# M II May 2016

Q.5 a) Show that the plane ${2x-2y+ z+12=0}$ touches the sphere ${x^2+y^2+z^2-2x-4y+2z-3=0}$. Also find the point of contact.

Solution : The sphere has center at the point ${(1,2,-1)}$ and its radius is equal to

${\sqrt {(1)^2 + (2)^2 + (-1)^2 - (-3)} = \sqrt {1+4+1+3} = \sqrt {9} = 3 \ units}$

Length of perpendicular from the center to the plane will be

${\begin {vmatrix} \frac {2(1) -2 (2)+ 1(-1) =12}{\sqrt {2^2 + (-2)^2 + 1(1)^2}} \end {vmatrix} = \begin {vmatrix} \frac {2-4-1+12}{\sqrt {9}} \end {vmatrix} = \frac {9}{3} = 3 \ units}$

Thus the length of perpendicular from the center to the plane is equal to the radius of sphere. This proves that the plane touches the sphere.

The equation of line through the center and normal to the plane is given by

${\frac {x-1}{2} = \frac {y-2}{-2} = \frac {z+1}{1} = \lambda}$

Hence,

${x = 2 \lambda + 1, \ y= -2 \lambda +2, \ z= \lambda-1}$

If this point lies on the plane, its coordinates must satisfy the equation of the plane. Therefore,

${2(2 \lambda+1)-2 (-2 \lambda+2) + 1 (\lambda-1)+12 =0}$

Therefore,

${4 \lambda +2+4 \lambda -4+ \lambda -1+12=0, \ i.e. \ 9 \lambda+9 =0, \ i.e. \ \lambda =-1}$

Substituting ${\lambda}$ in the coordinates, we get the point of contact as

${(x,y,z) = \big ( 2(-1)+1, \ -2(-1)+2 , \ (-1)-1 \big ) = \big (-1,4,-2 \big )}$

Q.5 b) Find the equation of right circular cone passing through ${(2,-2,1)}$ with vertex at origin and axis parallel to the line

${\frac {x-2}{5} = \frac {y-1}{1} = \frac {z+2}{1}}$

Solution : Let the vertex ${(0,0,0)}$ be denoted by ${V}$ and the point ${(2,-2,1)}$ by ${A}$. So,

${\vec {VA} = 2 \hat i - 2 \hat j + \hat k}$

Since parallel lines have same direction ratios, the axis of the cone will also have direction ratios ${5,1,1}$. Let ${\vec d}$ be the vector along the axis, then

${\vec d = 5 \hat i + \hat j + \hat k}$

Let ${\alpha}$ be the semi-vertical angle of cone. By the definition of dot product of vectors,

${\vec {VA} \cdot \vec {d} = |\vec {VA}| |\vec d| cos (\alpha)}$

This gives

${(2 \hat i - 2 \hat j + \hat k) \cdot (5 \hat i + \hat j + \hat k) = \sqrt {2^2 + (-2)^2 + (1)^2} \sqrt {(5)^2 +(1)^2 + (1)^2} \ cos (\alpha)}$

${10-2+1 = \sqrt {9} \sqrt {27} cos (\alpha)}$

${cos (\alpha) = \frac {10-2+1}{ \sqrt {9} \sqrt {27}} = \frac {9}{3 \times 3 \sqrt 3} = \frac {1}{\sqrt {3}}}$

Having obtained ${cos (\alpha)}$, we can go for the equation of right circular cone. Note that knowing ${cosine}$ of the semi-vertical angle is sufficient.

Let ${P(x,y,z)}$ be any point on the cone. Then,

${\vec {VP} = x \hat i + y\hat j + z \hat k}$

Using the dot product of vectors, we can write

${\vec {VP} \cdot \vec d = |\vec {VP}| |\vec d| cos (\alpha)}$

${(x \hat i + y \hat j + z \hat k) \cdot (5 \hat i + \hat j + \hat k) = \sqrt {x^2 + y^2 + z^2} \sqrt {(5)^2 +(1)^2 + (1)^2} \times \frac {1}{\sqrt 3}}$

Taking the dot product and squaring both sides, we get

${(5x+y+z)^2 = (x^2+y^2+z^2) \times 27 \times \frac 1 3 = (x^2+y^2+z^2) \times 9}$

Expanding the bracket,

${25x^2+y^2+z^2+10xy+2yz+10zx = 9x^2+9y^2+9z^2}$

On further simplifying, the required equation of cone is

${16x^2-8y^2-8z^2+10xy + 2yz + 10zx =0}$

Q.5 c) Find the equation of right circular cylinder, whose axis is ${x=2y=-z}$ and radius is ${4}$ units.

