3 Marks Question - MATHS STD 12 Science Questions - Vidyadip

Questions

3 Marks Question

🎯

Test yourself on this topic

19 questions · timed · auto-graded

Question 13 Marks

Find the vector equation of the line passing through the point (1, 2, -4) and perpendicular to the two lines:$ \frac { x - 8 } { 3 } = \frac { y + 19 } { - 16 } = \frac { z - 10 } { 7 }$ and $ \frac { x - 15 } { 3 } = \frac { y - 29 } { 8 } = \frac { z - 5 } { - 5 }.$

Answer

We have to find the vector and cartesian equations of a line passing through (1, 2,-4) and perpendicular to the two lines $ \frac { x - 8 } { 3 } = \frac { y + 19 } { - 16 } = \frac { z - 10 } { 7 }$ and $ \frac { x - 15 } { 3 } = \frac { y - 29 } { 8 } = \frac { z - 5 } { - 5 }.$
Now, we know that any line through ( 1, 2, -4) can be written as
$ \frac { x - 1 } { a } = \frac { y - 2 } { b } = \frac { z + 4 } { c }$
where a,b,c are the direction ratios of line (i)
Now, the line (i) is perpendicular to the lines
$\frac { x - 8 } { 3 } = \frac { y + 19 } { - 16 } = \frac { z - 10 } { 7 }$
and $\frac { x - 15 } { 3 } = \frac { y - 29 } { 8 } = \frac { z - 5 } { - 5 }$
The direction ratios of the above lines are (3,-16,7) and (3,8,-5), respectively which are perpendicular to the Equation (i).
3a - 16b + 7c = 0 ...(ii)
and 3a + 8b - 5c = 0 ...(iii)
By cross-multiplication, we get
$\frac { a } { 80 - 56 } = \frac { b } { 21 + 15 } = \frac { c } { 24 + 48 }$
$\Rightarrow \frac { a } { 24 } = \frac { b } { 36 } = \frac { c } { 72 }$
$\Rightarrow \frac { a } { 2 } = \frac { b } { 3 } = \frac { c } { 6 } = \lambda ( \text { say } )$
$ \Rightarrow a = 2 \lambda , b = 3 \lambda , c = 6 \lambda$
The equation of the required line in cartesian form is
$\frac { x - 1 } { 2 \lambda } = \frac { y - 2 } { 3 \lambda } = \frac { z + 4 } { 6 \lambda } \text { or } \frac { x - 1 } { 2 } = \frac { y - 2 } { 3 } = \frac { z + 4 } { 6 }$ and in vector form is $ \vec { r } = ( \hat { i } + 2 \hat { j } - 4 \hat { k } ) + \lambda ( { 2 } \hat { i } + 3 \hat { j } + 6 \hat { k } ).$

View full question & answer→

Question 23 Marks

Find the shortest distance between lines $\vec{r}=6 \hat{i}+2 \hat{j}+2 \hat{k}+\lambda(\hat{i}-2 \hat{j}+2 \hat{k})$ and $\vec{r}=-4 \hat{i}-\hat{k}+\mu(3 \hat{i}-2 \hat{j}-2 \hat{k})$

