NCERT Solutions For Class 12 Maths Chapter 11 Exercise 11.3

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Chapter 11 Ex.11.3 Question 1

In each of the following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

(a) \(z = 2\)

(b) \(x + y + z = 1\)

(c) \({\rm{2x + 3y - z}} = 5\)

(d) \(5y + 8 = 0\)

Solution

(a) The equation of the plane is \(z = 2\) or \(0x + 0y + z = 2 \quad \ldots \left( 1 \right)\)

The direction ratios of normal are \(0,0\) and \(1\).

Therefore,

\[\sqrt {{0^2} + {0^2} + {1^2}} = 1\]

Dividing both sides of equation (1) by \(1,\) we obtain

\[0.x + 0.y + 1.z = 2\]

This is of the form \(lx + my + nz = d\), where \(l,\;m,\;n\) are the direction cosines of normal to the plane and \(d\) is the distance of the perpendicular drawn from the origin.

Hence, the direction cosines are \(0,0\) and \(1\) and the distance of the plane form the origin is \(2\) units.

(b) \(x + y + z = 1 \quad \ldots \left( 1 \right)\)

The direction ratios of normal are \(1,1\) and \(1\).

Therefore,

\[\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 3 \]

Dividing both sides of equation (1) by \(\sqrt 3 \) , we get

\[\frac{1}{{\sqrt 3 }}x + \frac{1}{{\sqrt 3 }}y + \frac{1}{{\sqrt 3 }}z = \frac{1}{{\sqrt 3 }}\]

This equation is one of the form \(lx + my + nz = d\), where \(l,\;m,\;n\) are direction cosines of normal to the plane and \(d\) is the distance of normal from the origin.

Hence, the direction cosines of the normal are \(\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}\) and \(\frac{1}{{\sqrt 3 }}\) and the distance of normal form the origin is \(\frac{1}{{\sqrt 3 }}\) units.

(c) \(2x + 3y - z = 5 \quad \ldots \left( 1 \right)\)

The direction ratios of normal are \(2,3\) and \( - 1\).

Therefore,

\[\sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( { - 1} \right)}^2}} = \sqrt {14} \]

Dividing both sides of equation (1) by \(\sqrt {14} \) , we get

\[\frac{2}{{\sqrt {14} }}x + \frac{3}{{\sqrt {14} }}y - \frac{1}{{\sqrt {14} }}z = \frac{5}{{\sqrt {14} }}\]

This equation is one of the form \(lx + my + nz = d\), where \(l,\;m,\;n\) are direction cosines of normal to the plane and \(d\) is the distance of normal from the origin.

Hence, the direction cosines of the normal are \(\frac{2}{{\sqrt {14} }},\frac{3}{{\sqrt {14} }}\) and \(\frac{{ - 1}}{{\sqrt {14} }}\) and the distance of normal form the origin is \(\frac{5}{{\sqrt {14} }}\) units.

(d) \(5y + 8 = 0\)

\[ \Rightarrow \; 0x + 5y + 0z = 8 \quad  \ldots \left( 1 \right)\]

The direction ratios of normal are \(0, - 5\) and \(0\).

Therefore,

\[\sqrt {{0^2} + {{\left( { - 5} \right)}^2} + {0^2}} = 5\]

Dividing both sides of equation (1) by \(5,\) we get

\[0x + y + 0z = \frac{8}{5}\]

This equation is one of the form \(lx + my + nz = d\), where \(l,\;m,\;n\) are direction cosines of normal to the plane and \(d\) is the distance of normal from the origin.

Hence, the direction cosines of the normal to the plane are \(0,1\) and \(0\) and the distance of normal form the origin is \(\frac{8}{5}\) units.

Chapter 11 Ex.11.3 Question 2

Find the vector equation of a plane which is at the distance of \(7\) units from the origin and normal to the vector \(3\hat i + 5\hat j - 6\hat k\).

Solution

The normal vector is, \(\vec n = 3\hat i + 5\hat j - 6\hat k\)

\[\begin{align}\hat n &= \frac{{\vec n}}{{\left| {\vec n} \right|}} = \frac{{3\hat i + 5\hat j - 6\hat k}}{{\sqrt {{{\left( 3 \right)}^2} + {{\left( 5 \right)}^2} + {{\left( 6 \right)}^2}} }}\\&= \frac{{3\hat i + 5\hat j - 6\hat k}}{{\sqrt {70} }}\end{align}\]

The equation of the plane with position vector \(\vec r\) is given by, \(\vec r.\hat n = d\)

Hence,

\[\vec r.\left( {\frac{{3\hat i + 5\hat j - 6\hat k}}{{\sqrt {70} }}} \right) = 7\]

Chapter 11 Ex.11.3 Question 3

Find the Cartesian equation of the following planes:

