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 between the lines (x-3)/3=(y-8)/(-1)=(z-3)/1 and (x+3)/(-3)=(y+7)/2=(z–6)/4 is

Solution

(d) Given lines are r =3i ̂+8j ̂+3k ̂+l(3i ̂-j ̂+k ̂) and r =-3i ̂-7j ̂+6k ̂+m(-3i ̂+2j ̂+4k ̂) Required shortest distance =(|(6i ̂+15j ̂-3k ̂ ).((3i ̂-j ̂+k ̂)×@(-3i ̂+2j ̂+4k ̂))|)/(|(3i ̂-j ̂+k ̂)×(-3i ̂+2j ̂+4k ̂)|) =(|(6i ̂+15j ̂-3k ̂ ).(-6i ̂-15j ̂+3k ̂ )|)/(|(-6i ̂-15j ̂+3k ̂)|) =(36+225+9)/√(36+225+9)=270/√270=√270=3√30

(d) Given lines are r =3i ̂+8j ̂+3k ̂+l(3i ̂-j ̂+k ̂) and r =-3i ̂-7j ̂+6k ̂+m(-3i ̂+2j ̂+4k ̂) Required shortest distance =(|(6i ̂+15j ̂-3k ̂ ).((3i ̂-j ̂+k ̂)×@(-3i ̂+2j ̂+4k ̂))|)/(|(3i ̂-j ̂+k ̂)×(-3i ̂+2j ̂+4k ̂)|) =(|(6i ̂+15j ̂-3k ̂ ).(-6i ̂-15j ̂+3k ̂ )|)/(|(-6i ̂-15j ̂+3k ̂)|) =(36+225+9)/√(36+225+9)=270/√270=√270=3√30

**Q2.**If a line makes an angle of Ï€/4 with the positive direction of each of x-axis and y-axis, then the angle that the line makes with the positive direction of the z-axis is

Solution

(c) Here l=cos Ï€/4,m=cosÏ€/4 Let the line make an angle ‘Î³’ with z-axis ∴l^2+m^2+n^2=1 ⇒cos^2Ï€/4+cos^2Ï€/4+cos^2Î³=1 ⇒1/2+1/2+cos^2Î³=1 ⇒2 cos^2Î³=0 ⇒cosÎ³=0 ⇒ Î³=Ï€/2

(c) Here l=cos Ï€/4,m=cosÏ€/4 Let the line make an angle ‘Î³’ with z-axis ∴l^2+m^2+n^2=1 ⇒cos^2Ï€/4+cos^2Ï€/4+cos^2Î³=1 ⇒1/2+1/2+cos^2Î³=1 ⇒2 cos^2Î³=0 ⇒cosÎ³=0 ⇒ Î³=Ï€/2

**Q3.**The length of projection of the line segment joining the points (1,0,-1) and (-1,2,2) on the plane x+3y-5z=6, is equal to

Solution

(d) Let A(1,0,-1),B(-1;2,2) Direction ratios of segment AB are <2,-2,-3> cosÎ¸=|2×1+3(-2)-5(-3)|/(√(1+9+25) √(4+4+9))=11/(√17 √35)=11/√595 Length of projection =(AB) sinÎ¸ =√((2)^2+(2)^2+(3)^2 )×√(1-121/595) =√17 √474/(√17 √35)=√(474/35) units

(d) Let A(1,0,-1),B(-1;2,2) Direction ratios of segment AB are <2,-2,-3> cosÎ¸=|2×1+3(-2)-5(-3)|/(√(1+9+25) √(4+4+9))=11/(√17 √35)=11/√595 Length of projection =(AB) sinÎ¸ =√((2)^2+(2)^2+(3)^2 )×√(1-121/595) =√17 √474/(√17 √35)=√(474/35) units

**Q4.**Distance of point P(p ⃗) from the plane r ∙n =0 is

Solution

(c) Let Q(q) be the foot of altitude drawn from ‘P’ to the plane r∙n=0 ⇒q-p=Î»n ⇒q=p+Î»n Also q∙n=0⇒ (p+Î»n)∙n=0 ⇒Î»=-(p∙n)/|n|^2 ⇒q -p =-((p∙n))/|n|^2 n Thus, required distance =|q -p|=(|p ∙n|)/(|n|)=|p ∙n ̂|