Answer

We know that shortest distance between lines with vector equations
$\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1} \text { and } \vec{r}=\vec{a}_{2}+\lambda \vec{b}_{2} \text { is }$
$\left|\frac{(\vec{ {b}_{1}} \times \vec{ {b}_{2}}) \cdot(\vec{ {a}_{2}}-\vec{ {a}_{1}})}{|\vec{ {b}_{1}} \times \vec{ {b}_{2}}|}\right|$
Given,
$\vec{\mathrm{r}}=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})+\lambda({1} \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+{2} \hat{k})$
Here, $\vec{a}_{1}=(6 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$ and $\vec{b}_{1}=(1 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+{2} \hat{k})$
and,
$\vec{\mathrm{r}}=(-4 \hat{\mathrm{i}}-\hat{\mathrm{k}})+\mu(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-2 \hat{\mathrm{k}})$
Here, $\vec{a}_{2}=(-4 \hat{\mathrm{i}}-\hat{\mathrm{k}})$ and $\vec{b}_{2}=(3 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-\hat{2})$
Now,
$\Rightarrow(\vec{a_{2}}-\vec{a_{1}})=(-4 \hat{i}-\hat{k})-(6 \hat{i}+2 \hat{j}+2 \hat{k}) $
$=((-4-6) \hat{\mathrm{i}}+(0-2) \hat{\mathrm{j}}+(-1-2) \hat{\mathrm{k}})$
$=(-10 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$
$\Rightarrow(\vec{b}_{1} \times \vec{b}_{2})=\left|\begin{array}{ccc} {\hat{\mathrm{i}}} & {\hat{\mathrm{j}}} & {\hat{\mathrm{k}}} \\ {1} & {-2} & {2} \\ {3} & {-2} & {-2} \end{array}\right|$
$\Rightarrow\hat{\mathrm{i}}[(-2 \times-2)-(-2 \times 2)]-\hat{\mathrm{j}}[(1 \times-2)-(3 \times 2)]+\hat{\mathrm{k}}[(1 \times-2)-(3 \times-2)]$
$\Rightarrow\hat{\mathrm{i}}[4+4]-\hat{\mathrm{j}}[-2-6]+\hat{\mathrm{k}}[-2+6]$
$\Rightarrow 8 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}$
Magnitude of $\vec{b}_{1} \times \vec{b}_{2}=|\vec{b}_{1} \times \vec{b}_{2}|=\sqrt{8^{2}+8^{2}+4^{2}}=\sqrt{64+64+16}$
= $\sqrt{144}$
= 12
Also,
$\Rightarrow\left(\vec{b}_{1} \times \vec{b}_{2}\right) \cdot(\vec{a}_{2}-\vec{a}_{1})=(8 \hat{\mathrm{i}}+8 \hat{\mathrm{j}}+4 \hat{\mathrm{k}}) \cdot(-10 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}-3 \hat{\mathrm{k}})$
= -80 + (-16) + (-12)
= -108
Shortest Distance = $\left|\frac{\left(\overrightarrow{\mathrm{b}}_{1} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}})}{|\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}|}\right|=\left|\frac{-108}{12}\right|=|-9|=9$
Hence, the shortest distance between the given two lines is 9.

View full question & answer→

Question 33 Marks

Find the angle between the lines whose direction ratios are a, b, c and b - c, c - a, a - b.

Answer

Direction ratios of one line are a, b, c
$\Rightarrow $ A vector along this line is ${\vec b_1} = a\hat i + b\hat j + c\hat k$
Direction ratios of second line are $b - c,c - a,a - b$
$\Rightarrow$ A vector along second line is ${\vec b_2} = \left( {b - c} \right)\hat i + \left( {c - a} \right)\hat j + \left( {a - b} \right)\hat k$
Let $\theta $ be the angle between the two lines, then
$\cos \theta = \frac{{\left| {{{\vec b}_1}.{{\vec b}_2}} \right|}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}} = \frac{{a\left( {b - c} \right) + b\left( {c - a} \right) + c\left( {a - b} \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {a - b} \right)}^2}} }}$
$= \frac{{ab - ac + bc - ab + ac - bc}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {a - b} \right)}^2}} }} = 0 = \cos {90^0}$
$\Rightarrow \theta = {90^0}$

View full question & answer→

Question 43 Marks

Find the angle between the lines $\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$ and $\frac{{x - 5}}{4} = \frac{{y - 2}}{1} = \frac{{z - 3}}{8}$.

Answer

$\frac{{x - 0}}{2} = \frac{{y - 0}}{2} = \frac{{z - 0}}{1}$
$\frac{{x - 5}}{4} = \frac{{y - 2}}{1} = \frac{{z - 3}}{8}$
$a_1 = 2, b_1 = 2, c_1 = 1$
$a_2 = 4, b_2 = 1, c_2 = 8$
$\cos \theta = \left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {a_1^2 + {b_1^2} + c_1^2} \sqrt {a_2^2 + b_2^2 + c_2^2} }}} \right|$
$ = \left| {\frac{{2(4) + 2(1) + 1(8)}}{{\sqrt {{2^2} + {2^2} + 1} \sqrt {{4^2} + {1^2} + {8^2}} }}} \right|$
$ = \left| {\frac{{8 + 2 + 8}}{{\sqrt 9 \sqrt {81} }}} \right|$
$ = \frac{{18}}{{27}}$
$ = \frac{2}{3}$
$\theta = {\cos ^{ - 1}}\left( {\frac{2}{3}} \right)$

View full question & answer→

Question 53 Marks

Find the angle between the lines $\frac{{x - 2}}{2} = \frac{{y - 1}}{5} = \frac{{z + 3}}{{ - 3}}$ and $\frac{{x + 2}}{{ - 1}} = \frac{{y - 4}}{8} = \frac{{z - 5}}{4}$.

