Who is this material for?

This material is prepared for class 12 CBSE students who want to gain more insight on the subject and score good marks in exams.

How can I use this material?

Students can use this material as a guide for the textbook and practice from 8 to improve their performance.

How are RD Sharma books better than NCERT?

NCERT books are good for basic knowledge but RD Sharma books are much more informative and contain a lot of important concepts.

Is the entire material available online?

Yes, students can access this material through courier 360s website using any device with a browser.

can I save preparation time through this material?

Yes, students can save a lot of time by preparing from this material as it contains questions and answers in the same place.

RD Sharma Class 12 Exercise 27.3 Straight line in space Solutions Maths - Download PDF Free Online

Schools
RD Sharma Solutions
RD Sharma Class 12 Exercise 27.3 Straight line in space Solutions Maths - Download PDF Free Online

RD Sharma Class 12 Exercise 27.3 Straight line in space Solutions Maths - Download PDF Free Online

Updated on 25 Jan 2022, 09:47 AM IST

The RD Sharma class 12 solution of Straight line in space exercise 27.3 is used by thousands of students and teachers for the practical knowledge of maths. The RD Sharma class 12th exercise 27.3 consists of 10 questions including subparts which explores vector and cartesian form, finding the point of intersection between the two lines.

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

Straight Line in Space Excercise: 27.3

Straight Line in Space exercise 27.3 question 1

Answer: The given line intersect at a point $(2, 6, 3).$

Hint: By putting the value of $\gamma$.

Given:$\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}$ …(i)
and $\frac{x-2}{2}=\frac{y-6}{3}=\frac{z-3}{4}$ …(ii)

Solution:$\frac{x}{1}=\frac{y-2}{2}=\frac{z+3}{3}=\gamma$ (say)
Then $\mathrm{x}=\gamma, y=2 \gamma+2, z=3 \gamma-3$
Substituting $x,y,z$ in eqn (ii)
$\frac{\gamma-2}{2}=\frac{2 \gamma+2-6}{3}=\frac{3 \gamma-3-3}{4}$
Taking first and second ratios, we get
$\begin{aligned} &\frac{\gamma-2}{2}=\frac{2 \gamma-4}{3} \\\\ &\Rightarrow 3 \gamma-6=4 \gamma-8 \\\\ &\Rightarrow \gamma=2 \end{aligned}$
Now taking second and third ratios, we get
$\begin{aligned} &\frac{2 \gamma-4}{3}=\frac{3 \gamma-6}{4} \\\\ &\Rightarrow 8 \gamma-16=9 \gamma-18 \\\\ &\Rightarrow \gamma=2 \end{aligned}$
The line intersects at a point $(\gamma, 2 \gamma+2,3 \gamma-3)$
taking $\gamma=2$
Points will be $(2, 6, 3).$

Straight Line in Space exercise 27.3 question 2

Answer: The given line don’t intersect each other.

Hint: By substituting $\lambda=-17 \text { and } \mu=-12$

Given:$\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5} \text { and } \frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}$

Solution: The co- ordinates of any point on the first line are given by
$\begin{aligned} &\frac{x-1}{3}=\frac{y+1}{2}=\frac{z-1}{5}=\lambda(\text { say }) \\ \end{aligned}$
$\begin{aligned} &\bullet\text {} -x=3 \lambda+1 \\\\ &\bullet y=2 \lambda-1 \\\\ &\bullet z=5 \lambda+1 \end{aligned}$
The co-ordinates of first line is $(3 \lambda+1,2 \lambda-1,5 \lambda+1)$
The co- ordinates of any point on the second line are given by
$\frac{x+2}{4}=\frac{y-1}{3}=\frac{z+1}{-2}=\mu(s a y)$
$\begin{aligned} &\text { } \bullet \mathrm{x}=4 \mu-2 \\\\ &\text { } \bullet \mathrm{y}=3 \mu+1 \\\\ &\text { } \bullet \mathrm{z}=-2 \mu-1 \end{aligned}$
The co-ordinates of second line is $(4 \mu-2,3 \mu+1,-2 \mu-1)$
If the line intersect, then they have a common point. So, for some value of λ and μ.
We must have,
$3 \lambda+1=4 \mu-2, \quad 2 \lambda-1=3 \mu+1, \quad 5 \lambda+1=-2 \mu-1$
$\begin{aligned} &\Rightarrow 3 \lambda-4 \mu=-3\quad \ldots(i)\\ &\Rightarrow 2 \lambda-3 \mu=2 \quad \ldots(ii)\\ &\Rightarrow 5 \lambda+2 \mu=-2\quad \ldots(iii) \end{aligned}$
Solving (i)and (ii), we get
$\begin{aligned} &\lambda=-17 \\ &\mu=-12 \end{aligned}$
Substituting $\lambda=-17 \text { and } \mu=-12$ in (iii)
$L.H.S$
$\begin{aligned} &=5 \lambda+2 \mu \\ &=5(-17)+2(-12) \end{aligned}$
$\begin{aligned} &=-85-24 \\ &=-109 \\ &\neq-2 \\ &\text { L.H.S } \neq R . H . S \end{aligned}$

