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Prove the lines are mutually perpendicular
lines through the points are perpendicular to the line through the points
lines through the points are parallel to the line through the points
The angle between the pair of lines given in vector and cartesian form
Equation of line passing through the points
Perpendicularity of two lines
Straight Line in Space Exercise 27.2 Question 1
Answer: three direction cosines i.e. a, b, c are perpendicular to each otherStraight Line in Space Exercise 27.2 Question 2
Answer : the lines are perpendicular to the given points showedStraight Line in Space Exercise 27.2 Question 3
Answer: The given two lines will be parallel through the two points equal to ‘0’ showed;
Given: Show that the line through the points$(4,7,8) and (2,3,4)$ is parallel to the line through the points $(-1,-2,1) and (1,2,5)$
Hint: for showing parallel v1 and v2 must be equal to ‘0’
Solution:
$\begin{aligned} &\vec{v}_{1}=(4-2) \hat{i}+(7-3) \hat{j}+(8-4) \hat{k} \\ &\overrightarrow{v_{1}}=2 \hat{i}+4 \hat{j}+4 \hat{k} \\ &\overline{v_{2}}=(1+1) \hat{i}+(2+2) \hat{j}+(5-1) \hat{k} \\ &\overline{v_{2}}=2 \hat{i}+4 \hat{j}+4 \hat{k} \end{aligned}$
Now, $\begin{aligned} \cos \theta=\cos \theta &=\frac{\overrightarrow{v_{1}} \overline{v_{2}}}{\left|\overrightarrow{v_{1}}\right|\left|\overrightarrow{v_{2}}\right|} \\ \end{aligned}$
$\begin{aligned} &=\frac{4+16+16}{\sqrt{36} \sqrt{36}} \\ \end{aligned}$
$\begin{aligned} =& \frac{36}{36} \\ \end{aligned}$
$\begin{aligned} &=1 \\ \end{aligned}$
$\begin{aligned} \cos \theta &=1 \\ \end{aligned}$
$\begin{aligned} \cos \theta=\cos 0^{\circ} \\ \end{aligned}$
$\begin{aligned} \theta=0^{\circ} \\ \end{aligned}$
$\begin{aligned} \mathrm{v}_{1} \| \mathrm{v}_{2} \text { (proved) } \end{aligned}$
So the two lines are parallel to each other
Straight Line in Space Exercise 27.2 Question 4
Answer : the Cartesian equation which is parallel to given lines will be $\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$Straight Line in Space Exercise 27.2 Question 5
Answer : the given lines $\frac{x-5}{7}=\frac{y+2}{-5}=\frac{z}{1}$ and $\frac{x}{1}=\frac{y}{2}=\frac{z}{3}$ are perpendicular to each otherStraight Line in Space Exercise 27.2 Question 6
Answer: the line joining by the point $(2,1,1)$ is perpendicular to the line determined by the other two pointsSo the line joining to the point $(2,1,1)$ is perpendicular to the line joining by the determined points $(3,5,-1) and (4,3,-1)$
Straight Line in Space Exercise 27.2 Question 7
Answer: the equation parallel to x-axis will be $\frac{x}{1}=\frac{y}{0}=\frac{3}{0}$Straight Line in Space Exercise 27.2 Question 8(i)
Answer: the angles between the lines will be $0^{0}$The angle between the lines will be $0^{0}$
Straight Line in Space Exercise 27.2 Question 8(ii)
Answer: the angle between the points of lines will be $\cos ^{-1}\frac{19}{21}$Straight Line in Space Exercise 27.2 Question 8(iii)
Answer: the angle between the points of lines will be $\frac{\pi }{3}$Straight Line in Space Exercise 27.2 Question 9(i)
Answer: the angle between the given pairs of lines will be $\cos ^{-1}=\frac{8}{5\sqrt{3}}$Straight Line in Space Exercise 27.2 Question 9(ii)
Answer: the angle between the given pairs of a line is $\cos ^{-1}=\frac{10}{9\sqrt{22}}$Straight Line in Space Exercise 27.2 Question 9(iii)
Answer: the angle between the lines given in the question is $\cos ^{-1}\frac{11}{14}$Solution:
$$$ \begin{aligned} &\frac{-(x-5)}{-2}=\frac{y-(-3)}{1}=\frac{-(3-1)}{3} \\ &\frac{x-b}{2}=\frac{y+3}{1}=\frac{(3+1)}{3} \end{aligned}$
$Direction \: \: vector \vec{a}=2 \hat{i}+\hat{j}-3 \hat{k}$
Again:
$\frac{x-0}{3}=\frac{(y-1)}{ 2}=\frac{(z+5)}{-1}$
$\begin{aligned} Direction \: \: vector \vec{b}=3 \hat{i}+2 \hat{j}-\hat{k} \end{aligned}$
$\begin{aligned} &\cos \theta=\frac{6+2+3}{\left|\sqrt{2^{2}+1^{2}+(-3)^{2}}\right| \sqrt{3^{2}+(-2)^{2}+(-1)^{2}} \mid} \\ \end{aligned}$
$\frac{11}{\sqrt{14} \sqrt{14}}=\frac{11}{14} \\$
$\theta=\cos ^{-1} \frac{11}{14}$