Answer

${\overrightarrow b _1} = 2\hat i + 5\hat j - 3\hat k$
${\overrightarrow b _2} = \hat i + 8\hat j + 4\hat k$
$\cos \theta = \left| {\frac{{{{\overrightarrow b }_1} \cdot {{\overrightarrow b }_2}}}{{\left| {{{\overrightarrow b }_1}} \right|\left| {{{\overrightarrow b }_2}} \right|}}} \right|$
$ = \left| {\frac{{2( - 1) + 5(8) + ( - 3)(4)}}{{\sqrt {38} \sqrt {81} }}} \right|$
$ = \frac{{26}}{{9\sqrt {38} }}$
$\theta = {\cos ^{ - 1}}\left( {\frac{{26}}{{9\sqrt {38} }}} \right)$

View full question & answer→

Question 63 Marks

Find the angle between the pair of lines
$\vec r = (3\hat i + \hat j - 2\hat k) + \lambda (\hat i - \hat j -2\hat k)$ and $\vec r = (2\hat i - \hat j - 56\hat k) + \mu (3\hat i - 5\hat j - 4\hat k)$

Answer

Let $\theta $ be the angle between the given lines
${\vec b_1} = \hat i - \hat j + 2\hat k$ and ${\vec b_2} = 3\hat i - 5\hat j - 4\hat k$
$\cos \theta = \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|\left| {{{\vec b}_2}} \right|}}} \right|$
$\left| {\frac{{\left( {\hat i - \hat j - 2\hat k} \right).\left( {3\hat i - 5\hat j - 4\hat k} \right)}}{{\left| {\hat i - \hat j - 2\hat k} \right|\left| {3\hat i - 5\hat j - 4\hat k} \right|}}} \right|$
$=|\frac{3+5+8}{\sqrt{1^2+1^2+2^2}\sqrt{3^2+5^2+4^2}}|=\frac{16}{\sqrt6\sqrt{50}}=\frac{16}{\sqrt{2\times3}\sqrt{2\times25}}$
$= \frac{16}{10\times\sqrt3}$
$=\frac{8}{5\sqrt3}$
$\Rightarrow\theta=\cos^{-1}\left(\frac{8}{5\sqrt3}\right)$

View full question & answer→

Question 73 Marks

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

Answer

We know that
If $\theta$ is the acute angle between $\vec{\mathrm{r}}=\vec{\mathrm{a}_{1}}+\lambda \vec{\mathrm{b}_{1}}$ and $\vec{\mathrm{r}}=\vec{\mathrm{a}_{2}}+\mu \vec{\mathrm{b}_{2}}$ , then
$\cos \theta=\left|\begin{array}{c} \frac {{\vec{b}_{1} \cdot \vec{b}_{2}}} {{|\vec {b}_{1}||\vec {b}_{2}}|} \end{array}\right|$ .......(i)
Given that, $\vec{\mathrm{r}}=2 \hat{\mathrm{i}}-5 \hat{\mathrm{j}}+\hat{\mathrm{k}}+\lambda(3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}})$ and $\vec{\mathrm{r}}=7 \hat{\mathrm{i}}-6 \hat{\mathrm{k}}+\mu(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$
Here, $\vec{b}_{1}=3 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+6 \hat{\mathrm{k}}$ and $\vec{b}_{2}=\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}$
So, from (i), we have
$\cos \theta=\left|\frac{(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})}{|\hat{3} \hat{i}+2 \hat{j}+6 \hat{k}| \cdot|\hat{i}+2 \hat{j}+2 \hat{k}|}\right|$ ......(ii)
$\Rightarrow\left|{a}{\hat{i}}+{\mathrm{b} \hat{\mathrm{j}}}+\mathrm{c} \hat{\mathrm{k}}\right|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}+\mathrm{c}^{2}}$
$\Rightarrow |3 \hat{i}+2 \hat{j}+6 \hat{k}|$ = $\sqrt{3^{2}+2^{2}+6^{2}}=\sqrt{9+4+36}=\sqrt{49}=7$
And $|\hat{i}+2 \hat{j}+2 \hat{k}|=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{1+4+4}=\sqrt{9}=3$
Now, $\left(a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\right) \cdot\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right)=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}$
$\Rightarrow (3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})$ = $3 \times 1+2 \times 2+6 \times 2=3+4+12=19$
⇒ By (ii), we have
$\Rightarrow \cos \theta=\frac{19}{7 \times 3}=\frac{19}{21}$
$\Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right)$

