12th NCERT Three Dimensional Geometry Exercise Miscellineous Questions 23
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## Exercise 11.1

Question (1)

Show that the line joining the origin to the point (2, 1, 1) is perpendicular to the line determined by the points (3, 5, -1), (4, 3, -1).

Solution

The direction vector of line joining origin and point P(2, 1, 1) $\overrightarrow {OP} = (2,1,1) - (0,0,0) = (2,1,1)$ The direction vector of line joining point A(3, 5, -1) and B(4, 3, -1) $\overrightarrow {AB} = (4,3, - 1) - (3,5, - 1) = (1, - 2,0)$ $\begin{array}{l}\overrightarrow {OP} \cdot \overrightarrow {AB} = (2,1,1) \cdot (1, - 2,0)\\\quad \quad \quad = 2 - 2 + 0 = 0\end{array}$ $\overrightarrow {OP} \bot \overrightarrow {AB}$ So the lines containing these vectors are also perpendicular.+

Question (2)

If l1, m1,n1 and l2, m2, n2 are the direction cosines of two mutually perpendicular lines, show that the direction cosines of the line perpendicular to both of these are m1 n2 - m2 n1 , n1 l2 - n2 l1 , l1 m2 - l2 m1 .

Solution

If vector a and vector b are two distinct vectors, and vector c is other vector perpendicular to both vectors a and b, then $\overrightarrow c = \overrightarrow a \times \overrightarrow b$
If l1, m1,n1 and l2, m2, n2 are the direction cosines of the two lines. the direction cosines of the line perpendicular to both will be given by $\overrightarrow c = \overrightarrow a \times \overrightarrow b$ $\left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\{{l_1}}&{{m_1}}&{{n_1}}\\{{l_2}}&{{m_2}}&{{n_2}}\end{array}} \right| = \left( {{m_1}{n_2} - {m_2}{n_1}} \right)\widehat i - \left( {{l_1}{n_2} - {l_2}{n_1}} \right)\widehat j + \left( {{l_1}{n_2} - {l_2}{n_1}} \right)\widehat k$ $\left( {{m_1}{n_2} - {m_2}{n_1}} \right)\widehat i + \left( {{l_2}{n_1} - {l_1}{n_2}} \right)\widehat j + \left( {{l_1}{n_2} - {l_2}{n_1}} \right)\widehat k$ So cosine direction of the line perpendicular to both lines is $\left( {{m_1}{n_2} - {m_2}{n_1}} \right),\left( {{l_2}{n_1} - {l_1}{n_2}} \right),\left( {{l_1}{n_2} - {l_2}{n_1}} \right)$

Question (3)

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

Solution

$Let\;\overrightarrow x = a\widehat i + b\widehat j + c\widehat k,and\quad \overrightarrow y = (b - c)\widehat i + (c - a)\widehat j + (a - b)\widehat k$ Let θ be the angle between two vectors x and y $\cos \theta = \frac{{\overrightarrow {x \cdot } \overrightarrow y }}{{\left| {\overrightarrow x \left| {\overrightarrow y } \right|} \right|}} = \frac{{\left( {a\widehat i + b\widehat j + c\widehat k} \right) \cdot \left( {(b - c)\widehat i + (c - a)\widehat j + (a - b)\widehat k} \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{(b - c)}^2} + {{(c - a)}^2} + {{(a - b)}^2}} }}$ $\cos \theta = \frac{{ab - ac + bc - ba + ac - bc}}{{\sqrt {{a^2} + {b^2} + {c^2}} \sqrt {{{(b - c)}^2} + {{(c - a)}^2} + {{(a - b)}^2}} }} = 0$ $\theta = {\cos ^{ - 1}}0 = {90^0}$ So angle between the lines is 90°

Question (4)

Find the equation of the line parallel to x-axis and passing through the origin.