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

(b) \(\vec r.\left( {2\hat i + 3\hat j - 4\hat k} \right) = 1\)

(c) \(\vec r.\left[ {\left( {s - 2t} \right)\hat i + \left( {3 - t} \right)\hat j + \left( {2s + t} \right)\hat k} \right] = 15\)

Solution

(a) Given equation of the plane is

\[\vec r.\left( {\hat i + \hat j - \hat k} \right) = 2 \qquad \ldots \left( 1 \right)\]

For any arbitrary point, \(P\left( {x,y,z} \right)\) on the plane, position vector \(\vec r\) is given by, \(\vec r = x\hat i + y\hat j - z\hat k\)

Putting the values of \(\vec r\) in equation \(\left( 1 \right)\), we get

\[\begin{align}&\left( {x\hat i + y\hat j - z\hat k} \right).\left( {\hat i + \hat j - \hat k} \right) = 2\\& \Rightarrow \; x + y - z = 2\end{align}\]

(b) \(\vec r.\left( {2\hat i + 3\hat j - 4\hat k} \right) = 1 \qquad \ldots \left( 1 \right)\)

For any arbitrary point \(P\left( {x,y,z} \right)\) on the plane, position vector \(\vec r\) is given by, \(\vec r = x\hat i + y\hat j - z\hat k\)

Putting the values of \(\overrightarrow r \) in equation \(\left( 1 \right)\), we get

\[\begin{align}&\left( {x\hat i + y\hat j - z\hat k} \right).\left( {2\hat i + 3\hat j - 4\hat k} \right) = 1\\& \Rightarrow 2x + 3y - 4z = 1\end{align}\]

(c) \(\vec r.\left[ {\left( {s - 2t} \right)\hat i + \left( {3 - t} \right)\hat j + \left( {2s + t} \right)\hat k} \right] = 15 \quad \ldots \left( 1 \right)\)

For any arbitrary point, \(P\left( {x,y,z} \right)\) on the plane, position vector \(\vec r\) is given by,

\(\vec r = x\hat i + y\hat j - z\hat k\)

Putting the values of \(\vec r\) in equation \(\left( 1 \right)\), we get

\[\begin{align}&\left( {x\hat i + y\hat j - z\hat k} \right).\left[ {\left( {s - 2t} \right)\hat i + \left( {3 - t} \right)\hat j + \left( {2s + t} \right)\hat k} \right] = 15\\& \Rightarrow\; \left( {s - 2t} \right)x + \left( {3 - t} \right)y + \left( {2s + t} \right)z = 15\end{align}\]

Chapter 11 Ex.11.3 Question 4

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

(a) \(2x + 3y + 4z-12 = 0\)

(b) \(3y + 4z-6 = 0\)

(c) \(x + y + z = 1\)

(d) \(5y + 8 = 0\)

Solution

(a) Let the coordinates of the foot of perpendicular \(P\) from the origin to the plane be \(\left( {{x_1},{y_1},{z_1}} \right)\)

\[2x + 3y + 4z--12 = 0 \qquad  \ldots \left( 1 \right)\]

The direction ratios of normal are \(2,3\) and \({\rm{4}}\)

Therefore,

\[\sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} = \sqrt {29} \]

Dividing both sides of equation (1) by \(\sqrt {29} \), we get

\[\frac{2}{{\sqrt {29} }}x + \frac{3}{{\sqrt {29} }}y + \frac{4}{{\sqrt {29} }}z = \frac{{12}}{{\sqrt {29} }} \qquad \ldots \left( 2 \right)\]

This equation is one of the form \(lx + my + nz = d\), where \(l,\;m,\;n\) are direction cosines of normal to the plane and \(d\) is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by \(\left( {ld,md,nd} \right)\)

Hence, the coordinates of the foot of the perpendicular are

\[\begin{align}&\left( {\frac{2}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }},\frac{3}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }},\frac{4}{{\sqrt {29} }} \times \frac{{12}}{{\sqrt {29} }}} \right)\\& \Rightarrow \;\left( {\frac{{24}}{{29}},\frac{{36}}{{29}},\frac{{48}}{{29}}} \right)\end{align}\]

(b) Let the coordinates of the foot of perpendicular \(P\) from the origin to the plane be\(\left( {{x_1},{y_1},{z_1}} \right)\)

\[3y + 4z--6 = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]

The direction ratios of the normal are \(0,3\) and \(4\).