View full question & answer→

Question 83 Marks

Find the shortest distance between the lines whose vector equations are $\vec r = \hat i + 2\hat j + 3\hat k$ + $\lambda \left( {\hat i - 3\hat j + 2\hat k} \right)$ and $\vec r = 4\hat i + 5\hat j + 6\hat k$ + $\mu \left( {2\hat i + 3\hat j + \hat k} \right)$

Answer

On comparing the given equations with: $\overrightarrow{r}=\overrightarrow{{{a}_{1}}}+\lambda \overrightarrow{{{b}_{1}}},$ and $\overrightarrow{r}=\overrightarrow{{{a}_{2}}}+\mu \overrightarrow{{{b}_{2}}}$
, we get:
$\overrightarrow {{a_1}} = \hat i + 2\hat j + 3\hat k,\overrightarrow {{b_1}} = \hat i - 3\hat j + 2\hat k,$
and $\overrightarrow {{a_2}} = 4\hat i + 5\hat j + 6\hat k,\overrightarrow {{b_2}} = 2\hat i + 3\hat j + \hat k$
$\therefore S.D = \left| {\frac{{(\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} ).(\overrightarrow {{a_2}} - \overrightarrow {{a_2}} )}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}}} \right|$
$= \left| {\frac{{( - 9\hat i + 3\hat j + 9\hat k).(3\hat i + 3\hat j + 3\hat k)}}{{\sqrt {171} }}} \right|$
$= \left| {\frac{{ - 27 + 9 + 27}}{{3\sqrt {19} }}} \right| = \left| {\frac{9}{{3\sqrt {19} }}} \right| = \frac{3}{{\sqrt {19} }}$

View full question & answer→

Question 93 Marks

Find the shortest distance between the lines

$\vec r = \left( {\hat i + 2\hat j + \hat k} \right) + \lambda \left( {\hat i - \hat j + \hat k} \right)$

$\vec r = \left( {2\hat i - \hat j - \hat k} \right) + \mu \left( {2\hat i + \hat j + 2\hat k} \right)$

Answer

${\vec a_1} = \hat i + 2\hat j + \hat k,{\vec b_1} = \hat i - \hat j + \hat k$

${\vec a_2} = 2\hat i - \hat j - \hat k,{\vec b_2} = 2\hat i + \hat j + 2\hat k$

$d = \left| {\frac{{\left( {{{\vec a}_2} - {{\vec a}_1}} \right).\left( {{{\vec b}_1} \times {{\vec b}_2}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|$

${\vec a_2} - \vec a{_1} = \hat i - 3\hat j - 2\hat k$

${\vec b_1} \times {\vec b_2} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 1&{ - 1}&1 \\ 2&1&2 \end{array}} \right|\\ $

$=\hat i(-2-1)-\hat j(2-2)+\hat k(1+2)$

$ = - 3\hat i + 3\hat k$

$d = \left| {\frac{{\left( {\hat i - 3\hat j - 2\hat k} \right).\left( { - 3\hat i + 3\hat k} \right)}}{{\left| { - 3\hat i + 3\hat k} \right|}}} \right|$

$ = \left| {\frac{{ - 3 - 6}}{{\sqrt {9 + 9} }}} \right| = \left| {\frac{9}{{3\sqrt 2 }}} \right| = \frac{3}{{\sqrt 2 }}$

View full question & answer→

Question 103 Marks

Find the values of p so that the lines $\frac{{1 - x}}{3} = \frac{{7y - 14}}{{2p}} = \frac{{z - 3}}{2}$ and $\frac{{7 - 7x}}{{3p}} = \frac{{y - 5}}{1} = \frac{{6 - z}}{5}$ are at right angles.

