12th NCERT Three Dimensional Geometry Exercise 11.3 Questions 14
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## Exercise 11.3 Questions 14

Question (1)

In each of the following cases, determine the direction cosine of the normal to the plane and the distance from the origin.
(a) z = 2

Solution

The standard cartesian equation of the plane is ax + by + cz + d = 0. Then (a, b, c) is the normal vector of the plane denoted by vector n.
The direction cosine of normal is a unit vector of normal. $\widehat n = \frac{{\overrightarrow n }}{{\left| {\overrightarrow n } \right|}}$ The perpendicular distance of the plane from the origin is given by $p = \frac{{\left| d \right|}}{{\left| {\overrightarrow n } \right|}}$
z = 2 ⇒ z - 2 = 0
Comparing to standard form we get, a = 0, b = 0, c = 1, d = -2 So the normal n = ( a, b, c) = (0,0,1)
So cosine direction is 0, 0, 1.
The perpandicular distance from the origin is 2.
(b) x + y+ z = 1

Solution

Comparing to standard form we get, a = 1, b = 1, c = 1, d = - 1 So the normal n = ( a, b, c) = (1,1,1)
$\left| {\overrightarrow n } \right| = \sqrt {{1^2} + {1^2} + {1^2}} = \sqrt {1 + 1 + 1} = \sqrt 3$ So direction cosine of the normal is $\widehat n = \frac{{\overrightarrow n }}{{\left| {\overrightarrow n } \right|}} = \frac{{(1,1,1)}}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }},\frac{1}{{\sqrt 3 }}$ And perpendicular distance p is, $p = \frac{{\left| d \right|}}{{\left| {\overrightarrow n } \right|}} = \frac{{\left| { - 1} \right|}}{{\sqrt 3 }} = \frac{1}{{\sqrt 3 }}$ (c) 2x + 3y - z = 5

Solution

Comparing to standard form we get, a = 2, b = 3, c = -1, d = -5 So the normal n = ( a, b, c) = (2,3,-1)
$\left| {\overrightarrow n } \right| = \sqrt {{2^2} + {3^2} + - {1^2}} = \sqrt {4 + 9 + 1} = \sqrt {14}$ So direction cosine of the normal is $\widehat n = \frac{{\overrightarrow n }}{{\left| {\overrightarrow n } \right|}} = \frac{{(2,3, - 1)}}{{\sqrt {14} }} = \frac{2}{{\sqrt {14} }},\frac{3}{{\sqrt {14} }},\frac{{ - 1}}{{\sqrt {14} }}$ And perpendicular distance p is,$p = \frac{{\left| d \right|}}{{\left| {\overrightarrow n } \right|}} = \frac{{\left| { - 5} \right|}}{{\sqrt {14} }} = \frac{5}{{\sqrt {14} }}$ (d) 5y + 8 = 0

Solution

Comparing to standard form we get, a = 0, b = 5, c = 0, d = 8 So the normal n = ( a, b, c) = (0,5,0)
$\left| {\overrightarrow n } \right| = \sqrt {{0^2} + {5^2} + {0^2}} = \sqrt {0 + 25 + 0} = \sqrt {25} = 5$ So direction cosine of the normal is $\widehat n = \frac{{\overrightarrow n }}{{\left| {\overrightarrow n } \right|}} = \frac{{(0,5, - 0)}}{5} = 0,1,0$ And perpendicular distance p is,$p = \frac{{\left| d \right|}}{{\left| {\overrightarrow n } \right|}} = \frac{{\left| 8 \right|}}{5} = \frac{8}{5}$

Question (2)

