 2[\overrightarrow{\mathrm{a} 2 \mathrm{~b} 1 \mathrm{~b} 2}]=[\overrightarrow{\mathrm{a} 1 \mathrm{~b} 1 \mathrm{~b} 2}]$
$a \cdot(b \times c)=\left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=0$ [ If a,b,c are coplanar ]
$\begin{aligned} &\Rightarrow \quad[\overrightarrow{a 1}\overrightarrow{b1}\overrightarrow{b2}]=2[\overrightarrow{a 2} \overrightarrow{b1}\overrightarrow{b 2}] \\ & \end{aligned}$
$\Rightarrow \quad k=2$
So , the value of $k= 2$

Straight Line in Space Exercise Fill in the blanks question 16

Answer : $\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}$
Hint : $\frac{x}{a}+\frac{y}{b}=1$ where a and b are non zero.
Given : The equation of x-axis in symmetric form.
Solution : It should be noted that the symmetric form of the equation
of a line does not make it possible to find the slope of a line
∴ x in symmetric form
$\frac{x-0}{1}=\frac{y-0}{0}=\frac{z-0}{0}$

Straight Line in Space Exercise Fill in the blanks question 17

Answer : $y =0 , z = 0$
Hint : x will not be zero in its coordinates.
Given : The equation of x-axis in unsymmetrical form
Solution : $y =0 , z = 0$
Point ( x, 0, 0 ) points on x-axis.

Straight Line in Space Exercise Fill in the blanks question 18

Answer : $3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}+\lambda(-2 \hat{\imath}-5 \hat{\jmath}+13 \hat{k})$
Hint : Use the vector equation
Given : (3 , 4 , -7) and ( 1 , -1 , 6)
Solution : position vector of (3 , 4 , -7) is
$\vec{a}=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}$
And position vector of ( 1 , -1 , 6) is
$\vec{b}=\hat{\imath}-\hat{\jmath}+6 \hat{k}$
Also,
The vector equation of a line which passes through two
Points whose position vectors are $\vec{a} \text { and } \vec{b}$ is
$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
Hence , required equation of line is
$\begin{aligned} &\Rightarrow \vec{r}=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}+\lambda[\hat{\imath}-\hat{\jmath}+6 \hat{k}-(3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k})] \\ & \end{aligned}$
$\Rightarrow \vec{r}=3 \hat{\imath}+4 \hat{\jmath}-7 \hat{k}+\lambda(-2 \hat{\imath}-5 \hat{\jmath}+13 \hat{k})$

Straight Line in Space Exercise Fill in the blanks Question 19

Answer : $(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k})$
Hint : Compare the equation with the formula
Given : $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
Solution : We have given line as
$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$
By comparing with equation
$\frac{x-x 1}{a}=\frac{y-y 1}{b}=\frac{z-z 1}{c}$
We get given line passes through the point (x1 , x2, x3)
i.e. (5 , -4 , 6) and direction ratios are (a ,b ,c) i.e. ( 3 , 7 , 2)
Now , we can write vector equation of the line as
$\vec{A}=(5 \hat{\imath}-4 \hat{\jmath}+6 \hat{k})+\lambda(3 \hat{\imath}+7 \hat{\jmath}+2 \hat{k})$

Straight Line in Space Exercise Fill in the blanks Question 20

Answer : Perpendicular
Hint : Angle should be $90^{\circ}$
Given : The line of shortest distance between two skew-lines is
-------- to both the lines.
Solution : The line of shortest distance between two skew-lines is
perpendicular to both the lines.
Two or more lines which have no intersections but are not parallel are called skew lines

Class 12, mathematics, chapter 27, Straight Line in Space consists of 5 exercises, ex 27.1 to ex 27.5. There are 20 questions in the Chapter 27 FBQ section. The concepts present in this section are vector equations, cartesian equations, direction ratios, the shortest distance between the lines, direction cosines, equation of a plane, angle between two lines, etc. Even though the chapter is a little easy, when the students are asked these questions in the Fill in the Blanks format, they tend to get confused. And this is why the role of the RD Sharma class 12 Chapter 27 FBQ solution book is inevitable.

All the answers provided in the RD Sharma Class 12th Chapter 27 FBQ book are framed by a team of mathematical experts. They have contributed their knowledge and skill for the welfare of the students. However, only with a good amount of practice, this section can be attended with full confidence. Here is where the additional practice questions present in the Class 12 RD Sharma Chapter 27 FBQ Solution play a major role. The more sums students work out, the better they understand the concept and increase the speed in completing the sums.

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