Alternative English Year - 2017 Full Marks : 100
Ncert Solutions for Class 12 Maths Three Dimensional Geometry
Exercixe 11.3
1. In each of the following cases ,determine the direction cosine of the nornal tothe plane and the distance from the origin.
(a) \(z=2\) (b) x + y+z = 1
(c) 2x + 3y - z = 5 (d) 5y +8 =0
Sol:(a) \(z=2\)
Dividing both sides of the equation by \(\frac{1}{\sqrt{0^{2}+1^{2}+0^{2}}}\) = \(\frac{1}{\sqrt{1}}\) we get
\(\dfrac{z}{\sqrt{1}}=\dfrac{2}{\sqrt{1}}\)
z = 2
\(\therefore\) Comparing with the equation lx + my +nz= d we get
l=0, m =0, n=1,d=2
(b) x+y+z = 1
dividing both side of the equation by \(\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}\) = \(\frac{1}{\sqrt{3}}\)
We get
\(\dfrac{1}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}y+\dfrac{1}{\sqrt{3}}z=\dfrac{1}{\sqrt{3}}\)
\(\therefore\) Comparing with the equation lx + my +nz= d we get
\(l=\dfrac{1}{\sqrt{3}}, m =\dfrac{1}{\sqrt{3}}, n=\dfrac{1}{\sqrt{3}},d=\dfrac{1}{\sqrt{3}} \)
(c) 2x + 3y -z = 5
Dividing both sides of the equation by \(\frac{1}{\sqrt{2^{2}+3^{2}+(-1)^{2}}}\) = \(\frac{1}{\sqrt{14}}\) we get
\(\dfrac{2}{\sqrt{14}}x+\dfrac{3}{\sqrt{14}}y+\dfrac{-1}{\sqrt{14}}z=\dfrac{5}{\sqrt{14}}\)
\(\therefore\) Comparing with the equation lx + my +nz= d we get
\(l=\dfrac{2}{\sqrt{14}}, m =\dfrac{3}{\sqrt{14}}, n=\dfrac{-1}{\sqrt{14}},d=\dfrac{5}{\sqrt{14}} \)
(d) 5y +8 =0
Dividing both sides of the equation by \(\frac{1}{\sqrt{0^{2}+5^{2}+0^{2}}}\) = \(\frac{1}{\sqrt{5}}\) we get
\(\dfrac{5}{\sqrt{5}}y+\dfrac{8}{\sqrt{5}}=\dfrac{0}{\sqrt{5}}\)
y = -8
\(\therefore\) Comparing with the equation lx + my +nz= d we get
l=0, m =1, n=0,d=-8
2. Find the vector equation of a plain which is at a distance of 7 unit from the origin and normal to the vector \(3 \hat i+ 5 \hat j- 6\hat k .\)
Sol:
Vector the equation of a plane normal to the vector \(3\hat i+`5\hat j-6\hat k\) and at a distance of 7 unit from the origin is given by
\(\vec r.\hat n\) = d
\(\hat r.\). \(\dfrac{3\hat i+5\hat j-6\hat k}{\sqrt{3^{2}+5^{2}+6^{2}}}\) = 7
\(\implies\)\(\vec r\).\(\dfrac{3\hat i+5\hat j-6\hat k}{\sqrt{70}}\) = 7
3. find the Cartesian equation of the following planes :
(a) \(\vec r.(\hat i +\hat j-\hat k)=2\) (b) \(\vec r.(2 \hat i + 3 \hat j - 4 \hat k)=2\)
(c) \(\vec r.[(s-2t) \hat i +(3-t) \hat j + (2s+t) \hat k)]=15\)
Sol:
(a.)\(\vec r\).\(\hat i+\hat j +\hat k\)
let \(\vec r\). = \(x\hat i+y\hat j+z\hat k\)
\(\vec n\).= \(\hat i+\hat j+\hat k\) (l=1, m=1, k=-1)
\(\therefore\)the required cartesian equation of the plane is given by
(\(x\hat i+y\hat j+z\hat k\)).(\(\hat i+\hat j+\hat k\)) = 2
\(\implies\) x+y-z =2
(b).
\(\vec r\).\(2\hat i+3\hat j-4\hat k\)= 1
let \(\vec r\)=\(x\hat i+y\hat j+z\hat k\)
\(\vec n\)=\(l\hat i+m\hat j+n\hat k\)=\(2\hat i+3\hat j-4\hat k\)
putting the value of \(\vec r\) in the given equation we get
(\(x\hat i+ y\hat j+z\hat k\)).(\(2\hat i+3\hat j-4\hat k\)) = 1
2x+3y-4z = 1
this the required cartesian equation of of yhe plane.
(c).
Let \(\vec r\) = \(x\hat i+y\hat j+z\hat k\)
putting the value of \(\vec r\) in the given equation we get
(\(x\hat i+y\hat j+z\hat k\)).[(s-2t)\(\hat i\)+(3-t)\(\hat j\)+(2s+t)\(\hat k\)] = 15
\(\implies\) (s-2t)x+(3-t)y+(2s+t)y = 15
(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
Sol:
(a).2x+3y+4z-12 = 0
let the \(x_{1},y_{1},z_{1}\) be the coordinates of the foot of the perpendicular p from the origin to the plane.
The idstance ratio of the line OP is thus \( x_{1},y_{1},z_{1}\).
Therefore ,equation of the plane in normal from can be written as
\(\dfrac{2}{\sqrt{29}}x+\dfrac{3}{\sqrt{29}}y+\dfrac{4}{\sqrt{29}}z=\dfrac{12}{\sqrt{29}}\).....(1)
where \(\dfrac{2}{\sqrt{29}}, \dfrac{3}{\sqrt{29}}, \dfrac{4}{\sqrt{29}}\) are the direction cosines of OP
Since direction ratios and direction cosines of a line are proportional, so we have
\(\dfrac{x_{1}}{\dfrac{2}{\sqrt{29}}}=\dfrac{y_{1}}{\dfrac{3}{\sqrt{29}}}=\dfrac{z_{1}}{\dfrac{4}{\sqrt{29}}}=k\)
\(\implies\) \(x_{1}\) =\(\dfrac{2k}{\sqrt{29}}\), \(y_{1}\) =\(\dfrac{3k}{\sqrt{29}}\), \(z_{1}\) =\(\dfrac{4k}{\sqrt{29}}\)
Putting these value in the equation (1) we get
\begin{align}
\dfrac{2}{\sqrt{29}}\dfrac{2k}{\sqrt{29}}+\dfrac{3}{\sqrt{29}}\dfrac{3k}{\sqrt{29}}+\dfrac{4}{\sqrt{29}}\dfrac{4k}{\sqrt{29}}&=\dfrac{12}{\sqrt{29}}\\
\\
\dfrac{4k}{29} + \dfrac{9k}{29} +\dfrac{16k}{29} &= \dfrac{12}{\sqrt{29}}\\
\\
\dfrac{29k}{29} &= \dfrac{12}{\sqrt{29}}\\
\\
k&=\dfrac{12}{\sqrt{29}}\\
\end{align}
Hence putting the value of k in \(x_{1}\), \(y_{1}\) and \(z_{1}\) we get the foot of the perpendicular to be \(\left(\dfrac{24}{29},\dfrac{36}{29},\dfrac{48}{29}\right)\)
(b) 3x + 4z -6=0
3x + 4z =6
let the \(x_{1},y_{1},z_{1}\) be the coordinates of the foot of the perpendicular p from the origin to the plane.
The idstance ratio of the line OP is thus \( x_{1},y_{1},z_{1}\).