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

Solution

The vectar equation of the plane whose perpendicular distance from the origin is p is given by $\overrightarrow r \cdot \widehat n = p,where\;\widehat n = \frac{{\overrightarrow n }}{{\left| {\overrightarrow n } \right|}}$
In a given sum p = 7. $\overline n = 3\hat i + 5\hat j - 6\hat k\quad \left| {\overrightarrow n } \right| = \sqrt {{3^2} + {5^2} + - {6^2}} = \sqrt {9 + 25 + 36} = \sqrt {70}$ $\widehat n = \frac{{3\hat i + 5\hat j - 6\hat k}}{{\sqrt {70} }}$ The equation of the plane whose perpendicular distance is 7 unit is $\overrightarrow r \cdot \left( {\frac{{3\hat i + 5\hat j - 6\hat k}}{{\sqrt {70} }}} \right) = 7$

Question (3)

Find the cartesian equation of the following planes : (a) $\overrightarrow r \cdot \left( {\widehat i + \widehat j - \widehat k} \right) = 2$

Solution

$\vec r\cdot\left( {\hat i + \hat j - \hat k} \right) = 2$ $\left( {x\widehat i + y\widehat j + z\widehat k} \right)\cdot\left( {\hat i + \hat j - \hat k} \right) = 2$ $x + y - z = 2$ (b) $\overrightarrow r \cdot \left( {2\widehat i + 3\widehat j - 4\widehat k} \right) = 1$

Solution

$\overrightarrow r \cdot \left( {2\widehat i + 3\widehat j - 4\widehat k} \right) = 1$ $\left( {x\widehat i + y\widehat j + z\widehat k} \right)\cdot\left( {2\hat i + 3\hat j - 4\hat k} \right) = 1$ $2x + 3y - 4z = 1$ (c) $\overrightarrow r \cdot \left[ {\left( {s - 2t)\widehat i + (3 - t)\widehat j + (2s + t)\widehat k} \right)} \right] = 15$

Solution

$\overrightarrow r \cdot \left[ {\left( {s - 2t)\widehat i + (3 - t)\widehat j + (2s + t)\widehat k} \right)} \right] = 15$ $\left( {x\widehat i + y\widehat j + z\widehat k} \right)\cdot\left[ {\left( {s - 2t)\hat i + (3 - t)\hat j + (2s + t)\hat k} \right)} \right] = 15$ $(s - 2t)x + (3 - t)y + (2s + t)z = 15$

Question (4)

In the following cases, find the coordinates of foot of the perpendicular drawn from the origin.
(a) 2x + 3y + 4z - 12 = 0

Solution Let the equation plane 'p' be 2x + 3y +4z - 12 = 0. so normal of the plane 'p' is n = ( 2, 3, 4),
Origin is not on the plane . Let M with position vector m be the foot of perpendicular drawn from origin to the plane 'p'
The normal is always perpendicular to plane. So vector OM and normal are parallel to each other.
So the direction vector of line OM "b"is same as the normal 'n'.
$\overrightarrow b = \overrightarrow n = 2\widehat i + 3\widehat j + 4\widehat k$ The equation of line OM passing through origin and having direction vector 'b' ig given by $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$ $\overrightarrow r = \left( {0\widehat i + 0\widehat j + 0\widehat k} \right) + \lambda \left( {2\widehat i + 3\widehat j + 4\widehat k} \right)$ $\overrightarrow r = \left( {2\lambda \widehat i + 3\lambda \widehat j + 4\lambda \widehat k} \right)$ The point M is on the line , so will satisfy the equation of a line. $\overline m = \left( {2\lambda ,3\lambda ,4\lambda } \right)$ The point M is also on the plane , so it will satisfy the equation of the plane . $2(2\lambda ) + 3(3\lambda ) + 4(4\lambda ) - 12 = 0$ $4\lambda + 9\lambda + 16\lambda = 12$ $29\lambda = 12 \Rightarrow \lambda = \frac{{12}}{{29}}$ Replacing the value of λ in the vector m, we get coordinate of M which is the foot of the perpendicular from origin on the plane. $\overline m = \left( {\frac{{24}}{{29}},\frac{{36}}{{29}},\frac{{48}}{{29}}} \right)$ (b) 3y + 4z - 6 = 0

