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.**A plane passing through (1, 1, 1) cuts positive direction of co-ordinate axes at A,B and C, then the volume of tetrahedron OABC satisfies

Solution

(b) Let the equation of the plane be x/a+y/b+z/c=1 ⇒1/a+1/b+1/c=1 ⇒ Volume of tetrahedron OABC=V=1/6 (abc) Now (abc)^(1/3)≥3/(1/a+1/b+1/c)≥3 (G.M.≥H.M.) ⇒abc≥27 ⇒V≥9/2

(b) Let the equation of the plane be x/a+y/b+z/c=1 ⇒1/a+1/b+1/c=1 ⇒ Volume of tetrahedron OABC=V=1/6 (abc) Now (abc)^(1/3)≥3/(1/a+1/b+1/c)≥3 (G.M.≥H.M.) ⇒abc≥27 ⇒V≥9/2

**Q2.**The direction cosines of a line satisfy the relations Î»(l+m)=n and mn+nl+lm=0. The value of Î», for which the two lines are perpendicular to each other, is

Solution

(b) Eliminating n, we get Î»(l+m)^2+lm=0 ⇒(Î»l^2)/m^2 +(2Î»+1)l/m+Î»=0 ⇒(l_1 l_2)/(m_1 m_2 )=1 (product of roots l_1/m_1 and l_2/m_2 ) Where l_1/m_1 and l_2/m_2 are the roots of this equation, further eliminating m, we get Î»l^2-ln-n^2=0 ⇒(l_1 l_2)/(n_1 n_2 )=-1/Î» Since the lines with direction cosines (l_1,m_1,n_1) and (l_2,m_2,n_2) are perpendicular, we have l_1 l_2+m_1 m_2+n_1 n_2=0 ⇒1+1-Î»=0 ⇒ Î»=2

(b) Eliminating n, we get Î»(l+m)^2+lm=0 ⇒(Î»l^2)/m^2 +(2Î»+1)l/m+Î»=0 ⇒(l_1 l_2)/(m_1 m_2 )=1 (product of roots l_1/m_1 and l_2/m_2 ) Where l_1/m_1 and l_2/m_2 are the roots of this equation, further eliminating m, we get Î»l^2-ln-n^2=0 ⇒(l_1 l_2)/(n_1 n_2 )=-1/Î» Since the lines with direction cosines (l_1,m_1,n_1) and (l_2,m_2,n_2) are perpendicular, we have l_1 l_2+m_1 m_2+n_1 n_2=0 ⇒1+1-Î»=0 ⇒ Î»=2

**Q3.**The plane 4x+7y+4z+81=0 is rotated through a right angle about its line of intersection with the plane 5x+3y+10z=25. The equation of the plane in its new position is

Solution

(a) The equation of the plane through the line of intersection of the plane 4x+7y+4z+81=0 and 5x+3y+10z=25 is (4x+7y+4z+81)+Î»(5x+3y+10z-25)=0 ⇒(4+5Î»)x+(7+3Î»)y+(4+10Î»)z+81-25Î»=0 (i) Which is perpendicular to 4x+7y+4z+81=0 ⇒4(4+5Î»)+7(7+3Î»)+4(4+10Î»)=0 ⇒81Î»+81=0 ⇒Î»=-1 Hence the place is x-4y+6z=106

(a) The equation of the plane through the line of intersection of the plane 4x+7y+4z+81=0 and 5x+3y+10z=25 is (4x+7y+4z+81)+Î»(5x+3y+10z-25)=0 ⇒(4+5Î»)x+(7+3Î»)y+(4+10Î»)z+81-25Î»=0 (i) Which is perpendicular to 4x+7y+4z+81=0 ⇒4(4+5Î»)+7(7+3Î»)+4(4+10Î»)=0 ⇒81Î»+81=0 ⇒Î»=-1 Hence the place is x-4y+6z=106

**Q4.**A line makes an angle Î¸ with each of the x- and z-axes. If the angle Î², which it makes with y-axis, is such that sin^2Î²=3 sin^2Î¸, then cos^2Î¸ equals

Solution

(c) Here sin^2Î²=3 sin^2Î¸ (i) By the question, cos^2Î¸+cos^2Î¸+cos^2Î²=1 (ii) ⇒ cos^2Î²=1-2 cos^2Î¸ (iii) Adding (i) and (iii), we get 1=1+3 sin^2Î¸-2 cos^2Î¸ ⇒1=1+3(1-cos^2Î¸ )-2 cos^2Î¸ ⇒5 cos^2Î¸=3 ⇒cos^2Î¸=3/5

(c) Here sin^2Î²=3 sin^2Î¸ (i) By the question, cos^2Î¸+cos^2Î¸+cos^2Î²=1 (ii) ⇒ cos^2Î²=1-2 cos^2Î¸ (iii) Adding (i) and (iii), we get 1=1+3 sin^2Î¸-2 cos^2Î¸ ⇒1=1+3(1-cos^2Î¸ )-2 cos^2Î¸ ⇒5 cos^2Î¸=3 ⇒cos^2Î¸=3/5

**Q6.**In a three dimensional co-ordinate system, P,Q and R are images of a point A(a,b,c) in the x-y, y-z and z-x planes, respectively. If G is the centroid of triangle PQR, then area of triangle AOG is (O is the origin)

Solution

(a) Point A is (a,b,c)⇒ Points P,Q,R are (a,b,-c),(-a,b,c) and (a,-b,c), respectively ⇒ Centroid of triangle PQR is (a/3,b/3,c/3)⇒G≡(a/3,b/3,c/3) ⇒A,O,G are collinear ⇒ area of triangle AOG is zero