Therefore ,equation of the plane in normal from can be written as
\(\dfrac{3}{\sqrt{25}}y + \dfrac{4}{\sqrt{25}}z = \dfrac{6}{\sqrt{25}}\).....(1)
where \(0, \dfrac{3}{\sqrt{25}} , \dfrac{4}{\sqrt{25}}\) are the direction cosines of OP
Since direction ratios and direction cosines of a line are proportional, so we have
\(\dfrac{y_{1}}{\dfrac{3}{\sqrt{25}}}=\dfrac{x_{1}}{0}=\dfrac{z_{1}}{\dfrac{4}{\sqrt{25}}}=k\)
\(\implies\) \(y_{1}\) =\(\dfrac{3k}{\sqrt{25}}\), \(x_{1}\) =0, \(z_{1}\) =\(\dfrac{4k}{\sqrt{25}}\)
Putting these value in the equation (1) we get
\begin{align}
\dfrac{3}{\sqrt{25}}\dfrac{3k}{\sqrt{25}}+0+\dfrac{4}{\sqrt{25}}\dfrac{4k}{\sqrt{25}}&=\dfrac{6}{\sqrt{25}}\\
\\
\dfrac{9k}{25} +\dfrac{16k}{25} &= \dfrac{6}{\sqrt{25}}\\
\\
\dfrac{25k}{25} &= \dfrac{6}{\sqrt{25}}\\
\\
k&=\dfrac{6}{\sqrt{25}}\\
\end{align}
Hence putting the value of k in \(x_{1}\), \(y_{1}\) and \(z_{1}\) we get the foot of the perpendicular to be \(\left(0,\dfrac{18}{25},\dfrac{24}{25}\right)\)
(c)
x+y+z=1
Let the \(x_{1},y_{1},z_{1}\) be the coordinates of the foot of the perpendicular p from the origin to the plane.
The distance ratio of the line OP is thus \( x_{1},y_{1},z_{1}\).
Therefore ,equation of the plane in normal from can be written as
\(\dfrac{1}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}y+\dfrac{1}{\sqrt{3}}z=\dfrac{1}{\sqrt{3}}\).....(1)
where \(\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}\) are the direction cosines of OP
Since direction ratios and direction cosines of a line are proportional, so we have
\(\dfrac{x_{1}}{\dfrac{1}{\sqrt{3}}}=\dfrac{y_{1}}{\dfrac{1}{\sqrt{3}}}=\dfrac{z_{1}}{\dfrac{1}{\sqrt{3}}}=k\)
\(\implies\) \(x_{1}\) =\(\dfrac{k}{\sqrt{3}}\), \(y_{1}\) =\(\dfrac{k}{\sqrt{3}}\), \(z_{1}\) =\(\dfrac{k}{\sqrt{3}}\)
Putting these value in the equation (1) we get
\begin{align}
\dfrac{1}{\sqrt{3}}\dfrac{k}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}\dfrac{k}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}\dfrac{k}{\sqrt{3}}&=\dfrac{1}{\sqrt{3}}\\
\\
\dfrac{k}{3} + \dfrac{k}{3} +\dfrac{k}{3} &= \dfrac{1}{\sqrt{3}}\\
\\
\dfrac{3k}{3} &= \dfrac{1}{\sqrt{3}}\\
\\
k&=\dfrac{1}{\sqrt{3}}\\
\end{align}
Hence putting the value of k in \(x_{1}\), \(y_{1}\) and \(z_{1}\) we get the foot of the perpendicular to be \(\left(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}\right)\)
(d) 5y+8=0
Let the \(x_{1},y_{1},z_{1}\) be the coordinates of the foot of the perpendicular p from the origin to the plane.
The distance ratio of the line OP is thus \( x_{1},y_{1},z_{1}\).
Therefore ,equation of the plane in normal from can be written as
\(\dfrac{0}{\sqrt{25}}x+\dfrac{5}{\sqrt{25}}y+\dfrac{0}{\sqrt{25}}z=\dfrac{1}{\sqrt{25}}\)
y=\(\dfrac{-8}{5}\).....(1)
where 0, 1, 0 are the direction cosines of OP.
Since direction ratios and direction cosines of a line are proportional, so we have
\(\dfrac{x_1}{0}=\dfrac{y_1}{1}=\dfrac{z_1}{0}=k\)
\(\implies\) \(x_{1}\) =0, \(y_{1}\) =k, \(z_{1}\) =0
Putting these value in the equation (1) we get
\begin{align}
0.0+1.k+0.0&=\dfrac{-8}{5}\\
\\
k &= \dfrac{-8}{5}\\
\end{align}
Hence putting the value of k in \(x_{1}\), \(y_{1}\) and \(z_{1}\) we get the foot of the perpendicular to be ( 0, \(\frac{-8}{5}\), 0).
(5).Find the vector and cartesian equation of the planes .
(a). that passes through the point (1,0,-2) and the normal to the plane is \(\hat i + \hat j - \hat k.\)
(b) that passes through the point (1,4,6) and the normal vector to the plane is \(\hat i - 2\hat j+\hat k.\)
Sol: Equation of a line passing through a point \(P(x_{1}, y_{1}, z_{1})\) and perpendicular to a vector \(\vec N\) in vector form is given by $$ (\vec r - \vec a). \vec N = 0 $$
where \(\vec a\) is the position vector of P.
and in cartesian form it is given by $$ a(x -x_{1}) + b(y - y_{1} + c(z-z_{1})=0 $$
where a, b and c are direction ratios of the perpendicular.
(a) Here , \(P(x_{1}, y_{1}, z_{1})\) = (1, 0, -2)
So, position vector \(\vec a = \hat i + 0\hat j - 2\hat k=\hat i -2\hat k\)
and \(\vec N = \hat i +\hat j -\hat k\)
Thus the required equation in vector form is
\([\vec r -(\hat i-2\hat k)].(\hat i +\hat j -\hat k) =0\)
d.r"s of \(\vec N\) are a =1, b=1, c=-1
\(\therefore\) Required equation in Cartesian form is
\begin{align}
1(x-1) +1(y-0) + (-1)(z-(-2)) =0\\
x-1 + y -(z+2)=0\\
x-1+y-z-2=0\\
x+y-z=3\\
\end{align}
(b) Here , \(P(x_{1}, y_{1}, z_{1})\) = (1, 4, 6)
So, position vector \(\vec a = \hat i + 4\hat j - 6\hat k\)
and \(\vec N = \hat i -2\hat j +\hat k\)
Thus the required equation in vector form is
\([\vec r -(\hat i + 4\hat j - 6\hat k)].(\hat i -2\hat j +\hat k) =0\)
d.r"s of \(\vec N\) are a =1, b=-2, c=-1
\(\therefore\) Required equation in Cartesian form is
\begin{align}
1(x-1) +(-2)(y-4) + 1(z-6) =0\\
x-1 -2y +8 +z-6=0\\
x+-2y+z+1=0\\
\end{align}
(6). Find the equation of the planes that passes through three point .
(a). (1,1,-1), (6,4,-5), (-4,-2,3)
(b). (1,1,0), (1.2,1), (-2,2,-1)
Sol:
(a) The points are collinear so infinite no of planes are possible. Do it yourself.
