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Maths 2 Marks

Three Dimensional Geometry · Gujarat Board (GSEB) STD 12 Science (GSEB - English Medium). 15 questions.

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Question 12 Marks
If the lines $\frac{{x - 1}}{{ - 3}} = \frac{{y - 2}}{{2k}} = \frac{{z - 3}}{2}$ and $\frac{{x - 1}}{{3k}} = \frac{{y - 1}}{1} = \frac{{z - 6}}{{ - 5}}$ are perpendicular, find the value of 'k'.
Answer
Given: Equation of one line is $\frac{{x - 1}}{{ - 3}} = \frac{{y - 2}}{2k} = \frac{{z - 3}}{2}$ 

Direction ratios of this line are its denominators, i.e., $ - 3,2k,2 = {a_1},{b_1},{c_1}$ 

$\therefore$ A vector along this line is ${\vec b_1} = - 3\hat i + 2k\hat j + 2\hat k$

Again, equation of second line is $\frac{{x - 1}}{{3k}} = \frac{{y - 1}}{1} = \frac{{z - 6}}{{ - 5}}$

Direction ratios of this line are its denominators, i.e., $3k,1, - 5 = {a_2},{b_2},{c_2}$ 

$\therefore$A vector along this line is ${\vec b_2} = 3k\hat i + \hat j - 5\hat k$

Since these given lines are perpendicular.

$\therefore {\vec b_1}.{\vec b_2} = {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$

$\Rightarrow \left( { - 3} \right)\left( {3k} \right) + \left( {2k} \right)\left( 1 \right) + 2\left( { - 5} \right) = 0$$ \Rightarrow - 9k + 2k - 10 = 0$

$ \Rightarrow - 7k = 10 \Rightarrow k = \frac{{ - 10}}{7}$

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Question 22 Marks
Find the equation of a line parallel to x-axis and passing through the origin.
Answer
We know that a unit vector along the x-axis is $\hat i = \hat i + 0\hat j + 0\hat k$ 
$\therefore$ Direction cosines of the x-axis are coefficients of $\hat i,\hat j,\hat k$ in the unit vector
i.e., 1, 0, 0 $ = l,m,n$
$\therefore$ Equation of the required line passing through the origin (0, 0, 0) and parallel to the x-axis is $\frac{{x - 0}}{1} = \frac{{y - 0}}{0} = \frac{{z - 0}}{0} \Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{0}$ 
Vector equation of the required line is $\vec r = \vec a + \lambda \vec b$ 
$ \Rightarrow \vec r = \vec 0 + \lambda \hat i$ [$\vec a = \vec 0$ and $\vec b = \vec i$]
$ \Rightarrow \vec r = \lambda \hat i$
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Question 32 Marks
The Cartesian equation of a line is $\frac{{x - 5}}{3} = \frac{{y + 4}}{7} = \frac{{z - 6}}{2}$. Write its vector form.
Answer
$\vec r = \vec a + \lambda \vec b$

$\vec a = 5\hat i - 4\hat j + 6\hat k$

$\vec b = 3\hat i + 7\hat j + 2\hat k$

$\vec r = \left( {5\hat i - 4\hat j + 6\hat k} \right) + \lambda \left( {3\hat i + 7\hat j + 2\hat k} \right)$