Therefore,

\[\sqrt {{0^2} + {3^2} + {4^2}} = 5\]

Dividing both sides of equation (1) by \(5,\) we get

\[0x + \frac{3}{5}y + \frac{4}{5}z = \frac{6}{5}\]

This equation is one of the form \(lx + my + nz = d\), where \(l,\;m,\;n\) are direction cosines of normal to the plane and \(d\) is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by \(\left( {ld,md,nd} \right)\)

Hence, the coordinates of the foot of the perpendicular are

\[\begin{align}&\left( {0 \times \frac{6}{5},\frac{3}{5} \times \frac{6}{5},\frac{4}{5} \times \frac{6}{5}} \right)\\& \Rightarrow \ \left( {0,\frac{{18}}{{25}},\frac{{24}}{{25}}} \right)\end{align}\]

(c) Let the coordinates of the foot of perpendicular \(P\) from the origin to the plane be

\[x + y + z = 1 \qquad \ldots \left( 1 \right)\]

The direction ratios of the normal are \(1,1\) and \(1\).

Therefore,

\[\sqrt {{{\left( 1 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} = \sqrt 3 \]

Dividing both sides of equation (1) by \(\sqrt 3 \), we get,

\[\frac{1}{{\sqrt 3 }}x + \frac{1}{{\sqrt 3 }}y + \frac{1}{{\sqrt 3 }}z = \frac{1}{{\sqrt 3 }}\]

This equation is one of the form \(lx + my + nz = d\), where \(l,\;m,\;n\) are direction cosines of normal to the plane and \(d\) is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by \(\left( {ld,md,nd} \right)\)

Hence, the coordinates of the foot of the perpendicular are

\[\begin{align}&\left( {0 \times \frac{6}{5},\frac{3}{5} \times \frac{6}{5},\frac{4}{5} \times \frac{6}{5}} \right)\\& \Rightarrow\; \left( {0,\frac{{18}}{{25}},\frac{{24}}{{25}}} \right)\end{align}\]

(d) Let the coordinates of the foot of perpendicular \(P\) from the origin to the plane be

\[\begin{align}&5y + 8 = 0\\&\Rightarrow\;0x--5y + 0z = 8 \qquad  \ldots \left( 1 \right) \end{align}\]

The direction ratios of the normal are \(0, - 5\) and \(0\).

Therefore,

\[\sqrt {{0^2} + {{\left( { - 5} \right)}^2} + 0} = 5\]

Dividing both sides of equation (1) by \(5,\) we obtain

\[0x--y + 0z = \frac{8}{5}\]

This equation is one of the form \(lx + my + nz = d\), where \(l,\;m,\;n\) are direction cosines of normal to the plane and \(d\) is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by \(\left( {ld,md,nd} \right)\)

Hence, the coordinates of the foot of the perpendicular are

\[\begin{align}&\left( {0 \times \frac{8}{5}, - 1 \times \frac{8}{5},0 \times \frac{8}{5}} \right)\\&\Rightarrow \;\left( {0, - \frac{8}{5},0} \right)\end{align}\]

Chapter 11 Ex.11.3 Question 5

Find the vector and Cartesian equation of the planes

(a) that passes through the point \(\left( {{\rm{1}},0, - {\rm{2}}} \right)\) and the normal to the plane is \(\hat i + \hat j - \hat k\)

(b) that passes through the point \(\left( {{\rm{1}},{\rm{4}},6} \right)\)and the normal vector to the plane is .

Solution

(a) The position vector of point \(\left( {{\rm{1}},0, - {\rm{2}}} \right)\) is \(\vec a = \hat i - 2\hat k\)

The normal vector \(\vec N\) perpendicular to the plane is \(\vec N = \hat i + \hat j - \hat k\)

The vector equation of the plane is given by,

\[\begin{align}& \Rightarrow \left( {\vec r - \vec a} \right).\vec N = 0\\ &\Rightarrow \left[ {\vec r - \left( {\vec i - 2\hat k} \right)} \right].\left( {\hat i + \hat j - \hat k} \right) = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}\]

Since, \(\vec r\) is the positive vector of any point \(P\left( {x,y,z} \right)\)in the plane.

Hence,

\(\vec r = x\hat i + y\hat j + z\hat k\)

Thus, equation (1) becomes

\[\begin{align} &\Rightarrow \left[ {\left( {x\hat i + y\hat j + z\hat k} \right) - \left( {\hat i - 2\hat k} \right)} \right].\left( {\hat i + \hat j - \hat k} \right) = 0\\ &\Rightarrow \left[ {\left( {x - 1} \right)\hat i + y\hat j + \left( {z + 2} \right)\hat k} \right].\left( {\hat i + \hat j - \hat k} \right) = 0\\&\Rightarrow \left( {x - 1} \right) + y - \left( {z + 2} \right) = 0\\ &\Rightarrow x + y - z - 3 = 0\\ &\Rightarrow x + y - z = 3\end{align}\]

(b) The position vector of point \(\left( {{\rm{1}},{\rm{4}},6} \right)\) is \(\vec a = \hat i + 4\hat j + 6\hat k\)

The normal vector \(\vec N\) perpendicular to the plane is \(\vec N = \hat i - 2\hat j + \hat k\)

The vector equation of the plane is given by,

\[\begin{align} &\Rightarrow \left( {\vec r - \vec a} \right).\vec N = 0\\ &\Rightarrow \left[ {\vec r - \left( {\hat i + 4\hat j + 6\hat k} \right)} \right].\left( {\hat i - 2\hat j + \hat k} \right) = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}\]

Since, \(\vec r\) is the positive vector of any point \(P\left( {x,y,z} \right)\)in the plane.

