These NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Ex 11.2 Questions and Answers are prepared by our highly skilled subject experts. https://mcq-questions.com/ncert-solutions-for-class-12-maths-chapter-11-ex-11-2/
NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Exercise 11.2
Ex 11.2 Class 12 NCERT Solutions 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.
Solution:
Consider the lines with direction cosines
\(\frac{12}{13}, \frac{-3}{13}, \frac{-4}{13} \text { and } \frac{4}{13}, \frac{12}{13}, \frac{3}{13}\)
The lines with direction cosines l1, m1, n1 and l2, m2, n2 are perpendicular if
l1l2 + m1m2 + n1n2 = 0
l1l2 + m1m2 + n1n2
= \(\left(\frac{12}{13} \times \frac{4}{13}\right)+\left(\frac{-3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{3}{13}\right)\)
= \(\frac{48}{169}-\frac{36}{169}-\frac{12}{169}=0\)
∴ The lines are perpendicular.
Consider the lines with direction cosines
are mutually perpendicular.
Class 12 Math Ex 11.2 NCERT Solutions 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).
Solution:
Let A (1, – 1, 2), B(3, 4, – 2), C (0, 3, 2) and D(3, 5, 6) be the points
Direction ratios of AB = 3 – 1, 4 – 1, – 2 – 2 = 2, 5, – 4
Direction ratios of CD = 3 – 0, 5 – 3, 6 – 2 = 3, 2, 4
a1a2 + b1b2 + c1c2 = 2(3) + 5(2) + (- 4) (4) = 6 + 10 – 16 = 0
Hence AB ⊥ CD.
Ex 11.2 12 Class NCERT Solutions 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) and (1, 2, 5).
Solution:
Let A(4, 7, 8), B (2, 3, 4), C (- 1, – 2, 1) and D(1, 2, 5) be the points
Direction ratios of AB = 2 – 4, 3 – 7, 4 – 8 = – 2, – 4, – 4
Direction ratios of CD = 1 – 1, 2 – 2, 5 – 1 = 2, 4, 4
Question 4.
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector \(3\hat { i } +2\hat { j } -2\hat { k }\)
Solution:
Let \(\vec{a}\) be the position vector of the point (1, 2, 3) and \(\vec{b}\) be the vector \(3 \hat{i}+2 \hat{j}-2 \hat{k}\)
∴ \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)
The vector equation of the line is \(\vec{r}\) = \(\vec{a}\) + λ\(\vec{b}\)
\(\vec{r}\) = (∴ \(\vec{a}=\hat{i}+2 \hat{j}+3 \hat{k}\)) + λ(\(3\hat { i } +2\hat { j } -2\hat { k }\))
Question 5.
Find the equation of the line in vector and in cartesian form that passes through the point with position vector \(2\hat { i } -\hat { j } +4\hat { k }\) and is in the direction \(\hat { i } +2\hat { j } -\hat { k }\).
Solution:
Let \(\vec{a}\) be the position vector of the point \(\vec{a}\) = \(\vec{a}=2 \hat{i}-\hat{j}+4 \hat{k}\)
\(\vec{b}=\hat{i}+2 \hat{j}-\hat{k}\)
The vector equation of a line passing through the point with position vector \(\vec{a}\) and parallel to \(\vec{b}\) is \(\vec{r}\) = \(\vec{a}+\lambda \vec{b}\)
Hence the equation of the line is
\(\vec{r}=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})\)
The cartesian equation is
\(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}\)
Another Method:
The point is (2, – 1, 4)
The direction ratios of the line are 1, 2, – 1
The equation of the line passing through (x1, y1, z1) and having direction ratios a, b,c is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
Hence cartesian equation is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
The vector equation is
\(\vec{r}=(2 \hat{i}-\hat{j}+4 \hat{k})+\lambda(\hat{i}+2 \hat{j}-\hat{k})\)
Question 6.
Find the cartesian equation of the line which passes through the point (- 2, 4, -5) and parallel to the line given by \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}\)
Solution:
The cartesian equation of a line passing through the point (x1, y1, z1) and parallel to the line with direction ratios a, b, c is
\(\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}\)
Hence the cartesian equation of the required line is \(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)
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.
Solution:
The equation of the line is \(\frac{x-5}{3}=\frac{y+4}{7}=\frac{z-6}{2}\)
The required line passes through the point (5, – 4, 6) and is parallel to the vector \(3 \hat{i}+7 \hat{j}+2 \hat{k}\). Let \(\vec{r}\) be the position vector of any point on the line, then the vector equation of the line is
\(\vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})\)
Question 8.
Find the vector and the cartesian equations of the lines that passes through the origin and (5, – 2, 3).
Solution:
Let \(\vec{a}\) and \(\vec{b}\) be the position vectors of the point A (0, 0,0) and B (5, – 2, 3)
Question 9.
Find the vector and cartesian equations of the line that passes through the points (3, – 2, – 5), (3, – 2, 6).
