**Q1.**The plane 2x-1+Î»y+3Î»z=0 passes through the intersection of the planes:

Solution

We know that the equation of a plane passing through the intersection of the planes a_1 x+b_1 y+c_1 z+d_1=0 And a_2 x+b_2 y+c_2 z+d_2=0 is (a_1 x+b_1 y+c_1 z+d_1 )+Î»(a_2 x+b_2 y+c_2 z+d_2 )=0 Where Î» is constant Thus, the equation of plane 2x-(1+Î»)y+3Î»z=0 can be rewritten as (2x-y)+Î»(-y+3z)=0 So, it is clear that the equation of plane passes through the intersection of the planes 2x-y=0 and y-3z=0 .

We know that the equation of a plane passing through the intersection of the planes a_1 x+b_1 y+c_1 z+d_1=0 And a_2 x+b_2 y+c_2 z+d_2=0 is (a_1 x+b_1 y+c_1 z+d_1 )+Î»(a_2 x+b_2 y+c_2 z+d_2 )=0 Where Î» is constant Thus, the equation of plane 2x-(1+Î»)y+3Î»z=0 can be rewritten as (2x-y)+Î»(-y+3z)=0 So, it is clear that the equation of plane passes through the intersection of the planes 2x-y=0 and y-3z=0 .

**Q2.**The position vector of the point where the line

Solution

r ⃗=(i ̂-j ̂+k ̂ )+t(i ̂+j ̂-k ̂) =(1+t) i ̂-(1-t) j ̂+(1-t)k ̂ Also r ⃗∙(i ̂+j ̂+k ̂ )=5 ⇒(1+t)-(1-t)+(1-t)=5 ⇒1+t=5 ⇒t=4 ∴ r ⃗=(1+4) i ̂-(1-4) j ̂+(1-4) k ̂=5i ̂+3j ̂-3k ̂

r ⃗=(i ̂-j ̂+k ̂ )+t(i ̂+j ̂-k ̂) =(1+t) i ̂-(1-t) j ̂+(1-t)k ̂ Also r ⃗∙(i ̂+j ̂+k ̂ )=5 ⇒(1+t)-(1-t)+(1-t)=5 ⇒1+t=5 ⇒t=4 ∴ r ⃗=(1+4) i ̂-(1-4) j ̂+(1-4) k ̂=5i ̂+3j ̂-3k ̂

**Q3.**The locus of a point which moves so that the difference of the squares of its distances from two given points is constant, is a

Solution

Let the position vectors of the given points A and B be a ⃗ and b ⃗ respectively and that of the variable point P be r ⃗. It is given that PA^2-PB^2=k (Constant) ⇒|A ⃗P|^2-|B ⃗P|^2=k ⇒|r ⃗-a ⃗ |^2-|r ⃗-b ⃗ |^2=k ⇒{|r ⃗ |^2+|a ⃗ |^2-2 r ⃗∙a ⃗ }-{|r ⃗ |^2+|b ⃗ |^2-2 r ⃗∙b ⃗ }=k ⇒2r ⃗∙(b ⃗-a ⃗ )=k+|b ⃗ |^2-|a ⃗ |^2 ⇒r ⃗∙(b ⃗-a ⃗ )=Î», where, Î»=1/2 {k+|b ⃗ |^2-|a ⃗ |^2 } Clearly, it represents a plane .

Let the position vectors of the given points A and B be a ⃗ and b ⃗ respectively and that of the variable point P be r ⃗. It is given that PA^2-PB^2=k (Constant) ⇒|A ⃗P|^2-|B ⃗P|^2=k ⇒|r ⃗-a ⃗ |^2-|r ⃗-b ⃗ |^2=k ⇒{|r ⃗ |^2+|a ⃗ |^2-2 r ⃗∙a ⃗ }-{|r ⃗ |^2+|b ⃗ |^2-2 r ⃗∙b ⃗ }=k ⇒2r ⃗∙(b ⃗-a ⃗ )=k+|b ⃗ |^2-|a ⃗ |^2 ⇒r ⃗∙(b ⃗-a ⃗ )=Î», where, Î»=1/2 {k+|b ⃗ |^2-|a ⃗ |^2 } Clearly, it represents a plane .