You can prove the points to be non collinear in the same way as Question No. 4 solution in Exercise 11.1
(b) Equation of a plane in Cartesian form passing through three non collinear point \((x_{1}, y_{1}, z_{1})\), \((x_{2}, y_{2}, z_{2})\), and \((x_{3}, y_{3}, z_{3})\) is given by
$$ \left|\begin{array}{ccc}
x-x_{1}& y-y_{1} & z-z_{1}\\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2} - z_{1}\\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}\\
\end{array}
\right| = 0 $$
So Equation of the plane passing through (1, 1 ,0), (1,2,1) and (-2,2,-1) is given by
$$ \left|\begin{array}{ccc}
x-1& y-1 & z-0\\
1-1 & 2-1 & 1-0\\
-2-1 & 2-1 & -1-0\\
\end{array}
\right| = 0$$
$$ \left|\begin{array}{ccc}
x-1& y-1 & z\\
0 & 1 & 1\\
-3 & 1 & -1\\
\end{array}
\right| = 0$$
\begin{align}
(x-1)[1(-1)-1\times1]-(y-1)[0(-1)-(-3)1] +z(0\times1 - (-3)1]&=0\\
(x-1)(-1-1) -(y-1) (3) +z(3)&=0\\
-2x +2 -3y+3 +3z &=0\\
-2x -3y +3z +5 &=0\\
2x+3y-3z &=5\\
\end{align}
(7). Find the intercepts cut off by the plane 2x+y-z=5
Sol: The equation Ax+By+Cz = Dcan be written inintercept from as \(\dfrac{x}{a}\)+ \(\dfrac{y}{b}\)+ \(\dfrac{z}{c}\)=1when a,b,and c are intercept made on the x,y and z axes respective
Comparing 2x + y -z + 5 with Ax + By + Cz =D
We get
A= 2
B = 1
C = -1
a = \(\frac{-D}{A}\) = \(\frac{5}{2}\) = \(\frac{5}{2}\)
b = \(\frac{-D}{B}\) = \(\frac{5}{1}\) = 5
c = \(\frac{-D}{C}\) + \(\frac{5}{-1}\) = -5
\(\therefore\) the intercept made by plasne on the x,y,and z.
(8).Find the equation of the plane with intercept 3 on the y- axis and parallel to ZOX plane .
(9). Find equatrion of the plane through the intersection of the planes 3x-y+2z-4=0 and x+y+z=0 and the point (2,2,1).
Sol:
The equaion of the plane through the a intersection of the plane 3x-y+27+4 = 0 and x+y+z-2 = 4 is given by
(3x-y+2z-4)+ \(\lambda\)(x+y+z-2)= 0 ..........(1)
Since,it passes through (2,2,1) we can substitute x = 2,y = 2 and z=2 in above equation .Thus,we get \(3\times{2}\) - \(2+2\times{1}\)+ \(\lambda\) (2+2+1-2) = 0
\(\implies\) 6-2+2-4+ \(3\lambda\) = 0
\(\implies\) \(3\lambda\) = 2
\(\lambda\) = \(\frac{-2}{3}\)
putting \(\lambda\) = \(\frac{-2}{3}\) in equation we get
3x-y+2z- \(\frac{-2}{3}\)(x+y+z-2) =0
\(\implies\) 9x-3y+6z-12-2x-2y-2z+4 = 0
\(\implies\) 7x-5y+4Z-8 = 0
(10). Find the vector equation of the plane passing through the intersection of the planes
\(\vec r.(2 \hat i+ 2 \hat j- 3 \hat k)=7\), \(\vec r.(2 \hat i + 5\hat j + 3 \hat k)=9\) and through the point (2,1,3).
Sol: Equation of a plane passing through the intersection of the plane \(\vec r\).\(\vec n_{1}\) and \(\vec r\).\(\vec n_{2}\) = d_2 since by \(\vec r\).(\(\vec n_{1}\)+\(\lambda\)\(\vec n_{2}\) = d_1+xd_2
Here, \(\vec n_{1}\) = \(2\hat i\)+\(2\hat j\)-\(3\hat k\)
\(\vec n_{2}\) = \(2\hat i\)+\(5\hat j\)+\(3\hat k\)
Hence,the required equation is given by
\(2\hat i+2\hat j-3\hat k+\lambda(2\hat i+5\hat j+3\hat k)\) = 7+9 \(\lambda\)................(1)
Let \(\vec r =x\hat i+y\hat j+z\hat k\)
\(x\hat i+y\hat j+z\hat k\).\(2\hat i+5\hat j-3\hat k+\lambda (2\hat i+5\hat j+3\hat k)\) =7+9\ (\lambda\).......(2)
Since, it passes throught the point (2,1,3) it must satisfy equation (2)
\(\therefore\)(2+2x)2+(2+5x)1+(-3+3\(\lambda\))3=7+9\(\lambda\)
\(\implies\)4+4x +2+5x-9+9\(\lambda\)=7+9\(\lambda\)
\(\implies\)9\(\lambda\) = 7+9-4-2
9\(\lambda\) = 10
\(\lambda\) = \(\frac{10}{9}\)
putting the value of \(\lambda\) in (1)
we get.
\begin{align}
\vec r.[2\hat i+2\hat j-3\hat k+\frac{10}{9}(2\hat i+2\hat j+3\hat k)] &= 7+9\times\frac{10}{9}\\
\vec r.[2\hat i+2\hat j-3\hat k+\frac{10}{9}(2\hat i+2\hat j+3\hat k)] &= 17\\
\vec r.[18\hat i+18\hat j-27\hat 20\hat i+50\hat j+30\hat k ]&= 17\times{9}\\
\vec r.[38\hat i+68\hat j+3\hat k]&= 135\\
\end{align}
(11).Find the equation of the plane through the lline of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0.
Sol:
Equation of a plane passing through the line of intersection of the plane x+y+z = 1 and 2x+3y+4z = 5 is given by
\begin{align}
(x+y+z - 1) + \lambda(2x+3y+ 4z -5)&=0\\
\implies (1+2\lambda)x+(1+3\lambda)y+(1+4\lambda)z-(1+5\lambda) &= 0 .... (i)
\end{align}
The direction plane ratio of the abone plave plane is (1+2\lambda),(1+3\lambda),and (1+4\lambda)
The required plane is perpendicular to x - y +z = 0.
Hence the some of the product of their respective direction ratios will be equal to zero
\begin{align}
i.e\ \ \ a_1a_2+b_1b_2+c_1c_2 &=0\\
\implies\ \ (1+2\lambda)1+(1+3\lambda)(-1)+(1+4\lambda)1 &=0\\
\implies\ \ 1+2\lambda-1-3\lambda +1+4\lambda &= 0\\
\implies\ \ 3\lambda +1 &= 0\\
\implies\ \ \lambda &= \frac{-1}{3} \\
\text{Putting the value of }\lambda \text{we get}\\
(x+y+z - 1) + \frac{-1}{3}(2x+3y+ 4z -5)&=0\\
\implies x-z+2 = 0\\
\end{align}
This is the required equation .