This shows that the given line will not intersect each other.

Straight Line in Space exercise 27.3 question 3

Answer: The answer is $\left(\frac{1}{2}, \frac{-1}{2}, \frac{-3}{2}\right)$

Hint: $\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda \text { and } \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu(\text { say })$

Given:$\frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7} \text { and } \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}$

Solution: we have,
$\begin{gathered} \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda(s a y) \\ x=3 \lambda-1, y=5 \lambda-3, z=7 \lambda-5 \end{gathered}$
So, the co-ordinates of a general point on this line are $(3 \lambda-1,5 \lambda-3,7 \lambda-5)$
The equation of second line is given
$\begin{aligned} &\frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu(\text { say }) \\ &\mathrm{x}=\mu+2, \quad y=3 \mu+4, \quad z=5 \mu+6 \end{aligned}$
So, the co-ordinates of general point on this line are $(\mu+2,3 \mu+4,5 \mu+6)$
If the line intersect, then they have a common point. So, for some value of λ and μ,we must have
$\begin{aligned} &3 \lambda-1=\mu+2, \quad 5 \lambda-3=3 \mu+4, \quad 7 \lambda-5=5 \mu+6\\ &3 \lambda-\mu=3\ldots(i)\\ &5 \lambda-3 \mu=7\ldots(ii)\\ &7 \lambda-5 \mu=11 \quad \ldots(i i i) \end{aligned}$
Solving the first two equations (i) and (ii), we get
$\begin{aligned} &\lambda=\frac{1}{2} \\\\ &\mu=\frac{-3}{2} . \end{aligned}$
Since,$\lambda=\frac{1}{2} \text { and } \mu=\frac{-3}{2}$ satisfying the eqn(iii) $7 \lambda-5 \mu=11$

Hence,the given lines intersect each other.
When $\lambda=\frac{1}{2} \text { in }(3 \lambda-1,5 \lambda-3,7 \lambda-5)$
the co-ordinates of the required point of intersection are $\left(\frac{1}{2}, \frac{-1}{2}, \frac{-3}{2}\right)$

Straight Line in Space exercise 27.3 question 4

Answer: The answer of the given equation is $(10,14,4).$

Hint: By substituting the value of λ and μ.

Given:$A(0,-1,-1), B(4,5,1) , C(3,9,4) \; and \; D(-4,4,4).$

Solution: The co-ordinates of any point on the line $AB$ are given by
$\begin{aligned} &\frac{x-0}{4-0}=\frac{y+1}{5+1}=\frac{z+1}{1+1}=\lambda \\\\ &\Rightarrow x=4 \lambda \\\\ &\Rightarrow y=6 \lambda-1 \\\\ &\Rightarrow z=2 \lambda-1 \end{aligned}$
The co-ordinates of general point on $AB$ are
$(4 \lambda, 6 \lambda-1,2 \lambda-1)$
The co-ordinates of any point on the line $CD$ are given by
$\begin{aligned} &\frac{x-3}{-4-3}=\frac{y-9}{4-9}=\frac{z-4}{4-4}=\mu \\\\ &\Rightarrow x=-7 \mu+3 \\\\ &\Rightarrow y=-5 \mu+9 \\\\ &\Rightarrow z=4 \end{aligned}$
The co-ordinates of general point on $CD$ are
$(-7 \mu+3,-5 \mu+9,4)$
If the line intersect, then they have a common point. So, for some value of λ and μ.
We must have,
$\begin{aligned} &4 \lambda=-7 \mu+3, \quad 6 \lambda-1=-5 \mu+9,2 \lambda-1=4\\\\ &4 \lambda+7 \mu=3 \quad \ldots(i)\\\\ &6 \lambda+5 \mu=10 \quad \ldots(ii)\\\\ &\lambda=\frac{5}{2} \quad \cdots(iii) \end{aligned}$
Solving (ii) and (iii), we get
$\begin{aligned} &\lambda=\frac{5}{2} \\\\ &\mu=-1 \end{aligned}$
Substituting$\lambda=\frac{5}{2} \text { and } \mu=-1$ in (i),we get
$\begin{aligned} &\text { L.H.S } \\\\ &=4 \lambda-7 \mu \\\\ &=4\left(\frac{5}{2}\right)-7(1) \\\\ &=3 \\ &\text { R.H.S } \end{aligned}$
Since $\lambda=\frac{5}{2} \text { and } \mu=1$ satisfying eqn(iii),
Hence, the given line intersect.
Substituting the value of λ and μ in the Co- ordinates of a general point on the line $AB$ and $CD$, we get
$\begin{aligned} &x=10 \\ &y=14 \\ &z=4 \end{aligned}$

