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****Q1. **If the foot of the perpendicular from the origin to a plane is (a, b, c), then equation of the plane is
Solution

Let P be the foot of the perpendicular from the origin on the plane, then direction ratios ofOP, the normal to the plane area-0,b-0,c-0 ie,a,b,c. Also, since, it passes through (a,b,c), the equation of the plane is
a(x-a)+b(y-b)+c(z-c)=0
⇒ ax+by+cz=a^2+b^2+c^2

**Q2.**
The point on the line x-21=y+3-2=z+5-2 at a distance of 6 from the point (2, -3, -5) is
Solution

DC’s 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
∴ Point is (2+r/3,-3-2r/3,-5-2r/3)
∴r=±6

**Q3. **The distance of the point P(1,2,3) from the line which passes through the point A4,2,2and parallel to the vector 2i+3j+6k is
**Q4. **
Equation of plane passing through the points (2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x+6y+6z-1=0, is
Solution

Equation of a plane passing through (2, 2, 1) is
a(x-2)+b(y-2)+c(z-1)=0 … (i)
This passes through (9, 3, 6) and is perpendicular to 2x+6y+6z-1=0
∴7a+b+5c=0 and, 2a+6b+6c=0
Solving these two by cross-multiplication, we get
a/(-24)=b/(-32)=c/40⇒a/(-3)=b/(-4)=c/5
Substituting the values of a,b,c in (i), we get
3x+4y-5z-9=0 as the required plane

**Q5.**The vector equation of the plane through the point 2, 1, -1 and passing through the line of intersection of the plane ri+3 j-k=0 and rj+2 k=0, is
Solution

The vector equation of a plane through the line of intersection of the planes r ⃗∙(i ̂+3j ̂-k ̂ )=0 and r ⃗∙(j ̂+2k ̂ )=0 can be written as
{r ⃗∙(i ̂+3j ̂+-k ̂ )}+Î»{r ⃗∙(j ̂+2k ̂ )}=0 …(i)
This passes through 2i ̂+j ̂-k ̂
∴(2i ̂+j ̂-k ̂ )∙(i ̂+3j ̂-k ̂ )+Î»(2i ̂+j ̂-k ̂ )∙(j ̂+2k ̂ )=0
⇒(2+3+1)+Î»(0+1-2)=0⇒Î»=6
Putting the value of Î» in (i), we get the equation of the required plane as
r ⃗∙(i ̂+9j ̂+11k ̂ )=0

**Q6. **The equation of the plane passing through the points (0, 1, 2) and (-1, 0, 3) and perpendicular to the plane 2x+3y+z=5 is
Solution

Any plane passing through (0, 1, 2) is
a(x-0)+b(y-1)+c(z-2)=0
⟹ax+b(y-1)+c(z-2)=0 ….(i)
Since, it is passing through (-1,0,3), we get
-a-b+c=0 …..(ii)
Also, Eq. is perpendicular to 2x+3y+z=5
∴2a+3b+c=0 ….(iii)
On solving Eqs. (ii) and (iii), we get
a/(-4)=b/3=c/(-1)
∴From Eq. (i)
-4x+3(y-1)-1(z-2)=0
⟹4x-3y+z+1=0

**Q7.**The centre of sphere passes through four points (0, 0, 0), (0,2,0), (1,0, 0)and (0, 0, 4)
is
Solution

Let the equation of sphere passing through (0, 0, 0) be
x^2+y^2+z^2+2ux+2vy+2wz=0
Also, it passes through (0, 2, 0), (1, 0, 0), (0, 0, 4) respectively are
4+4v=0
⟹v=-1
1+2u=0⟹u=-1/2
and 16+8w⟹w=-2
∴ Centre is (-u,-v,-w)=(1/2,1,2)

**Q8.**A plane Ï€ makes intercepts 3 and 4 respectively on z-axis and x-axis. If Ï€ is parallel to y-axis, then its equation is
Solution

Given, a=4,c=3
Equation of the plane Ï€ is
x/4+y/b+z/3=1
Since, Ï€ is parallel to y-axis
∴Coefficient of y=0 ie,1/b=0
Thus, the equation of plane Ï€ is
x/4+z/3=1
⟹3x+4z-12=0

**Q9.**The length of the perpendicular from the origin to the plane passing through three non-collinear points a, b, c is
Solution

The vector equation of the plane passing through points a ⃗,b ⃗,c ⃗ is
r ⃗∙(a ⃗×b ⃗+b ⃗×c ⃗+c ⃗×a ⃗ )=[a ⃗ b ⃗ c ⃗ ]
Therefore, the length of the perpendicular from the origin to this plane is given by

