Q1. The distance between the points (1,4,5) and (2,2,3) is
Solution
The distance between given points
=√((2-1)^2+(2-4)^2+(3-5)^2 )
=√(1+4+4)=3
Q2.The distance of the point of intersection of the line
(x-2)/3=(y+1)/4=(z-2)/12 and the plane x-y+z=5 from the point (-1,-5,-10)is
Solution
Given line is
(x-2)/3=(y+1)/4=(z-2)/12=k (say)
Any point on the line is (3k+2,4k-1,12k+2)
This point lies on the plane x-y+z=5
∴3k+2-(4k-1)+12k+2=5
⟹11k=0 ⟹k=0
∴ Intersection point is (2,-1,2)
∴ Distance, between points(2-1,2) and (-1,-5,-10)
=√((-1-2)^2+(-5+1)^2+(-10-2)^2 )
=√(9+16+144)=13
Q3. The triangle formed by the points (0, 7, 10), (-1,6,6),(-4,9,6) is
Solution
Let A(0,7,10),B(-1,6,6) and C(-4,9,6)
Then, AB=√((-1-0)^2+(6-7)^2+(6-10)^2 )
=√(1+1+16)=√18=3√2
BC=√((-4+1)^2+(9-6)^2+(6-6)^2 )
=√(9+9+0)=√18=3√2
AC=√((-4-0)^2+(9-7)^2+(6-10)^2 )
=√(16+4+16)=√36=6
Clearly, AC^2=AB^2+BC^2
Hence, triangle is right angled. Also, AB=BC
∴ Triangle is right angled isosceles
Q4. The equation of the plane containing the lines r ⃗=(a1 ) ⃗+λ (a2 ) ⃗ and r ⃗=(a2 ) ⃗+λ (a1 ) ⃗, is-
- None of these
- [r ⃗ (a_1 ) ⃗ (a_2 ) ⃗ ]=(a_1 ) ⃗∙(a_2 ) ⃗
- [r ⃗ (a_1 ) ⃗ (a_2 ) ⃗ ]=0
Solution
Q5.The lines r ⃗=a ⃗+λ(b ⃗×c ⃗ ) and r ⃗=b ⃗+μ(c ⃗×a ⃗ ) will intersect, if
Solution
The lines r ⃗=a ⃗+λ(b ⃗×c ⃗ ) and r ⃗=b ⃗+μ(c ⃗×a ⃗ ) pass through points a ⃗ and b ⃗ respectively and are parallel to vectors b ⃗×c ⃗ and c ⃗×a ⃗ respectively. Therefore, they will intersect, if
a ⃗-b ⃗,b ⃗×c ⃗ and c ⃗×a ⃗ are coplanar
⇒(a ⃗-b ⃗ )∙{(b ⃗×c ⃗ )×(c ⃗×a ⃗ )}=0
⇒(a ⃗-b ⃗ )∙{[b ⃗ c ⃗ a ⃗ ] c ⃗-[b ⃗ c ⃗ c ⃗ ] a ⃗ }=0
⇒(a ⃗-b ⃗ )∙c ⃗[b ⃗ c ⃗ a ⃗ ]=0
⇒a ⃗∙c ⃗-b ⃗∙c ⃗=0⇒a ⃗∙c ⃗=b ⃗∙c ⃗
Q6. The equation of the plane passing through the origin and containing the line
(x-1)/5=(y-2)/4=(z-3)/5 is
Solution
The equation of the plane through given line is
a(x-1)+b(y-2)+c(z-3)=0 …(i)
Since, the straight line lie on the plane.
