5 Marks Questions

Question 15 Marks

Find the shortest distance between the lines whose vector equations are $ \vec r = \left( {1 - t} \right)\hat i + (t - 2)\hat j + (3 - 2t)\hat k$ and $\vec r = \left( {s + 1} \right)\hat i + (2s - 1)\hat j - (2s + 1)\hat k$

Answer

We have,
$\vec r=\hat i - 2\hat j + 3\hat k + t( - \hat i + \hat j - 2\hat k)$
$\vec r = \hat i - \hat j - \hat k + s(\hat i + 2\hat j - 2\hat k)$
${\vec a_1} = \hat i - 2\hat j + 3\hat k$
${\vec b_1} = - \hat i + \hat j - 2\hat k$
${\vec a_2} = \hat i - \hat j - \hat k$
${\vec b_2} = \hat i + 2\hat j - 2\hat k$
${\vec a_2} - {\vec a_1} = \hat j - 4\hat k$
${\vec b_1} \times {\hat b_2} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ { - 1}&1&{ - 2} \\ 1&2&{ - 2} \end{array}} \right|$
$ = 2\hat i - 4\hat j - 3\hat k$
$\left| {{{\vec b}_1} \times {{\vec b}_2}} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 3} \right)}^2}} $
$ = \sqrt {29} $
$d = \left| {\frac{{\left( {{{\vec a}_2} - {{\vec a}_1}} \right)\left( {{b_1} \times {b_2}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|$
$ = \frac{8}{{\sqrt {29} }}$

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Question 25 Marks

Find the shortest distance between the lines $\frac{{x + 1}}{7} = \frac{{y + 1}}{{ - 6}} = \frac{{z + 1}}{1}$ and $\frac{{x - 3}}{1} = \frac{{y - 5}}{{ - 2}} = \frac{{z - 7}}{1}$

Answer

On comparing the given equations with:
in the Cartesian form two lines $\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}$ and $\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}$
We get;
${x_1} = - 1,{y_1} = - 1,{z_1} = - 1;{a_1} = 7,{b_1} = - 6,{c_1} = 1$ and ${x_2} = 3,{y_2} = 5,{z_2} = 7;{a_2} = 1,{b_2} = - 2,{c_2} = 1$
now, the shortest distance the line is given by:
$S.D. = \frac{{\left| {\begin{array}{*{20}{c}} {{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}} \\ {{a_1}}&{{b_1}}&{{c_1}} \\ {{a_2}}&{{b_2}}&{{c_2}} \end{array}} \right|}}{{\sqrt {{{({b_1}{c_2} - {b_2}{c_1})}^2} + {{({c_1}{a_2} - {c_2}{a_1})}^2} + {{({a_1}{b_2} - {a_2}{b_1})}^2}} }}$
$S.D. = \frac{1}{{\sqrt {116} }}\left| {\begin{array}{*{20}{c}} {3 - ( - 1)}&{5 - ( - 1)}&{7 - ( - 1)} \\ 7&{ - 6}&1 \\ 1&{ - 2}&1 \end{array}} \right|$
$ = \frac{1}{{\sqrt {116} }}\left| {\begin{array}{*{20}{c}} 4&6&8 \\ 7&{ - 6}&1 \\ 1&{ - 2}&1 \end{array}} \right|$
$= \frac{1}{{\sqrt {116} }}[ - 16 - 36 - 64]$
$= \frac{1}{{\sqrt {116} }}\left| { - 116} \right| = \sqrt {116} = 2\sqrt {29}$

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Question 35 Marks

Find the direction cosines of the sides of the triangle whose vertices are $(3, 5, -4), (-1, 1, 2)$ and $(-5, -5, -2).$

Answer

The direction cosines of the two points passing through $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ is given by
$(x_2 - x_1), (y_{2 }- y_1), (z_{2 }- z_1)$
And the direction cosines of the line AB is $\frac{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)}{\mathrm{AB}}, \frac{\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)}{\mathrm{AB}}, \frac{\left(\mathrm{z}_{2}-\mathrm{z}_{1}\right)}{\mathrm{AB}}$
Where AB = $\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}}$