Hence, $AB$ and $CD$ intersect at point $(10, 14, 4).$

Straight Line in Space exercise 27.3 question 5

Answer: The answer of the given question is $(4,0,-1).$

Hint: Equate the coefficient of vector equation of line.

Given: vector $r=(\hat{\imath}+\hat{\jmath}-\hat{k})+\lambda(3 \hat{\imath}-\hat{\jmath}) \text { and vector } r=(4 \hat{\imath}-\hat{k})+\mu(2 \hat{\imath}+3 \hat{k})$

Solution: General points on the lines are
$(1+3 \lambda) \hat{\imath}+(1-\lambda) \hat{\jmath}-\hat{k} \text { and }(4+2 \mu) \hat{\imath}+(3 \mu-1) \hat{k}$
Lines intersect if
$\begin{aligned} &(1+3 \lambda)=(4+2 \mu)\\\\ &1-\lambda=0\\\\ &\text { and } 3 \mu-1=-1 \end{aligned}$
From (ii) and (iii), $\lambda=1, \mu=0$
Substituting in equation (i)
Since,$1+3(1)=4+2(0)$ is true.
Hence, the lines intersect.
Point of intersection is :$4 \hat{\imath}-\hat{k} \text { or }(4,0,-1)$

Straight Line in Space exercise 27.3 question 6(i)

Answer: The answer of the given question is that the line do not intersect each other.

Hint: Assume $\lambda=\lambda_{1} \text { and } \mu=\mu_{1}$

Given: $\vec{r}=(\hat{\imath}-\widehat{\jmath})+\lambda(2 \hat{\imath}+\hat{k}) \text { and } \vec{r}=(2 \hat{\imath}-\hat{\jmath})+\mu(\hat{\imath}+\hat{\jmath}-\hat{k})$

Solution: let’s first assume that the line intersect for $\lambda=\lambda_{1} \text { and } \mu=\mu_{1}$
Then, $\left(2 \lambda_{1}+1\right) \hat{\imath}-\hat{\jmath}+\lambda_{1} \hat{k}=\left(2+\mu_{1}\right) \hat{\imath}+\left(\mu_{1}-1\right) \hat{\jmath}-\mu_{1} \hat{k}$
Equations we get,
$\begin{aligned} &\left(2 \lambda_{1}+1\right)=\left(2+\mu_{1}\right) &\ldots(i)\\\\ &-1=\mu_{1}-1 &\cdots(ii)\\\\ &\lambda_{1}=-\mu_{1} &\ldots(iii) \end{aligned}$
Solving equation (ii) and (iii), we get $\lambda_{1}=\mu_{1}=0$
Which doesn’t satisfy the equation (i) which is a contradiction.
Thus, the above lines are skew lines i.e. they neither intersect nor parallel to each other.

Straight Line in Space exercise 27.3 question 6(ii)

Answer: The given line do not intersect.

Hint: If $\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=0$ then the lines intersect each other.

