Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.

**Q1.**The shortest distance from the plane 12x+y+3z=327 to the sphere x^2+y^2+z^2+4x-2y-6z=155 is

**Q2.**The angle between i ̂ line of the intersection of the plane r ∙(i ̂+2j ̂+3k ̂ )=0 and r ∙(3i ̂+3j ̂+k ̂ )=0, is

Solution

(d) Line of intersection of r ∙(i ̂+2j ̂+3k ̂ )=0 and r ∙(3i ̂+3j ̂+k ̂ )=0 will be parallel to (3i ̂+3j ̂+k ̂ )×(i ̂+2j ̂+3k ̂ ), i.e., 7i ̂-8j ̂+3k ̂ If the required angle is Î¸, then cosÎ¸=7/√(49+64+9)=7/√122

(d) Line of intersection of r ∙(i ̂+2j ̂+3k ̂ )=0 and r ∙(3i ̂+3j ̂+k ̂ )=0 will be parallel to (3i ̂+3j ̂+k ̂ )×(i ̂+2j ̂+3k ̂ ), i.e., 7i ̂-8j ̂+3k ̂ If the required angle is Î¸, then cosÎ¸=7/√(49+64+9)=7/√122

**Q4.**Let PM be the perpendicular from the point P(1,2,3) to the x-y plane. If (OP) makes an angle Î¸ with the positive direction of the z-axis and (OM) makes an angle Ï• with the positive direction of x-axis, where O is the origin and Î¸ and Ï• are acute angles, then

Solution

(d) If P be (x,y,z), then from the figure x=r sinÎ¸ cosÏ•,y=r sinÎ¸ sinÏ• and z=r cosÎ¸ 1=r sinÎ¸ cosÏ•,2=r sinÎ¸ sinÏ• and 3=r cosÎ¸ ⇒1^2+2^2+3^2=r^2 ⇒r=±√14 ∴ sinÎ¸ cosÏ•=1/√14,sinÎ¸ sinÏ•=2/√14 and cosÎ¸=3/√14 (neglecting negative sign as Î¸ and Ï• are acute) (sinÎ¸ sinÏ•)/(sinÎ¸ cosÏ• )=2/1 ⇒tanÏ•=2 Also, tanÎ¸=√5/3

(d) If P be (x,y,z), then from the figure x=r sinÎ¸ cosÏ•,y=r sinÎ¸ sinÏ• and z=r cosÎ¸ 1=r sinÎ¸ cosÏ•,2=r sinÎ¸ sinÏ• and 3=r cosÎ¸ ⇒1^2+2^2+3^2=r^2 ⇒r=±√14 ∴ sinÎ¸ cosÏ•=1/√14,sinÎ¸ sinÏ•=2/√14 and cosÎ¸=3/√14 (neglecting negative sign as Î¸ and Ï• are acute) (sinÎ¸ sinÏ•)/(sinÎ¸ cosÏ• )=2/1 ⇒tanÏ•=2 Also, tanÎ¸=√5/3

**Q5.**The equation of the line x+y+z-1=0 and 4x+y-2z+2=0 written in the symmetrical form is

Solution

(b) x+y+z-1=0 4x+y-2z+2=0 Therefore, the line is along the vector (i ̂+j ̂+k ̂ )×(4i ̂+j ̂-2k ̂ )=3i ̂-6j ̂+3k ̂ Let z=k. Then x=k-1 and y=2-2k Therefore, (k-1,2-2k,k) is any point on the line Hence, (-1,2,0),(0,0,1) and (-1/2,1,1/2) are the points on the line

(b) x+y+z-1=0 4x+y-2z+2=0 Therefore, the line is along the vector (i ̂+j ̂+k ̂ )×(4i ̂+j ̂-2k ̂ )=3i ̂-6j ̂+3k ̂ Let z=k. Then x=k-1 and y=2-2k Therefore, (k-1,2-2k,k) is any point on the line Hence, (-1,2,0),(0,0,1) and (-1/2,1,1/2) are the points on the line

**Q6.**Consider a set of points R in the space which is at a distance of 2 units from the line x/1=(y-1)/(-1)=(z+2)/2 between the planes x-y+2z+3=0 and x-y+2z-2=0

Solution

(b) Distance between the planes is h=5/√6 Also the figure formed is cylinder, whose radius is r=2 units Hence, the volume of the cylinder is Ï€r^2 h=Ï€(2)^2.5/√6=20Ï€/√6 cubic units Also the curved surface area is 2Ï€rh=2Ï€(2).5/√6=20Ï€/√6

(b) Distance between the planes is h=5/√6 Also the figure formed is cylinder, whose radius is r=2 units Hence, the volume of the cylinder is Ï€r^2 h=Ï€(2)^2.5/√6=20Ï€/√6 cubic units Also the curved surface area is 2Ï€rh=2Ï€(2).5/√6=20Ï€/√6

**Q7.**A rod of length 2 units whose one end is (1,0,-1) and other end touches the plane x-2y+2z+4=0, then

Solution

(d) The rod sweeps out the figure which is a cone The distance of point A(1,0,-1) from the plane is (|1-2+4|)/√9=1 unit The slant height l of the cone is 2 units Then the radius of the base of the cone is √(l^2-1)=√(4-1)=√3 Hence, the volume of the cone is Ï€/3 (√3)^2.1=Ï€ cubic units Area of the circle on the plane which the of traces is 3Ï€ Also, the centre of the circle is Q(x,y,z). Then (x-1)/1=(y-0)/(-2)=(z+1)/2=(-(1-0-2+4))/(1^2+(-2^2)+2^2 ), or Q(x,y,z)≡(2/3,2/3,(-5)/3)

(d) The rod sweeps out the figure which is a cone The distance of point A(1,0,-1) from the plane is (|1-2+4|)/√9=1 unit The slant height l of the cone is 2 units Then the radius of the base of the cone is √(l^2-1)=√(4-1)=√3 Hence, the volume of the cone is Ï€/3 (√3)^2.1=Ï€ cubic units Area of the circle on the plane which the of traces is 3Ï€ Also, the centre of the circle is Q(x,y,z). Then (x-1)/1=(y-0)/(-2)=(z+1)/2=(-(1-0-2+4))/(1^2+(-2^2)+2^2 ), or Q(x,y,z)≡(2/3,2/3,(-5)/3)

**Q8.**Consider the planes 3x-6y+2z+5=0 and 4x-12y+3z=3. The plane 67x-162y+47z+44=0 bisects the angle between the given planes which

Solution

(b) 3x-6y+2z+5=0 (i) -4x+12y-3z+3=0 (ii) Bisectors are (3x-6y+2z+5)/√(9+36+4)=±(-4x+12y-3z+3)/√(16+144+9) The plane which bisects the angle between the planes that contains the origin 13(3x-6y+2z+5)=7(-4x+12y-3z+3) 67x-162y+47z+44=0 (iii) Further, 3×(-4)+(-6)(12)+2×(-3)<0 Hence, the origin lies in acute angle

(b) 3x-6y+2z+5=0 (i) -4x+12y-3z+3=0 (ii) Bisectors are (3x-6y+2z+5)/√(9+36+4)=±(-4x+12y-3z+3)/√(16+144+9) The plane which bisects the angle between the planes that contains the origin 13(3x-6y+2z+5)=7(-4x+12y-3z+3) 67x-162y+47z+44=0 (iii) Further, 3×(-4)+(-6)(12)+2×(-3)<0 Hence, the origin lies in acute angle