(c) Let Q(q) be the foot of altitude drawn from ‘P’ to the plane r∙n=0 ⇒q-p=Î»n ⇒q=p+Î»n Also q∙n=0⇒ (p+Î»n)∙n=0 ⇒Î»=-(p∙n)/|n|^2 ⇒q -p =-((p∙n))/|n|^2 n Thus, required distance =|q -p|=(|p ∙n|)/(|n|)=|p ∙n ̂|

**Q5.**The three planes 4y+6z=5;2x+3y+5z=5 and 6x+5y+9z=10

Solution

(b) Plane passing through the line of intersection if planes 4y+6z=5 and 2x+3y+5z=5 is (4y+6z-5)+Î»(2x+3y+5z-5)=0, or 2Î»x+(3Î»+4)y+(5Î»+6)z-5Î»-5=0 Clearly, for Î»=-3, we get the plane 6x+5y+9z=10 Hence, the given three planes have common line of intersection

(b) Plane passing through the line of intersection if planes 4y+6z=5 and 2x+3y+5z=5 is (4y+6z-5)+Î»(2x+3y+5z-5)=0, or 2Î»x+(3Î»+4)y+(5Î»+6)z-5Î»-5=0 Clearly, for Î»=-3, we get the plane 6x+5y+9z=10 Hence, the given three planes have common line of intersection

**Q6.**Equation of the plane containing the straight linex/2=y/3=z/4 and perpendicular to the plane containing the straight linesx/3=y/4=z/2 and x/4=y/2=z/3 is

Solution

(c) Equation of plane containing the linex/2=y/3=z/4 is a(x-0)+b(y-0)+c(z-0)=0 …(i) and 2a+3b+4c=0 ….(ii) Another equation of the plane containing the other two lines is a_1 (x-0)+b_1 (y-0)+c_1 (z-0)=0 ….(iii) Also, 3a_1+4b_1+2c_1=0 and 4a_1+2b_1+3c_1=0 on solving we get a_1/8=b_1/(-1)=c/(-10) ∴ Eq. (iii) becomes 8x-y-10c=0 …(iv) Since, the plane (i) is perpendicular to the plane (ii) ∴ 8a-b-10c=0 …(v) On solving Eqs. (ii) and (v), we get a/(-26)=b/52=c/(-26) or a/1=b/(-2)=c/1 ∴ From Eq. (i) x-2y+z=0 Alternate Let a ⃗=2i ̇ ̂+3j ̇ ̂+2k ̂ , b ⃗=3i ̇ ̂+4j ̇ ̂+2k ̂ and c ⃗=4i ̇ ̂+2j ̇ ̂+3k ̂ a ⃗×(b ⃗×c ⃗ )=(a ⃗∙c ⃗ ) b ⃗-(a ⃗∙b ⃗ ) c ⃗ =26(-i ̇ ̂+2j ̇ ̂-k ̂ ) ⟹ Direction ratio of normal to the required plane (passing through origin ) is 1,-2,1 ⟹ Equation of required plane is x-2y+z=0

(c) Equation of plane containing the linex/2=y/3=z/4 is a(x-0)+b(y-0)+c(z-0)=0 …(i) and 2a+3b+4c=0 ….(ii) Another equation of the plane containing the other two lines is a_1 (x-0)+b_1 (y-0)+c_1 (z-0)=0 ….(iii) Also, 3a_1+4b_1+2c_1=0 and 4a_1+2b_1+3c_1=0 on solving we get a_1/8=b_1/(-1)=c/(-10) ∴ Eq. (iii) becomes 8x-y-10c=0 …(iv) Since, the plane (i) is perpendicular to the plane (ii) ∴ 8a-b-10c=0 …(v) On solving Eqs. (ii) and (v), we get a/(-26)=b/52=c/(-26) or a/1=b/(-2)=c/1 ∴ From Eq. (i) x-2y+z=0 Alternate Let a ⃗=2i ̇ ̂+3j ̇ ̂+2k ̂ , b ⃗=3i ̇ ̂+4j ̇ ̂+2k ̂ and c ⃗=4i ̇ ̂+2j ̇ ̂+3k ̂ a ⃗×(b ⃗×c ⃗ )=(a ⃗∙c ⃗ ) b ⃗-(a ⃗∙b ⃗ ) c ⃗ =26(-i ̇ ̂+2j ̇ ̂-k ̂ ) ⟹ Direction ratio of normal to the required plane (passing through origin ) is 1,-2,1 ⟹ Equation of required plane is x-2y+z=0