Solution

The direction cosine of x axis. is (1, 0, 0), line is parallel to x axis so its direction cosine is (1, 0, 0). The line is passes through origin. Its vector equation is given by $\overrightarrow r = 0\widehat i + 0\widehat j + 0\widehat k + \lambda \left( {\widehat i + 0\widehat j + 0\widehat k} \right)$ The cartesian equation is given by $\frac{x}{1} = \frac{y}{0} = \frac{z}{0}$

Question (5)

If the coordinates of the points A,B, C, Dbe (1, 2, 3), (4, 5, 7), (-4, 3, -6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

Solution

The direction vector of line AB = (4, 5, 7)- (1, 2, 3) = (3, 3, 4) $\left| {\overrightarrow {AB} } \right| = \sqrt {{3^2} + {3^2} + {4^2}} = \sqrt {9 + 9 + 16} = \sqrt {34}$ The direction vector of line CD = (2, 9, 2) - (-4, 3, -6) = (6, 6, 8) $\left| {\overrightarrow {CD} } \right| = \sqrt {{6^2} + {6^2} + {8^2}} = \sqrt {36 + 36 + 64} = \sqrt {136} = 2\sqrt {34}$ Let angle between AB and CD be θ, then $\cos \theta = \frac{{\overrightarrow {AB \cdot } \overrightarrow {CD} }}{{\left| {\overrightarrow {AB} \left| {\overrightarrow {CD} } \right|} \right|}} = \frac{{\left( {3\widehat i + 3\widehat j + 4\widehat k} \right) \cdot \left( {6\widehat i + 6\widehat j + 8\widehat k} \right)}}{{\sqrt {34} 2\sqrt {34} }}$ $\cos \theta = \frac{{18 + 18 + 32}}{{2(34)}} = \frac{{68}}{{68}} = 1$ $\theta = {\cos ^{ - 1}}1 = {0^0}$

Question (6)

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.

Solution

${\frac{{x - 1}}{{ - 3}} = \frac{{y - 2}}{{2k}} = \frac{{z - 3}}{2}}$ ${\overrightarrow {{b_1}} = - 3\hat i + 2k\hat j + 2\hat k}$] $\frac{{x - 1}}{{3k}} = \frac{{y - 1}}{1} = \frac{{z - 6}}{{ - 5}}$ $\overrightarrow {{b_2}} = 3k\hat i + \hat j - 5\hat k$ As the lines are perpendicular, dot product of directions is zero. $\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} = 0$ $\left( { - 3\widehat i + 2k\widehat j + 2\widehat k} \right) \cdot \left( {3k\widehat i + \widehat j - 5\widehat k} \right) = 0$ $- 9k + 2k - 10 = 0$ $- 7k = 10 \Rightarrow k = \frac{{ - 10}}{7}$

Question (7)

Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane $\overrightarrow r \cdot \left( {\widehat i + 2\widehat j - 5\widehat k} \right) + 9 = 0$.

Solution

The vector equation of the plane is $\overrightarrow r \cdot \left( {\widehat i + 2\widehat j - 5\widehat k} \right) + 9 = 0$. So normal of the plane is $\overrightarrow n = \left( {\hat i + 2\hat j - 5\hat k} \right)$ The normal is perpendicular to the plane, the line is also perpendicular to plane so normal and line are perpendicular to each other. So direction vector of the line is equal to normal $\overrightarrow b = \overrightarrow n = \left( {\hat i + 2\hat j - 5\hat k} \right)\ The line is passing through ( 1, 2, 3). The vector equation of the line will be \[\overrightarrow r = \overrightarrow a + \lambda \overline b ,\lambda \in R$ $\overrightarrow r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda \left( {\hat i + 2\hat j - 5\hat k} \right)$

Question (8)

Find the equation of the plane passing through (a, b, c) and parallel to the plane $\overrightarrow r \cdot \left( {\widehat i + \widehat j + \widehat k} \right) = 2$

Solution

The equation of the plane is $\overrightarrow r \cdot \left( {\widehat i + \widehat j + \widehat k} \right) = 2$ $\overrightarrow n = \left( {\hat i + \hat j + \hat k} \right)$ The required plane is parallel to given plane, so normal of requried plane will be same as of given plane. $\overrightarrow n = \left( {\hat i + \hat j + \hat k} \right)$ The plane is passing throgh (a, b, c) , so the equation of the plane will be $\overrightarrow r \cdot \overrightarrow n = \overrightarrow a \cdot \overrightarrow n$ $\vec r\cdot\left( {\hat i + \hat j + \hat k} \right) = \left( {a\hat i + b\hat j + c\hat k} \right) \cdot \left( {\hat i + \hat j + \hat k} \right)$ $x + y + z = a + b + c$

Question (9)