Hence,

\(\vec r = x\hat i + y\hat j + z\hat k\)

Thus, equation (1) becomes

\[\begin{align}& \Rightarrow \left[ {\left( {x\hat i + y\hat j + z\hat k} \right) - \left( {\hat i + 4\hat j + 6\hat k} \right)} \right].\left( {\hat i - 2\hat j + \hat k} \right) = 0\\ &\Rightarrow \left[ {\left( {x - 1} \right)\hat i + \left( {y - 4} \right)\hat j + \left( {z - 6} \right)\hat k} \right].\left( {\hat i - 2\hat j + \hat k} \right) = 0\\& \Rightarrow \left( {x - 1} \right) - 2\left( {y - 4} \right) + \left( {z - 6} \right) = 0\\& \Rightarrow x - 2y + z + 1 = 0\end{align}\]

Chapter 11 Ex.11.3 Question 6

Find the equations of the planes that passes through the points.

(a) \(\left( {1,1, - 1} \right),\left( {6,4, - 5} \right),\left( { - 4,2,3} \right)\)

(b) \(\left( {1,1,0} \right),\left( {1,2,1} \right),\left( { - 2,2, - 1} \right)\)

Solution

(a) The given points are \(A\left( {1,1, - 1} \right),B\left( {6,4, - 5} \right)\) and\(C\left( { - 4,2,3} \right)\).

\[\begin{align}\left( {\begin{array}{*{20}{c}}1&1&{ - 1}\\6&4&{ - 5}\\{ - 4}&{ - 2}&3\end{array}} \right) &= \left( {12 - 10} \right) - \left( {18 - 20} \right) - \left( { - 12 + 16} \right)\\ &= 2 + 2 - 4\\& = 0\end{align}\]

Since, the points are collinear, there will be infinite number of planes passing through the given points.

(b) The given points are \(A\left( {1,1,0} \right),B\left( {1,2,1} \right)\) and \(C\left( { - 2,2, - 1} \right)\).

\[\begin{align}\left( {\begin{array}{*{20}{c}}1&1&0\\1&2&1\\{ - 2}&2&{ - 1}\end{array}} \right) &= \left( { - 2 - 2} \right) - \left( {2 + 2} \right)\\ &= - 8\\ &\ne 0\end{align}\]

Thus, a plane will pass through the points.

The equation of the plane through the points \(\left( {{x_1},{y_1},{z_1}} \right),\left( {{x_2},{y_2},{z_2}} \right)\) and \(\left( {{x_3},{y_3},{z_3}} \right)\)is given by

\[\begin{align}& \Rightarrow \left( {\begin{array}{*{20}{c}}{x - {x_1}}&{y - {y_1}}&{z - {z_1}}\\{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\end{array}} \right) = 0\\ &\Rightarrow \left( {\begin{array}{*{20}{c}}{x - 1}&{y - 1}&z\\0&1&1\\{ - 3}&1&{ - 1}\end{array}} \right) = 0\\ &\Rightarrow \left( { - 2} \right)\left( {x - 1} \right) - 3\left( {y - 1} \right) + 3z = 0\\& \Rightarrow - 2x + 2 - 3y + 3 + 3z = 0\\& \Rightarrow - 2x - 3y + 3z + 5 = 0\\ &\Rightarrow 2x + 3y - 3z = 5\end{align}\]

Chapter 11 Ex.11.3 Question 7

Find the intercepts cut off by the plane \(2x + y - z = 5\)

Solution

\(2x + y - z = 5\)

Dividing both sides of equation by \(5\), we get

\[\begin{align} \Rightarrow \frac{2}{5}x + \frac{y}{5} - \frac{z}{5} = 1\\ \Rightarrow \frac{x}{{\frac{5}{2}}} + \frac{y}{5} + \frac{z}{{ - 5}} = 1\end{align}\]

The equation of a plane in intercept form is \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\) , where \(a,b\) and \(c\) are intercepts cut off by the plane at \(x,y\) and \(z - \)axes respectively.

Hence, for the given equation,

\(a = \frac{5}{2},\;b = 5\) and \(c = - 5\)

Thus, the intercepts cut off by plane are \(\frac{5}{2},5\) and \( - 5\).