Given:$\frac{x-1}{2}=\frac{y+1}{3}=z \text { and } \frac{x+1}{5}=\frac{y-2}{1} ; z=2$

Solution:$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-0}{1} \text { and } \frac{x+1}{5}=\frac{y-2}{1}=\frac{z-2}{0}$
The first line passes through the point (1,-1,0) and has direction ratio proportional to 2,3,1 and its vector equation is
$\vec{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}}$ …(i)
Here,$\begin{aligned} &\overrightarrow{a_{1}}=\hat{i}-\hat{j}+0 \widehat{k} \\\\ &\overrightarrow{b_{1}}=2 \hat{i}+3 \hat{j}+\widehat{k} \end{aligned}$
Also, the second line passes through the point $(-1,2,2)$ and has direction ratios proportional to $5,1,0.$
Its vector equation is
$\vec{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}}$ ...........(ii)
Here,
$\begin{aligned} &\overrightarrow{a_{2}}=-\hat{i}+2 \hat{j}+2 \widehat{k} \\\\ &\overrightarrow{b_{2}}=5 \hat{i}+\hat{j}+0 \widehat{k} \end{aligned}$
Now,
$\overrightarrow{a_{2}}-\overrightarrow{a_{1}}=-2 \hat{i}+3 \hat{j}+2 \widehat{k}$
And
$\begin{aligned} &\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\left|\begin{array}{lll} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 1 \\ 5 & 1 & 0 \end{array}\right| \\\\ &=-\hat{\imath}+\overline{5}-13 \hat{k} \\\\ &\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)=(-2 \hat{i}+3 \hat{j}+2 \widehat{k}) \cdot(-\hat{i}+5 \hat{j}-13 \widehat{k}) \\ &=2+15-26 \\ &=-9 . \end{aligned}$
We observe , $\left(\overrightarrow{a_{2}}-\overrightarrow{a_{1}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \neq 0$
Thus,the given lines don't intersect.

Straight Line in Space exercise 27.3 question 6(iii)

Answer: The line intersect at the point of intersection is $(4,0,-1).$

Hint: Equate the coefficient of vector equation of line.

Given:$\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0} \text { and } \frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}$

Solution: co-ordinate of first equation of line
$\begin{aligned} &\frac{x-1}{3}=\frac{y-1}{-1}=\frac{z+1}{0}=\lambda(\text { let })\quad \dots (i)\\\\ &\mathrm{x}=3 \lambda+1, y=-\lambda+1, z=-1 \end{aligned}$
Co-ordinate of second equation of line
$\begin{aligned} &\frac{x-4}{2}=\frac{y-0}{0}=\frac{z+1}{3}=\mu(\text { let })\quad\dots(ii)\\\\ &x=2 \mu+4, y=0, z=3 \mu-1 \end{aligned}$
If the lines intersect, then they have a common point for some value of λ and μ.
$\begin{aligned} &\text { So, } 3 \lambda+1=2 \mu+4\quad\dots(iii)\\\\ &-\lambda+1=0 \Rightarrow \lambda=1\quad\dots(iv)\\\\ &3 \mu-1=-1 \Rightarrow \mu=0\quad\dots(v)\\\\ &\lambda=1 \text { and } \mu=0 \end{aligned}$
Satisfy equation (iii). So, the given line intersect and the point of intersection is $(4,0,-1).$

Straight Line in Space exercise 27.3 question 6(iv)

Answer: The line intersect each other and the point of intersection is $(1,3,2).$

Hint: Equate the coefficient of vector equation of line.

Given: Two equation of line.

Solution: The equation of the first line,
$\begin{aligned} &\frac{x-5}{4}=\frac{y-7}{4}=\frac{z+3}{-5}=\lambda(l e t) \\ \\&x=4 \lambda+5, \quad y=4 \lambda+7, \quad z=-5 \lambda-3 \end{aligned}$
So, the co-ordinates of a general point on this line are $(4 \lambda+5,4 \lambda+7,-5 \lambda-3)$
The equation of the second line,
$\begin{aligned} &\frac{x-8}{7}=\frac{y-4}{1}=\frac{z-5}{3}=\mu(\text { let }) \\\\ &\mathrm{x}=7 \mu+8, \quad y=\mu+4, \quad z=3 \mu+5 \end{aligned}$
So, the co-ordinates of a general point on this line are ($(7 \mu+8, \mu+4,3 \mu+5)$
If the lines intersect, then they have a common point for some value of λ and μ
We have,
$\begin{aligned} &4 \lambda+5=7 \mu+8,4 \lambda+7=\mu+4,-5 \lambda-3=3 \mu+5\\ &4 \lambda-7 \mu=3\quad\dots(i)\\\\ &\mu=4 \lambda+3\quad\dots(ii)\\\\\ &-5 \lambda-3 \mu=8\quad\dots(iii)\\\\ \end{aligned}$
Putting the value of μ from equation (ii) to (i), we get
$\begin{aligned} &4 \lambda-7 \mu=3 \\\\ &4 \lambda-7(4 \lambda+3)=3 \\\\ &4 \lambda-28 \lambda-21=3 \end{aligned}$
$\begin{aligned} &-24 \lambda=24 \\\\ &\lambda=-1 \end{aligned}$
Now putting the value of λ in equation (ii), we get
$\begin{aligned} &\mu=4 \lambda+3 \\\\ &\mu=4(-1)+3 \\\\ &\mu=-1 \end{aligned}$
As we can see by putting the value of λ and μ in equation (iii), that it satisfy the equation
Check, $\begin{aligned} &\text { }-5 \lambda-3 \mu=8 \\ \end{aligned}$
$-5(-1)-3(-1)=8 \\\\$
$5+3=8 \$
$8=8$
$L.H.S.=R.H.S.$
Hence intersection point exists or line of intersects.
Thus, point of intersection $(4 \lambda+5,4 \lambda+7,-5 \lambda-3)$
$\begin{aligned} &=(4(-1)+5,4(-1)+7,-5(-1)-3) \\\\ &=(1,3,2) \end{aligned}$