Find the shortest distance between lines $\overrightarrow r = \left( {6\widehat i + 2\widehat j + 2\widehat k} \right) + \lambda \left( {\widehat i - 2\widehat j + 2\widehat k} \right)and\overrightarrow {\;r} = \left( { - 4\widehat i - \widehat k} \right) + \mu \left( {3\widehat i - 2\widehat j - 2\widehat k} \right)$

Solution

$\vec r = \left( {6\hat i + 2\hat j + 2\hat k} \right) + \lambda \left( {\hat i - 2\hat j + 2\hat k} \right)$ $\overrightarrow {{a_1}} = 6\hat i + 2\hat j + 2\hat k\quad \overrightarrow {{b_1}} = \widehat i - 2\widehat j + 2\widehat k$ $\overrightarrow {\;r} = \left( { - 4\hat i - \hat k} \right) + \mu \left( {3\hat i - 2\hat j - 2\hat k} \right)$ $\overrightarrow {{a_2}} = - 4\hat i - \hat k\quad \overrightarrow {{b_2}} = 3\widehat i - 2\widehat j - 2\widehat k$ Calculate b1 × b2 $\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\1&{ - 2}&2\\3&{ - 2}&{ - 2}\end{array}} \right| = 8\widehat i + 8\widehat j + 4\widehat k \ne \overline 0$ As the cross product is non zero, lines are either intersecting or skew. $\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right| = \sqrt {{8^2} + {8^2} + {4^2}} = \sqrt {64 + 64 + 16} = \sqrt {144} = 12$ $\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = \left( { - 4\hat i - \hat k} \right) - \left( {6\hat i + 2\hat j + 2\hat k} \right) = \left( { - 10\hat i - 2\hat j - 3\hat k} \right)$ $\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = \left( { - 10\hat i - 2\hat j - 3\hat k} \right) \cdot \left( {8\widehat i + 8\widehat j + 4\widehat k} \right)$ $\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = - 80 - 16 - 12 = - 108 \ne 0$ So lines are skew. the shortest distance between skew lines is p, $p = \frac{{\left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right)} \right|}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}} = \frac{{\left| { - 108} \right|}}{{12}} = 9$ So the shortest distance between two skew lies is 9 unit.

Question (10)

Find the coordinates of point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ- plane.

Solution

The cartesian equation of line passing through two points (x1, y1, z1), (x2, y2, z2) is given by $\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}} = \lambda$ $\frac{{x - 5}}{{3 - 5}} = \frac{{y - 1}}{{4 - 1}} = \frac{{z - 6}}{{1 - 6}} = \lambda$ $\frac{{x - 5}}{{ - 2}} = \frac{{y - 1}}{3} = \frac{{z - 6}}{{ - 5}} = \lambda$ $x = - 2\lambda + 5,y = 3\lambda + 1,z = - 5\lambda + 6 - - - (1)$ The line crosses YZ plane. so x coordinate =0. $0 = - 2\lambda + 5\; \Rightarrow \lambda = \frac{5}{2}$ Replacing the value of λ in equation (1), we get $y = 3\lambda + 1,\quad z = - 5\lambda + 6$ $y = \frac{{15}}{2} + 1 = \frac{{17}}{2}\;\;z = \frac{{ - 25}}{2} + 6 = \frac{{ - 13}}{2}$ So coordinate of the point where line crosses YZ plane $\left( {x,y,z} \right) = \left( {0,\frac{{17}}{2},\frac{{ - 13}}{2}} \right)$

Question (11)

Find the coordinates of point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX- plane.

Solution

The cartesian equation of line passing through two points (x1, y1, z1), (x2, y2, z2) is given by $\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}} = \lambda$ $\frac{{x - 5}}{{3 - 5}} = \frac{{y - 1}}{{4 - 1}} = \frac{{z - 6}}{{1 - 6}} = \lambda$ $\frac{{x - 5}}{{ - 2}} = \frac{{y - 1}}{3} = \frac{{z - 6}}{{ - 5}} = \lambda$ $x = - 2\lambda + 5,y = 3\lambda + 1,z = - 5\lambda + 6 - - - (1)$ The line crosses ZX plane. so y coordinate =0. $0 = 3\lambda + 1\; \Rightarrow \lambda = \frac{{ - 1}}{3}$ Replacing the value of λ in equation (1), we get $x = - 2\lambda + 5,\quad z = - 5\lambda + 6$ $x = \frac{2}{3} + 5 = \frac{{17}}{3},\quad z = \frac{5}{3} + 6 = \frac{{23}}{3}$ So coordinate of the point where line crosses ZX plane $\left( {x,y,z} \right) = \left( {\frac{{17}}{3},0,\frac{{23}}{3}} \right)$

Question (12)

Find the coordinates of the point where the line through (3, -4, -5)and (2, -3, 1) crosses the plane 2x + y + z = 7.

