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

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## 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
Let the Directions cosines given for lines L1, L2 and L3.
Let α be the angle between L1, and L2 similarly for L2 and L3 and L3 and L1 are perpendicular. 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).
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).
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 $$3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}$$ 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 $$\hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}}$$
Let $$\bar{a}=2 i-j+4 \bar{k} \text { and } \bar{b}=i+2 j-\bar{k}$$
Required vector equation is $$\overline{\mathbf{r}}=\overline{\mathbf{a}}+\lambda \overline{\mathbf{b}}$$ Question 6.
Find the cartesian equation of the line which passes through the point (- 2, 4, – 5) and parallel to the line given by
$$\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}$$
$$\bar{a}=-2 i+4 j-5 \bar{k} \text { and } \bar{b}=3 i+5 j+6 \bar{k}$$ equation of a line through (-2, 4, -5) and is parallel to the vector $$3 i+5 j+6 \bar{k}$$
$$\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}$$ 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.
$$\bar{a}=5 i-4 j+6 \bar{k}, \bar{b}=3 i+7 j+2 \bar{k}$$
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).
Equation of a line through two points is  Question 9.
Find the vector and the cartesian equations of the line that passes through the points
(3, – 2, – 5), (3, – 2, 6). 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}
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 (ii) $$\overrightarrow{\mathbf{r}}=3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \text {and }\overrightarrow{\mathbf{r}}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-56\hat{\mathbf{k}}+\mu(3 \hat{\mathbf{i}}-5 \hat{\mathbf{j}}-4 \hat{\mathbf{k}})$$
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  Question 11.
Find the angle between the following pair of lines:

(i) $$\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3} \text { and } \frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}$$
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  (ii) $$\frac{x}{2}=\frac{y}{2}=\frac{z}{1} \text { and } \frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}$$
Dr’ of lines are (2,2,1), {4,1,8}
Let θ be the angle between them 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  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.
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 ‘⊥’ 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}  Question 15.
Find the shortest distance between the lines
$$\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \text { and } \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}$$ 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}   \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} 