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Question 42 Marks
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
Find the equation of the line in cartesian form that passes through the point (-2, 4, -5) and parallel to the line given by
$\frac{{x + 3}}{3} = \frac{{y - 4}}{5} = \frac{{z + 8}}{6}$
is given by:
$\frac{{x - {x_1}}}{l} = \frac{{y - {y_1}}}{m} = \frac{{z - {z_1}}}{n}$ ${x_1} = - 2,{x_2} = 4,{x_3} = - 5$ And l = 3 , m = 5 and n = 6 .
Therefore,
$\frac{{x + 2}}{3} = \frac{{y - 4}}{5} = \frac{{z + 5}}{6}$
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Question 52 Marks
Find the equation of the line in cartesian form that passes through the point with position vector $2\hat i - \hat j + 4\hat k$ and is in the direction $\hat i + 2\hat j - \hat k$.
Answer
We have ,
$\overrightarrow a = 2\hat i - \hat j + 4\hat k$ and $\overrightarrow b = \hat i + 2\hat j - \hat k$, then, its Cartesian equation is given by :
$\frac{{x - {x_1}}}{l} = \frac{{y - {y_1}}}{m} = \frac{{z - {z_1}}}{n}$
Here, ${x_1} = 2,{x_2} = - 1,{x_3} = 4$ And l = 1 , m = 2 and n = -1 .
Therefore , $\frac{{x - 2}}{1} = \frac{{y + 1}}{2} = \frac{{z - 4}}{{ - 1}}$
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Question 62 Marks
Find the equation of the line which passes through the point (1, 2, 3) and is parallel to the vector $3\hat i + 2\hat j - 2\hat k$.
Answer
The equation of the line which passes through the point (1, 2, 3) and is parallel to the vector
$3\hat i + 2\hat j - 2\hat k$ , let vector $\overrightarrow a = \widehat i + \widehat j + \widehat k$ and vector $\overrightarrow b = 3\hat i + 2\hat j - 2\hat k$,
the equation of line is :
$\overrightarrow a + \lambda \overrightarrow b \\ $ $= (\widehat i + \widehat j + \widehat k) + \lambda (3\hat i + 2\hat j - 2\hat k)$
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Question 72 Marks
Show that the line through points (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (1, 2, 5).
Answer
We know that direction ratios of the line joining the points A (4, 7, 8) and B (2, 3, 4) are ${x_2} - {x_1}, {y_2} - {y_1},{z_2} - {z_1}$ 
$ \Rightarrow 2 - 4,3 - 7,4 - 8$
$ \Rightarrow - 2, - 4, - 4 = {a_1},{b_1},{c_1}$ (say)
Again direction ratios of the line joining the points $C\left( { - 1, - 2,1} \right)$ and D (1, 2, 5) are $x_2-x_1,y_2-y_1,z_2-z_1$
$\Rightarrow 1 - \left( { - 1} \right),2 - \left( { - 2} \right),5 - 1$
$ \Rightarrow 2,4,4 = {a_2},{b_2},{c_2}$(say)
For the lines AB and CD, $\frac{{{a_1}}}{{{a_2}}} = \frac{{ - 2}}{2},\frac{{{b_1}}}{{{b_2}}} = \frac{{ - 4}}{4},\frac{{{c_1}}}{{{c_2}}} = \frac{{ - 4}}{4} = - 1$ 
Since, $\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$
Therefore, line AB is parallel to line CD.
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Question 82 Marks
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
We know that direction ratios of the line joining the points $A\left( {1, - 1,2} \right)$ and $B\left( {3,4, - 2} \right)$are ${x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}$
$ \Rightarrow 3 - 1,4 - \left( { - 1} \right), - 2 - 2$
$ \Rightarrow 2,5, - 4 = {a_1},{b_1},{c_1}$
Again, direction ratios of the line joining the points C (0, 3, 2) and D (3, 5, 6) are ${x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}$
$ \Rightarrow 3 - 0,5 - 3,6 - 2$
$ \Rightarrow 3,2,4 = {a_2},{b_2},{c_2}$(say)
For lines AB and CD, ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 2 \times 3 + 5 \times 2\left( { - 4} \right) \times 4 = 6 + 10 - 16 = 0$
Since, it is 0, therefore, line AB is perpendicular to line CD.
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Question 92 Marks
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.
Answer
Equation of one line $\frac{{x - 5}}{7} = \frac{{y + 2}}{{ - 5}} = \frac{z}{1}$ 
$\therefore$ Direction ratios of this line are $7, - 5,1 = {a_1},{b_1},{c_1}$
$\Rightarrow {\vec b_1} = 7\hat i - 5\hat j + \hat k$
Again equation of another line $\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$ 
$\therefore$ Direction ratios of this line are 1, 2, 3 $ = {a_2},{b_2},{c_2}$
$ \Rightarrow \vec b = \hat i + 2\hat j + 3\hat k$
Now ${\vec b_1}.{\vec b_2} = {a_1}{a_2} + {b_1}{b_2} = {c_1}{c_2}\, = 7 \times 1 + \left( { - 5} \right) \times 2 + 1 \times 3$$ = 7 - 10 + 3 = 0$ 
Hence, the given two lines are perpendicular to each other.
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Question 102 Marks
Find the direction cosines of a line which makes equal angles with the co-ordinate axes.
Answer
Let a line make equal angles $\alpha ,\alpha ,\alpha $ with the co-ordinate axes.
$\therefore$Direction cosines of the line are $\cos \alpha ,\cos \alpha ,\cos \alpha $…(i)
$\therefore {\cos ^2}\alpha + {\cos ^2}\alpha + {\cos ^2}\alpha = 1\,\left[ {\because {{\cos }^2}\alpha + {{\cos }^2}\beta + {{\cos }^2}\gamma = 1} \right]$
$ \Rightarrow 3{\cos ^2}\alpha = - 1$
$ \Rightarrow {\cos ^2}\alpha = \frac{1}{3}$
$ \Rightarrow \cos \alpha = \pm \frac{1}{{\sqrt 3 }}$
Putting $\cos \alpha = \pm \frac{1}{{\sqrt 3 }}$ in eq. (i), direction cosines of the required line making equal angles with the co-ordinate axes are $\pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}, \pm \frac{1}{{\sqrt 3 }}$ .