Solution

The cartesian equation of line passing through two points (x1, y1, z1), (x2, y2, z2) is given by $\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}} = \lambda$ $\frac{{x - 3}}{{2 - 3}} = \frac{{y + 4}}{{ - 3 + 4}} = \frac{{z + 5}}{{1 + 5}} = \lambda$ $\frac{{x - 3}}{{ - 1}} = \frac{{y + 4}}{1} = \frac{{z + 5}}{6} = \lambda$ $x = - \lambda + 3,\quad y = \lambda - 4,\quad z = 6\lambda - 5 - - - (1)$ Let M (x, y, z) be the cordinate of the point where line crosses the given plane. So ordinate of M is $\left( {x,y,z} \right) = \left( { - \lambda + 3,\;\lambda - 4,\;6\lambda - 5} \right)$ M is on the plane, so will satisfy the equation of the plane. $2x + y + z = 7$ $2( - \lambda + 3) + \lambda - 4 + \;6\lambda - 5 = 7$ $- 2\lambda + 6 + \lambda - 4 + \;6\lambda - 5 = 7$ $5\lambda = 10\quad \Rightarrow \lambda = 2$ Replacing the value of λ in equation (1), we get $\left( {x,y,z} \right) = \left( { - \lambda + 3,\;\lambda - 4,\;6\lambda - 5} \right)$ $= (1, - 2,7)$ So the cordinate of the plane crosses the plane is ( 1, -2, 7)

Question (13)

Find the equation of the plane passing through the points (-1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

Solution

The equations of the planes are x + 2y +3z = 5 and 3x + 3y + z = 0. So there normal vectors are $\overrightarrow {{n_1}} = \widehat i + 2\widehat j + 3\widehat k\quad \overrightarrow {{n_2}} = 3\widehat i + 3\widehat j + \widehat k$ The required plane is perpendicular to both planes. So normal of required plane is $\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}}$ $= \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\1&2&3\\3&3&1\end{array}} \right|$ $= - 7\widehat i + 8\widehat j - 3\widehat k$ The equation of the plane passing through (-1, 3, 2) is gien by $\overrightarrow r \cdot \overrightarrow n = \overrightarrow a \cdot \overrightarrow n$ $\overrightarrow r \cdot \left( { - 7\widehat i + 8\widehat j - 3\widehat k} \right) = \left( { - \widehat i + 3\widehat j + 2\widehat k} \right) \cdot \left( { - 7\widehat i + 8\widehat j - 3\widehat k} \right)$ $- 7x + 8y - 3z = 7 + 24 - 6$ $7x - 8y + 3z + 25 = 0$

Question (14)

If the points (1, 1, p) and (-3, 0, 1) are equidistant from the plane $\overrightarrow r \cdot \left( {3\widehat i + 4\widehat j - 12\widehat k} \right) + 13 = 0$, then find the value of p.

Solution

Lrt p1 and p2 be the perpendicular distance of the plane from the points (1, 1, p) and (-3, 0, 1) respectively. $\overrightarrow r \cdot \left( {3\widehat i + 4\widehat j - 12\widehat k} \right) + 13 = 0$ $3x + 4y - 12z + 13 = 0$ Comparing to standard form the plane we a = 3, b = 4, c = -12, and d = 13. $p = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$ (x1, y1, z1) =( 1, 1, p) ${p_1} = \frac{{\left| {3(1) + 4(1) - 12(p) + 13} \right|}}{{\sqrt {{3^2} + {4^2} + - {{12}^2}} }}$ ${p_1} = \frac{{\left| {20 - 12p} \right|}}{{13}}$ (x2, y2, z2) =( -3, 0, 1) ${p_2} = \frac{{\left| {a{x_2} + b{y_2} + c{z_2} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$ ${p_2} = \frac{{\left| {3( - 3) + 4(0) - 12(1) + 13} \right|}}{{\sqrt {{3^2} + {4^2} + - {{12}^2}} }}$ ${p_2} = \frac{{\left| { - 8} \right|}}{{13}} = \frac{8}{{13}}$ The points are equdistance from planes p1 = p2 $\frac{{\left| {20 - 12p} \right|}}{{13}} = \frac{8}{{13}}$ $\Rightarrow 20 - 12p = 8,\quad 20 - 12p = - 8$ $\Rightarrow 12p = 12,\quad \;\quad 12p = 28$ $\Rightarrow p = 1,\quad \quad p = \frac{{28}}{{12}} = \frac{7}{3}$