Answer

Given: Equation of one line $\frac{{1 - x}}{3} = \frac{{7y - 14}}{{2p}} = \frac{{z - 3}}{2} \Rightarrow \frac{{ - \left( {x - 1} \right)}}{3} = \frac{{7\left( {y - 2} \right)}}{{2p}} = \frac{{z - 3}}{2}$
$\Rightarrow \frac{{ \left( {x - 1} \right)}}{-3} = \frac{{y - 2}}{{\frac{{2p}}{7}}} = \frac{{z - 3}}{2}$
$\therefore$ Direction ratios of this line are $- 3,\frac{{2p}}{7},2 = {a_1},{b_1},{c_1}$ (say)
Again, equation of another line is $\frac{{7 - 7x}}{{3p}} = \frac{{y - 5}}{1} = \frac{{6 - z}}{5} = \frac{{ - 7\left( {x - 1} \right)}}{{3p}} = \frac{{y - 5}}{1} = \frac{{ - \left( {z - 6} \right)}}{5}$
$\Rightarrow \frac{{x - 1}}{{\frac{{ - 3p}}{7}}} = \frac{{y - 5}}{1} = \frac{{z - 6}}{{ - 5}}$
$\therefore$ Direction ratios of this line are $\frac{{ - 3p}}{7},1 - 5 = {a_2},{b_2},{c_2}$ (say)
Since these two lines are perpendicular.
Therefore, ${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$
$\Rightarrow \left( { - 3} \right)\left( {\frac{{ - 3p}}{7}} \right) + \left( {\frac{{2p}}{7}} \right)\left( 1 \right) + \left( 2 \right)\left( { - 5} \right) = 0 \Rightarrow \frac{{9p}}{7} + \frac{{2p}}{7} - 10 = 0$
$\Rightarrow \frac{{11p}}{7} = 10 \Rightarrow p = \frac{{70}}{{11}}$

View full question & answer→

Question 113 Marks

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

We know that
If $l_1, m_1, n_1$ and $l_2, m_2, n_2$ are the direction cosines of two lines; and $\theta$ is the acute angle between the two lines; then $\cos \theta = |l_1l_2 + m_1m_2 + n_1n_2|$
If two lines are perpendicular, then the angle between the two is $\theta = 90^\circ$
$\Rightarrow$ For perpendicular lines, $| l_1l_2 + m_1m_2 + n_1n_2 | = \cos 90^\circ = 0,$ i.e.
$| l_1l_2 + m_1m_2 + n_1n_2 | = 0$
So, in order to check if the three lines are mutually perpendicular, we compute |$ l_1l_2 + m_1m_2 + n_1n_2 $| for all the pairs of the three lines.
Now let the direction cosines of $L_1, L_2$ and $L_3$ be $l_1, m_1, n_1; l_2, m_2, n_2$ and $l_3, m_3, n_3.$
First, consider
$\left|1_{1} 1_{2}+\mathrm{m}_{1} \mathrm{m}_{2}+\mathrm{n}_{1} \mathrm{n}_{2}\right| = |\left(\frac{12}{13} \times \frac{4}{13}\right)+\left(\frac{-3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{3}{13}\right)| = \frac{48}{13}+\left(\frac{-36}{13}\right)+\left(\frac{-12}{13}\right)$
$= \frac{48+(-48)}{13}=0$
$\Rightarrow L_{1 } \perp L_2 ……(i)$
Next, We take,
$\left|1_{2} 1_{3}+\mathrm{m}_{2} \mathrm{m}_{3}+\mathrm{n}_{2} \mathrm{n}_{3}\right| = |\left(\frac{4}{13} \times \frac{3}{13}\right)+\left(\frac{12}{13} \times \frac{-4}{13}\right)+\left(\frac{3}{13} \times \frac{12}{13}\right)| = \frac{12}{13}+\left(\frac{-48}{13}\right)+\frac{36}{13}$
$= \frac{48+(-48)}{13}=0$
$\Rightarrow L_{2 }\perp L_3 …(ii)$
Thereafter,
$\left|1_{3} 1_{1}+\mathrm{m}_{3} \mathrm{m}_{1}+\mathrm{n}_{3} \mathrm{n}_{1}\right| = |\left(\frac{3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{-3}{13}\right)+\left(\frac{12}{13} \times \frac{-4}{13}\right)| = \frac{36}{13}+\frac{12}{13}+\left(\frac{-48}{13}\right)$
= $\frac{48+(-48)}{13}=0$
$\Rightarrow L_{1 } \perp L_3 …(iii)$
$\therefore$ By $(i), (ii)$ and $(iii),$ we conclude ,
$L_1, L_2$ and $L_3$ are mutually perpendicular.