Solution : The axis of cylinder is ${x=2y=-z}$. Making the adjustments to bring these equations into general equations of a line in 3 dimensions, we get,

${\frac {x}{1} = \frac {y}{1/2} = \frac {z}{-1}}$

The direction ratios of the axis are ${1, \frac 1 2}$ and ${-1}$. Now, we need to find a point on the axis. (We can arbitrarily choose any value of ${x}$ and get ${y}$ and ${z}$ from above equations. The most trivial value of ${x}$ is ${0}$. This gives ${y=0}$ and ${z=0}$.) So, we got a point on the axis, which is the origin itself, ${O(0,0,0)}$.

Let ${P(x,y,z)}$ be any point on the cylinder. We drop a perpendicular from ${P}$ on the axis. Let the foot be ${M}$. ${PM}$ has length ${4}$ units, because it is the radius of the cylinder.

${\Delta OPM}$ is a right angled triangle, with ${\angle M = 90^o}$. Let ${\angle POM}$ be ${\theta}$. By property of right angled triangles,

${OP^2 = OM^2 + PM^2}$

${\therefore x^2+y^2+z^2 = OM^2 + 4^2}$

${cos (\theta) = \frac {OM}{OP}}$, so ${OM = OP cos (\theta)}$. Once we get ${OM}$ in terms of ${x,y,z}$, we can go ahead with above equation.

Let ${\vec d}$ be the vector along the axis of the cylinder (which has direction ratios ${1, \frac 1 2, -1}$). Multiplying each direction ratio by same non-zero number will still give us direction ratios. So, ${\vec d = 2 \hat i + \hat j - \hat k}$ (We multiplied by ${2}$, so the fraction ${1/2}$ is gone, which could have possibly made the task tougher.)

The unit vector along the axis,

${\hat d = \frac {\vec d}{|\vec d|} = \frac {2 \hat i + \hat j - \hat k}{\sqrt {2^2 + 1^2 + (-1)^2}} = \frac 2 3 \hat i + \frac 1 3 \hat j - \frac 2 3 \hat k}$.

We now consider the dot product ${\vec {OP} \cdot \hat d}$.

${\vec {OP} \cdot \hat d = |\vec {OP}| |\hat d| cos (\theta) = OP \times 1 \times cos (\theta) = OP \ cos (\theta) = OM}$

${\therefore \ OM = \vec {OP} \cdot {\hat d} = (x \hat i + y \hat j + z \hat k) \cdot \left (\frac 2 3 \hat i + \frac 1 3 \hat j - \frac 2 3 \hat k \right ) = \frac {2x}{3} + \frac {y}{3} - \frac {2z}{3}}$

Using this in ${OP^2 = OM^2 + PM^2}$,

${x^2+y^2+z^2 = \left (\frac {2x}{3} + \frac {y}{3} - \frac {2z}{3} \right)^2 + 16}$

On expanding and simplifying,

${x^2 + y^2 + z^2 -16 = \frac 1 9 \left (2x+y-2z \right)^2}$

${\therefore 9x^2+9y^2 + 9z^2 - 144 = 4x^2+y^2+4z^2 + 4xy - 4yz - 8zx}$

${\therefore 5x^2+8y^2+5z^2 - 4xy + 4yz + 8zx - 144 =0}$

This is the required equation.

Q.6 a) Find the equation of the sphere which has its center at ${(2,3,-1)}$ and touches the line

${\frac {x+1}{-5}= \frac {y-8}{3} = \frac {z-4}{4}}$

Solution : The tangent line is

${\frac {x+1}{-5}= \frac {y-8}{3} = \frac {z-4}{4} = \lambda, \ \lambda \in \mathbb R}$

Any point on this line can be represented by the coordinates ${(-5 \lambda -1, \ 3 \lambda +8, \ 4 \lambda + 4)}$.

Let ${C (2,3,-1)}$ be the center of sphere and ${M}$ be the point of contact. Then,

${\vec {CM} = (-5 \lambda -1 -2)\hat i + (3 \lambda + 8 - 3) \hat j + (4 \lambda + 4 - \ -1) \hat k }$

${=(-5 \lambda - 3) \hat i + (3 \lambda + 5) \hat j + (4 \lambda + 5) \hat k}$

Let ${\vec d}$ be the vector along the line. Its components are the direction ratios of the line, i.e. ${\vec d = -5 \hat i + 3 \hat j + 4 \hat k}$.

Since ${CM}$ is perpendicular to the line,

${\vec {CM} \cdot \vec d =0}$

${\therefore \left [(-5 \lambda - 3) \hat i + (3 \lambda + 5) \hat j + (4 \lambda + 5) \hat k \right ] \cdot (-5 \hat i + 4 \hat j + 3 \hat k) = 0}$

${\therefore 25 \lambda + 15 + 9 \lambda + 15 + 16 \lambda + 20 = 0}$

${\therefore 50 \lambda + 50 = 0}$

${\therefore \lambda = -1}$

So, coordinates of ${M}$ are ${(-5 \times -1 -1, \ 3 \times -1 + 8, \ 4 \times -1 + 4) = (4,5,0)}$.