(12). Find the angle between the plane whose vecrtor equation are
\(\vec r.(2 \hat i+2 \hat j -3 \hat k)=5\) and \(\vec r.(3 \hat i-3 \hat j+5 \hat k)=3\)
Sol: Angle between two plane
\(\vec r.\vec n_{1}=d_{1}\) and \(\vec r.\vec n_{2}= d_{2}\) is given by
\(\cos\theta =\left|\dfrac{\vec n_{1}.\vec n_{2}}{|\vec n_{1}|.|\vec n_{2}|}\right|\)
Here, \(\vec n_{1}=2\hat i + 2\hat j -3\hat k\)
\(\vec n_{2}=3\hat i - 3\hat j +5\hat k\)
\begin{align}
\vec n_{1}. \vec n_{2} &=(2\hat i + 2\hat j -3\hat k).(3\hat i -3\hat j + 5\hat k)\\
&=2.3 + 2(-3) + (-3)5\\
&=6-6-15\\
&=-15\\
|\vec n_{1}|. |\vec n_{2}|&=|2\hat i +2\hat j-3\hat k|. |3\hat i -3\hat j + 5\hat k|\\
&=\sqrt{2^2 + 2^2 +(-3)^2}. \sqrt{3^2 + (-3)^2 + 5^2}\\
&=\sqrt{4+4+9}\sqrt{9+9+25}\\
&=\sqrt{17}\sqrt{43}\\
\therefore \cos \theta &= \left| \frac{-15}{\sqrt{17}\sqrt{43}}\right| \\
\\
\implies \theta &= cos^{-1} \frac{15}{\sqrt{731}}\\
\end{align}
(13). In the following cases, determine whetherthe given plnes are parallel or perpindicular, and in cases they are nither, findthe angle 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+x-4=0
Sol: Two planes \( a_{1}x+b_{1}y +c_{1}z+d_{1}\) = 0 and \( a_{2}x+b_{2}y +c_{2}z+d_{2}\) = 0 are parallel if \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}\)
(a)
Here, \(a_{1} =7, b_{1} =5, c_{1}=6\)
\(a_{2} =3, b_{2}=-1, c_{2}=-10\)
Since, it is neither parallel nor perpendicular, the angle between the planes is given as
\begin{align}
\cos \theta &= \left|\dfrac{a_{1}a_{2}+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2 + c_1^2}\sqrt{a_2^2 +b_2^2+c_2^2}}\right|\\
&=\left|\dfrac{7.3 +5(-1) +6(-10)}{\sqrt{7^2+5^2+6^2}\sqrt{3^2 +(-1)^2+(-10)^2}}\right|\\
&=\left|\dfrac{-44}{\sqrt{110}\sqrt{110}}\right|\\
&=\dfrac{44}{110}\\
&=\dfrac{2}{5}\\
\theta &= \cos^{-1}\frac{2}{5}\\
\end{align}
(b)Here, \(a_{1} =2, b_{1} =1, c_{1}=3\)
\(a_{2} =1, b_{2}=-2, c_{2}=0\)
We have
\begin{align}
a_1a_2 + b_1b_2 + c_1c_2 &= 2\times 1+ 1\times(-2) +3\times 0\\
&= 2-2\\
&=0\\
\end{align}
Hence the two planes are perpendicular.
(c)
Here, \(a_{1} =2, b_{1} =-2, c_{1}=4\)
\(a_{2} =3, b_{2}=-3, c_{2}=6\)
We have
\(\dfrac{a_1}{a_2}=\dfrac{2}{3}, \ \ \dfrac{b_1}{b_2}=\dfrac{-2}{-3}=\dfrac{2}{3}, \ \ \ \dfrac{c_1}{c_2}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\therefore\) \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\)
Hence, the two planes are parallel.
(d)
Here, \(a_{1} =2, b_{1} =-1, c_{1}=3\)
\(a_{2} =2, b_{2}=-1, c_{2}=3\)
We have
\(\dfrac{a_1}{a_2}=\dfrac{2}{2}=1, \ \ \dfrac{b_1}{b_2}=\dfrac{-1}{-1}=1, \ \ \ \dfrac{c_1}{c_2}=\dfrac{3}{3}=1\)
\(\therefore\) \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\)
Hence, the two planes are parallel.
(e)
Here, \(a_{1} =4, b_{1} =8, c_{1}=1\)
\(a_{2} =0, b_{2}=1, c_{2}=1\)
Since, it is neither parallel nor perpendicular, the angle between the planes is given as
\begin{align}
\cos \theta &= \left|\dfrac{a_{1}a_{2}+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2 + c_1^2}\sqrt{a_2^2 +b_2^2+c_2^2}}\right|\\
&=\left|\dfrac{ 4\times 0+8\times 1+ 1\times 1}{\sqrt{4^2+8^2+1^2}\sqrt{0^2 +1^2+1^2}}\right|\\
&=\left|\dfrac{9}{\sqrt{81}\sqrt{2}}\right|\\
&=\dfrac{9}{9\sqrt{2}}\\
&=\dfrac{1}{\sqrt{2}}\\
\theta &= \cos^{-1}\frac{1}{\sqrt2}\\
&=45^o
\end{align}
(14) In the following cases find the distance of each of the given point from the crossponding given plane .
point plane
(000) 3x-4y+12z=3
(3,-2,1) 2x-y+2z+3=0
(2,3,-5) x+2y-2z+3=9
(-6,0,0) 2x-3y+6z-2=0
Sol:
(a) Here \(\vec a = 0\hat i +0\hat j + 0\hat k\)
(c) 2x + 3y - z = 5 (d) 5y +8 =0
Sol:(a) \(z=2\)
Dividing both sides of the equation by \(\frac{1}{\sqrt{0^{2}+1^{2}+0^{2}}}\) = \(\frac{1}{\sqrt{1}}\) we get
\(\dfrac{z}{\sqrt{1}}=\dfrac{2}{\sqrt{1}}\)
z = 2
\(\therefore\) Comparing with the equation lx + my +nz= d we get
l=0, m =0, n=1,d=2
(b) x+y+z = 1
dividing both side of the equation by \(\frac{1}{\sqrt{1^{2}+1^{2}+1^{2}}}\) = \(\frac{1}{\sqrt{3}}\)
We get
\(\dfrac{1}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}y+\dfrac{1}{\sqrt{3}}z=\dfrac{1}{\sqrt{3}}\)
\(\therefore\) Comparing with the equation lx + my +nz= d we get
\(l=\dfrac{1}{\sqrt{3}}, m =\dfrac{1}{\sqrt{3}}, n=\dfrac{1}{\sqrt{3}},d=\dfrac{1}{\sqrt{3}} \)
(c) 2x + 3y -z = 5
Dividing both sides of the equation by \(\frac{1}{\sqrt{2^{2}+3^{2}+(-1)^{2}}}\) = \(\frac{1}{\sqrt{14}}\) we get
\(\dfrac{2}{\sqrt{14}}x+\dfrac{3}{\sqrt{14}}y+\dfrac{-1}{\sqrt{14}}z=\dfrac{5}{\sqrt{14}}\)
\(\therefore\) Comparing with the equation lx + my +nz= d we get
\(l=\dfrac{2}{\sqrt{14}}, m =\dfrac{3}{\sqrt{14}}, n=\dfrac{-1}{\sqrt{14}},d=\dfrac{5}{\sqrt{14}} \)
(d) 5y +8 =0
Dividing both sides of the equation by \(\frac{1}{\sqrt{0^{2}+5^{2}+0^{2}}}\) = \(\frac{1}{\sqrt{5}}\) we get
\(\dfrac{5}{\sqrt{5}}y+\dfrac{8}{\sqrt{5}}=\dfrac{0}{\sqrt{5}}\)
y = -8
\(\therefore\) Comparing with the equation lx + my +nz= d we get
l=0, m =1, n=0,d=-8
2. Find the vector equation of a plain which is at a distance of 7 unit from the origin and normal to the vector \(3 \hat i+ 5 \hat j- 6\hat k .\)
Sol:
Vector the equation of a plane normal to the vector \(3\hat i+`5\hat j-6\hat k\) and at a distance of 7 unit from the origin is given by
\(\vec r.\hat n\) = d
\(\hat r.\). \(\dfrac{3\hat i+5\hat j-6\hat k}{\sqrt{3^{2}+5^{2}+6^{2}}}\) = 7
\(\implies\)\(\vec r\).\(\dfrac{3\hat i+5\hat j-6\hat k}{\sqrt{70}}\) = 7
3. find the Cartesian equation of the following planes :
(a) \(\vec r.(\hat i +\hat j-\hat k)=2\) (b) \(\vec r.(2 \hat i + 3 \hat j - 4 \hat k)=2\)
(c) \(\vec r.[(s-2t) \hat i +(3-t) \hat j + (2s+t) \hat k)]=15\)
Sol:
(a.)\(\vec r\).\(\hat i+\hat j +\hat k\)
let \(\vec r\). = \(x\hat i+y\hat j+z\hat k\)
\(\vec n\).= \(\hat i+\hat j+\hat k\) (l=1, m=1, k=-1)
\(\therefore\)the required cartesian equation of the plane is given by
(\(x\hat i+y\hat j+z\hat k\)).(\(\hat i+\hat j+\hat k\)) = 2
\(\implies\) x+y-z =2
(b).