**Q10. **If P3, 2, -4, Q(5, 4, -6) and R(9, 8, -10) are collinear, then R divides PQ in the ratio

Solution

Suppose R divides PQ in the ratio Î»∶1. Then, the coordinates of R are
((5 Î»+3)/(Î»+1),(4 Î»+2)/(Î»+1),(-6 Î»-4)/(Î»+1))
But, the coordinates of R are given as (9,8,-10)
∴(5 Î»+3)/(Î»+1)=9,(4 Î»+2)/(Î»+1)=8 and (-6 Î»-4)/(Î»+1)=-10
⇒Î»=-3/2
Hence, R divides PQ externally in the ratio 3 : 2

**Q1.**If the foot of the perpendicular from the origin to a plane is (a, b, c), then equation of the plane is

Solution

Let P be the foot of the perpendicular from the origin on the plane, then direction ratios ofOP, the normal to the plane area-0,b-0,c-0 ie,a,b,c. Also, since, it passes through (a,b,c), the equation of the plane is a(x-a)+b(y-b)+c(z-c)=0 ⇒ ax+by+cz=a^2+b^2+c^2

Let P be the foot of the perpendicular from the origin on the plane, then direction ratios ofOP, the normal to the plane area-0,b-0,c-0 ie,a,b,c. Also, since, it passes through (a,b,c), the equation of the plane is a(x-a)+b(y-b)+c(z-c)=0 ⇒ ax+by+cz=a^2+b^2+c^2

**Q2.**The point on the line x-21=y+3-2=z+5-2 at a distance of 6 from the point (2, -3, -5) is

Solution

DC’s 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 ∴ Point is (2+r/3,-3-2r/3,-5-2r/3) ∴r=±6

DC’s 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 ∴ Point is (2+r/3,-3-2r/3,-5-2r/3) ∴r=±6

**Q3.**The distance of the point P(1,2,3) from the line which passes through the point A4,2,2and parallel to the vector 2i+3j+6k is

**Q4.**Equation of plane passing through the points (2, 2, 1), (9, 3, 6) and perpendicular to the plane 2x+6y+6z-1=0, is

Solution

Equation of a plane passing through (2, 2, 1) is a(x-2)+b(y-2)+c(z-1)=0 … (i) This passes through (9, 3, 6) and is perpendicular to 2x+6y+6z-1=0 ∴7a+b+5c=0 and, 2a+6b+6c=0 Solving these two by cross-multiplication, we get a/(-24)=b/(-32)=c/40⇒a/(-3)=b/(-4)=c/5 Substituting the values of a,b,c in (i), we get 3x+4y-5z-9=0 as the required plane

Equation of a plane passing through (2, 2, 1) is a(x-2)+b(y-2)+c(z-1)=0 … (i) This passes through (9, 3, 6) and is perpendicular to 2x+6y+6z-1=0 ∴7a+b+5c=0 and, 2a+6b+6c=0 Solving these two by cross-multiplication, we get a/(-24)=b/(-32)=c/40⇒a/(-3)=b/(-4)=c/5 Substituting the values of a,b,c in (i), we get 3x+4y-5z-9=0 as the required plane

**Q5.**The vector equation of the plane through the point 2, 1, -1 and passing through the line of intersection of the plane ri+3 j-k=0 and rj+2 k=0, is

Solution

The vector equation of a plane through the line of intersection of the planes r ⃗∙(i ̂+3j ̂-k ̂ )=0 and r ⃗∙(j ̂+2k ̂ )=0 can be written as {r ⃗∙(i ̂+3j ̂+-k ̂ )}+Î»{r ⃗∙(j ̂+2k ̂ )}=0 …(i) This passes through 2i ̂+j ̂-k ̂ ∴(2i ̂+j ̂-k ̂ )∙(i ̂+3j ̂-k ̂ )+Î»(2i ̂+j ̂-k ̂ )∙(j ̂+2k ̂ )=0 ⇒(2+3+1)+Î»(0+1-2)=0⇒Î»=6 Putting the value of Î» in (i), we get the equation of the required plane as r ⃗∙(i ̂+9j ̂+11k ̂ )=0

The vector equation of a plane through the line of intersection of the planes r ⃗∙(i ̂+3j ̂-k ̂ )=0 and r ⃗∙(j ̂+2k ̂ )=0 can be written as {r ⃗∙(i ̂+3j ̂+-k ̂ )}+Î»{r ⃗∙(j ̂+2k ̂ )}=0 …(i) This passes through 2i ̂+j ̂-k ̂ ∴(2i ̂+j ̂-k ̂ )∙(i ̂+3j ̂-k ̂ )+Î»(2i ̂+j ̂-k ̂ )∙(j ̂+2k ̂ )=0 ⇒(2+3+1)+Î»(0+1-2)=0⇒Î»=6 Putting the value of Î» in (i), we get the equation of the required plane as r ⃗∙(i ̂+9j ̂+11k ̂ )=0

**Q6.**The equation of the plane passing through the points (0, 1, 2) and (-1, 0, 3) and perpendicular to the plane 2x+3y+z=5 is