Distance ${CM}$ is the radius of the sphere and using distance formula,

${CM^2 = (4-2)^2 + (5-3)^2 + (0+1)^2 = 2^2+2^2+1^2 = 9}$

The required equation of sphere in center-radius form is

${(x-2)^2 + (y-3)^2 + (z+1)^2 = 9}$

Q.6 b) Find the equation of the cone with vertex at ${(1,2,-3)}$, semi-vertical angle ${cos^{-1} \frac {1}{\sqrt 3}}$ and the line

${\frac {x-1}{1}= \frac {y-2}{2} + \frac {z+1}{-1}}$

as the axis of the cone.

Solution : Let ${V (1,2,-3)}$ be the vertex of cone and ${\alpha}$ be the semi-vertical angle. Let ${P (x,y,z)}$ be any point on the cone. Since the axis of the cone is ${\frac {x-1}{1}= \frac {y-2}{2} + \frac {z+1}{-1}}$, the direction ratios are ${1,2,-1}$. Let ${\vec d}$ be the vector along the axis. Then,

${\vec d = \hat i + 2 \hat j - \hat k}$

Using the definition of dot product,

${\vec {VX} \cdot \vec d = |\vec {VX}| |\vec d| cos (\alpha)}$

${\therefore \left ( (x-1)\hat i + (y-2) \hat j + (z+3) \hat k \right ) \cdot (\hat i + 2 \hat j - \hat k)}$

${= \sqrt {(x-1)^2 + (y-2)^2 + (z+3)^2} \times \sqrt {1^2 + 2^2 + (-3)^2} \times \frac {1}{\sqrt 3}}$

${\therefore x-1 + 2 (y-2) + (-1) (z+3) = \sqrt {(x-1)^2 + (y-2)^2 + (z+3)^2} \times \sqrt {1+4+1} \times \frac {1}{\sqrt 3}}$

Squaring both the sides,

${\left ( x-1+2y-4-z-3 \right)^2 = \left (x^2-2x+1+y^2-4y+4+z^2+6z+9 \right ) \times 6 \times \frac 1 3}$

LHS is ${(x+2y-z-8)^2}$. Let’s expand it separately :

${LHS = (x+2y)^2 + (-z-8)^2 + 2 (x+2y)(-z-8)}$

${= x^2 + 4xy + 4y^2 + z^2 +16z + 64 - 2zx - 16x-2yz -16y}$

${RHS = 2x^2 + 2y^2 + 2z^2 - 4x - 8y + 12z + 28}$

Equating 2 sides and taking all terms on one side,

${x^2-2y^2+z^2+2xz+2yz-4xy+12x+8y-4z-36=0}$

This is the required equation.

Q.7 a)  Evaluate the following over the region ${y \ge x^2, x \ge 1}$.

${\iint \frac {1}{x^4+y^2} dxdy}$

Solution The region of integration is plotted below :

Clearly, to the left of line ${x=1}$, the region is absent, hence the lower limit on ${x}$ will be ${1}$. The upper limit will be ${\infty}$. No restriction on the highest value of ${x}$.

Let’s now consider a strip parallel to ${Y}$ axis. It will first intersect the region at ${y=x^2}$. So, the lower limit on ${y}$ is ${x^2}$. The upper limit will be ${\infty}$ because no curve bounds the region in that direction. Hence, given integral will be

${\int \limits_{x=1}^{\infty} \int \limits_{y=x^2}^{\infty} \frac {1}{x^4+y^2} dydx = I \ say}$

Since inner limits are of ${y}$, let’s first integrate w.r.t. ${y}$.

${I = \int \limits_{x=1}^{\infty} \left [\frac {1}{x^2} tan^{-1} \left (\frac {y}{x^2} \right) \right]_{x^2}^{\infty}dx}$

${= \int \limits_{x=1}^{\infty} \frac {1}{x^2} \left[ \frac {\pi}{2} - \frac {\pi}{4} \right] dx}$

${= \frac {\pi}{4} \int \limits_1^{\infty} x^{-2}dx}$

${= \frac {\pi}{4} \left [\frac {x^{-2+1}}{-2+1} \right ]_{1}^{\infty}}$

${= \frac {\pi}{4} \times \frac {1}{-1} \times \left[\frac 1 x \right]_1^{\infty}}$

${= \frac {\pi}{4} \times -1 \times [0 - 1]}$

${= \frac {\pi}{4}}$