\(\vec r\).\(2\hat i+3\hat j-4\hat k\)= 1
let \(\vec r\)=\(x\hat i+y\hat j+z\hat k\)
\(\vec n\)=\(l\hat i+m\hat j+n\hat k\)=\(2\hat i+3\hat j-4\hat k\)
putting the value of \(\vec r\) in the given equation we get
(\(x\hat i+ y\hat j+z\hat k\)).(\(2\hat i+3\hat j-4\hat k\)) = 1
2x+3y-4z = 1
this the required cartesian equation of of yhe plane.
(c).
Let \(\vec r\) = \(x\hat i+y\hat j+z\hat k\)
putting the value of \(\vec r\) in the given equation we get
(\(x\hat i+y\hat j+z\hat k\)).[(s-2t)\(\hat i\)+(3-t)\(\hat j\)+(2s+t)\(\hat k\)] = 15
\(\implies\) (s-2t)x+(3-t)y+(2s+t)y = 15
(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
Sol:
(a).2x+3y+4z-12 = 0
let the \(x_{1},y_{1},z_{1}\) be the coordinates of the foot of the perpendicular p from the origin to the plane.
The idstance ratio of the line OP is thus \( x_{1},y_{1},z_{1}\).
Therefore ,equation of the plane in normal from can be written as
\(\dfrac{2}{\sqrt{29}}x+\dfrac{3}{\sqrt{29}}y+\dfrac{4}{\sqrt{29}}z=\dfrac{12}{\sqrt{29}}\).....(1)
where \(\dfrac{2}{\sqrt{29}}, \dfrac{3}{\sqrt{29}}, \dfrac{4}{\sqrt{29}}\) are the direction cosines of OP
Since direction ratios and direction cosines of a line are proportional, so we have
\(\dfrac{x_{1}}{\dfrac{2}{\sqrt{29}}}=\dfrac{y_{1}}{\dfrac{3}{\sqrt{29}}}=\dfrac{z_{1}}{\dfrac{4}{\sqrt{29}}}=k\)
\(\implies\) \(x_{1}\) =\(\dfrac{2k}{\sqrt{29}}\), \(y_{1}\) =\(\dfrac{3k}{\sqrt{29}}\), \(z_{1}\) =\(\dfrac{4k}{\sqrt{29}}\)
Putting these value in the equation (1) we get
\begin{align}
\dfrac{2}{\sqrt{29}}\dfrac{2k}{\sqrt{29}}+\dfrac{3}{\sqrt{29}}\dfrac{3k}{\sqrt{29}}+\dfrac{4}{\sqrt{29}}\dfrac{4k}{\sqrt{29}}&=\dfrac{12}{\sqrt{29}}\\
\\
\dfrac{4k}{29} + \dfrac{9k}{29} +\dfrac{16k}{29} &= \dfrac{12}{\sqrt{29}}\\
\\
\dfrac{29k}{29} &= \dfrac{12}{\sqrt{29}}\\
\\
k&=\dfrac{12}{\sqrt{29}}\\
\end{align}
Hence putting the value of k in \(x_{1}\), \(y_{1}\) and \(z_{1}\) we get the foot of the perpendicular to be \(\left(\dfrac{24}{29},\dfrac{36}{29},\dfrac{48}{29}\right)\)
(b) 3x + 4z -6=0
3x + 4z =6
let the \(x_{1},y_{1},z_{1}\) be the coordinates of the foot of the perpendicular p from the origin to the plane.
The idstance ratio of the line OP is thus \( x_{1},y_{1},z_{1}\).
Therefore ,equation of the plane in normal from can be written as
\(\dfrac{3}{\sqrt{25}}y + \dfrac{4}{\sqrt{25}}z = \dfrac{6}{\sqrt{25}}\).....(1)
where \(0, \dfrac{3}{\sqrt{25}} , \dfrac{4}{\sqrt{25}}\) are the direction cosines of OP
Since direction ratios and direction cosines of a line are proportional, so we have
\(\dfrac{y_{1}}{\dfrac{3}{\sqrt{25}}}=\dfrac{x_{1}}{0}=\dfrac{z_{1}}{\dfrac{4}{\sqrt{25}}}=k\)
\(\implies\) \(y_{1}\) =\(\dfrac{3k}{\sqrt{25}}\), \(x_{1}\) =0, \(z_{1}\) =\(\dfrac{4k}{\sqrt{25}}\)
Putting these value in the equation (1) we get
\begin{align}
\dfrac{3}{\sqrt{25}}\dfrac{3k}{\sqrt{25}}+0+\dfrac{4}{\sqrt{25}}\dfrac{4k}{\sqrt{25}}&=\dfrac{6}{\sqrt{25}}\\
\\
\dfrac{9k}{25} +\dfrac{16k}{25} &= \dfrac{6}{\sqrt{25}}\\
\\
\dfrac{25k}{25} &= \dfrac{6}{\sqrt{25}}\\
\\
k&=\dfrac{6}{\sqrt{25}}\\
\end{align}
Hence putting the value of k in \(x_{1}\), \(y_{1}\) and \(z_{1}\) we get the foot of the perpendicular to be \(\left(0,\dfrac{18}{25},\dfrac{24}{25}\right)\)
(c)
x+y+z=1
Let the \(x_{1},y_{1},z_{1}\) be the coordinates of the foot of the perpendicular p from the origin to the plane.
The distance ratio of the line OP is thus \( x_{1},y_{1},z_{1}\).
Therefore ,equation of the plane in normal from can be written as
\(\dfrac{1}{\sqrt{3}}x+\dfrac{1}{\sqrt{3}}y+\dfrac{1}{\sqrt{3}}z=\dfrac{1}{\sqrt{3}}\).....(1)
where \(\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}\) are the direction cosines of OP
Since direction ratios and direction cosines of a line are proportional, so we have
\(\dfrac{x_{1}}{\dfrac{1}{\sqrt{3}}}=\dfrac{y_{1}}{\dfrac{1}{\sqrt{3}}}=\dfrac{z_{1}}{\dfrac{1}{\sqrt{3}}}=k\)
\(\implies\) \(x_{1}\) =\(\dfrac{k}{\sqrt{3}}\), \(y_{1}\) =\(\dfrac{k}{\sqrt{3}}\), \(z_{1}\) =\(\dfrac{k}{\sqrt{3}}\)
Putting these value in the equation (1) we get
\begin{align}
\dfrac{1}{\sqrt{3}}\dfrac{k}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}\dfrac{k}{\sqrt{3}}+\dfrac{1}{\sqrt{3}}\dfrac{k}{\sqrt{3}}&=\dfrac{1}{\sqrt{3}}\\
\\
\dfrac{k}{3} + \dfrac{k}{3} +\dfrac{k}{3} &= \dfrac{1}{\sqrt{3}}\\
\\
\dfrac{3k}{3} &= \dfrac{1}{\sqrt{3}}\\
\\
k&=\dfrac{1}{\sqrt{3}}\\
\end{align}
Hence putting the value of k in \(x_{1}\), \(y_{1}\) and \(z_{1}\) we get the foot of the perpendicular to be \(\left(\dfrac{1}{3},\dfrac{1}{3},\dfrac{1}{3}\right)\)
(d) 5y+8=0
Let the \(x_{1},y_{1},z_{1}\) be the coordinates of the foot of the perpendicular p from the origin to the plane.