View full question & answer→

Question 123 Marks

Show that the points (2, 3, 4), (-1, -2, 1), (5, 8, 7) are collinear.

Answer

The given points are $A\left( {2,3,4} \right),B\left( { - 1, - 2,1} \right)$and $C\left( {5,8,7} \right)$
$\therefore$ Direction ratios of the line joining A and B are
$- 1 - 2, - 2 - 3,1 - 4\,\left[ {\because {x_2} - {x_1},{y_2} - {y_1},{z_2} - {z_1}} \right]$ …(i)
$ \Rightarrow - 3, - 5, - 3 = {a_1},{b_1},{c_1}$ (say)
Again Direction ratios of the line joining B and C are
$5 - \left( { - 1} \right),8 - \left( { - 2} \right),7 - 1 = 6,10,6 = {a_2},{b_2},{c_2}$ (say) ….(ii)
From eq. (i) and (ii),
$\frac{{ - 3}}{6} = \frac{{ - 1}}{2},\frac{{ - 5}}{{10}} = \frac{{ - 1}}{2},\frac{{ - 3}}{6} = \frac{{ - 1}}{2}$
$\Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}$
Therefore, AB is parallel to BC.
But point B is common to both AB and BC. Hence points A, B, C are collinear.

View full question & answer→

Question 133 Marks

If a line has the direction ratios $-18, 12, -4$ then what are its direction cosines $?$

Answer

Here, $a = -18, b = 12, c = -4$
$a^{2 }+ b^{2 }+ c^2 = (-18)^2 + (12)^2 + (-4)^2$
$= 484$
Therefore, direction cosines are,
$\left( { - \frac{{18}}{{484}},\frac{{12}}{{484}}, - \frac{4}{{484}}} \right)$

View full question & answer→

Question 143 Marks

Find the shortest distance between the lines $l_1$ and $l_2$ whose vector equations are
$\vec r = \hat i + \hat j + \lambda (2\hat i - \hat j + \hat k) ...(1) $
and $\vec r = 2\hat i + \hat j - \hat k + \mu (3\hat i - 5\hat j + 2\hat k) ...(2)$

Answer

${\vec a_1} = \hat i + \hat j,{\vec b_1} = 2\hat i - \hat j + \hat k$
${\vec a_2} = 2\hat i + \hat j - \hat k,{\vec b_2} = 3\hat i - 5\hat j + 2\hat k$
${\vec a_2} - {\vec a_1} = \hat i - \hat k$
${\vec b_1} \times {\vec b_2} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&{ - 1}&1 \\ 3&{ - 5}&2 \end{array}} \right|$
$=\hat i (-2+5)-\hat j(4-3)+\hat k(-10+3)$
$ = 3\hat i - \hat j - 7\hat k$
$\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|=\sqrt {9+1+49}= \sqrt {59} $
Also, $(\vec b_1×\vec b_2).(\vec a_2-\vec a_1)=(3\hat i-\hat j-7 \hat k)(\hat i-\hat k)=3+7+0=10$
$d=\left|\frac{(\vec b_1×\vec b_2).(\vec a_2-\vec a_1)}{|\vec b_1×\vec b_2|}\right|=\frac{10}{\sqrt{59}}$

View full question & answer→

Question 153 Marks

Find the angle between the pair of lines given by $\vec r = 3\hat i + 2\hat j - 4\hat k + \lambda (\hat i + 2\hat j + 2\hat k)$ and $\vec r = 5\hat i - 2\hat j + \mu (3\hat i + 2\hat j + 6\hat k)$