Solution Let the equation plane 'p' be 3y + 4z - 6 = 0. so normal of the plane 'p' is n = ( 0, 3, 4),
Origin is not on the plane . Let M with position vector m be the foot of perpendicular drawn from origin to the plane 'p'
The normal is always perpendicular to plane. So vector OM and normal are parallel to each other.
So the direction vector of line OM "b"is same as the normal 'n'.
$\overrightarrow b = \overrightarrow n = 0\widehat i + 3\widehat j + 4\widehat k$ The equation of line OM passing through origin and having direction vector 'b' ig given by $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$ $\overrightarrow r = \left( {0\widehat i + 0\widehat j + 0\widehat k} \right) + \lambda \left( {0\widehat i + 3\widehat j + 4\widehat k} \right)$ $\overrightarrow r = \left( {0\lambda \widehat i + 3\lambda \widehat j + 4\lambda \widehat k} \right)$ The point M is on the line , so will satisfy the equation of a line. $\overline m = \left( { 3\lambda ,4\lambda } \right)$ The point M is also on the plane , so it will satisfy the equation of the plane . $3(3\lambda ) + 4(4\lambda ) - 6 = 0$ $9\lambda + 16\lambda = 6$ $25\lambda = 6 \Rightarrow \lambda = \frac{{6}}{{25}}$ Replacing the value of λ in the vector m, we get coordinate of M which is the foot of the perpendicular from origin on the plane. $\overline m = \left( {0,\frac{{18}}{{25}},\frac{{24}}{{25}}} \right)$ (c) x + y + z = 1

Solution Let the equation plane 'p' be x + y + z - 1 = 0. so normal of the plane 'p' is n = ( 1, 1, 1),
Origin is not on the plane . Let M with position vector m be the foot of perpendicular drawn from origin to the plane 'p'
The normal is always perpendicular to plane. So vector OM and normal are parallel to each other.
So the direction vector of line OM "b"is same as the normal 'n'.
$\overrightarrow b = \overrightarrow n = 1\widehat i + 1\widehat j + 1\widehat k$ The equation of line OM passing through origin and having direction vector 'b' ig given by $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$ $\overrightarrow r = \left( {0\widehat i + 0\widehat j + 0\widehat k} \right) + \lambda \left( {\widehat i + \widehat j + \widehat k} \right)$ $\overrightarrow r = \left( {\lambda \widehat i + \lambda \widehat j + \lambda \widehat k} \right)$ The point M is on the line , so will satisfy the equation of a line. $\overline m = \left( {\lambda ,\lambda ,\lambda } \right)$ The point M is also on the plane , so it will satisfy the equation of the plane . $(\lambda ) + (\lambda ) + (\lambda ) - 1 = 0$ $\lambda + \lambda + \lambda = 1$ $3\lambda = 1 \Rightarrow \lambda = \frac{{1}}{{3}}$ Replacing the value of λ in the vector m, we get coordinate of M which is the foot of the perpendicular from origin on the plane. $\overline m = \left( {\frac{{1}}{{3}},\frac{{1}}{{3}},\frac{{1}}{{3}}} \right)$ (d) 5y + 8 = 0

