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**Q1. **The shortest distance between the lines
(x-2)/3=(y+3)/4=(z-1)/5 and
(x-5)/1=(y-1)/2=(z-6)/3 is
**Q2.**
The equation of the plane through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x+6y+6z-1=0, is
Solution

B

**Q3. **The equation of the plane containing the line (x+1)/-3=(y-3)/2=(z+2)/1 and the point 0, 7, -7, is
Solution

The position vector of any point on the given line is
i ̂+j ̂+Î»(2i ̂+j ̂+4k ̂ )=(2Î»+1) i ̂+(Î»+1) j ̂+4 Î» k ̂
Clearly, this point lies on the plane r ⃗∙(i ̂+2j ̂-k ̂ )=3
Hence, the plane r ⃗∙(i ̂+2j ̂-k ̂ )=3 contains the given line

**Q4. **f the direction ratio of two lines are given by
3lm-4ln+mn=0 and
l+2m+3n=0, then the angle between the line is
Solution

Given, 3lm-4ln+mn=0 …..(i)
and l+2m+3n=0 …(ii)
From Eq. (ii), l=-(2m+3n) putting in Eq. (i)
-3(2m+3n)m+4(2m+3n)n+mn=0
⟹ -〖6m〗^2+〖12n〗^2=0
⟹m=±√2 n
Now,m=√2 n
⟹l=-(2√2 n+3n)=-(2√2+3)n
∴l:m:n=-(3+2√2)n:√2 n:n
=-(3+2√2):√2:1
Also,m=-√2 n⟹l=-(-2√2+3)n
∴l:m:n=-(3-2√2)n:-√2:n
=-(3-2√2):-√2:1
=cosÎ¸
=((3+2√2)(3-2√2)+(√2)(-√2)+1∙1)/(√((3+2√2 )^2+(√2)^2+1^2 ) √((3-2√2)^2+(-√2)^2+1^2 ))
=0
⟹ Î¸=Ï€/2

**Q5.**The planex2+y3+z4=1 cuts the coordinate axes in A,B,C then the area of the ∆ABC
Solution

Equation of the plane is x/2+y/3+z/4=1
Here,a=2,b=3,c=4
∴Area of ∆ABC=1/2 √(a^2 b^2+b^2 c^2+c^2 a^2 )
=1/2 √(2^2∙3^2+3^2∙4^2+4^2∙2^2 )
=1/2 √244=√61 sq units

**Q6. **The equation to the straight line passing through the points (4, -5, -2) and (-1, 5, 3) is
**Q7.**A straight line r=a+Î» b meets the plane rn=0 in P. The position vector of P is

**Q1.**The shortest distance between the lines (x-2)/3=(y+3)/4=(z-1)/5 and (x-5)/1=(y-1)/2=(z-6)/3 is

**Q2.**The equation of the plane through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x+6y+6z-1=0, is

Solution

B

B

**Q3.**The equation of the plane containing the line (x+1)/-3=(y-3)/2=(z+2)/1 and the point 0, 7, -7, is

Solution

The position vector of any point on the given line is i ̂+j ̂+Î»(2i ̂+j ̂+4k ̂ )=(2Î»+1) i ̂+(Î»+1) j ̂+4 Î» k ̂ Clearly, this point lies on the plane r ⃗∙(i ̂+2j ̂-k ̂ )=3 Hence, the plane r ⃗∙(i ̂+2j ̂-k ̂ )=3 contains the given line

The position vector of any point on the given line is i ̂+j ̂+Î»(2i ̂+j ̂+4k ̂ )=(2Î»+1) i ̂+(Î»+1) j ̂+4 Î» k ̂ Clearly, this point lies on the plane r ⃗∙(i ̂+2j ̂-k ̂ )=3 Hence, the plane r ⃗∙(i ̂+2j ̂-k ̂ )=3 contains the given line

**Q4.**f the direction ratio of two lines are given by 3lm-4ln+mn=0 and l+2m+3n=0, then the angle between the line is

Solution

Given, 3lm-4ln+mn=0 …..(i) and l+2m+3n=0 …(ii) From Eq. (ii), l=-(2m+3n) putting in Eq. (i) -3(2m+3n)m+4(2m+3n)n+mn=0 ⟹ -〖6m〗^2+〖12n〗^2=0 ⟹m=±√2 n Now,m=√2 n ⟹l=-(2√2 n+3n)=-(2√2+3)n ∴l:m:n=-(3+2√2)n:√2 n:n =-(3+2√2):√2:1 Also,m=-√2 n⟹l=-(-2√2+3)n ∴l:m:n=-(3-2√2)n:-√2:n =-(3-2√2):-√2:1 =cosÎ¸ =((3+2√2)(3-2√2)+(√2)(-√2)+1∙1)/(√((3+2√2 )^2+(√2)^2+1^2 ) √((3-2√2)^2+(-√2)^2+1^2 )) =0 ⟹ Î¸=Ï€/2