Question (15)

Find the equation of the plane passing through the line of intersection of the planes $\overrightarrow r \cdot \left( {\widehat i + \widehat j + \widehat k} \right) = 1\;and\;\overrightarrow r \cdot \left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0$ and parallel to x-axis.

Solution

The vector equation of first plane is $\vec r\cdot\left( {\hat i + \hat j + \hat k} \right) = 1$ So its cartesian equation is x + y + z - 1 = 0 The vector equation of the second plane is $\vec r\cdot\left( {2\hat i + 3\hat j - \hat k} \right) + 4 = 0$ So its cartesian equation is 2x + 3y - z +4 = 0
The equation of required plane which passes through the intersection of two planes is given by. $(1 + 2\lambda )x + (1 + 3\lambda ) + (1 + 4\lambda ) - 1 + 4\lambda = 0 - - - (1)$ $(1 + 2\lambda )x + (1 + 3\lambda ) + (1 + 4\lambda ) - 1 + 4\lambda = 0$ $so\;\overrightarrow n = (1 + 2\lambda )\widehat i + (1 + 3\lambda )\widehat j + (1 + 4\lambda )\widehat k$ The plane is parallel to x axis. So normal is perpendicular to x axis. Direction of x axis is (1, 0, 0). The dot product of normal and direction of x axis is zero. $\overrightarrow n \cdot \overrightarrow b = 0$ $\left( {(1 + 2\lambda )\widehat i + (1 + 3\lambda )\widehat j + (1 + 4\lambda )\widehat k} \right) \cdot \left( {\widehat i} \right) = 0$ $1 + 2\lambda = 0\quad \Rightarrow \lambda = \frac{{ - 1}}{2}$ Replacing the value λin equation (1) , we get $(x + y + z - 1) + \frac{{ - 1}}{2}(2x + 3y - z + 4) = 0$ $2x + 2y + 2z - 2 - 2x - 3y + z - 4 = 0$ $- y + 3z - 6 = 0$ $y - 3z + 6 = 0$

Question (16)

If O be the origin and the coordinates of P be (1, 2, -3) then find the equation of the plane passing through P and perpendicular to OP.

Solution

Let P(1, 2, -3) and origin is (0, 0, 0). Direction vector of OP = (1, 2, -3). The plane is perpendicular to OP. normal is perpendicular to plane. So OP and normal are parallel to each other. $\overrightarrow n = \widehat i + 2\widehat j - 3\widehat k$ The plane is passing through P, the equation of the plane will be $\overrightarrow r \cdot \overrightarrow n = \overrightarrow p \cdot \overrightarrow n$ $\overrightarrow r \cdot \left( {\widehat i + 2\widehat j - 3\widehat k} \right) = \left( {\widehat i + 2\widehat j - 3\widehat k} \right) \cdot \left( {\widehat i + 2\widehat j - 3\widehat k} \right)$ $x + 2y - 3z = 1 + 4 + 9$ $x + 2y - 3z - 14 = 0$

Question (17)

Find the equation of the plane which contains the line of intersection of the planes $\overrightarrow r \cdot \left( {\widehat i + 2\widehat j + 3\widehat k} \right) - 4 = 0\;and\;\overrightarrow r \cdot \left( {2\widehat i + \widehat j - \widehat k} \right) + 5 = 0$ and which is perpendicular to plane $\overrightarrow r \cdot \left( {5\widehat i + 3\widehat j - 6\widehat k} \right) + 8 = 0$