∴ DR’s of the plane is perpendicular to the line ie,
5a+4b+5c=0 …(ii)
Since, the plane passes through (0, 0, 0), we get
-a-2b-3c=0
⟹a+2b+3c=0 ….(iii)
On solving Eqs. (ii) and (ii), we get
a/2=b/(-10)=c/6
From Eq. (i),
2(x-1)-10(y-2)+6(z-3)=0
⟹2x-10y+6z=0
⟹x-5y+3z=0
Q7.If the plane 3x+y+2z+6=0 is parallel to the line
(3x-1)/2b=3-y=(z-1)/a, then the value of 3a+3b is
Solution
Given line can be rewritten as
(x-1/3)/(2b/3)=(y-3)/(-1)=(z-1)/a
Given plane3x+y+2z+6=0 is parallel to the above line
∴2b/3∙3+1∙(-1)+2∙a=0
⟹2a+2b=1
⟹3a+3b=3/2
Q8.If the direction cosines of two lines are such that l+m+n=0,l^2+m^2+n^2=0,
then the angle between them is
Solution
Given, l+m+n=0 ….(i)
and l^2+m^2-n^2=0 …..(ii)
∴ l^2+m^2-(-l-m)^2=0
⟹2lm=0
⟹l=0 or m=0
if l=0,then n=-m
⟹l:m:n=0:1:-1
and if m=0, then n=-1
⟹l:m:n=1:0:-1
∴cos〖θ=(0+0+1)/(√(0+1+1) √(0+1+1))=1/2〗
⟹ θ=Ï€/3
Q9.The equation of the plane through the point (1, 2,3) and parallel to the plane x+2y+5z=0 is
Solution
Equation of plane passing through the point (1, 2, 3) is
A(x-1)+B(y-2)+C(z-3)=0 …(i)
Since, plane (i) is parallel to plane x+2y+5z=0
⇒A=1,B=2,C=5
Putting these values in Eq. (i), we get
(x-1)+2(y-2)+5(z-3)=0 is the required plane
Q10. The line drown from (4,-1,2) the point (-3,2,3) meets a plane at right angle at the point (-10,5,4), then the equation of plane is
Solution
Since, the line passing through the points (4,-1,2) and (-3,2,3). So, the DR’s of the line is (4+3,-1-2,2-3)ie,(7,-3,-1)
Since, the line is perpendicular to the plane therefore DR’s of this line is proportional to the normal of the plane.
∴ Required equation plane is
7(x+10)-3(y-5)-1(z-4)=0
⟹7x-3y-z+89=0
Q1. The distance between the points (1,4,5) and (2,2,3) is
Solution
The distance between given points =√((2-1)^2+(2-4)^2+(3-5)^2 ) =√(1+4+4)=3
The distance between given points =√((2-1)^2+(2-4)^2+(3-5)^2 ) =√(1+4+4)=3
Q2.The distance of the point of intersection of the line
(x-2)/3=(y+1)/4=(z-2)/12 and the plane x-y+z=5 from the point (-1,-5,-10)is
Solution
Given line is (x-2)/3=(y+1)/4=(z-2)/12=k (say) Any point on the line is (3k+2,4k-1,12k+2) This point lies on the plane x-y+z=5 ∴3k+2-(4k-1)+12k+2=5 ⟹11k=0 ⟹k=0 ∴ Intersection point is (2,-1,2) ∴ Distance, between points(2-1,2) and (-1,-5,-10) =√((-1-2)^2+(-5+1)^2+(-10-2)^2 ) =√(9+16+144)=13
Given line is (x-2)/3=(y+1)/4=(z-2)/12=k (say) Any point on the line is (3k+2,4k-1,12k+2) This point lies on the plane x-y+z=5 ∴3k+2-(4k-1)+12k+2=5 ⟹11k=0 ⟹k=0 ∴ Intersection point is (2,-1,2) ∴ Distance, between points(2-1,2) and (-1,-5,-10) =√((-1-2)^2+(-5+1)^2+(-10-2)^2 ) =√(9+16+144)=13