The distance ratio of the line OP is thus \( x_{1},y_{1},z_{1}\).
Therefore ,equation of the plane in normal from can be written as
\(\dfrac{0}{\sqrt{25}}x+\dfrac{5}{\sqrt{25}}y+\dfrac{0}{\sqrt{25}}z=\dfrac{1}{\sqrt{25}}\)
y=\(\dfrac{-8}{5}\).....(1)
where 0, 1, 0 are the direction cosines of OP.
Since direction ratios and direction cosines of a line are proportional, so we have
\(\dfrac{x_1}{0}=\dfrac{y_1}{1}=\dfrac{z_1}{0}=k\)
\(\implies\) \(x_{1}\) =0, \(y_{1}\) =k, \(z_{1}\) =0
Putting these value in the equation (1) we get
\begin{align}
0.0+1.k+0.0&=\dfrac{-8}{5}\\
\\
k &= \dfrac{-8}{5}\\
\end{align}
Hence putting the value of k in \(x_{1}\), \(y_{1}\) and \(z_{1}\) we get the foot of the perpendicular to be ( 0, \(\frac{-8}{5}\), 0).
(5).Find the vector and cartesian equation of the planes .
(a). that passes through the point (1,0,-2) and the normal to the plane is \(\hat i + \hat j - \hat k.\)
(b) that passes through the point (1,4,6) and the normal vector to the plane is \(\hat i - 2\hat j+\hat k.\)
Sol: Equation of a line passing through a point \(P(x_{1}, y_{1}, z_{1})\) and perpendicular to a vector \(\vec N\) in vector form is given by $$ (\vec r - \vec a). \vec N = 0 $$
where \(\vec a\) is the position vector of P.
and in cartesian form it is given by $$ a(x -x_{1}) + b(y - y_{1} + c(z-z_{1})=0 $$
where a, b and c are direction ratios of the perpendicular.
(a) Here , \(P(x_{1}, y_{1}, z_{1})\) = (1, 0, -2)
So, position vector \(\vec a = \hat i + 0\hat j - 2\hat k=\hat i -2\hat k\)
and \(\vec N = \hat i +\hat j -\hat k\)
Thus the required equation in vector form is
\([\vec r -(\hat i-2\hat k)].(\hat i +\hat j -\hat k) =0\)
d.r"s of \(\vec N\) are a =1, b=1, c=-1
\(\therefore\) Required equation in Cartesian form is
\begin{align}
1(x-1) +1(y-0) + (-1)(z-(-2)) =0\\
x-1 + y -(z+2)=0\\
x-1+y-z-2=0\\
x+y-z=3\\
\end{align}
(b) Here , \(P(x_{1}, y_{1}, z_{1})\) = (1, 4, 6)
So, position vector \(\vec a = \hat i + 4\hat j - 6\hat k\)
and \(\vec N = \hat i -2\hat j +\hat k\)
Thus the required equation in vector form is
\([\vec r -(\hat i + 4\hat j - 6\hat k)].(\hat i -2\hat j +\hat k) =0\)
d.r"s of \(\vec N\) are a =1, b=-2, c=-1
\(\therefore\) Required equation in Cartesian form is
\begin{align}
1(x-1) +(-2)(y-4) + 1(z-6) =0\\
x-1 -2y +8 +z-6=0\\
x+-2y+z+1=0\\
\end{align}
(6). Find the equation of the planes that passes through three point .
(a). (1,1,-1), (6,4,-5), (-4,-2,3)
(b). (1,1,0), (1.2,1), (-2,2,-1)
Sol:
(a) The points are collinear so infinite no of planes are possible. Do it yourself.
You can prove the points to be non collinear in the same way as Question No. 4 solution in Exercise 11.1
(b) Equation of a plane in Cartesian form passing through three non collinear point \((x_{1}, y_{1}, z_{1})\), \((x_{2}, y_{2}, z_{2})\), and \((x_{3}, y_{3}, z_{3})\) is given by
$$ \left|\begin{array}{ccc}
x-x_{1}& y-y_{1} & z-z_{1}\\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2} - z_{1}\\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}\\
\end{array}
\right| = 0 $$
So Equation of the plane passing through (1, 1 ,0), (1,2,1) and (-2,2,-1) is given by
$$ \left|\begin{array}{ccc}
x-1& y-1 & z-0\\
1-1 & 2-1 & 1-0\\
-2-1 & 2-1 & -1-0\\
\end{array}
\right| = 0$$
$$ \left|\begin{array}{ccc}
x-1& y-1 & z\\
0 & 1 & 1\\
-3 & 1 & -1\\
\end{array}
\right| = 0$$
\begin{align}
(x-1)[1(-1)-1\times1]-(y-1)[0(-1)-(-3)1] +z(0\times1 - (-3)1]&=0\\
(x-1)(-1-1) -(y-1) (3) +z(3)&=0\\
-2x +2 -3y+3 +3z &=0\\
-2x -3y +3z +5 &=0\\
2x+3y-3z &=5\\
\end{align}
(7). Find the intercepts cut off by the plane 2x+y-z=5
Sol: The equation Ax+By+Cz = Dcan be written inintercept from as \(\dfrac{x}{a}\)+ \(\dfrac{y}{b}\)+ \(\dfrac{z}{c}\)=1when a,b,and c are intercept made on the x,y and z axes respective
Comparing 2x + y -z + 5 with Ax + By + Cz =D
We get
A= 2
B = 1
C = -1
a = \(\frac{-D}{A}\) = \(\frac{5}{2}\) = \(\frac{5}{2}\)
b = \(\frac{-D}{B}\) = \(\frac{5}{1}\) = 5
c = \(\frac{-D}{C}\) + \(\frac{5}{-1}\) = -5
\(\therefore\) the intercept made by plasne on the x,y,and z.
Ncert Solutions for Class 12 Maths Three Dimensional Geometry
Previous Exercises
(9). Find equatrion of the plane through the intersection of the planes 3x-y+2z-4=0 and x+y+z=0 and the point (2,2,1).
Sol:
The equaion of the plane through the a intersection of the plane 3x-y+27+4 = 0 and x+y+z-2 = 4 is given by
(3x-y+2z-4)+ \(\lambda\)(x+y+z-2)= 0 ..........(1)
Since,it passes through (2,2,1) we can substitute x = 2,y = 2 and z=2 in above equation .Thus,we get \(3\times{2}\) - \(2+2\times{1}\)+ \(\lambda\) (2+2+1-2) = 0
\(\implies\) 6-2+2-4+ \(3\lambda\) = 0
\(\implies\) \(3\lambda\) = 2
\(\lambda\) = \(\frac{-2}{3}\)
putting \(\lambda\) = \(\frac{-2}{3}\) in equation we get
3x-y+2z- \(\frac{-2}{3}\)(x+y+z-2) =0
\(\implies\) 9x-3y+6z-12-2x-2y-2z+4 = 0
\(\implies\) 7x-5y+4Z-8 = 0
(10). Find the vector equation of the plane passing through the intersection of the planes
\(\vec r.(2 \hat i+ 2 \hat j- 3 \hat k)=7\), \(\vec r.(2 \hat i + 5\hat j + 3 \hat k)=9\) and through the point (2,1,3).