Chapter 11 Ex.11.3 Question 8

Find the equation of the plane with intercept on the \(y\)-axis and parallel to \(ZOX\) plane.

Solution

The equation of the plane \(ZOX\) is \(Y = 0\)

Any plane parallel to it is of the form, \(y = a\)

Since the \(y\)-intercept of the plane is \(3\),

Therefore, \(a = 3\)

Hence, the equation of the required plane is \(y = 3\).

Chapter 11 Ex.11.3 Question 9

Find the equation of the plane through the intersection of the planes \(3x - y + 2z - 4 = 0\) and \(x + y + z - 2 = 0\) and the point \(\left( {2,2,1} \right) \).

Solution

The equation of the given plane through the intersection of the planes \(3x - y + 2z - 4 = 0\) and \(x + y + z - 2 = 0\) is given by

\[\left( {3x - y + 2z - 4} \right) + \alpha \left( {x + y + z - 2} \right) = 0;\;\;\alpha \in R \qquad \ldots \left( 1 \right)\]

This plane passes through the point \(\left( {2,2,1} \right)\).

Hence, this point will satisfy equation

\[\begin{align}& \Rightarrow \left( {3 \times 2 - 2 + 2 \times 1 - 4} \right) + \alpha \left( {2 + 2 + 1 - 2} \right) = 0\\ &\Rightarrow 2 + 3\alpha = 0\\ &\Rightarrow \alpha = \frac{{ - 2}}{3}\end{align}\]

Putting \(\alpha = \frac{{ - 2}}{3}\) in equation (1), we get

\[\begin{align} &\Rightarrow \left( {3x - y + 2z - 4} \right) - \frac{2}{3}\left( {x + y + z - 2} \right) = 0\\& \Rightarrow 3\left( {3x - y + 2z - 4} \right) - 2\left( {x + y + z - 2} \right) = 0\\& \Rightarrow 9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0\\ &\Rightarrow 7x - 5y + 4z - 8 = 0\end{align}\]

Chapter 11 Ex.11.3 Question 10

Find the vector equation of the plane passing through the intersection of the planes \(\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 7,\;\vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) = 9\) and through the point \(\left( {2,1,3} \right)\).

Solution

The equations of the planes are \(\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 7\) and \(\vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) = 9\)

Hence,

\[\begin{align} \Rightarrow \vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) - 7 = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\ \Rightarrow \vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) - 9 = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

Equation of the required plane is given by

\(\left[ {\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) - 7} \right]{\rm{ }} + \lambda \left[ {\vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) - 9} \right] = 0;\;\;\lambda \in R\)

Therefore,

\[\begin{align} &\Rightarrow \vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) - 7{\rm{ }} + \lambda \vec r.\left( {2\hat i + 5\hat j + 3\hat k} \right) - 9\lambda = 0\\ &\Rightarrow \vec r.\left[ {\left( {2\hat i + 2\hat j - 3\hat k} \right) + \lambda \left( {2\hat i + 5\hat j + 3\hat k} \right)} \right] = 9\lambda + 7\\ &\Rightarrow \vec r.\left[ {\left( {2 + 2\lambda } \right)\hat i + \left( {2 + 5\lambda } \right)\hat j + \left( {3\lambda - 3} \right)\hat k]} \right] = 9\lambda + 7\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}\]

The plane passes through the point \(\left( {2,1,3} \right)\)

Hence, its position vector is given by, \(\vec r = 2\hat i + \hat j + 3\hat k\)

Putting in equation (3), we get

\[\begin{align} &\Rightarrow \left( {2\hat i + \hat j + 3\hat k} \right).\left[ {\left( {2 + 2\lambda } \right)\hat i + \left( {2 + 5\lambda } \right)\hat j + \left( {3\lambda - 3} \right)\hat k]} \right] = 9\lambda + 7\\& \Rightarrow 2\left( {2 + 2\lambda } \right) + \left( {2 + 5\lambda } \right) + 3\left( {3\lambda - 3} \right) = 9\lambda + 7\\ &\Rightarrow 4 + 4\lambda + 2 + 5\lambda + 9\lambda - 9 - 9\lambda - 7 = 0\\& \Rightarrow 9\lambda - 10 = 0\\ &\Rightarrow \lambda = \frac{{10}}{9}\end{align}\]

Putting \(\lambda = \frac{{10}}{9}\) in equation (3), we get

\[\begin{align} &\Rightarrow \vec r.\left( {\frac{{38}}{9}\hat i + \frac{{68}}{9}\hat j + \frac{3}{9}\hat k} \right) = 17\\& \Rightarrow \vec r.\left( {38\hat i + 68\hat j + 3\hat k} \right) = 153\end{align}\]

Chapter 11 Ex.11.3 Question 11

Find the equation of the plane through the line of intersection of the planes \(x + y + z = 1\) and \(2x + 3y + 4z = 5\) which is perpendicular to the plane \(x - y + z = 0\).