**Q7.**The value of k such that (x-4)/1=(y-2)/1=(z-k)/2 lies in the plane 2x-4y+z=7, is

Solution

Since, line lies in a plane, it means point (4,2,k) lies in a plane. ∴8-8+k=7 ⟹k=7

Since, line lies in a plane, it means point (4,2,k) lies in a plane. ∴8-8+k=7 ⟹k=7

**Q8.**If lines x=y=z and x=y/2=z/3, and third line passing through (1, 1, 1) form a triangle of area √6 units, then point of intersection of third line with second line will be

Solution

(b) Let any point on second line be (Î»,2Î»,3Î») cosÎ¸=6/√42,sinÎ¸=√6/√42 Î”_OAB=1/2 (OA)OB sinÎ¸=1/2 √3 Î»√14×√6/√42=√6 ⇒Î»=2 So B is (2, 4, 6)

(b) Let any point on second line be (Î»,2Î»,3Î») cosÎ¸=6/√42,sinÎ¸=√6/√42 Î”_OAB=1/2 (OA)OB sinÎ¸=1/2 √3 Î»√14×√6/√42=√6 ⇒Î»=2 So B is (2, 4, 6)

**Q9.**The point on the line (x-2)/1=(y+3)/(-2)=(z+5)/(-2) at a distance of 6 from the point (2,-3,-5) is

Solution

(b) Direction cosines of the given line are 1/3,-2/3,-2/3 Hence, the equation of line can be point in the form (x-2)/(1/3)=(y+3)/(-2/3)=(z+5)/(-2/3)=r Therefore, any point on the line is (2+r/3,-3-2r/3,-5-2r/3), where r=±6 Points are (4,-7,-9) and (0,1,-1)

(b) Direction cosines of the given line are 1/3,-2/3,-2/3 Hence, the equation of line can be point in the form (x-2)/(1/3)=(y+3)/(-2/3)=(z+5)/(-2/3)=r Therefore, any point on the line is (2+r/3,-3-2r/3,-5-2r/3), where r=±6 Points are (4,-7,-9) and (0,1,-1)

**Q10.**A sphere of constnat radius 2k passes through the origin and meets the axes in A,B and C. The locus of a centroid of the tetrahedron OABC is

Solution

(b) Let the equation of the sphere be x^2+y^2+z^2-ax-by-cz=0. This meets the axes at A(a,0,0),B(0,b,0) and C(0,0,c) Let (Î±,Î²,Î³) be the coordinares of the centroid of the tetrahedron OABC. Then a/4=Î±,b/4=Î²,c/4=Î³ ⇒a=4Î±,b=4Î²,c=4Î³ Now, radius of the sphere =2k ⇒1/2 √(a^2+b^2+c^2 )=2k ⇒a^2+b^2+c^2=16k^2 ⇒16(Î±^2+Î²^2+Î³^2 )=16k^2 Hence, the locus of (Î±,Î²,Î³) is (x^2+y^2+z^2 )=k^2

(b) Let the equation of the sphere be x^2+y^2+z^2-ax-by-cz=0. This meets the axes at A(a,0,0),B(0,b,0) and C(0,0,c) Let (Î±,Î²,Î³) be the coordinares of the centroid of the tetrahedron OABC. Then a/4=Î±,b/4=Î²,c/4=Î³ ⇒a=4Î±,b=4Î²,c=4Î³ Now, radius of the sphere =2k ⇒1/2 √(a^2+b^2+c^2 )=2k ⇒a^2+b^2+c^2=16k^2 ⇒16(Î±^2+Î²^2+Î³^2 )=16k^2 Hence, the locus of (Î±,Î²,Î³) is (x^2+y^2+z^2 )=k^2