Solution:
Let \(\vec{a}\) and \(\vec{b}\) be the position vector of the point A (3, -2, -5) and B (3, -2, 6)
\(\vec{a}=3 \hat{i}-2 \hat{j}-5 \hat{k}\) and \(\vec{b}=3 \hat{i}-2 \hat{j}+6 \hat{k}\)
∴ \(\vec{b}-\vec{a}=0 \hat{i}+0 \hat{j}+11 \hat{k}\)
Let \(\vec{r}\) be the position vector of any point on the line. Then the vector equation of the line is \(\vec{r}\) = \(\vec{a}\) + λ(\(\vec{b}\) – \(\vec{a}\))
\(\vec{r}\) = \((3 \hat{i}-2 \hat{j}-5 \hat{k})+\lambda(11 \hat{k})\)
The cartesian equation is
\(\frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{11}\)
Question 10.
Find the angle between the following pair of lines
(i) \(\overrightarrow { r } =2\hat { i } -5\hat { j } +\hat { k } +\lambda (3\hat { i } +2\hat { j } +6\hat { k } )\)
\(and\quad \overrightarrow { r } =7\hat { i } -6\hat { j } +\mu (\hat { i } +2\hat { j } +2\hat { k } )\)
(ii) \(\overrightarrow { r } =3\hat { i } +\hat { j } -2\hat { k } +\lambda (\hat { i } -\hat { j } -2\hat { k } )\)
\(\overrightarrow { r } =2\hat { i } -\hat { j } -56\hat { k } +\mu (3\hat { i } -5\hat { j } -4\hat { k } )\)
Solution:
Question 11.
Find the angle between the following pair of lines
(i) \(\frac{x-2}{2}=\frac{y-1}{5}=\frac{z+3}{-3}\) and \(\frac{x+2}{-1}=\frac{y-4}{8}=\frac{z-5}{4}\)
(ii) \(\frac{x}{2}=\frac{y}{2}=\frac{z}{1} \text { and } \frac{x-5}{4}=\frac{y-2}{1}=\frac{z-3}{8}\)
Solution:
i. The direction ratios of the first line are 2, 5, – 3 and the direction ratios of the second line are -1, 8,4
Let θ be the acute angle between the lines, then
ii. The direction ratios of the first line are 2, 2, 1 and the direction ratios of the second line are 4, 1, 8 Let θ be the acute angle between the lines.
Question 12.
Find the values of p so that the lines
\(\frac{1-x}{3}=\frac{7 y-14}{2 p}=\frac{z-3}{2}\) and \(\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}\) are at right angles
Solution:
The given lines are
Question 13.
Show that the 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 other
Solution:
The direction ratios of the first line are 7, – 5, 1 and the direction ratios of second line are 1, 2, 3
a1a2 + b1b2 + c1c2 = (7)(1) + (-5)(2) + (1)(3) = 7 – 10 + 3 = 0
∴ The lines are perpendicular to each other.
Question 14.
Find the shortest distance between the lines
\(\overrightarrow { r } =(\hat { i } +2\hat { j } +\hat { k } )+\lambda (\hat { i } -\hat { j } +\hat { k } )\) and \(\overrightarrow { r } =(2\hat { i } -\hat { j } -\hat { k } )+\mu (2\hat { i } +\hat { j } +2\hat { k } )\)
Solution:
\(\overrightarrow { r } =(\hat { i } +2\hat { j } +\hat { k } )+\lambda (\hat { i } -\hat { j } +\hat { k } )\) … (1)
\(\overrightarrow { r } =(2\hat { i } -\hat { j } -\hat { k } )+\mu (2\hat { i } +\hat { j } +2\hat { k } )\) … (2)
Question 15.
Find the shortest distance between the lines \(\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}\) and \(\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}\)
Solution:
Shortest distance between the lines
Question 16.
Find the distance between die lines whose vector equations are:
\(\overrightarrow { r } =(\hat { i } +2\hat { j } +3\hat { k) } +\lambda (\hat { i } -3\hat { j } +2\hat { k } )\) and \(\overrightarrow { r } =(4\hat { i } +5\hat { j } +6\hat { k) } +\mu (2\hat { i } +3\hat { j } +\hat { k } )\)
Solution:
Question 17.
Find the shortest distance between the lines whose vector equations are
\(\overrightarrow { r } =(1-t)\hat { i } +(t-2)\hat { j } +(3-2t)\hat { k }\) and \(\overrightarrow { r } =(s+1)\hat { i } +(2s-1)\hat { j } -(2s+1)\hat { k }\)
Solution:
The given equation can be reduced as
\(\vec{r}=(\hat{i}-2 \hat{j}+3 \hat{k})+t(-\hat{i}+\hat{j}-2 \hat{k})\)
and \(\vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2 \hat{j}-2 \hat{k})\)
Comparing with the standard equation, we get