Here $A = (3, 5, -4)$	$B = (-1, 1, 2)$	$C = (-5, -5, -2)$
$B = (-1, 1, 2)$	$C = (-5, -5, -2)$	$A = (3, 5, -4)$
Direction ratios of $AB$	Direction ratio of BC	Direction ratios of CA
$(-1-3), (1 - 5), (2 - (-4))$	$(-5 + 1), (-5 - 1), (-2, -2)$	$(3 + 5), (5 + 5), (-4 + 2)$
$= -4, -4, 6$	$= (-4, -6, -4)$	$= (8, 10, -2)$
$\sqrt{\mathrm{AB}}=\sqrt{(68)}=2 \sqrt{17}$	$\sqrt{\mathrm{BC}}=\sqrt{(68)}=2 \sqrt{17}$	$\sqrt{C A}=\sqrt{(168)}=2 \sqrt{42}$
Direction Cosines of $AB$ are	Direction Cosines of $BC$ are	Direction Cosines of $CA$ are
$\frac{-4}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}, \frac{6}{2 \sqrt{17}}$	$\frac{-4}{2 \sqrt{17}}, \frac{-6}{2 \sqrt{17}}, \frac{-4}{2 \sqrt{17}}$	$\frac{8}{2 \sqrt{42}}, \frac{10}{2 \sqrt{42}}, \frac{-2}{2 \sqrt{42}}$
$\frac{-2}{\sqrt{17}}, \frac{-2}{\sqrt{17}}, \frac{3}{\sqrt{17}}$	$\frac{-2}{\sqrt{17}}, \frac{-3}{\sqrt{17}}, \frac{-2}{\sqrt{17}}$	$\frac{4}{\sqrt{42}}, \frac{5}{\sqrt{42}}, \frac{-1}{\sqrt{42}}$

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Question 45 Marks

Find the vector and Cartesian equations of the line through the point (5, 2, -4) and which is parallel to the vector $3\hat i + 2\hat j - 8\hat k$.

Answer

$\vec a = 5\hat i + 2\hat j - 4\hat k,\vec b = 3\hat i + 2\hat j - 8\hat k$
Vector equation of line is
$\vec r = \vec a + \lambda \vec b$
$ = 5\hat i + 2\hat j - 4\hat k + \lambda (3\hat i + 2\hat j - 8\hat k)$
Cartesian equation is
$x\hat i + y\hat j + z\hat k = 5\hat i + 2\hat j - 4\hat k + \lambda (3\hat i + 2\hat j - 8\hat k)$
$\Rightarrow x\hat i + y\hat j + z\hat k = (5 + 3\lambda )\hat i + (2 +2\lambda )\hat j + ( - 4 - 8\lambda )\hat k$
$\Rightarrow x = 5 + 3\lambda ,y = 2 + 2\lambda ,z = - 4 - 8\lambda $
$\Rightarrow\frac{{x - 5}}{3} = \frac{{y - 2}}{2} = \frac{{z + 4}}{{ - 8}}$$=\lambda$
Therefore, required equation is,
$\frac{x-5}{3}=\frac{y-2}{2}=\frac{z+4}{-8}$

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Question 55 Marks

Find the distance between the lines $l_1$ and $l_2$ given by
$\vec r = \hat i + 2\hat j - 4\hat k + \lambda (2\hat i + 3\hat j + 6\hat k)$
and $\vec r = 3\hat i + 3\hat j - 5\hat k + \mu (2\hat i + 3\hat j + 6\hat k)$

Answer

${\vec a_1} = \hat i + 2\hat j - 4\hat k$
${\vec a_2} = 3\hat i + 3\hat j - 5\hat k$
${\vec b_1} = {\vec b_2}=\vec b$$=2\hat i+3\hat j+6\hat k$
$|\vec b|=\sqrt{4+9+36}=\sqrt {49}$
Hence lines are parallel.
${\vec a_2} - {\vec a_1} = 2\hat i + \hat i - \hat k$
$\vec b \times \left( {{{\vec a}_2} - {{\vec a}_1}} \right) = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k} \\ 2&3&6 \\ 2&1&{ - 1} \end{array}} \right|$
$=-9\vec i+14\vec j-4\vec k$
$d = \left| {\frac{{\vec b \times \left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {\vec b} \right|}}} \right|$
$ = \left| {\frac{{ - 9\hat i + 14\hat j - 4\hat k}}{{\sqrt {49} }}} \right|$
$=\frac{{\sqrt {81+196+16} }}{{\sqrt {49} }} = \frac{{\sqrt {293} }}{7}$

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Maths 5 Marks Questions

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