2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2

Students can Download Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Questions and Answers, Notes Pdf, 2nd PUC Maths Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and score more marks in your examinations.

Karnataka 2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2

2nd PUC Maths Three Dimensional Geometry NCERT Text Book Questions and Answers Ex 11.2

Question 1
Show that the three lines with direction cosines
\(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} ; \frac{4}{13}, \frac{12}{13}, \frac{3}{13} ; \frac{3}{13}, \frac{-4}{13}, \frac{12}{13}\) are mutually perpendicular
Answer:
Let the Directions cosines given for lines L1, L2 and L3.
Let α be the angle between L1, and L2

2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.1
similarly for L2 and L3 and L3 and L1 are perpendicular.

KSEEB Solutions

a

Question 2.
Show that the line through the points (1, -1, 2), (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Answer:
Let A (1, -1, 2), B (3, 4, -2), C (0, 3, 2), D (3, 5, 6)
Dr’ of AB = (2,5,-4)
Dr’of CD = (3,2, 4)
a1a2 + b1b2 + C1C2 = 2 x 3 + 5 x 2+-4 x 4 = 0
hence the lines are perpendicular.

Question 3.
Show that the line through the points (4,7, 8), (2, 3, 4) is parallel to the line through the points (- 1, – 2, 1), (1, 2, 5).
Answer:
A (4,7,8), B (2,3,4), C (-1, -2,1), D (1,2,5)
Dr’ of AB (2-4,3-7,4-8), (-2, -4, -4)
Dr’ of CD (1+1, 2+2, 5-1) (2,4,4)
since Dr’ of AB and Dr’ of CD are proportional
\(\frac{-2}{2}=\frac{-4}{4}=\frac{-4}{4} \quad \mathrm{AB} \| \mathrm{CD}\)

Question 4.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}\)
Answer:

2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.2

Question 5.
Find the equation of the line in vector and in cartesian form that passes through the point with position vector \(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}\) and is in the direction \(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}\)
Answer:
Let \(\bar{a}=2 i-j+4 \bar{k} \text { and } \bar{b}=i+2 j-\bar{k}\)
Required vector equation is \(\overline{\mathbf{r}}=\overline{\mathbf{a}}+\lambda \overline{\mathbf{b}}\)
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.3

Question 6.
Find the cartesian equation of the line which passes through the point (- 2, 4, – 5) and parallel to the line given by
\(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
Answer:
\(\bar{a}=-2 i+4 j-5 \bar{k} \text { and } \bar{b}=3 i+5 j+6 \bar{k}\) equation of a line through (-2, 4, -5) and is parallel to the vector \(3 i+5 j+6 \bar{k}\)
\(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)

KSEEB Solutions

Question 7.
The cartesian equation of a line is
\(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\) .
write its vector form.
Answer:
\(\bar{a}=5 i-4 j+6 \bar{k}, \bar{b}=3 i+7 j+2 \bar{k}\)
equation of a line through the point with p.v
\(5i-4j+6\bar { k } and\quad is\quad parallel\quad to\quad 3i+7j+2\bar { k } is\quad \overrightarrow { { r } } =(5{ i }-4{ j }+6\overline { { k } } )+\lambda (3{ i }+7{ j }+2\overline { { k } } )\)

Question 8.
Find the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3).
Answer:
Equation of a line through two points is
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.4

KSEEB Solutions

Question 9.
Find the vector and the cartesian equations of the line that passes through the points
(3, – 2, – 5), (3, – 2, 6).
Answer:
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.5

Question 10.
Find the angle between the following pairs of lines:
(i) \(\begin{aligned}&\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}-5\hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \text { and }\\&\overrightarrow{\mathbf{r}}=7 \hat{\mathbf{i}}-6 \hat{\mathbf{k}}+\mu(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\end{aligned}\)
Answer:
The given lines are respectively parallel to
\(\overline{\mathrm{b}}_{1}=3 \mathrm{i}+2 \mathrm{j}+6 \overline{\mathrm{k}} \text { and } \mathrm{i}+2 \mathrm{j}+2\overline{\mathrm{k}}=\overline{\mathrm{b}}_{2}\)
Let θ be the angle between them
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.6

(ii) \(\overrightarrow{\mathbf{r}}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \text {and }\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-56\hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})\)
Answer:
The given lines are respectively parallel to
\(\overline{\mathrm{b}}_{1}=\mathrm{i}-\mathrm{j} 2 \overline{\mathrm{k}} \text { and } \overline{\mathrm{b}}_{2}=3 \mathrm{i}-5 \mathrm{j}-4 \overline{\mathrm{k}}\)
Let θ be the angle between them
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.7
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.8

Question 11.
Find the angle between the following pair of lines:

(i) \(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3} \text { and } \frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\)
Answer:
The given lines are respectively parallel to
\(\overline{\mathrm{b}}_{1}=2 \mathrm{i}+5 \mathrm{j}-3 \overline{\mathrm{k}} \text { and } \overline{\mathrm{b}}_{2}=-\mathrm{i}+8 \mathrm{j}+4 \overline{\mathrm{k}}\)
Let θ be the angle between them
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.9

KSEEB Solutions

(ii) \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1} \text { and } \frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
Answer:
Dr’ of lines are (2,2,1), {4,1,8}
Let θ be the angle between them
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.10

Question 12.
Find the values of p so that the lines 1-x
\(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2} \text { and }\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles
Answer:
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.11
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.12

Question 13.
Show that the lines
\(\frac { x-5 }{ 7 } =\frac { y+2 }{ -5 } =\frac { z }{ 1 } { and } \quad \frac { x }{ 1 } =\frac { y }{ 2 } =\frac { z }{ 3 } \) are perpendicular to each other.
Answer:
Dr’s are (7,-5,1), and (1, 2,3)
a1b1 + a2b+ a3b3
7 x 1- 5 x 2 + 1 x 3 ⇒ 10 – 10 = 0
The products of Dr’s are equal, lines are ‘⊥’

KSEEB Solutions

Question 14.
Find the shortest distance between the lines
\(\begin{aligned}&\overrightarrow{\mathbf{r}}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \text { and }\\&\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}+\mu(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})\end{aligned}\)
Answer:
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.13
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.14

Question 15.
Find the shortest distance between the lines
\(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Answer:
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.15

Question 16.
Find the shortest distance between the lines whose vector equations are
\(\begin{aligned}&\overrightarrow{\mathbf{r}}=(\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\lambda(\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}})\\&\text { and } \overrightarrow{\mathbf{r}}=(4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}+\mu(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+\hat{\mathbf{k}})\end{aligned}\)
Answer:
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.16
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.17

KSEEB Solutions

Question 17.
Find the shortest distance between the lines whose vector equations are
\(\begin{aligned}&\overrightarrow{\mathbf{r}}=(1-t) \hat{\mathbf{i}}+(t-2) \hat{\mathbf{j}}+(3-2 \mathbf{t}) \hat{\mathbf{k}} \text { and }\\&\overrightarrow{\mathbf{r}}=(\mathbf{s}+1) \hat{\mathbf{i}}+(2 \mathbf{s}-\mathbf{1}) \hat{\mathbf{j}}-(2 \mathbf{s}+\mathbf{1}) \hat{\mathbf{k}}\end{aligned}\)
Answer:
2nd PUC Maths Question Bank Chapter 11 Three Dimensional Geometry Ex 11.2.18

a