**Q4.**If the distance of the point (1, 1,1) from the origin is half is distance from the plane x+y+z+k=0, then k is equal to

Solution

Distance of a point (1,1,1) from x+y+z+k=0 is |(1+1+1+k)/√3|=|(3+k)/√3| According to question |(3+k)/√3|=±2√3 ⇒k=3,-9

Distance of a point (1,1,1) from x+y+z+k=0 is |(1+1+1+k)/√3|=|(3+k)/√3| According to question |(3+k)/√3|=±2√3 ⇒k=3,-9

**Q5.**A line AB in three-dimensional space makes angle 45° and 120° with the positive x-axis and the positive y-axis respectively. If AB makes an acute angle Î¸ with the positive z-axis, then Î¸ equals

Solution

cos^2 45°+cos^2 120°+cos^2 Î¸=1 ⟹1/2+1/4+cos^2 Î¸=1⟹cos^2 Î¸=1-3/4=1/4 ⟹cos〖Î¸=1/2〗 (∵ Î¸ is acute) ⟹ Î¸=60°

cos^2 45°+cos^2 120°+cos^2 Î¸=1 ⟹1/2+1/4+cos^2 Î¸=1⟹cos^2 Î¸=1-3/4=1/4 ⟹cos〖Î¸=1/2〗 (∵ Î¸ is acute) ⟹ Î¸=60°

**Q6.**-Find the direction ratio of 3-x1=y-25=2z1

Solution

Given equation of line is
(3-x)/1=(y-2)/5=(2z-3)/1
⟹ (x-3)/(-1)=(y-2)/5=(z-3/2)/(1/2)
∴ Direction ratios of line are-1,5,1/2

**Q7.**The position vectors of points A and B are i-j+3 k and 3 i+3 j+3 k respectively. The equation of a plane is r5 i+2 j-7 k+9=0. The points A and B=

Solution

The position vectors of two given points are a ⃗=i ̂-j ̂+3k ̂ and b ⃗=3i ̂+3j ̂+3k ̂ and the equation of the given plane is
r ⃗∙(5i ̂+2j ̂-7k ̂ )+9=0 or, r ⃗∙n ⃗+d=0
We have,
a ⃗∙n ⃗+d=(i ̂-j ̂+3k ̂ )∙(5i ̂+2j ̂-7k ̂ )+9
=5-2-21+9<0 and="" b="" d="(3i" i="" j="" k="" n="">
0
So, the points a ⃗ and b ⃗ are on the opposite sides of the plane

**Q8.**The position vector of a point at a distance of 311 units from i-j+2k on a line passing through the points i-j+2k and 3i+j+k iscc

Solution

The equation of a line passing through the points A(i ̂-j ̂+2 k ̂ ) and B(3i ̂+j ̂+k ̂ ) is given by r ⃗=(i ̂-j ̂+2k ̂ )+Î»(3i ̂+j ̂+k ̂ ) The position vector of a variable point P on the line, is (i ̂-j ̂+2k ̂ )+Î»(3i ̂+j ̂+k ̂ ) ∴A ⃗P=Î»(3i ̂+j ̂+k ̂ )⇒|A ⃗P|=|Î»|√11 Now, |Î»| √11=3√11,⇒Î»=±3 Thus, the position vectors of P are 10i ̂+2j ̂+5k ̂ and -8i ̂-4j ̂-k ̂

The equation of a line passing through the points A(i ̂-j ̂+2 k ̂ ) and B(3i ̂+j ̂+k ̂ ) is given by r ⃗=(i ̂-j ̂+2k ̂ )+Î»(3i ̂+j ̂+k ̂ ) The position vector of a variable point P on the line, is (i ̂-j ̂+2k ̂ )+Î»(3i ̂+j ̂+k ̂ ) ∴A ⃗P=Î»(3i ̂+j ̂+k ̂ )⇒|A ⃗P|=|Î»|√11 Now, |Î»| √11=3√11,⇒Î»=±3 Thus, the position vectors of P are 10i ̂+2j ̂+5k ̂ and -8i ̂-4j ̂-k ̂

**Q9.**The distance of the point (-1, -5, -10) from the point of intersection of the line x-23=y+14=z-212 and the plane x-y+z=5 is

Solution

Clearly point (2,-1,2) lies on the line as well as plane ∴ Required distance of point (-1,-5,-10) =√((-1-2)^2+(-5+1)^2+(-10-2)^2 ) =√(9+16+144) =√169=13

Clearly point (2,-1,2) lies on the line as well as plane ∴ Required distance of point (-1,-5,-10) =√((-1-2)^2+(-5+1)^2+(-10-2)^2 ) =√(9+16+144) =√169=13

**Q10.**The distance of the point (2, 3, -5) from the plane x+2y-2z=9 is

Solution

The distance of the point (2, 3, -5) from the plane x+2y-2z=9 is D=(|2(1)+2(3)-2(-5)-9|)/√(1^2+2^2+(-2)^2 ) =(|2+6+10-9|)/√(1+4+4)=3

The distance of the point (2, 3, -5) from the plane x+2y-2z=9 is D=(|2(1)+2(3)-2(-5)-9|)/√(1^2+2^2+(-2)^2 ) =(|2+6+10-9|)/√(1+4+4)=3