Solution

The equation of the plane through the intersection of the planes \(x + y + z = 1\) and \(2x + 3y + 4z = 5\)is

\[\begin{align}& \Rightarrow \left( {x + y + z - 1} \right) + \lambda \left( {2x + 3y + 4z - 5} \right) = 0\\ &\Rightarrow \left( {2\lambda + 1} \right)x + \left( {3\lambda + 1} \right)y + \left( {4\lambda + 1} \right)z - \left( {5\lambda + 1} \right) = 0 \qquad \ldots \left( 1 \right)\end{align}\]

The plane in equation (1) is perpendicular to the plane \(x - y + z = 0\)

Since the planes are perpendicular, \({a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\)

Here,

\({a_1} = \left( {2\lambda + 1} \right),\;{b_1} = \left( {3\lambda + 1} \right)\) and \({c_1} = \left( {4\lambda + 1} \right)\)

\({a_2} = 1,\;{b_2} = - 1\) and \({c_2} = 1\)

Hence,

\[\begin{align} &\Rightarrow \left( {2\lambda + 1} \right) \times 1 + \left( {3\lambda + 1} \right) \times \left( { - 1} \right) + \left( {4\lambda + 1} \right) \times 1 = 0\\& \Rightarrow 2\lambda + 1 - 3\lambda - 1 + 4\lambda + 1 = 0\\& \Rightarrow 3\lambda + 1 = 0\\ &\Rightarrow \lambda = \frac{{ - 1}}{3}\end{align}\]

Putting \(\lambda = \frac{{ - 1}}{3}\) in equation (1), we get

\[\begin{align}&\Rightarrow \frac{1}{3}x + \frac{1}{3}z + \frac{2}{3} = 0\\& \Rightarrow x - z + 2 = 0\end{align}\]

Chapter 11 Ex.11.3 Question 12

Find the angle between the planes whose vector equations are \(\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 5\) and \(\vec r.\left( {3\hat i - 3\hat j + 5\hat k} \right) = 3\).

Solution

The equations of the given planes are \(\vec r.\left( {2\hat i + 2\hat j - 3\hat k} \right) = 5\) and \(\vec r.\left( {3\hat i - 3\hat j + 5\hat k} \right) = 3\)

If \({\vec n_1}\) and \({\vec n_2}\) are normal to the planes, \(\vec r.{\vec n_1} = {d_1}\) and \(\vec r.{\vec n_2} = {d_2}\)

Then the angle between them \({\rm{\theta }}\) is given by,

\[{\rm{\theta }} = {\cos ^{ - 1}}\frac{{\left| {{{\vec n}_1}.{{\vec n}_2}} \right|}}{{\left| {{{\vec n}_1}} \right|\left| {{{\vec n}_2}} \right|}} \qquad \ldots \left( 1 \right)\]

Here,

\({\vec n_1} = 2\hat i + 2\hat j - 3\hat k\) and \({\vec n_2} = 3\hat i - 3\hat j + 5\hat k\)

Hence,

\[\begin{align}{{\vec n}_1}.{{\vec n}_2} &= \left( {2\hat i + 2\hat j - 3\hat k} \right).\left( {3\hat i - 3\hat j + 5\hat k} \right)\\ &= 2 \times 3 + 2 \times \left( { - 3} \right) + \left( { - 3} \right) \times 5\\ &= - 15\\\left| {{{\vec n}_1}} \right| &= \sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 3} \right)}^2}} = \sqrt {17} \\\left| {{{\vec n}_2}} \right| &= \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 5 \right)}^2}} = \sqrt {43}\end{align}\]

Substituting these values in equation (1), we obtain

\[\begin{align}{\rm{\theta }} &= {\cos ^{ - 1}}\frac{{\left| { - 15} \right|}}{{\left| {\sqrt {17} } \right|.\left| {\sqrt {43} } \right|}}\\& = {\cos ^{ - 1}}\frac{{15}}{{\sqrt {731} }}\end{align}\]

Chapter 11 Ex.11.3 Question 13

In the following cases, determine whether the given planes are parallel or perpendicular, and in case they are neither, find the angles between them.