Given, 3lm-4ln+mn=0 …..(i) and l+2m+3n=0 …(ii) From Eq. (ii), l=-(2m+3n) putting in Eq. (i) -3(2m+3n)m+4(2m+3n)n+mn=0 ⟹ -〖6m〗^2+〖12n〗^2=0 ⟹m=±√2 n Now,m=√2 n ⟹l=-(2√2 n+3n)=-(2√2+3)n ∴l:m:n=-(3+2√2)n:√2 n:n =-(3+2√2):√2:1 Also,m=-√2 n⟹l=-(-2√2+3)n ∴l:m:n=-(3-2√2)n:-√2:n =-(3-2√2):-√2:1 =cosÎ¸ =((3+2√2)(3-2√2)+(√2)(-√2)+1∙1)/(√((3+2√2 )^2+(√2)^2+1^2 ) √((3-2√2)^2+(-√2)^2+1^2 )) =0 ⟹ Î¸=Ï€/2

**Q5.**The planex2+y3+z4=1 cuts the coordinate axes in A,B,C then the area of the ∆ABC

Solution

Equation of the plane is x/2+y/3+z/4=1 Here,a=2,b=3,c=4 ∴Area of ∆ABC=1/2 √(a^2 b^2+b^2 c^2+c^2 a^2 ) =1/2 √(2^2∙3^2+3^2∙4^2+4^2∙2^2 ) =1/2 √244=√61 sq units

Equation of the plane is x/2+y/3+z/4=1 Here,a=2,b=3,c=4 ∴Area of ∆ABC=1/2 √(a^2 b^2+b^2 c^2+c^2 a^2 ) =1/2 √(2^2∙3^2+3^2∙4^2+4^2∙2^2 ) =1/2 √244=√61 sq units

**Q6.**The equation to the straight line passing through the points (4, -5, -2) and (-1, 5, 3) is

**Q7.**A straight line r=a+Î» b meets the plane rn=0 in P. The position vector of P is

Solution

The straight line r ⃗=a ⃗+Î» b ⃗ meets the plane r ⃗∙n ⃗=0 in P for which Î» is given by (a ⃗+Î» b ⃗ )∙n ⃗=0⇒Î»=-(a ⃗∙n ⃗)/(b ⃗∙n ⃗ ) Thus, the position vector of P is r ⃗=a ⃗((a ⃗∙n ⃗)/(b ⃗∙n ⃗ )) b ⃗ [Putting the value of Î» in r ⃗=a ⃗+Î»b ⃗ ]

The straight line r ⃗=a ⃗+Î» b ⃗ meets the plane r ⃗∙n ⃗=0 in P for which Î» is given by (a ⃗+Î» b ⃗ )∙n ⃗=0⇒Î»=-(a ⃗∙n ⃗)/(b ⃗∙n ⃗ ) Thus, the position vector of P is r ⃗=a ⃗((a ⃗∙n ⃗)/(b ⃗∙n ⃗ )) b ⃗ [Putting the value of Î» in r ⃗=a ⃗+Î»b ⃗ ]

**Q8.**Cosine of the angle between two diagonals of cube is equal to

Solution

Let OA,OB,OC be the sides of a cube such that OA=OB=OC=a ∴ Direction ratios of OE are (a-0,a-0)ie (a,a,a) ∴Direction cosines of AF are (1/√3,1/√3,1/√3) Similarly, direction of AF are (-1/√3,1/√3,1/√3). ∴ Angle between OE and AF is cos^(-1)〖[-1/√3∙1/√3+1/√3∙1/√3+1/√3∙1/√3] cos^(-1)(1/3) 〗

Let OA,OB,OC be the sides of a cube such that OA=OB=OC=a ∴ Direction ratios of OE are (a-0,a-0)ie (a,a,a) ∴Direction cosines of AF are (1/√3,1/√3,1/√3) Similarly, direction of AF are (-1/√3,1/√3,1/√3). ∴ Angle between OE and AF is cos^(-1)〖[-1/√3∙1/√3+1/√3∙1/√3+1/√3∙1/√3] cos^(-1)(1/3) 〗

**Q9.**The line segment adjoining the points A, B makes projection 1, 4, 3 on x, y, z-axes respectively. Then, the direction cosines of AB are

Solution

DC^' s of AB=1/√(1^2+4^2+3^2 ),4/√(1^2+4^2+3^2 ),3/√(1^2+4^2+3^2 ) =1/√26,4/√26,3/√26

DC^' s of AB=1/√(1^2+4^2+3^2 ),4/√(1^2+4^2+3^2 ),3/√(1^2+4^2+3^2 ) =1/√26,4/√26,3/√26

**Q10.**A variable plane moves so that sum of the reciprocals of its intercepts on the coordinate axes is 1/2 Then, the plane passes through

Solution

Let x/a+y/b+z/c=1 ….(i) Then,1/a+1/b+1/c=1/2 …(ii) On comparing Eqs. (i) and (ii), we get x=2,y=2,z=2

Let x/a+y/b+z/c=1 ….(i) Then,1/a+1/b+1/c=1/2 …(ii) On comparing Eqs. (i) and (ii), we get x=2,y=2,z=2