KC Sinha Solution Class 12 Chapter 29 3D Geometry : Straight Lines (त्रिविमीय ज्यामिति : सरल रेखा ) Exercise 29.1 (Q1-Q26)

  Exercise 29.1 

TYPE-I : दो रेखाओं के बीच के कोण पर आधारित प्रश्न :

Question 1

[निम्नलिखित प्रत्येक रेखा युग्म के बीच का कोण ज्ञात करें]

Find the angle between each of the following pair of lines :

(i)

$\vec{r}=5 \hat{i}-7 \hat{j}+\lambda(-\hat{i}+4 \hat{j}+2 \hat{k})$

$\vec{r}=-2 \hat{i}+k+\mu(3 \hat{i}+4 \hat{k})$

Sol :

(ii)

$\vec{r}=(2+s) \hat{i}+(s-1) \hat{j}+(2-3 s) \hat{k}$

$\vec{r}=(1-t) \hat{i}+(2 t+3) \hat{j}+\hat{k}$

Sol :

Question 2

[निम्नलिखित रेखा युग्मों के बीच का कोण ज्ञात कीजिए :]

Find the angle between the following pairs of lines :

(i)

$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-4}{6}$ तथा (and)

$x+1=\frac{y+2}{2}=\frac{z-4}{2}$

Sol :

(ii)

$\frac{x+1}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ तथा (and)

$\frac{x+1}{1}=\frac{y-4}{4}=\frac{z-5}{2}$

Sol :

Question 3

[निम्नलिखित रेखा युग्मों के बीच का कोण ज्ञात कीजिए ।]

Find the angle between the lines

(i) $\frac{x+4}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ and $\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$

(ii) $\frac{x-3}{1}=\frac{y-2}{2}=\frac{z-2}{-4}$ and $\frac{x-0}{3}=\frac{y-5}{2}=\frac{z+2}{-6}$

(iii) $\vec{r}=4 \hat{i}-\hat{j}+\lambda(\hat{i}+2 \hat{j}-2 \hat{k})$ and $\vec{r}=\hat{i}-\hat{j}+2 \hat{k}-\mu(2 \hat{i}+4 \hat{j}-4 \hat{k})$

Sol :

Question 4

Show that the line $\frac{x-3}{2}=\frac{y+1}{-3}=\frac{z-2}{4}$ is perpendicular to the line $\frac{x+2}{2}=\frac{y-4}{4}=\frac{z+5}{2}$

Sol :

TYPE-II

Question 5

(i) If cartesian equation of a line is $\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$, find its vector equation.

(ii) If the cartesian equations of a line are $\frac{3-x}{5}=\frac{y+4}{7}=\frac{2 z-6}{4}$, write the vector equation for the line.

Sol :

Question 6

Find the equation of the line passing through point $2 \hat{i}-\hat{j}+4 \hat{k}$ and parallel to vector $i+\hat{j}-2 \hat{k}$ in vector form as well as cartesian form.

Question 7

Find the equation of the line passing through points $\hat{i}-2 \hat{j}+\hat{k}$ and $-2 \hat{j}+3 \hat{k}$ in vector form and cartesian form.

Question 8

Find the equation of the line passing through point $(1,0,2)$ having direction ratios $3,-1,5$. Prove that this line passes through (4,-1,7)

Question 9

Find the equation of the line parallel to the line $\frac{x-2}{3}=\frac{y+1}{1}=\frac{z-7}{9}$ and passing through the point (3,0,5)

Sol :

Question 10

Find the vector equation of a line passing through a point with position vector $2 \hat{i}-\hat{j}+\hat{k}$ and parallel to the line joining the points with position vectors $-\hat{i}+4 \hat{j}+\hat{k}$ and $\hat{i}+2 \hat{j}+2 \hat{k}$. Also find the cartesian equation of this line.

Sol :

Question 11

A line passes through (2,-1,3) and is perpendicular to the lines $\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k})$ and $\vec{r}=(2 \hat{i}-\hat{j}-3 \hat{k})+\mu(\hat{i}+2 \hat{j}+2 \hat{k})$. Obtain its equation in vector and cartesian form.

Sol :

Question 12

Find the vector equation of a line parallel to the vector $2 \hat{i}-\hat{j}+2 \hat{k}$ and passing through a point A with position vector $3 \hat{i}+\hat{j}-k$.

Sol :

Question 13

Find the vector and cartesian equations of the line passing through the point $(2,1,3)$ and perpendicular to the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z-3}{3}$ and $\frac{x}{-3}=\frac{y}{2}=\frac{z}{5}$

Sol :

Question 14

The Cartesian equations of a line are :

$\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}$

Find a vector equation for the line.

Sol :

Question 15

Find the vector equation of a straight line which passes through the points whose position vectors are $\hat{i}-2 \hat{j}+\hat{k}$ and $3 \hat{k}-2 \hat{j}$.

Question 16

Find the vector equation of the straight line passing through the points :

(i) (1,1,0) and (0,1,1)

(ii) (-2,1,3) and (3,1,-2)

(iii) Find the cartesian equation of the line which passes through the point $(-2,4,-5)$ and is parallel to the line $\frac{x+3}{3}=\frac{4-y}{5}=\frac{z+8}{6}$.

Sol :

TYPE-III

Question 17

(i) Find the point on the line $\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}$ at a distance $3 \sqrt{2}$ from the point (1,2,3)

Sol :

(ii) Find the distance of the point $(1,-2,3)$ from the plane $x-y+z=5$ measured parallel to the line $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$.

Sol :

Question 18

Find the coordinates of the foot of perpendicular drawn from the point A(1,8,4) to the line joining B(0,-1,3) and C(2,-3,-1)

Sol :

Question 19

Find the image of the point (1,6,3) in the line $\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}$.

Sol :

Question 20

Find the perpendicular distance of the point $(1,0,0)$ from the line $\frac{x-1}{2}=\frac{y+1}{-3}=\frac{z+10}{8}$.

Also find the coordinates of the foot of the perpendicular.

Sol :

Question 21

Find the foot of perpendicular from (0,2,7) to line

$\frac{x+2}{-1}=\frac{y-1}{3}=\frac{z-3}{-2}$

Sol :

Question 22

(i) Find the foot and hence the length of perpendicular from $(5,7,3)$ to the line $\frac{x-15}{3}=\frac{y-29}{8}=\frac{z-5}{-5}$. Find also the equation of the perpendicular.

Sol :

(ii) Find the equation of the perpendicular drawn from $(2,4,-1)$ to the line $\frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9}$

Sol :

Question 23

Find two points on the line through the points A(1,2,3) and B(3,5,9) at a distance of 14 units from the mid point of AB

Sol :

Question 24

(i) Find the length of the perpendicular from point $(3,4,5)$ on the line $\frac{x-2}{2}=\frac{y-3}{5}=\frac{z-1}{3}$.

Sol :

(ii) Find the length and the foot of the perpendicular drawn from the point $(2,-1,5)$ to the line $\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}$

Sol :

TYPE-IV

Question 25

Find the shortest distance between the lines $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$ and $\frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}$. Find also its equation.

Sol :

Question 26

Find the shortest distance and equation of shortest distance between the lines $\vec{r}=3 \hat{i}+5 \hat{j}+7 \hat{k}+\lambda(\hat{i}-2 \hat{j}+\hat{k})$ and $\vec{r}=-\hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}-6 \hat{j}+\hat{k})$

Sol :