So the angle between the lines will be$\cos ^{-1}\frac{11}{14}$
Straight Line in Space Exercise 27.2 Question 9(iv)
Answer: the angle between the pairs of lines will be $\frac{\lambda }{2}$Straight Line in Space Exercise 27.2 Question 9(v)
Answer: the angle between the pairs of given line is $\cos ^{-1}\frac{4}{5\sqrt{6}}$Straight Line in Space Exercise 27.2 Question 10(i)
Answer: the angle between the pairs of lines will be $\cos ^{-1}\frac{1}{65}$Straight Line in Space Exercise 27.2 Question 10(ii)
Answer: the angle between the given pairs of lines will be $\cos ^{-1}\frac{2}{3}$Straight Line in Space Exercise 27.2 Question 10(iii)
Answer: the angle between the given points of lines will be $90^{0}$Straight Line in Space Exercise 27.2 Question 10(iv)
Answer: the angle between the given pairs of lines will be $90^{0}$Straight Line in Space Exercise 27.2 Question 11
Answer: the angle between the pairs given in the question $\cos ^{-1}\frac{2}{3}$So the angle between the pairs will be $\cos ^{-1}\frac{2}{3}$
Straight Line in Space Exercise 27.2 Question 12
Answer: the equation will be:$\frac{x-1}{4}=\frac{y-2}{2}=\frac{z+4}{3}$Straight Line in Space Exercise 27.2 Question 13
Answer: the equation of the lines will be $\vec{r}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda(2 \hat{i}+2 / 3 \hat{j}-3 \hat{k})$Straight Line in Space Exercise 27.2 Question 14
Answer: the equation of the line passing through the point will be $\therefore \vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}-5 \hat{k})$Straight Line in Space Exercise 27.2 Question 15
Answer: the equation of the line will be $\vec{r}=(2 \hat{i}+\hat{j}+3 \hat{k})+\lambda(4 \hat{i}-14 \hat{j}+8 \hat{k})$Straight Line in Space Exercise 27.2 Question 16
Answer: The equation will be $\begin{aligned} &\vec{r}=(\hat{i}+\hat{j}-3 \hat{k})+\lambda(4 \hat{i}-5 \hat{j}+\hat{k}) \\ \end{aligned}$Straight Line in Space Exercise 27.2 Question 17
Answer : The equation will be $\frac{x-1}{10}=\frac{y+1}{y}=\frac{3}{7}$Straight Line in Space Exercise 27.2 Question 18
Answer: The equation of the given question will be $\vec{r}=(\hat{i}+2 \hat{j}-4 \hat{k})+\lambda(2 \hat{i}+3 \hat{j}+6 \hat{k})$Straight Line in Space Exercise 27.2 Question 19
Answer: The two lines are perpendicular to each other(shown)Straight Line in Space Exercise 27.4 Question 12
Answer: the equation will be $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z+1}{3}$ which is parallel to $\vec{r}=(2 \hat{i}-\hat{j}-\hat{k})+\lambda(\hat{i}+2 \hat{j}+3 \hat{k})$So, the vector equation will be
$\frac{x-2}{1}= \frac{y+1}{2}=\frac{z+1}{3}$
Straight Line in Space Exercise 27.2 Question 21
Answer : the value of the $\lambda$ will be $\frac{10}{7}$Straight Line in Space Exercise 27.2 Question 22
Answer : the angle cannot be made between AB and CD because both are parallel to each otherCD and AB are parallel. So it cannot make any angle
Straight Line in Space Exercise 27.2 Question 23
Answer: the value $\lambda$ will be 1Straight Line in Space Exercise 27.2 Question 24
Answer : the direction cosines will be $\frac{2}{7},\frac{3}{7},\frac{-6}{7}$ and the equation will be $\vec{r}=\left ( \hat{i}-2\hat{j}+3\hat{k} \right )+\lambda \left ( 2\hat{i}+3\hat{j}-6\hat{k} \right )$So the answer will be
$\vec{r}=\hat{i}-2\hat{j}+3\hat{k} +\lambda \left ( 2\hat{i}+3\hat{j}-6\hat{k} \right )$ and cosines will be $\frac{2}{7},\frac{3}{7},\frac{-6}{7}$
Straight Line in Space Exercise 27.2 Question 25
Answer: the value of k will be ‘$2$’Straight Line in Space Exercise 27.2 Question 26
Answer: the required vector equation is $\vec{r}=(\hat{i}+\hat{j}+\hat{k})+\lambda(-4 \hat{i}+4 \hat{j}-\hat{k})$Class 12 RD Sharma chapter 27 exercise 27.2 material is prepared by a group of subject experts who have years of experience in the field. This material is updated to the latest version of the book. Students can use it as a guide for the textbook and prepare accordingly. Additionally, it follows the CBSE syllabus, which means that students can use it for their homework.
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