(a) \(7x + 5y + 6z + 30 = 0\) and \(3x - y - 10z + 4 = 0\)

(b) \(2x + y + 3z - 2 = 0\) and \(x - 2y + 5 = 0\)

(c) \(2x - 2y + 4z + 5 = 0\) and \(3x - 3y + 6z - 1 = 0\)

(d) \(2x - y + 3z - 1 = 0\) and \(2x - y + 3z + 3 = 0\)

(e) \(4x + 8y + z - 8 = 0\) and \(y + z - 4 = 0\)

Solution

The directions ratios of normal to be the plane \({L_1}:{a_1}x + {b_1}y + {c_1}z = 0\) are \({a_1},{b_1},{c_1}\) and \({L_2}:{a_2}x + {b_2}y + {c_2}z = 0\) are \({a_2},{b_2},{c_2}\)

If,

\[\begin{align}{L_1}\parallel {L_2};\;\; &\Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\\{L_1} \bot {L_2};\;\;& \Rightarrow {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\end{align}\]

The angle between \({L_1}\) and \({L_2}\) is given by

\[{\rm{\theta }} = {\cos ^{ - 1}}\left| {\frac{{\left| {{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \right|}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} .\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\]

(a) The equations of the planes are \(7x + 5y + 6z + 30 = 0\) and \(3x - y - 10z + 4 = 0\)

Here,

\({a_1} = 7,{b_1} = 5\) and \({c_1} = 6\)

\({a_2} = 3,{b_2} = - 1\) and \({c_2} = - 10\)

Hence,

\[\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 7 \times 3 + 5 \times \left( { - 1} \right) + 6 \times \left( { - 10} \right)\\&= - 44\\& \ne 0\end{align}\]

Therefore, the given planes are not perpendicular.

Also,

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{7}{3}\\\frac{{{b_1}}}{{{b_2}}} = \frac{5}{{ - 1}} = - 5\\\frac{{{c_1}}}{{{c_2}}} = \frac{6}{{ - 10}} = - \frac{3}{5}\end{align}\]

It can be seen that, \(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\)

Therefore, the given planes are not parallel.

The angle between them is given by,

\[\begin{align}{\rm{\theta }} &= {\cos ^{ - 1}}\left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} .\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\\ &= {\cos ^{ - 1}}\left| {\frac{{7 \times 3 + 5 \times \left( { - 1} \right) + 6 \times \left( { - 10} \right)}}{{\sqrt {{{\left( 7 \right)}^2} + {{\left( 5 \right)}^2} + {{\left( 6 \right)}^2}} .\sqrt {{{\left( 3 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 10} \right)}^2}} }}} \right|\\&= {\cos ^{ - 1}}\left| {\frac{{21 - 5 - 60}}{{\sqrt {110} .\sqrt {110} }}} \right|\\& = {\cos ^{ - 1}}\left| {\frac{{ - 44}}{{110}}} \right|\\ &= {\cos ^{ - 1}}\frac{2}{5}\end{align}\]

(b) The equations of the planes are \(2x + y + 3z - 2 = 0\) and \(x - 2y + 5 = 0\)

Here,

\({a_1} = 2,{b_1} = 1\) and \({c_1} = 3\)

\({a_2} = 1,{b_2} = - 2\) and \({c_2} = 0\)

Hence,

\[\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 2 \times 1 + 1 \times \left( { - 2} \right) + 3 \times 0\\ &= 2 - 2 + 0\\ &= 0\end{align}\]

Thus, the given planes are perpendicular to each other.

(c) The equations of the planes are \(2x - 2y + 4z + 5 = 0\) and \(3x - 3y + 6z - 1 = 0\)

Here,

\({a_1} = 2,{b_1} = - 2\) and \({c_1} = 4\)

\({a_2} = 3,{b_2} = - 3\) and \({c_2} = 6\)

Hence,

\[\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 2 \times 3 + \left( { - 2} \right) \times \left( { - 3} \right) + 4 \times 6\\& = 6 + 6 + 24\\ &= 36\\& \ne 0\end{align}\]

Thus, the given planes are not perpendicular to each other.

Also,

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{2}{3}\\\frac{{{b_1}}}{{{b_2}}} = \frac{{ - 2}}{{ - 3}} = \frac{2}{3}\\\frac{{{c_1}}}{{{c_2}}} = \frac{4}{6} = \frac{2}{3}\end{align}\]

It can be seen that, \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\)

Hence, the given planes are parallel to each other.

(d) The equations of the planes are \(2x - y + 3z - 1 = 0\) and \(2x - y + 3z + 3 = 0\)

Here,

\({a_1} = 2,{b_1} = - 1\) and \({c_1} = 3\)

\({a_2} = 2,{b_2} = - 1\) and \({c_2} = 3\)

Hence,

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{2}{2} &= 1\\\frac{{{b_1}}}{{{b_2}}} &= \frac{{ - 1}}{{ - 1}} = 1\\\frac{{{c_1}}}{{{c_2}}} &= \frac{3}{3} &= 1\end{align}\]

Therefore, \(\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\)

Hence, the given lines are parallel to each other.