Solution Let the equation plane 'p' be 5y + 8 = 0. so normal of the plane 'p' is n = ( 0, 5, 0),
Origin is not on the plane . Let M with position vector m be the foot of perpendicular drawn from origin to the plane 'p'
The normal is always perpendicular to plane. So vector OM and normal are parallel to each other.
So the direction vector of line OM "b"is same as the normal 'n'.
$\overrightarrow b = \overrightarrow n = 5\widehat j$ The equation of line OM passing through origin and having direction vector 'b' ig given by $\overrightarrow r = \overrightarrow a + \lambda \overrightarrow b$ $\overrightarrow r = \left( {0\widehat i + 0\widehat j + 0\widehat k} \right) + \lambda \left( {5\widehat j} \right)$ $\overrightarrow r = \left( { 5\lambda \widehat j } \right)$ The point M is on the line , so will satisfy the equation of a line. $\overline m = \left( {0 ,5\lambda ,0 } \right)$ The point M is also on the plane , so it will satisfy the equation of the plane . $5(5\lambda ) ) + 8 = 0$ $25\lambda = -8 \Rightarrow \lambda = \frac{{-8}}{{25}}$ Replacing the value of λ in the vector m, we get coordinate of M which is the foot of the perpendicular from origin on the plane. $\overline m = \left( {0,\frac{{ - 40}}{{25}},0} \right)$ $\overline m = \left( {0,\frac{{ - 8}}{5},0} \right)$

Question (5)

Find the vector and cartesian equations of the planes
(a) that passes through the point (1, 0, - 2) and normal to the plane is $\widehat i + \widehat j - \widehat k.$

Solution

The vector equation of plane with normal (n) and passing through the point A(a) is given by $\vec r\cdot\overrightarrow n = \overrightarrow a \cdot\overrightarrow n = d$
The vector equation of the plane passing through ( 1, 0,-2) with normal(n) is given by $\vec r\cdot\left( {\hat i + \hat j - \hat k} \right). = \left( {\hat i + 0\hat j - 2\hat k.} \right)\cdot\left( {\hat i + \hat j - \hat k} \right)$ $\vec r\cdot\left( {\hat i + \hat j - \hat k} \right) = 1 + 2$ $\vec r\cdot\left( {\hat i + \hat j - \hat k} \right) = 3$ The cartesian equation of the plane is $x + y - z = 3$ (b) that passes through the point (1, 4, 6) and normal vector to the plane is $\widehat i - 2\widehat j + \widehat k.$

Solution

The vector equation of the plane passing through ( 1, 0,-2) with normal(n) is given by $\vec r\cdot\left( {\hat i - 2\hat j + \hat k} \right) = \left( {\hat i + 4\hat j + 6\hat k.} \right)\cdot\left( {\hat i - 2\hat j + \hat k.} \right)$ $\vec r\cdot\left( {\hat i - 2\hat j + \hat k} \right) - \left( {\hat i + 4\hat j + 6\hat k.} \right)\cdot\left( {\hat i - 2\hat j + \hat k.} \right) = 0$ $\left[ {\vec r\cdot - \left( {\hat i + 4\hat j + 6\hat k.} \right)} \right]\cdot\left( {\hat i - 2\hat j + \hat k.} \right) = 0$ The cartesian equation of the plane will be given by $\vec r\cdot\left( {\hat i - 2\hat j + \hat k.} \right) = \left( {\hat i + 4\hat j + 6\hat k.} \right) \cdot \left( {\hat i - 2\hat j + \hat k.} \right)$ $x - 2y + z = 1 - 8 + 6$ $x - 2y + z + 1 = 0$

Question (6)

Find the equations of the planes that passes through three points. (a) (1, 1, -1), (6, 4, -5), (-4, -2, 3)

Solution

The equation of the plane passing through three points (x1, y1,z1), (x2, y2,z2),(x3, y3,z3)is given by $\left| {\begin{array}{*{20}{c}}{x - {x_1}}&{y - {y_1}}&{z - {z_1}}\\{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{x_3} - {x_1}}&{{y_3} - {y_1}}&{{z_3} - {z_1}}\end{array}} \right| = 0$
Here (x1, y1,z1) = (1, 1, -1), (x2, y2,z2) = (6, 4, -5),(x3, y3,z3) = (-4, -2, 3) So the equation of the plane will be $\left| {\begin{array}{*{20}{c}}{x - 1}&{y - 1}&{z + 1}\\{6 - 1}&{4 - 1}&{ - 5 + 1}\\{ - 4 - 1}&{ - 2 - 1}&{3 + 1}\end{array}} \right| = 0$ $\left| {\begin{array}{*{20}{c}}{x - 1}&{y - 1}&{z + 1}\\5&3&{ - 4}\\{ - 5}&{ - 3}&4\end{array}} \right| = 0$ $(x - 1)0 - (y - 1)0 + (z + 1)0 = 0$ which is true . So we don't get equation of the plane.
It indicates that three points are colliear, so there exists infinite planes passing throug given three points. So we does not get a single equation.
(b) (1, 1, 0),(1, 2, 1), (-2, 2, -1)