Sol: Equation of a plane passing through the intersection of the plane \(\vec r\).\(\vec n_{1}\) and \(\vec r\).\(\vec n_{2}\) = d_2 since by \(\vec r\).(\(\vec n_{1}\)+\(\lambda\)\(\vec n_{2}\) = d_1+xd_2
Here, \(\vec n_{1}\) = \(2\hat i\)+\(2\hat j\)-\(3\hat k\)
\(\vec n_{2}\) = \(2\hat i\)+\(5\hat j\)+\(3\hat k\)
Hence,the required equation is given by
\(2\hat i+2\hat j-3\hat k+\lambda(2\hat i+5\hat j+3\hat k)\) = 7+9 \(\lambda\)................(1)
Let \(\vec r =x\hat i+y\hat j+z\hat k\)
\(x\hat i+y\hat j+z\hat k\).\(2\hat i+5\hat j-3\hat k+\lambda (2\hat i+5\hat j+3\hat k)\) =7+9\ (\lambda\).......(2)
Since, it passes throught the point (2,1,3) it must satisfy equation (2)
\(\therefore\)(2+2x)2+(2+5x)1+(-3+3\(\lambda\))3=7+9\(\lambda\)
\(\implies\)4+4x +2+5x-9+9\(\lambda\)=7+9\(\lambda\)
\(\implies\)9\(\lambda\) = 7+9-4-2
9\(\lambda\) = 10
\(\lambda\) = \(\frac{10}{9}\)
putting the value of \(\lambda\) in (1)
we get.
\begin{align}
\vec r.[2\hat i+2\hat j-3\hat k+\frac{10}{9}(2\hat i+2\hat j+3\hat k)] &= 7+9\times\frac{10}{9}\\
\vec r.[2\hat i+2\hat j-3\hat k+\frac{10}{9}(2\hat i+2\hat j+3\hat k)] &= 17\\
\vec r.[18\hat i+18\hat j-27\hat 20\hat i+50\hat j+30\hat k ]&= 17\times{9}\\
\vec r.[38\hat i+68\hat j+3\hat k]&= 135\\
\end{align}
(11).Find the equation of the plane through the lline of intersection of the planes x+y+z=1 and 2x+3y+4z=5 which is perpendicular to the plane x-y+z=0.
Sol:
Equation of a plane passing through the line of intersection of the plane x+y+z = 1 and 2x+3y+4z = 5 is given by
\begin{align}
(x+y+z - 1) + \lambda(2x+3y+ 4z -5)&=0\\
\implies (1+2\lambda)x+(1+3\lambda)y+(1+4\lambda)z-(1+5\lambda) &= 0 .... (i)
\end{align}
The direction plane ratio of the abone plave plane is (1+2\lambda),(1+3\lambda),and (1+4\lambda)
The required plane is perpendicular to x - y +z = 0.
Hence the some of the product of their respective direction ratios will be equal to zero
\begin{align}
i.e\ \ \ a_1a_2+b_1b_2+c_1c_2 &=0\\
\implies\ \ (1+2\lambda)1+(1+3\lambda)(-1)+(1+4\lambda)1 &=0\\
\implies\ \ 1+2\lambda-1-3\lambda +1+4\lambda &= 0\\
\implies\ \ 3\lambda +1 &= 0\\
\implies\ \ \lambda &= \frac{-1}{3} \\
\text{Putting the value of }\lambda \text{we get}\\
(x+y+z - 1) + \frac{-1}{3}(2x+3y+ 4z -5)&=0\\
\implies x-z+2 = 0\\
\end{align}
This is the required equation .
(12). Find the angle between the plane whose vecrtor equation are
\(\vec r.(2 \hat i+2 \hat j -3 \hat k)=5\) and \(\vec r.(3 \hat i-3 \hat j+5 \hat k)=3\)
Sol: Angle between two plane
\(\vec r.\vec n_{1}=d_{1}\) and \(\vec r.\vec n_{2}= d_{2}\) is given by
\(\cos\theta =\left|\dfrac{\vec n_{1}.\vec n_{2}}{|\vec n_{1}|.|\vec n_{2}|}\right|\)
Here, \(\vec n_{1}=2\hat i + 2\hat j -3\hat k\)
\(\vec n_{2}=3\hat i - 3\hat j +5\hat k\)
\begin{align}
\vec n_{1}. \vec n_{2} &=(2\hat i + 2\hat j -3\hat k).(3\hat i -3\hat j + 5\hat k)\\
&=2.3 + 2(-3) + (-3)5\\
&=6-6-15\\
&=-15\\
|\vec n_{1}|. |\vec n_{2}|&=|2\hat i +2\hat j-3\hat k|. |3\hat i -3\hat j + 5\hat k|\\
&=\sqrt{2^2 + 2^2 +(-3)^2}. \sqrt{3^2 + (-3)^2 + 5^2}\\
&=\sqrt{4+4+9}\sqrt{9+9+25}\\
&=\sqrt{17}\sqrt{43}\\
\therefore \cos \theta &= \left| \frac{-15}{\sqrt{17}\sqrt{43}}\right| \\
\\
\implies \theta &= cos^{-1} \frac{15}{\sqrt{731}}\\
\end{align}
(13). In the following cases, determine whetherthe given plnes are parallel or perpindicular, and in cases they are nither, findthe angle 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+x-4=0
Sol: Two planes \( a_{1}x+b_{1}y +c_{1}z+d_{1}\) = 0 and \( a_{2}x+b_{2}y +c_{2}z+d_{2}\) = 0 are parallel if \(\dfrac{a_{1}}{a_{2}}=\dfrac{b_{1}}{b_{2}}=\dfrac{c_{1}}{c_{2}}\)
(a)
Here, \(a_{1} =7, b_{1} =5, c_{1}=6\)
\(a_{2} =3, b_{2}=-1, c_{2}=-10\)
Since, it is neither parallel nor perpendicular, the angle between the planes is given as
\begin{align}
\cos \theta &= \left|\dfrac{a_{1}a_{2}+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2 + c_1^2}\sqrt{a_2^2 +b_2^2+c_2^2}}\right|\\
&=\left|\dfrac{7.3 +5(-1) +6(-10)}{\sqrt{7^2+5^2+6^2}\sqrt{3^2 +(-1)^2+(-10)^2}}\right|\\
&=\left|\dfrac{-44}{\sqrt{110}\sqrt{110}}\right|\\
&=\dfrac{44}{110}\\
&=\dfrac{2}{5}\\
\theta &= \cos^{-1}\frac{2}{5}\\
\end{align}
(b)Here, \(a_{1} =2, b_{1} =1, c_{1}=3\)
\(a_{2} =1, b_{2}=-2, c_{2}=0\)
We have
\begin{align}
a_1a_2 + b_1b_2 + c_1c_2 &= 2\times 1+ 1\times(-2) +3\times 0\\
&= 2-2\\
&=0\\
\end{align}
Hence the two planes are perpendicular.
(c)
Here, \(a_{1} =2, b_{1} =-2, c_{1}=4\)
\(a_{2} =3, b_{2}=-3, c_{2}=6\)
We have
\(\dfrac{a_1}{a_2}=\dfrac{2}{3}, \ \ \dfrac{b_1}{b_2}=\dfrac{-2}{-3}=\dfrac{2}{3}, \ \ \ \dfrac{c_1}{c_2}=\dfrac{4}{6}=\dfrac{2}{3}\)
\(\therefore\) \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\)
Hence, the two planes are parallel.