(a) Point A is (a,b,c)⇒ Points P,Q,R are (a,b,-c),(-a,b,c) and (a,-b,c), respectively ⇒ Centroid of triangle PQR is (a/3,b/3,c/3)⇒G≡(a/3,b/3,c/3) ⇒A,O,G are collinear ⇒ area of triangle AOG is zero

**Q7.**The lines which intercept the skew lines y=mx,z=z;y=-mx,z=-c and the x-axis lie on the surface

Solution

(c) Equating of the planes through y=mx,z=c and y=-mx,z=-c are respectively, (y-mx)+Î»_1 (z-c)=0 (i) and (y+mx)+Î»_2 (z+c)=0 (ii) It meets at x-axis, i.e., y=0=z ∴ Î»_2=Î»_1 From (i) and (ii), (y-mx)/(z-c)=(y+mx)/(z+c) ∴cy=mzx

(c) Equating of the planes through y=mx,z=c and y=-mx,z=-c are respectively, (y-mx)+Î»_1 (z-c)=0 (i) and (y+mx)+Î»_2 (z+c)=0 (ii) It meets at x-axis, i.e., y=0=z ∴ Î»_2=Î»_1 From (i) and (ii), (y-mx)/(z-c)=(y+mx)/(z+c) ∴cy=mzx

**Q8.**Equation of a line in the plane Ï€≡2x-y+z-4=0 which is perpendicular to the line l whose equation is (x-2)/1=(y-2)/(-1)=(z-3)/(-2) and which passes through the point of intersection of l and Ï€ is

Solution

(b) Let direction ratios of the line be (a,b,c), then 2a-b+c=0 and a-b-2c=0 i.e, a/3=b/5=c/(-1) Therefore, direction ratios of the line are (3,5,-1) Any point on the given line is (2+Î»,2-Î»,3-2Î»), it lies on the given plane Ï€ if 2(2+Î»)-(2-Î»)+(3-2Î»)=4 ⇒4+2Î»-2+Î»+3-2Î»=4 ⇒Î»=-1 Therefore, the point of intersection of the line and the plane is (1,3, 5) Therefore, equation of the required line is (x-1)/3=(y-3)/5=(z-5)/(-1)

(b) Let direction ratios of the line be (a,b,c), then 2a-b+c=0 and a-b-2c=0 i.e, a/3=b/5=c/(-1) Therefore, direction ratios of the line are (3,5,-1) Any point on the given line is (2+Î»,2-Î»,3-2Î»), it lies on the given plane Ï€ if 2(2+Î»)-(2-Î»)+(3-2Î»)=4 ⇒4+2Î»-2+Î»+3-2Î»=4 ⇒Î»=-1 Therefore, the point of intersection of the line and the plane is (1,3, 5) Therefore, equation of the required line is (x-1)/3=(y-3)/5=(z-5)/(-1)

**Q9.**The Cartesian equation of the plane r =(1+Î»-Î¼) i ̂+(2-Î») j ̂+(3-2Î»+2Î¼)k ̂ is

Solution

(c) Given plane is r =(1+Î»-Î¼) i ̂+(2-Î») j ̂+(3-2Î»+2Î¼)k ̂ ⇒ r=(i ̂+2j ̂+3k ̂ )+Î»(i ̂-j ̂-2k ̂ )+Î¼(-i ̂+2k ̂) Which is a plane passing through a =i ̂+2j ̂+3k ̂ and parallel to the vectors b =i ̂-j ̂-2k ̂ and c ⃗=-i ̂+2k ̂ Therefore, it is perpendicular to the vector n =b ⃗×c =-2i ̂-k ̂ Hence, equation of plane is -2(x-1)+(0)(y-2)-(z-3)=0 or 2x+z=5

(c) Given plane is r =(1+Î»-Î¼) i ̂+(2-Î») j ̂+(3-2Î»+2Î¼)k ̂ ⇒ r=(i ̂+2j ̂+3k ̂ )+Î»(i ̂-j ̂-2k ̂ )+Î¼(-i ̂+2k ̂) Which is a plane passing through a =i ̂+2j ̂+3k ̂ and parallel to the vectors b =i ̂-j ̂-2k ̂ and c ⃗=-i ̂+2k ̂ Therefore, it is perpendicular to the vector n =b ⃗×c =-2i ̂-k ̂ Hence, equation of plane is -2(x-1)+(0)(y-2)-(z-3)=0 or 2x+z=5

**Q10.**The coordinates of the point P on the line r =(i ̂+j ̂+k ̂ )+Î»(-i ̂+j ̂-k ̂) which is nearest to the origin is

Solution

(a) Let the point P be (x,y,z), then the vector (xi ̂+yj ̂+zk ̂ ) will lie on the line ⇒(x-1) i ̂+(y-1) j ̂+(z-1) k ̂=-Î»i ̂+Î»j ̂-Î»k ̂ ⇒x=1-Î»,y=1+Î» and z=1-Î» Now point P is nearest to the origin ⇒D=(1-Î»)^2+(1+Î»)^2+(1-Î»)^2 ⇒dD/d=-4(1-Î»)+2(1+Î»)=0 ⇒ Î»=1/3 ⇒ The point is (2/3,4/3,2/3)

(a) Let the point P be (x,y,z), then the vector (xi ̂+yj ̂+zk ̂ ) will lie on the line ⇒(x-1) i ̂+(y-1) j ̂+(z-1) k ̂=-Î»i ̂+Î»j ̂-Î»k ̂ ⇒x=1-Î»,y=1+Î» and z=1-Î» Now point P is nearest to the origin ⇒D=(1-Î»)^2+(1+Î»)^2+(1-Î»)^2 ⇒dD/d=-4(1-Î»)+2(1+Î»)=0 ⇒ Î»=1/3 ⇒ The point is (2/3,4/3,2/3)