(e) The equations of the given planes are \(4x + 8y + z - 8 = 0\) and \(y + z - 4 = 0\)

Here,

\({a_1} = 4,{b_1} = 8\) and \({c_1} = 1\)

\({a_2} = 0,{b_2} = 1\) and \({c_2} = 1\)

Hence,

\[\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 4 \times 0 + 8 \times 1 + 1\\& = 9\\& \ne 0\end{align}\]

Thus, the given lines are not perpendicular to each other.

Also,

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{4}{0}\\\frac{{{b_1}}}{{{b_2}}} &= \frac{8}{1} = 8\\\frac{{{c_1}}}{{{c_2}}}& = \frac{1}{1} = 1\end{align}\]

It can be seen that, \(\frac{{{a_1}}}{{{a_2}}} \ne \frac{{{b_1}}}{{{b_2}}} \ne \frac{{{c_1}}}{{{c_2}}}\)

Thus, the given lines not parallel to each other.

The angle between the planes is given by,

\[\begin{align}{\rm{\theta }} &= {\cos ^{ - 1}}\left| {\frac{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}}}{{\sqrt {{a_1}^2 + {b_1}^2 + {c_1}^2} .\sqrt {{a_2}^2 + {b_2}^2 + {c_2}^2} }}} \right|\\ &= {\cos ^{ - 1}}\left| {\frac{{4 \times 0 + 8 \times 1 + 1 \times 1}}{{\sqrt {{{\left( 4 \right)}^2} + {{\left( 8 \right)}^2} + {{\left( 1 \right)}^2}} \times \sqrt {{{\left( 0 \right)}^2} + {{\left( 1 \right)}^2} + {{\left( 1 \right)}^2}} }}} \right|\\ &= {\cos ^{ - 1}}\left| {\frac{9}{{9\sqrt 2 }}} \right|\\ &= {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right)\\& = 45^\circ\end{align}\]

Chapter 11 Ex.11.3 Question 14

In the following cases, find the distance of each of the given points from the corresponding given plane.

Point Plane
(a) \(\left( {0,0,0} \right)\) \(3x - 4y + 12z = 3\)
(b) \(\left( {3, - 2,1} \right)\) \(2x - y + 2z + 3 = 0\)
(c) \(\left( {2,3, - 5} \right)\) \(x + 2y - 2z = 9\)
(d) \(\left( { - 6,0,0} \right)\) \(2x - 3y + 6z - 2 = 0\)

Solution

The distance between a point, \(P\left( {{x_1},{y_1},{z_1}} \right)\) and a plane \(Ax + By + Cz + D = 0\) is given by,

\[d = \left| {\frac{{A{x_1} + B{y_1} + C{z_1} + D}}{{\sqrt {{A^2} + {B^2} + {C^2}} }}} \right|\]

(a) The given point is \(\left( {0,0,0} \right)\) and the plane is \(3x - 4y + 12z - 3 = 0\)

Therefore,

\[\begin{align}d &= \left| {\frac{{3 \times 0 + \left( { - 4} \right) \times 0 + 12 \times 0 + \left( { - 3} \right)}}{{\sqrt {{{\left( { - 3} \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( {12} \right)}^2}} }}} \right|\\ &= \left| {\frac{{ - 3}}{{\sqrt {169} }}} \right|\\ &= \frac{3}{{13}}\end{align}\]

(b) The given point is \(\left( {3, - 2,1} \right)\) and the plane is \(2x - y + 2z + 3 = 0\)

Therefore,

\[\begin{align}d& = \left| {\frac{{2 \times 3 + \left( { - 1} \right) \times \left( { - 2} \right) + 2 \times 1 + 3}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( 2 \right)}^2}} }}} \right|\\ &= \left| {\frac{{13}}{{\sqrt 9 }}} \right|\\ &= \frac{{13}}{3}\end{align}\]

(c) The given point is \(\left( {2,3, - 5} \right)\) and the plane is \(x + 2y - 2z - 9 = 0\)

Therefore,

\[\begin{align}d &= \left| {\frac{{1 \times 2 + 2 \times 3 + \left( { - 2} \right) \times \left( { - 5} \right) + \left( { - 9} \right)}}{{\sqrt {{{\left( 1 \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} }}} \right|\\ &= \left| {\frac{9}{{\sqrt 9 }}} \right|\\ &= \frac{9}{3}\\&= 3\end{align}\]

(d) The given point is \(\left( { - 6,0,0} \right)\) and the plane is \(2x - 3y + 6z - 2 = 0\)

Therefore,

\[\begin{align}d &= \left| {\frac{{2 \times \left( { - 6} \right) + \left( { - 3} \right) \times 0 + 6 \times 0 + \left( { - 2} \right)}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( { - 3} \right)}^2} + {{\left( 6 \right)}^2}} }}} \right|\\ &= \left| {\frac{{ - 14}}{{\sqrt {49} }}} \right|\\ &= \frac{{14}}{7}\\ &= 2\end{align}\]