Solution

Let vector equation of the first plane is $\vec r\cdot\left( {\hat i + 2\hat j + 3\hat k} \right) - 4 = 0$ So cartesian equation of plane is x + 2y + 3z - 4 = 0
Let vector equation of the second plane is $\vec r\cdot\left( {2\hat i + \hat j - \hat k} \right) + 5 = 0$ Its cartesian equation is 2x + y - z + 5 = 0
The equation of plane which passes through the point of intersction of two plane is given by $\left( {x + 2y + 3z - 4} \right) + \lambda \left( {2x + y - z + 5} \right) = 0 - - - (1)$ $\left( {1 + 2\lambda } \right)x + \left( { 2 + \lambda } \right)y + \left( {3 - \lambda } \right)z - 4 + 5\lambda = 0$ $So\quad \overrightarrow n = \left( {1 + 2\lambda } \right)\widehat i + \left( { 2 + \lambda } \right)\widehat j + \left( {3 - \lambda } \right)\widehat k$ This required the plane is perpendicular to the third plane $\overrightarrow r \cdot \left( {5\widehat i + 3\widehat j - 6\widehat k} \right) + 8 = 0$ $\overrightarrow {{n_3}} = \left( {5\hat i + 3\hat j - 6\hat k} \right)$ $\overrightarrow n \cdot \overrightarrow {{n_3}} = 0$ $\left( {\left( {1 + 2\lambda } \right)\widehat i + \left( { 2 + \lambda } \right)\widehat j + \left( {3 - \lambda } \right)\widehat k} \right) \cdot \left( {5\hat i + 3\hat j - 6\hat k} \right) = 0$ $5 + 10\lambda + 6 + 3\lambda - 18 + 6\lambda = 0$ $19\lambda - 7 = 0\quad \Rightarrow \lambda = \frac{7}{{19}}$ Replacing the value of λ in equation (1) we get, $\left( {x + 2y + 3z - 4} \right) + \frac{7}{{19}}\left( {2x + y - z + 5} \right) = 0$ $19x + 38y + 57z - 76 + 14x + 7y - 7z + 35 = 0$ $33x + 45y + 50z - 41 = 0$

Question (18)

Find the distance of the point (-1, -5, -10) from the point of intersectio of the line $\overrightarrow r = \left( {2\widehat i - \widehat j + 2\widehat k} \right) + \lambda \left( {3\widehat i + 4\widehat j + 2\widehat k} \right)and \overrightarrow r \cdot \left( {\widehat i - \widehat j + \widehat k} \right) = 5$

Solution

Let M(m) be the point of intersection of a line and plane. The vector equation of the line is $\overrightarrow r = \left( {2\widehat i - \widehat j + 2\widehat k} \right) + \lambda \left( {3\widehat i + 4\widehat j + 2\widehat k} \right) \vec r = \left( {2 + 3\lambda } \right)\widehat i + \left( { - 1 + 4\lambda } \right)\widehat j + \left( {2 + 2\lambda } \right)\widehat k$ The point M is on the line , so coordinate of M is $\overline m = \left( {\left( {2 + 3\lambda } \right),\left( { - 1 + 4\lambda } \right),\left( {2 + 2\lambda } \right)} \right)$ the point M is on the plane ,$\vec r\cdot\left( {\hat i - \hat j + \hat k} \right) = 5$ $x - y + z - 5 = 0$ $\left( {2 + 3\lambda } \right) - \left( { - 1 + 4\lambda } \right) + \left( {2 + 2\lambda } \right) - 5 = 0$ $\begin{array}{l}2 + 3\lambda + 1 - 4\lambda + 2 + 2\lambda - 5 = 0\\\lambda = 0\end{array}$ Replacing the value of λ in coordinate of M we get, M( x, y, z) = (2, -1, 2) The coordinate of the point P is ( -1, -5, -10) By distance formula, $PM = \sqrt {{{(2 + 1)}^2} + {{( - 1 + 5)}^2} + {{(2 + 10)}^2}} = \sqrt {9 + 16 + 144} = \sqrt {169} = 13$ So distance of the point P from the point of intersection of line and plane is 13 unit.