Q3. The triangle formed by the points (0, 7, 10), (-1,6,6),(-4,9,6) is
Solution
Let A(0,7,10),B(-1,6,6) and C(-4,9,6) Then, AB=√((-1-0)^2+(6-7)^2+(6-10)^2 ) =√(1+1+16)=√18=3√2 BC=√((-4+1)^2+(9-6)^2+(6-6)^2 ) =√(9+9+0)=√18=3√2 AC=√((-4-0)^2+(9-7)^2+(6-10)^2 ) =√(16+4+16)=√36=6 Clearly, AC^2=AB^2+BC^2 Hence, triangle is right angled. Also, AB=BC ∴ Triangle is right angled isosceles
Let A(0,7,10),B(-1,6,6) and C(-4,9,6) Then, AB=√((-1-0)^2+(6-7)^2+(6-10)^2 ) =√(1+1+16)=√18=3√2 BC=√((-4+1)^2+(9-6)^2+(6-6)^2 ) =√(9+9+0)=√18=3√2 AC=√((-4-0)^2+(9-7)^2+(6-10)^2 ) =√(16+4+16)=√36=6 Clearly, AC^2=AB^2+BC^2 Hence, triangle is right angled. Also, AB=BC ∴ Triangle is right angled isosceles
Q4. The equation of the plane containing the lines r ⃗=(a1 ) ⃗+λ (a2 ) ⃗ and r ⃗=(a2 ) ⃗+λ (a1 ) ⃗, is
-
- None of these
- [r ⃗ (a_1 ) ⃗ (a_2 ) ⃗ ]=(a_1 ) ⃗∙(a_2 ) ⃗
- [r ⃗ (a_1 ) ⃗ (a_2 ) ⃗ ]=0
Solution
Q5.The lines r ⃗=a ⃗+λ(b ⃗×c ⃗ ) and r ⃗=b ⃗+μ(c ⃗×a ⃗ ) will intersect, if
Solution
The lines r ⃗=a ⃗+λ(b ⃗×c ⃗ ) and r ⃗=b ⃗+μ(c ⃗×a ⃗ ) pass through points a ⃗ and b ⃗ respectively and are parallel to vectors b ⃗×c ⃗ and c ⃗×a ⃗ respectively. Therefore, they will intersect, if a ⃗-b ⃗,b ⃗×c ⃗ and c ⃗×a ⃗ are coplanar ⇒(a ⃗-b ⃗ )∙{(b ⃗×c ⃗ )×(c ⃗×a ⃗ )}=0 ⇒(a ⃗-b ⃗ )∙{[b ⃗ c ⃗ a ⃗ ] c ⃗-[b ⃗ c ⃗ c ⃗ ] a ⃗ }=0 ⇒(a ⃗-b ⃗ )∙c ⃗[b ⃗ c ⃗ a ⃗ ]=0 ⇒a ⃗∙c ⃗-b ⃗∙c ⃗=0⇒a ⃗∙c ⃗=b ⃗∙c ⃗
The lines r ⃗=a ⃗+λ(b ⃗×c ⃗ ) and r ⃗=b ⃗+μ(c ⃗×a ⃗ ) pass through points a ⃗ and b ⃗ respectively and are parallel to vectors b ⃗×c ⃗ and c ⃗×a ⃗ respectively. Therefore, they will intersect, if a ⃗-b ⃗,b ⃗×c ⃗ and c ⃗×a ⃗ are coplanar ⇒(a ⃗-b ⃗ )∙{(b ⃗×c ⃗ )×(c ⃗×a ⃗ )}=0 ⇒(a ⃗-b ⃗ )∙{[b ⃗ c ⃗ a ⃗ ] c ⃗-[b ⃗ c ⃗ c ⃗ ] a ⃗ }=0 ⇒(a ⃗-b ⃗ )∙c ⃗[b ⃗ c ⃗ a ⃗ ]=0 ⇒a ⃗∙c ⃗-b ⃗∙c ⃗=0⇒a ⃗∙c ⃗=b ⃗∙c ⃗
Q6. The equation of the plane passing through the origin and containing the line
(x-1)/5=(y-2)/4=(z-3)/5 is
Solution
The equation of the plane through given line is a(x-1)+b(y-2)+c(z-3)=0 …(i) Since, the straight line lie on the plane. ∴ DR’s of the plane is perpendicular to the line ie, 5a+4b+5c=0 …(ii) Since, the plane passes through (0, 0, 0), we get -a-2b-3c=0 ⟹a+2b+3c=0 ….(iii) On solving Eqs. (ii) and (ii), we get a/2=b/(-10)=c/6 From Eq. (i), 2(x-1)-10(y-2)+6(z-3)=0 ⟹2x-10y+6z=0 ⟹x-5y+3z=0
