**Q.5 a)** Show that the plane touches the sphere . Also find the point of contact.

**Solution** : The sphere has center at the point and its radius is equal to

Length of perpendicular from the center to the plane will be

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

Hence,

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

Therefore,

Substituting in the coordinates, we get the point of contact as

**Q.5 b)**** **Find the equation of right circular cone passing through with vertex at origin and axis parallel to the line

**Solution** : Let the vertex be denoted by and the point by . So,

Since parallel lines have same direction ratios, the axis of the cone will also have direction ratios . Let be the vector along the axis, then

Let be the semi-vertical angle of cone. By the definition of dot product of vectors,

This gives

Having obtained , we can go for the equation of right circular cone. Note that knowing of the semi-vertical angle is sufficient.

Let be any point on the cone. Then,

Using the dot product of vectors, we can write

Taking the dot product and squaring both sides, we get

Expanding the bracket,

On further simplifying, the required equation of cone is

**Q.5 c)** Find the equation of right circular cylinder, whose axis is and radius is units.

**Solution** : The axis of cylinder is . Making the adjustments to bring these equations into general equations of a line in 3 dimensions, we get,

The direction ratios of the axis are and . Now, we need to find a point on the axis. (We can arbitrarily choose any value of and get and from above equations. The most trivial value of is . This gives and .) So, we got a point on the axis, which is the origin itself, .

Let be any point on the cylinder. We drop a perpendicular from on the axis. Let the foot be . has length units, because it is the radius of the cylinder.

is a right angled triangle, with . Let be . By property of right angled triangles,

, so . Once we get in terms of , we can go ahead with above equation.

Let be the vector along the axis of the cylinder (which has direction ratios ). Multiplying each direction ratio by same non-zero number will still give us direction ratios. So, (We multiplied by , so the fraction is gone, which could have possibly made the task tougher.)

The unit vector along the axis,

.

We now consider the dot product .

Using this in ,

On expanding and simplifying,

This is the required equation.

**Q.6 a)** Find the equation of the sphere which has its center at and touches the line

**Solution** : The tangent line is

Any point on this line can be represented by the coordinates .

Let be the center of sphere and be the point of contact. Then,

Let be the vector along the line. Its components are the direction ratios of the line, i.e. .

Since is perpendicular to the line,

So, coordinates of are .

Distance is the radius of the sphere and using distance formula,

The required equation of sphere in center-radius form is

**Q.6 b)** Find the equation of the cone with vertex at , semi-vertical angle and the line

as the axis of the cone.

**Solution** : Let be the vertex of cone and be the semi-vertical angle. Let be any point on the cone. Since the axis of the cone is , the direction ratios are . Let be the vector along the axis. Then,

Using the definition of dot product,

Squaring both the sides,

LHS is . Let’s expand it separately :

Equating 2 sides and taking all terms on one side,

This is the required equation.

**Q.7 a)** Evaluate the following over the region .

**Solution**** **The region of integration is plotted below :

Clearly, to the left of line , the region is absent, hence the lower limit on will be . The upper limit will be . No restriction on the highest value of .

Let’s now consider a strip parallel to axis. It will first intersect the region at . So, the lower limit on is . The upper limit will be because no curve bounds the region in that direction. Hence, given integral will be

Since inner limits are of , let’s first integrate w.r.t. .