Question (19)

Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes $\overrightarrow r \cdot \left( {\widehat i - \widehat j + 2\widehat k} \right) = 5\;and\;\overrightarrow r \cdot \left( {3\widehat i + \widehat j + \widehat k} \right) = 6$

Solution

The equations of the planes are $\overrightarrow r \cdot \left( {\widehat i - \widehat j + 2\widehat k} \right) = 5\;and\;\overrightarrow r \cdot \left( {3\widehat i + \widehat j + \widehat k} \right) = 6$ The line is parallel to two planes. So the line will be perpendicular to both the planes. So the direction vector b of the line is cross product of two normals. $\overrightarrow b = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}}$ $= \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\1&{ - 1}&2\\3&1&1\end{array}} \right|$ $= - 3\widehat i + 5\widehat j + 4\widehat k$ The equation of the line passing through (1, 2, 3) and with direction vector b is given by $\vec r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda ( - 3\widehat i + 5\widehat j + 4\widehat k)$

Question (20)

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

Solution

The equations or two lines are $\frac{{x - 8}}{3} = \frac{{y + 19}}{{ - 16}} = \frac{{z - 10}}{7}\;and\frac{{x - 15}}{3} = \frac{{y - 29}}{8} = \frac{{z - 5}}{{ - 5}}$ $\overrightarrow {{b_1}} = 3\widehat i - 16\widehat j + 7\widehat k,\quad \overrightarrow {{b_2}} = 3\widehat i + 8\widehat j - 5\widehat k$ The required line is perpendicular to both lines , so direction vector of required line is $\overrightarrow b = \overrightarrow {{b_1}} \times \overrightarrow {{b_2}}$ $= \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\3&{ - 16}&7\\3&8&{ - 5}\end{array}} \right|$ $= 24\widehat i + 36\widehat j + 72\widehat k$ The vector equation of line passing through(1,2,-4) is given by, $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b ,\lambda \in R\quad$ $\vec r = \left( {\hat i + 2\hat j - 4\hat k} \right) + {\lambda _1}(24\widehat i + 36\widehat j + 72\widehat k)$ $\vec r = \left( {\hat i + 2\hat j - 4\hat k} \right) + 12{\lambda _1}(2\widehat i + 3\widehat j + 6\widehat k)$ $\vec r = \left( {\hat i + 2\hat j - 4\hat k} \right) + \lambda (2\widehat i + 3\widehat j + 6\widehat k)$

Question (21)

Prove that if a plane has the intercepts a, b, c and is at a distance of p from the origin, then $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{{p^2}}}$

Solution

The plane has intercepts a, b, c. The equation of the plane in intercept form is given by ${\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1}$ ${\overrightarrow n = \left( {\frac{1}{a}\hat i + \frac{1}{b}\hat j + \frac{1}{c}\hat k} \right)}$ $\left| {\overrightarrow n } \right| = \sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}}$ The perpendicular distance p from the origin, is given by $p = \frac{{\left| d \right|}}{{\left| {\overrightarrow n } \right|}}$ $p = \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}$ $\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} = \frac{1}{p}$ Squaring on both the sides, we get $\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{{p^2}}}$ Choose the correct answer .

Question (22)

Distance between the two planes 2x + 3y +4z = 4 and 4x + 6y + 8z = 12 is.
(a) 2 units   (b) 4 units   (c) 8 units   (d) 2/√29 units.

Solution

The equation of the planes are 2x + 3y + 4z = 4 , and 4x + 6y + 8z = 12.
Direction ratio of the normals of the plane are in propertion, so planes are parallel. (2x + 3y + 4z = 4)×2 ⇒ 4x + 6y + 8z = 8. 4x + 6y + 8z = 12. ${\overrightarrow n = \left( {4\hat i + 6\hat j + 8\hat k} \right)}$ $\left| {\overrightarrow n } \right| = \sqrt {{4^2} + {6^2} + {8^2}} = \sqrt {16 + 36 + 64} = \sqrt {116} = 2\sqrt {29}$ The perpendicular distance between two parallel planes is $p = \frac{{\left| {{d_1} - {d_2}} \right|}}{{\left| {\overrightarrow n } \right|}} = \frac{{\left| {12 - 8} \right|}}{{2\sqrt {29} }} = \frac{4}{{2\sqrt {29} }} = \frac{2}{{\sqrt {29} }}$ So D is the correct option.

Question (23)

The planes 2x - y + 4z = 5 and 5x - 2.5y + 10z = 6 are
(a) Perpendicular   (b) Parallel   (c) intersects y- axis   (d) passes through (0,0,5/4)

Solution

The equations planes are 2x - y + 4z = 5 and 5x - 2.5y + 10z = 6 ,
The direction ratios of their normals are in properation, so planes are parallel.
So B is the correct option.