The equation of the plane through given line is a(x-1)+b(y-2)+c(z-3)=0 …(i) Since, the straight line lie on the plane. ∴ DR’s of the plane is perpendicular to the line ie, 5a+4b+5c=0 …(ii) Since, the plane passes through (0, 0, 0), we get -a-2b-3c=0 ⟹a+2b+3c=0 ….(iii) On solving Eqs. (ii) and (ii), we get a/2=b/(-10)=c/6 From Eq. (i), 2(x-1)-10(y-2)+6(z-3)=0 ⟹2x-10y+6z=0 ⟹x-5y+3z=0
Q7.If the plane 3x+y+2z+6=0 is parallel to the line
(3x-1)/2b=3-y=(z-1)/a, then the value of 3a+3b is
Solution
Given line can be rewritten as (x-1/3)/(2b/3)=(y-3)/(-1)=(z-1)/a Given plane3x+y+2z+6=0 is parallel to the above line ∴2b/3∙3+1∙(-1)+2∙a=0 ⟹2a+2b=1 ⟹3a+3b=3/2
Given line can be rewritten as (x-1/3)/(2b/3)=(y-3)/(-1)=(z-1)/a Given plane3x+y+2z+6=0 is parallel to the above line ∴2b/3∙3+1∙(-1)+2∙a=0 ⟹2a+2b=1 ⟹3a+3b=3/2
Q8.If the direction cosines of two lines are such that l+m+n=0,l^2+m^2+n^2=0,
then the angle between them is
Solution
Given, l+m+n=0 ….(i) and l^2+m^2-n^2=0 …..(ii) ∴ l^2+m^2-(-l-m)^2=0 ⟹2lm=0 ⟹l=0 or m=0 if l=0,then n=-m ⟹l:m:n=0:1:-1 and if m=0, then n=-1 ⟹l:m:n=1:0:-1 ∴cos〖θ=(0+0+1)/(√(0+1+1) √(0+1+1))=1/2〗 ⟹ θ=Ï€/3
Given, l+m+n=0 ….(i) and l^2+m^2-n^2=0 …..(ii) ∴ l^2+m^2-(-l-m)^2=0 ⟹2lm=0 ⟹l=0 or m=0 if l=0,then n=-m ⟹l:m:n=0:1:-1 and if m=0, then n=-1 ⟹l:m:n=1:0:-1 ∴cos〖θ=(0+0+1)/(√(0+1+1) √(0+1+1))=1/2〗 ⟹ θ=Ï€/3
Q9.The equation of the plane through the point (1, 2,3) and parallel to the plane x+2y+5z=0 is
Solution
Equation of plane passing through the point (1, 2, 3) is A(x-1)+B(y-2)+C(z-3)=0 …(i) Since, plane (i) is parallel to plane x+2y+5z=0 ⇒A=1,B=2,C=5 Putting these values in Eq. (i), we get (x-1)+2(y-2)+5(z-3)=0 is the required plane
Equation of plane passing through the point (1, 2, 3) is A(x-1)+B(y-2)+C(z-3)=0 …(i) Since, plane (i) is parallel to plane x+2y+5z=0 ⇒A=1,B=2,C=5 Putting these values in Eq. (i), we get (x-1)+2(y-2)+5(z-3)=0 is the required plane
Q10. The line drown from (4,-1,2) the point (-3,2,3) meets a plane at right angle at the point (-10,5,4), then the equation of plane is
Solution
Since, the line passing through the points (4,-1,2) and (-3,2,3). So, the DR’s of the line is (4+3,-1-2,2-3)ie,(7,-3,-1) Since, the line is perpendicular to the plane therefore DR’s of this line is proportional to the normal of the plane. ∴ Required equation plane is 7(x+10)-3(y-5)-1(z-4)=0 ⟹7x-3y-z+89=0
Since, the line passing through the points (4,-1,2) and (-3,2,3). So, the DR’s of the line is (4+3,-1-2,2-3)ie,(7,-3,-1) Since, the line is perpendicular to the plane therefore DR’s of this line is proportional to the normal of the plane. ∴ Required equation plane is 7(x+10)-3(y-5)-1(z-4)=0 ⟹7x-3y-z+89=0