(d)
Here, \(a_{1} =2, b_{1} =-1, c_{1}=3\)
\(a_{2} =2, b_{2}=-1, c_{2}=3\)
We have
\(\dfrac{a_1}{a_2}=\dfrac{2}{2}=1, \ \ \dfrac{b_1}{b_2}=\dfrac{-1}{-1}=1, \ \ \ \dfrac{c_1}{c_2}=\dfrac{3}{3}=1\)
\(\therefore\) \(\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}=\dfrac{c_1}{c_2}\)
Hence, the two planes are parallel.
(e)
Here, \(a_{1} =4, b_{1} =8, c_{1}=1\)
\(a_{2} =0, b_{2}=1, c_{2}=1\)
Since, it is neither parallel nor perpendicular, the angle between the planes is given as
\begin{align}
\cos \theta &= \left|\dfrac{a_{1}a_{2}+b_1b_2+c_1c_2}{\sqrt{a_1^2+b_1^2 + c_1^2}\sqrt{a_2^2 +b_2^2+c_2^2}}\right|\\
&=\left|\dfrac{ 4\times 0+8\times 1+ 1\times 1}{\sqrt{4^2+8^2+1^2}\sqrt{0^2 +1^2+1^2}}\right|\\
&=\left|\dfrac{9}{\sqrt{81}\sqrt{2}}\right|\\
&=\dfrac{9}{9\sqrt{2}}\\
&=\dfrac{1}{\sqrt{2}}\\
\theta &= \cos^{-1}\frac{1}{\sqrt2}\\
&=45^o
\end{align}
(14) In the following cases find the distance of each of the given point from the crossponding given plane .
point plane
(000) 3x-4y+12z=3
(3,-2,1) 2x-y+2z+3=0
(2,3,-5) x+2y-2z+3=9
(-6,0,0) 2x-3y+6z-2=0
Sol:
(a) Here \(\vec a = 0\hat i +0\hat j + 0\hat k\)
\(\vec N = 3\hat i - 4\hat j + 2 \hat k\)
The distance of the poin (0,0,0) from the given plane 3x -4y +12z = 3 is given by
\(\left|\dfrac{\vec a. \vec N - d}{|\vec N|}\right|\)
=\(\left|\dfrac{(0\hat i +0\hat j + 0\hat k). (3\hat i - 4\hat j + 2 \hat k) - 3}{\sqrt{3^2+(-4)^2+12^2}}\right|\)
=\(\left|\dfrac{0.3-0.4+0.2-3}{\sqrt{169}}\right|\)
=\(\left|\dfrac{-3}{13}\right|\)
=\(\dfrac{3}{13}\)
The distance of the poin (0,0,0) from the given plane 3x -4y +12z = 3 is given by
\(\left|\dfrac{\vec a. \vec N - d}{|\vec N|}\right|\)
=\(\left|\dfrac{(0\hat i +0\hat j + 0\hat k). (3\hat i - 4\hat j + 2 \hat k) - 3}{\sqrt{3^2+(-4)^2+12^2}}\right|\)
=\(\left|\dfrac{0.3-0.4+0.2-3}{\sqrt{169}}\right|\)
=\(\left|\dfrac{-3}{13}\right|\)
=\(\dfrac{3}{13}\)
(b)
Here \(\vec a = 3\hat i -2\hat j + \hat k\)
\(\vec N = 2\hat i - \hat j + 2 \hat k\)
The distance of the poin (3,-2,1) from the given plane 2x -y +2z + =0 is given by
\(\left|\dfrac{\vec a. \vec N - d}{|\vec N|}\right|\)
=\(\left|\dfrac{(3\hat i -2\hat j + \hat k). (2\hat i - \hat j + 2 \hat k) + 3}{\sqrt{2^2+(-1)^2+2^2}}\right|\)
=\(\left|\dfrac{3.2 + (-2)(-1)4+1.2-3}{\sqrt{169}}\right|\)
=\(\left|\dfrac{10 +3}{\sqrt {9}}\right|\)
=\(\left|\dfrac{13}{13}\right|\)
=\(\dfrac{13}{3}\)
\(\vec N = 2\hat i - \hat j + 2 \hat k\)
The distance of the poin (3,-2,1) from the given plane 2x -y +2z + =0 is given by
\(\left|\dfrac{\vec a. \vec N - d}{|\vec N|}\right|\)
=\(\left|\dfrac{(3\hat i -2\hat j + \hat k). (2\hat i - \hat j + 2 \hat k) + 3}{\sqrt{2^2+(-1)^2+2^2}}\right|\)
=\(\left|\dfrac{3.2 + (-2)(-1)4+1.2-3}{\sqrt{169}}\right|\)
=\(\left|\dfrac{13}{13}\right|\)
=\(\dfrac{13}{3}\)
(c)
Here \(\vec a = 2\hat i +3\hat j - 5\hat k\)
\(\vec N = 1\hat i + 2\hat j - 2\hat k\)
The distance of the poin (2,3,-5) from the given plane x +2y -2z =9 is given by
\(\left|\dfrac{\vec a. \vec N - d}{|\vec N|}\right|\)
=\(\left|\dfrac{(2\hat i +3\hat j - 5\hat k). ( 1\hat i + 2\hat j - 2\hat k) + 9}{\sqrt{1^2+2^2+(-2)^2}}\right|\)
=\(\left|\dfrac{2.1+3.2+(-5)(-2)-9}{\sqrt{169}}\right|\)
=\(\left|\dfrac{18-9}{\sqrt {9}}\right|\)
=\(\left|\dfrac{9}{3}\right|\)
=3
(d)
Here \(\vec a = -6\hat i +0\hat j - 0\hat k\)
\(\vec N = 2\hat i -3\hat j - 6\hat k\)
The distance of the poin (-6,0,0) from the given plane 2x -3y +6z - 2 = 0 is given by
\(\left|\dfrac{\vec a. \vec N - d}{|\vec N|}\right|\)
=\(\left|\dfrac{(-6\hat i +0\hat j - 0\hat k). ( 2\hat i -3\hat j - 6\hat k) + 9}{\sqrt{2^2+(-3)^2+6^2}}\right|\)
=\(\left|\dfrac{(-6).2+0(-3)+0.6-2}{\sqrt{4+9+36}}\right|\)
=\(\left|\dfrac{-12-2}{\sqrt {49}}\right|\)
=\(\left|\dfrac{-14}{7}\right|\)
=\(\dfrac{14}{7}\)
=2
\(\vec N = 1\hat i + 2\hat j - 2\hat k\)
The distance of the poin (2,3,-5) from the given plane x +2y -2z =9 is given by
\(\left|\dfrac{\vec a. \vec N - d}{|\vec N|}\right|\)
=\(\left|\dfrac{(2\hat i +3\hat j - 5\hat k). ( 1\hat i + 2\hat j - 2\hat k) + 9}{\sqrt{1^2+2^2+(-2)^2}}\right|\)
=\(\left|\dfrac{2.1+3.2+(-5)(-2)-9}{\sqrt{169}}\right|\)
=\(\left|\dfrac{9}{3}\right|\)
=3
(d)
Here \(\vec a = -6\hat i +0\hat j - 0\hat k\)
\(\vec N = 2\hat i -3\hat j - 6\hat k\)
The distance of the poin (-6,0,0) from the given plane 2x -3y +6z - 2 = 0 is given by
\(\left|\dfrac{\vec a. \vec N - d}{|\vec N|}\right|\)
=\(\left|\dfrac{(-6\hat i +0\hat j - 0\hat k). ( 2\hat i -3\hat j - 6\hat k) + 9}{\sqrt{2^2+(-3)^2+6^2}}\right|\)
=\(\left|\dfrac{(-6).2+0(-3)+0.6-2}{\sqrt{4+9+36}}\right|\)
=\(\left|\dfrac{-14}{7}\right|\)
=\(\dfrac{14}{7}\)
=2
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