Solution

Here (x1, y1,z1) = (1, 1, 0), (x2, y2,z2) = (1, 2, 1),(x3, y3,z3) = (-2, 2, -1) So the equation of the plane will be $\left| {\begin{array}{*{20}{c}}{x - 1}&{y - 1}&{z - 0}\\{1 - 1}&{2 - 1}&{1 - 0}\\{ - 2 - 1}&{2 - 1}&{ - 1 - 0}\end{array}} \right| = 0$ $\left| {\begin{array}{*{20}{c}}{x - 1}&{y - 1}&z\\0&1&1\\{ - 3}&1&{ - 1}\end{array}} \right| = 0$ $(x - 1)( - 2) - (y - 1)(3) + (z)3 = 0$ $- 2x + 2 - 3y + 3 + 3z = 0$ $- 2x - 3y + 3z + 5 = 0$ $2x + 3y - 3z - 5 = 0$

Question (7)

Find the intersepts cut off by the plane 2x + y - z = 5

Solution

When equation of the plane is written in the form $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ then a, b and c are the intercepts cut by the plane .
The cartesian equation of the plane is 2x + y - z = 5.
To convert into intercept form we will divide by 5, and we will get $\frac{{2x}}{5} + \frac{y}{5} - \frac{z}{5} = 1$ $\frac{x}{{5/2}} + \frac{y}{5} + \frac{z}{{ - 5}} = 1$ So the intercepts cut are 5/2, 5, -5.

Question (8)

Find the equation of the plane with intercept 3 on y axis and parallel to ZOX plane.

Solution

The plane intersects y axis.And parallel to ZOX plane it means it is not intersecting x and z axis.
So it will have only y intercept. and y intercept b = 3.
So the equation of the plane will be y / 3 = 1
y = 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 (2, 2, 1).

Solution

The equation of the plane passing through the intrsection of two planes α1 and α2 is given by ${\alpha _1} + \lambda {\alpha _2} = 0,\lambda \in R$
The equation of the plane through the intersection of the planes 3x - y + 2z - 4 = 0 and x + y + z - 2 = 0 will be given by $\left( {3x - y + 2z - 4} \right) + \lambda \left( {x + y + z - 2} \right) = 0$ The plane passes through (2, 2, 1).So the point will satisfy the equation of the plane. $\left( {3x - y + 2z - 4} \right) + \lambda \left( {x + y + z - 2} \right) = 0 - - - (1)$ $2 + 3\lambda = 0 \Rightarrow \lambda = \frac{{ - 2}}{3}$ Replacing the value of λ in equation (1) we get. $\left( {3x - y + 2z - 4} \right) + \frac{{ - 2}}{3}\left( {x + y + z - 2} \right) = 0 - - - (1)$ $9x - 3y + 6z - 12 - 2x - 2y - 2z + 4 = 0$ $7x - 5y + 4z - 8 = 0$

Question (10)