 $$ 

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Question 112 Marks
If a line makes angles ${90^0},{135^0},{45^0}$ with the x, y and z - axes respectively, find its direction cosines.
Answer
Here $\alpha = {90^0},\beta = {135^0}$ and $\gamma = {45^0}$ 

Since direction cosines of a line making angles $\alpha ,\beta ,\gamma $ with the x, y and z - axes respectively are $\cos \alpha ,\cos \beta ,\cos \gamma $ 

Therefore, the direction cosines of the required line are:

$\cos {90^0} = 0;\cos {135^0} = \frac{{ - 1}}{{\sqrt 2 }},\cos {45^0} = \frac{1}{{\sqrt 2 }}$

$\left[ {\because \cos {{135}^0} = \cos \left( {{{180}^0} - {{45}^0}} \right) = - \cos {{45}^0} = \frac{{ - 1}}{{\sqrt 2 }}} \right]$

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Question 122 Marks
Find the angle between the pair of lines 
$\frac{x+3}{3}=\frac{y-1}{5}=\frac{z+3}{4}$ 
$\frac{x+1}{1}=\frac{y-4}{1}=\frac{z-5}{2}$
Answer
The direction ratios of the first line are (3, 5, 4) and the direction ratios of the second line are (1, 1, 2).
If $\theta$ is the angle between them, then
$\cos \theta=\left|\frac{3.1+5.1+4.2}{\sqrt{3^{2}+5^{2}+4^{2}} \sqrt{1^{2}+1^{2}+2^{2}}}\right|$ = $\frac{16}{\sqrt{50} \sqrt{6}}=\frac{16}{5 \sqrt{2} \sqrt{6}}=\frac{8 \sqrt{3}}{15}$ 
Hence, the required angle is cos-1 $\left(\frac{8 \sqrt{3}}{15}\right)$
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Question 142 Marks
Find the direction cosines of the line passing through the two points (-2, 4, -5) and (1, 2, 3).
Answer
Let the given points be P(-2, 4, -5)  and Q(1, 2, 3). Then,
$\left| {PQ} \right| = \sqrt {{{\left( {1 + 2} \right)}^2} + {{\left( {2 - 4} \right)}^2} + {{\left( {3 + 5} \right)}^2}} $
$ = \sqrt {9 + 4 + 64} $
$ = \sqrt {77} $
The direction cosines of the line joining two points is
$\left( {\frac{{1 + 2}}{{\sqrt {77} }},\frac{{2 - 4}}{{\sqrt {77} }},\frac{{3 + 5}}{{\sqrt {77} }}} \right)$
= $\left( {\frac{3}{{\sqrt {77} }},\frac{{ - 2}}{{\sqrt {77} }},\frac{8}{{\sqrt {77} }}} \right)$
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Question 152 Marks
If a line makes angle 90°, 60° and 30° with the positive direction of x, y and z-axis respectively, find its direction cosines.
Answer
Let the direction cosine's of the lines be l, m, n. Then l = cos 90o = 0, m = cos 60o = $\frac{1}{2}$, n = cos 30o = $\frac{\sqrt{3}}{2}$.
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