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Solution
Given (3,4,-1)and (-1,2,3) are the end points of diameter of sphere ∴Radius=1/2 (length of the diameter) =1/2 √((3+1)^2+(4-2)^2+(-1-3)^2 ) =3

Q4. The equation of the plane in which the lines (x-5)/4=(y-7)/4=(z+3)/(-5) and (x-8)/7=(y-4)/1=(z-5)/3 lie, is
•  17x-47y+24z+172=0
•  17x+47y-24z+172=0
•  17x+47y+24z+172=0
•  17x-47y-24z+172=0
Solution
The equation of plane, in which the line (x-5)/4=(y-7)/4=(z+3)/(-5) lies is A(x-5)+B(y-7)+C(z+3)=0 …(i) Where A,B and C are the direction ratios of the plane. Since, the first line lie on the plane ∴ Direction ratios of normal to the plane is perpendicular to the direction ratios of line ie, 4A+4B-5C=0 ….(ii) Since, line (x-8)/7=(y-4)/1=(z-5)/3 lies in this plane. The direction ratios is also perpendicular to this line ∴7A+B+3C=0 …(iii) From Eqs. (ii) and (iii), we get A/17=B/(-47)=C/(-24) ∴ The required equation of plane is 17(x-5)-47(y-7)+(-24)(z+3)=0 ⇒ 17x-47y-24z+172=0

Q5.The equation of the plane passing through a point A(2,-1,3) and parallel to the vectors a ⃗=(3,0,-1) and b ⃗=(-3,2,2) is
•  2x-3y+6z-25=0
•  2x-3y+6z-25=0
•  3x-2y+6z-25=0
•  3x-2y+6z+25=0
Solution
The equation of any plane through (2,-1,3) is a(x-2)+b(y+1)+c(z-3)=0 …(i) Since, Eq. (i), is parallel to a ⃗ and b ⃗ ∴3a+0b-c=0 and-3a+2b+2c=0 ⟹a/2=b/(3-6)=c/6=k [say] ⟹a=2k,b=-3k,c=6k On putting the values of a,b and c in Eq. (i), we get 2k(x-2)-3k(y+1)+6k(z-3)=0 ⟹2x-3y+6z-25=0

Q6. Foot of the perpendicular from B(-2,1,4) to the plane is (3, 1, 2). Then, the equation of the plane is
•  4x-2y=11
•  5x-2z=11
• 5x-2y=10
•  5x+2z=11
Solution
Given, A(3,1,2) be the foot of the perpendicular from B(-2,1,4)on the plane, then direction ratios of BA, which is the normal to plane are (3+2,1-1,2-4)ie, (5,0,-2) ∴ The equation of plane is 5(x-3)+0(y-1)-2(z-2)=0 ⟹5x-2z=11

Q7.Let P(-7,1,-5) be a point on a plane and let Obe the origin. If OP is normal to the plane, then the equation of the plane is
•  7x+y+5z+73=0
•  7x+y-5z+73=0
•  7x-y+5z+75=0
•  7x-y-5z+75=0
Solution
Equation of any plane passing through (-7,1,-5) is a(x+7)+b(y-1)+c(z+5)=0 …..(i) The DR’s of normal to above plane are a=-7,b=1,c=-5 ∴ From Eq. (i) we get -7(x+7)+1(y-1)-5(z+1)=0 ⟹7x-y+5z+75=0

Q8.The value of Î» for which the lines (x-1)/1=(y-2)/Î»=(z+1)/(-1) and (x+1)/(-Î»)=(y+1)/2=(z-2)/1 are perpendicular to each other is
•  0
•  1
•  -1
•  None of these
Solution
The given lines are parallel to the vectors (b1 ) ⃗=i ̂+Î»j ̂-k ̂ and (b2 ) ⃗=-Î»i ̂+2j ̂+k ̂ respectively. The lines will be perpendicular to each other, if (b1 ) ⃗∙(b2 ) ⃗=0⇒-Î»+2Î»-1=0⇒Î»=1

Q9.Radius of the circle (r^2 ) ⃗+r ⃗∙(2i ̇ ̂-2j ̇ ̂-4k ̂ )-19=0 and r ⃗∙(i ̇ ̂-2j ̇ ̂+2k ̂ )+8=0
•  3
•  2
•  5
•  4
Solution
Required circle is intersection of sphere x^2+y^2+z^2+2x-2y-4z-19=0 and plane x-2y+2z+8=0 Centre of sphere is (-1,1,2) P=length of the perpendicular from, (-1,1,2) to the plane =(-1-2+4+8)/√(1+4+4) =9/3=3 R=radius of sphere =√(1+1+4+19)=5 Radius of the circle=√(R^2-P^2 ) =√(25-9)=4

Q10. The foot of the perpendicular drawn from a point with position vector i ̂+4 k ̂ on the line joining the points having position vectors as -11 j ̂+3 k ̂ and 2 i ̂-3 j ̂+k ̂ has the position vector
•  4 i ̂+5 j ̂+5 k ̂
•  4 i ̂+5 j ̂-5 k ̂
•  5 i ̂+4 j ̂-5 k ̂
• 4 i ̂-5 j ̂+5 k ̂ ̂
Solution
Let AB be the given line and the let its direction cosines of AB be l,m,n. Then, Projection of AB on x-axis =AB l=12 (given) Projection of AB on y-axis =AB m=4 (given) Projection of AB on z-axis =AB n=3 (given) ∴(AB)^2 (l^2+m^2+n^2 )=〖12〗^2+4^2+3^2⇒AB=13 Hence, direction cosines of AB are 12/13,4/13,3/13

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