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

Solution

First convert equation of plane in cartesian form. The equation of the plane passing through the intrsection of two planes α1 and α2 is given by ${\alpha _1} + \lambda {\alpha _2} = 0,\lambda \in R$
$\overrightarrow r \cdot \left( {2\widehat i + 2\widehat j - 3\widehat k} \right) = 7$, $2x + 2y - 3z - 7 = 0$ $\overrightarrow r \cdot \left( {2\widehat i + 5\widehat j + 3\widehat k} \right) = 9$ $2x + 5y + 3z - 9 = 0$ The equation of the required plane which passes through the intersection of two planes is given by $\left( {2x + 2y - 3z - 7} \right) + \lambda \left( {2x + 5y + 3z - 9} \right) = 0 - - - (1)$ The plane passes through (2, 1, 3).So the point will satisfy the equation of the plane. $\left( {4 + 2 - 9 - 7} \right) + \lambda \left( {4 + 5 + 9 - 9} \right) = 0$ $- 10 + 9\lambda = 0,\; \Rightarrow \lambda = \frac{{10}}{9}$ Replacing the value of λ in equation (1) we get. $\left( {2x + 2y - 3z - 7} \right) + \frac{{10}}{9}\left( {2x + 5y + 3z - 9} \right) = 0 - - - (1)$ $18x + 18y - 27z - 63 + 20x + 50y + 30z - 90 = 0$ $38x + 68y + 3z - 153 = 0$ The vector equation of the plane is $\overrightarrow r \cdot \left( {38\widehat i + 68\widehat j + 3\widehat k} \right) = 153$

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 line of intersection of the planes x + y+ z = 1 and 2x + 3y + 4z = 5 will be given by, $\left( {x + y + z - 1} \right) + \lambda \left( {2x + 3y + 4z - 5} \right) = 0 - - - (1)$ $\left( {1 + 2\lambda } \right)x + \left( {1 + 3\lambda } \right)y + \left( {1 + 4\lambda } \right)z - 1 - 5\lambda = 0$ So normal of the plane is $\overrightarrow n = \left( {1 + 2\lambda } \right)\widehat i + \left( {1 + 3\lambda } \right)\widehat j + \left( {1 + 4\lambda } \right)\widehat k$ The required plane is perpendicular to the plane x - y + z = 0.whose normal $\overrightarrow {{n_3}} = \widehat i - \widehat j + \widehat k$ As vectors are perpendicular dot product is zero. $\overrightarrow {{n_3}} \cdot \overline n = 0$ $\left( {\widehat i - \widehat j + \widehat k} \right) \cdot \left( {\left( {1 + 2\lambda } \right)\widehat i + \left( {1 + 3\lambda } \right)\widehat j + \left( {1 + 4\lambda } \right)\widehat k} \right) = 0$ $1 + 2\lambda - 1 - 3\lambda + 1 + 4\lambda = 0$ $3\lambda + 1 = 0\; \Rightarrow \lambda = \frac{{ - 1}}{3}$ Replacing the value of λ in equation (1) we get. $\left( {x + y + z - 1} \right) + \frac{{ - 1}}{3}\left( {2x + 3y + 4z - 5} \right) = 0 - - - (1)$ $3x + 3y + 3z - 3 - 2x - 3y - 4z + 5 = 0$ $x - z + 2 = 0$

Question (12)

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

Solution

Let θ be the angle between two planes. then $\cos \theta = \frac{{\overrightarrow {{n_1}} \cdot \overrightarrow {{n_2}} }}{{\left| {\overline {{n_1}} } \right|\left| {\overline {{n_2}} } \right|}}$ $\left| {\overrightarrow {{n_1}} } \right| = \sqrt {{2^2} + {2^2} + - {3^2}} = \sqrt {4 + 4 + 9} = \sqrt {17}$ $\overrightarrow {{n_2}} = 3\hat i - 3\hat j + 5\hat k$ $\left| {\overrightarrow {{n_2}} } \right| = \sqrt {{3^2} + {3^2} + {5^2}} = \sqrt {9 + 9 + 25} = \sqrt {43}$ $\cos \theta = \left| {\frac{{\overrightarrow {{n_1}} \cdot \overrightarrow {{n_2}} }}{{\left| {\overline {{n_1}} } \right|\left| {\overline {{n_2}} } \right|}}} \right|$ $\cos \theta = \left| {\frac{{\left( {2\hat i + 2\hat j - 3\hat k} \right) \cdot \left( {3\hat i - 3\hat j + 5\hat k} \right)}}{{\sqrt {17} \sqrt {43} }}} \right|$ $\cos \theta = \left| {\frac{{6 - 6 - 15}}{{\sqrt {17} \sqrt {43} }}} \right|$ $\cos \theta = \frac{{15}}{{\sqrt {731} }} \Rightarrow \theta = {\cos ^{ - 1}}\left( {\frac{{15}}{{\sqrt {731} }}} \right)$