Answer

Given that, ${\vec b_1} = \hat i + 2\hat j + 2\hat k$
${\vec b_2} = 3\hat i + 2\hat j + 6\hat k$
$\cos \theta = \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|\left| {{{\vec b}_2}} \right|}}} \right| = \frac{{19}}{{21}}$
$=\left| \frac{(\hat i +2\hat j+2\hat k).(3\hat i+2\hat j+6\hat k)}{\sqrt{1+4+4}\sqrt{9+4+36}}\right|$
$ =\frac{{3+4+12}}{3×7}$
$=\frac{19}{21}$
$\therefore \theta=cos^{-1}\frac{19}{21}$

View full question & answer→

Question 163 Marks

Show that the points A (2, 3, -4), B (1, -2, 3) and C (3, 8, -11) are collinear.

Answer

Direction ratios of line joining A and B are
(1 - 2, -2 - 3, 3 + 4) i.e., -1, -5, 7.
The direction ratios of line joining B and C are
(3 -1, 8 + 2, -11- 3), i.e., 2, 10, - 14.
It is clear that direction ratios of AB and BC are proportional, hence, AB is parallel to BC. But point B is common to both AB and BC. Therefore, A, B, C are collinear points.

View full question & answer→

Question 173 Marks

If the line has direction ratios 2, -1, -2 determine its direction Cosines.

Answer

a = 2, b = -1, c = -2

$\sqrt{a^2+b^2+c^2} = \sqrt{2^2+(-1)^2+(-2)^2}=\sqrt{4+1+4}=3$

Therefore, direction cosines are, $\frac{2}{3},\frac{-1}{3},\frac{-2}{3}$

View full question & answer→

Question 183 Marks

Find value of p when given lines $\frac{1-x}{3}=\frac{7 y-14}{3 p}=\frac{z-3}{2}$ and $\frac{7-7 x}{3 p}=\frac{y-5}{1}=\frac{6-z}{5}$ are perpendicular.

Answer

Write the lines in standard form $\frac{x-x_1}{a}$ $=\frac{y-y_1}{b}=\frac{z-z_1}{c}:$
$\therefore$ $L_1: \frac{x-1}{-3}=\frac{y-2}{3 p / 7}=\frac{z-3}{2}$
⟹ D.R.s are $\left(-3, \frac{3 p}{7}, 2\right)$
$\therefore$ $L_2: \frac{x-1}{-3 p / 7}=\frac{y-5}{1}=\frac{z-6}{-5}$
⟹ D.R.s are $\left(-\frac{3 p}{7}, 1,-5\right)$
Condition for Perpendicularity $\left(a_1 a_2+b_1 b_2+c_1 c_2=0\right):$
$(-3)\left(-\frac{3 p}{7}\right)+\left(\frac{3 p}{7}\right)(1)+$$(2)(-5)=0$
$=\frac{9 p}{7}+\frac{3 p}{7}-10=0$
$\Longrightarrow \frac{12 p}{7}=10$
$\therefore$ $12 p=70$
$\Longrightarrow p=\frac{70}{12}$
$=\frac{35}{6}$

View full question & answer→

Question 193 Marks

Show that the line passing through the points (4, 7, 8) (2, 3, 4) is parallel to the passing through the points (-1, -2, 1), (1, 2, 5)

Answer

Direction Ratios (D.R.s) of Line 1 $\left(L_1\right): a_1=x_2-x_1$
$=2-4$
$=-2$
$\therefore$ $b_1=y_2-y_1$
= $3-7$
$=-4$
$\therefore$ $c_1=z_2-z_1$
$=4-8$
$=-4$
$\therefore$ D.R.s of $L_1:(-2,-4,-4)$
Direction Ratios (D.R.s) of Line 2 $\left(L_2\right): a_2=1-(-1)$
$=2 b_2$
$\therefore$ $b_2=2-(-2)$
$=4 c_2$
= $5-1$
$=4$
$\therefore$ D.R.s of $L_2:(2,4,4)$
Check for Parallelism: Lines are parallel if $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
$\frac{-2}{2}=-1 ;$
$\frac{-4}{4}=$ -1;
$\therefore$ $\frac{-4}{4}=-1$ Since the ratios are equal, the lines are parallel.

View full question & answer→

Generate Paper Free All Three Dimensional Geometry topics

3 Marks Question - MATHS STD 12 Science Questions - Vidyadip