Solution

Any plane passing through (0, 1, 2) is a(x-0)+b(y-1)+c(z-2)=0 ⟹ax+b(y-1)+c(z-2)=0 ….(i) Since, it is passing through (-1,0,3), we get -a-b+c=0 …..(ii) Also, Eq. is perpendicular to 2x+3y+z=5 ∴2a+3b+c=0 ….(iii) On solving Eqs. (ii) and (iii), we get a/(-4)=b/3=c/(-1) ∴From Eq. (i) -4x+3(y-1)-1(z-2)=0 ⟹4x-3y+z+1=0

Any plane passing through (0, 1, 2) is a(x-0)+b(y-1)+c(z-2)=0 ⟹ax+b(y-1)+c(z-2)=0 ….(i) Since, it is passing through (-1,0,3), we get -a-b+c=0 …..(ii) Also, Eq. is perpendicular to 2x+3y+z=5 ∴2a+3b+c=0 ….(iii) On solving Eqs. (ii) and (iii), we get a/(-4)=b/3=c/(-1) ∴From Eq. (i) -4x+3(y-1)-1(z-2)=0 ⟹4x-3y+z+1=0

**Q7.**The centre of sphere passes through four points (0, 0, 0), (0,2,0), (1,0, 0)and (0, 0, 4) is

Solution

Let the equation of sphere passing through (0, 0, 0) be x^2+y^2+z^2+2ux+2vy+2wz=0 Also, it passes through (0, 2, 0), (1, 0, 0), (0, 0, 4) respectively are 4+4v=0 ⟹v=-1 1+2u=0⟹u=-1/2 and 16+8w⟹w=-2 ∴ Centre is (-u,-v,-w)=(1/2,1,2)

Let the equation of sphere passing through (0, 0, 0) be x^2+y^2+z^2+2ux+2vy+2wz=0 Also, it passes through (0, 2, 0), (1, 0, 0), (0, 0, 4) respectively are 4+4v=0 ⟹v=-1 1+2u=0⟹u=-1/2 and 16+8w⟹w=-2 ∴ Centre is (-u,-v,-w)=(1/2,1,2)

**Q8.**A plane Ï€ makes intercepts 3 and 4 respectively on z-axis and x-axis. If Ï€ is parallel to y-axis, then its equation is

Solution

Given, a=4,c=3 Equation of the plane Ï€ is x/4+y/b+z/3=1 Since, Ï€ is parallel to y-axis ∴Coefficient of y=0 ie,1/b=0 Thus, the equation of plane Ï€ is x/4+z/3=1 ⟹3x+4z-12=0

Given, a=4,c=3 Equation of the plane Ï€ is x/4+y/b+z/3=1 Since, Ï€ is parallel to y-axis ∴Coefficient of y=0 ie,1/b=0 Thus, the equation of plane Ï€ is x/4+z/3=1 ⟹3x+4z-12=0

**Q9.**The length of the perpendicular from the origin to the plane passing through three non-collinear points a, b, c is

Solution

The vector equation of the plane passing through points a ⃗,b ⃗,c ⃗ is r ⃗∙(a ⃗×b ⃗+b ⃗×c ⃗+c ⃗×a ⃗ )=[a ⃗ b ⃗ c ⃗ ] Therefore, the length of the perpendicular from the origin to this plane is given by

The vector equation of the plane passing through points a ⃗,b ⃗,c ⃗ is r ⃗∙(a ⃗×b ⃗+b ⃗×c ⃗+c ⃗×a ⃗ )=[a ⃗ b ⃗ c ⃗ ] Therefore, the length of the perpendicular from the origin to this plane is given by

**Q10.**If P3, 2, -4, Q(5, 4, -6) and R(9, 8, -10) are collinear, then R divides PQ in the ratio

Solution

Suppose R divides PQ in the ratio Î»∶1. Then, the coordinates of R are ((5 Î»+3)/(Î»+1),(4 Î»+2)/(Î»+1),(-6 Î»-4)/(Î»+1)) But, the coordinates of R are given as (9,8,-10) ∴(5 Î»+3)/(Î»+1)=9,(4 Î»+2)/(Î»+1)=8 and (-6 Î»-4)/(Î»+1)=-10 ⇒Î»=-3/2 Hence, R divides PQ externally in the ratio 3 : 2

Suppose R divides PQ in the ratio Î»∶1. Then, the coordinates of R are ((5 Î»+3)/(Î»+1),(4 Î»+2)/(Î»+1),(-6 Î»-4)/(Î»+1)) But, the coordinates of R are given as (9,8,-10) ∴(5 Î»+3)/(Î»+1)=9,(4 Î»+2)/(Î»+1)=8 and (-6 Î»-4)/(Î»+1)=-10 ⇒Î»=-3/2 Hence, R divides PQ externally in the ratio 3 : 2