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

Solution

Let θ be the angle between two planes. then $\cos \theta = \frac{{\overrightarrow {{n_1}} \cdot \overrightarrow {{n_2}} }}{{\left| {\overline {{n_1}} } \right|\left| {\overline {{n_2}} } \right|}}$ $\overrightarrow {{n_1}} = 7\hat i + 5\hat j + 6\hat k$ $\left| {\overrightarrow {{n_1}} } \right| = \sqrt {{7^2} + {5^2} + {6^2}} = \sqrt {49 + 25 + 36} = \sqrt {110}$ $\overrightarrow {{n_2}} = 3\hat i - \hat j - 10\hat k$ $\left| {\overrightarrow {{n_2}} } \right| = \sqrt {{3^2} + - {1^2} + - {{10}^2}} = \sqrt {9 + 1 + 100} = \sqrt {110}$ $\cos \theta = \left| {\frac{{\overrightarrow {{n_1}} \cdot \overrightarrow {{n_2}} }}{{\left| {\overline {{n_1}} } \right|\left| {\overline {{n_2}} } \right|}}} \right|$ $\cos \theta = \left| {\frac{{\left( {7\hat i + 5\hat j + 6\hat k} \right) \cdot \left( {3\hat i - \hat j - 10\hat k} \right)}}{{\sqrt {110} \sqrt {110} }}} \right|$ $\cos \theta = \left| {\frac{{21 - 5 - 60}}{{110}}} \right|$ $\cos \theta = \frac{{44}}{{110}} = \frac{2}{5} \Rightarrow \theta = {\cos ^{ - 1}}\left( {\frac{2}{5}} \right)$ (b) 2x + y + 3z - 2 = 0 and x - 2y + 5 = 0

Solution

$\overrightarrow {{n_1}} = \left( {2\widehat i + \widehat j + 3\widehat k} \right),\overrightarrow {{n_2}} = \left( {\widehat i - 2\widehat j} \right)$ $\overrightarrow {{n_1}} \cdot \overrightarrow {{n_2}} = \left( {2\widehat i + \widehat j + 3\widehat k} \right) \cdot \left( {\widehat i - 2\widehat j} \right) = 2 - 2 = 0$ So planes are perpendicular to each other.
(c) 2x - 2y + 4z + 5 = 0 and 3x - 3y + 6z - 1 = 0

Solution

$\overrightarrow {{n_1}} = \left( {2\widehat i - 2\widehat j + 4\widehat k} \right),\overrightarrow {{n_2}} = \left( {\widehat {3i} - 3\widehat j + 6\widehat k} \right)$ The direction cosine of the plane are in proportion, so planes are parallel.
(d) 2x - y + 3z - 1 = 0 and 2x - y + 3z + 3 = 0

Solution

$\overrightarrow {{n_1}} = \left( {2\widehat i - \widehat j + 3\widehat k} \right),\overrightarrow {{n_2}} = \left( {2\widehat i - \widehat j + 3\widehat k} \right)$ The direction cosine of the plane are in proportion, so planes are parallel.
(e) 4x + 8y + z - 8 = 0 and y + z - 4 = 0