Straight Line in Space exercise 27.3 question 7

Answer: The point of intersection is $(-1,-6,-12)$

Hint: Equate the coefficient of vector equation of line.

Given: $\vec{r}=3 \hat{\imath}+2 \hat{\jmath}-4 \hat{k}+\lambda(\hat{\imath}+2 \hat{\jmath}+2 \hat{k}) \text { and } \vec{r}=5 \hat{\imath}-2 \hat{\jmath}+\mu(3 \hat{\imath}+2 \hat{\jmath}+6 \hat{k})$

Solution: let the lines are
$\begin{aligned} &M: \vec{r}=\overrightarrow{3 t}+\overrightarrow{2 J}-\overrightarrow{4 k}+\lambda(\vec{\imath}+2 \vec{\jmath}+2 \vec{k}) \ \\\\ &N: \vec{r}=\overrightarrow{5 \imath}-\overrightarrow{2 \jmath}+\mu(\overrightarrow{3 l}+\overrightarrow{2 J}+\overrightarrow{6 k}) \end{aligned}$
Co-ordinate of any random point on $M$ are $\mathrm{P}(3+\lambda, 2+2 \lambda,-4+2 \lambda)$ and on $N$ are $\mathrm{Q}(5+3 \mu,-2+2 \mu, 6 \mu)$
If the lines Mand N intersect then ,they must have a common point on them i.e. P and Q must coincide for some values of λ and μ.
Now,
$\begin{aligned} &3+\lambda=5+3 \mu\quad \text {... (i)}\\\\ &2+2 \lambda=-2+2 \mu \quad \text {... (ii)}\\\\ &-4+2 \lambda=6 \mu\quad \text {... (iii)} \end{aligned}$
Solving (i) and (ii) ,we get
$\begin{aligned} &\lambda=-4 \\\\ &\mu=-2 \end{aligned}$
Substitute the value in equation (iii),
$\begin{aligned} &-4+2(-4)=6(-2) \\\\ &-4=-12+8 \\\\ &-4=-4 \end{aligned}$
So, the given line intersect each other.
Now, point of intersection is$P(-1,-6,-12).$

The Class 12 RD Sharma chapter 27 exercise 27.3 solution is a profoundly pursued book that students and instructors acclaim in the whole country. The RD Sharma solution is productive all-around out and is additionally exam-situated, straightforward, and fundamental. What's more, you will discover an assortment of tips and deceives to expand your critical thinking speed. The RD Sharma class 12th exercise 27.3 can be utilized for self-practice and assessment of imprints.

The following are the advantages of studying from the RD Sharma class 12th exercise 27.3 material:

RD Sharma class 12 solutions chapter 27 exercise 27.3 material is prepared by a group of subject experts who have immense knowledge about CBSE question paper patterns. Each answer goes through a series of quality checks to ensure that the best information is available.
Moreover RD Sharma class 12th exercise 27.3 material is updated to the latest version and follows the CBSE syllabus which means that students can use it to clear their doubts from class lectures and prepare for exams simultaneously.
Students can use RD Sharma class 12th exercise 27.3 material as a guide for the textbook and save time in their preparation. This is an efficient and paperless alternative to education which is beneficial in the long term.
Students can access RD Sharma class 12th exercise 27.3 material from career 360s website by searching the book name and exercise number. Additionally the entire material is made available for free to help students excel in their exams.

RD Sharma Chapter-wise Solutions