Solution

$\overrightarrow {{n_1}} = 4\hat i + 8\hat j + \hat k$ $\left| {\overrightarrow {{n_1}} } \right| = \sqrt {{4^2} + {8^2} + {1^2}} = \sqrt {16 + 64 + 1} = \sqrt {81} = 9$ $\overrightarrow {{n_2}} = \hat j + \hat k$ $\left| {\overrightarrow {{n_2}} } \right| = \sqrt {{1^2} + {1^2}} = \sqrt {1 + 1} = \sqrt 2$ $\cos \theta = \left| {\frac{{\overrightarrow {{n_1}} \cdot \overrightarrow {{n_2}} }}{{\left| {\overline {{n_1}} } \right|\left| {\overline {{n_2}} } \right|}}} \right|$ $\cos \theta = \left| {\frac{{\left( {4\hat i + 8\hat j + \hat k} \right) \cdot \left( {\hat j + \hat k} \right)}}{{9\sqrt 2 }}} \right|$ $\cos \theta = \left| {\frac{{8 + 1}}{{9\sqrt 2 }}} \right|$ $\cos \theta = \frac{9}{{9\sqrt 2 }} = \frac{1}{{\sqrt 2 }} \Rightarrow \theta = {\cos ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4}$

Question (14)

In the following cases, find the distance of each of the given points from the corresponding given plane. (a) Point (0, 0, 0) plane 3x - 4y + 12z = 3

Solution

The equation of the plane is 3x - 4y + 12z -3 = 0 .
comparing to standard equation of the plane we get, a = 3, b = -4, c = 12, and d = -3.
The point (x1, y1, z1 ) = ( 0, 0, 0).
The perpedicular distance of plane from the point be p, then $p = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$ $p = \frac{{\left| {3(0) - 4(0) + 12(0) - 3} \right|}}{{\sqrt {{3^2} + - {4^2} + {{12}^2}} }} = \left| {\frac{3}{{\sqrt {169} }}} \right| = \frac{3}{{13}}$ (b) point ( 3, -2, 1) plane 2x - y + 2z + 3 = 0

Solution

The equation of the plane is 2x - y + 2z + 3 = 0 .
comparing to standard equation of the plane we get, a = 2, b = -1, c = 2, and d = 3.
The point (x1, y1, z1 ) = ( 3, -2, 1).
The perpedicular distance of plane from the point be p, then $p = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$
$p = \frac{{\left| {2(3) - ( - 2) + 2(1) + 3} \right|}}{{\sqrt {{2^2} + - {1^2} + {2^2}} }} = \left| {\frac{{13}}{{\sqrt 9 }}} \right| = \frac{{13}}{3}$ (c) (2, 3, -5) plane x + 2y - 2z = 9.

Solution

The equation of the plane is x + 2y - 2z - 9 = 0 .
comparing to standard equation of the plane we get, a = 1, b = 2, c = -2, and d = -9.
The point (x1, y1, z1 ) = ( 2, 3, -5).
The perpedicular distance of plane from the point be p, then $p = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$ $p = \frac{{\left| {(2) + 2(3) - 2( - 5) - 9} \right|}}{{\sqrt {{1^2} + {2^2} + - {2^2}} }} = \left| {\frac{9}{{\sqrt 9 }}} \right| = \frac{9}{3} = 3$ (d) (-6, 0, 0) plane 2x - 3y + 6z - 2 = 0.

Solution

The equation of the plane is 2x - y + 2z + 3 = 0 .
comparing to standard equation of the plane we get, a = 2, b = -3, c = 6, and d = -2.
The point (x1, y1, z1 ) = ( -6, 0, 0).
The perpedicular distance of plane from the point be p, then $p = \frac{{\left| {a{x_1} + b{y_1} + c{z_1} + d} \right|}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}$ $p = \frac{{\left| {2( - 6) - 3(0) + 6(0) - 2} \right|}}{{\sqrt {{2^2} + - {3^2} + {6^2}} }} = \left| {\frac{{ - 14}}{{\sqrt {49} }}} \right| = \frac{{14}}{7} = 2$