Important topics for Maths has been designed in such a way that it offers very practical and application-based learning to further make it easier for students to understand every concept or topic by correlating it with the day-to-day experiences.
Q1. A tetrahedron has vertices O(0,0,0),A(1,2,1),B(2,1,3) and C(-1,1,2), then angle between faces OAB and ABC will be:
Solution
Q2.Let the equations of a line and a plane be (x+3)/2=(y-4)/3=(z+5)/2 and 4x-2y-z=1, respectively, the
Solution
4(2)-2(3)-1(2)=0 Also, point (-3,4,-5) does not lie on the plane Therefore, the line is parallel to the plane
4(2)-2(3)-1(2)=0 Also, point (-3,4,-5) does not lie on the plane Therefore, the line is parallel to the plane
Q3. The length of the perpendicular from the origin to the plane passing through the point a and containing the line r =b +λc ⃗is
Q5.The intercept made by the plane r ∙n =q on the x-axis is
Solution
x intercept is say x_1 ⇒ Plane passes through it ∴x_1 i ̂∙n ⃗=q ⇒x_1=q/(i ̂∙n ⃗ )
x intercept is say x_1 ⇒ Plane passes through it ∴x_1 i ̂∙n ⃗=q ⇒x_1=q/(i ̂∙n ⃗ )
Q6. What is the nature of the intersection of the set of planes x+ay+(b+c)z+d=0,x+by+(c+a)z+d=0 and x+cy+(a+b)z+d=0?
Q8.The line through i ̂+3j ̂+2k ̂ and ⊥ to the line r =(i ̂+2j ̂-k ̂ )+λ(2i ̂+j ̂+k ̂) and r =(2i ̂+6j ̂+k ̂ )+μ(i ̂+2j ̂+3k ̂) is
Solution
The required line passes through the point i ̂+3j ̂+2k ̂ and is perpendicular to the lines r ⃗=(i ̂+2j ̂-k ̂ )+λ(2i ̂+j ̂+k ̂) and r ⃗=(2i ̂+6j ̂+k ̂ )+μ(i ̂+2j ̂+3k ̂); therefore it is parallel to the vector b ⃗=(2i ̂+6j ̂+k ̂ )×μ(i ̂+2j ̂+3k ̂ )=(i ̂-5j ̂+3k ̂) Hence, the equation of the required line is r ⃗=(i ̂+3j ̂+2k ̂ )+λ(i ̂-5j ̂+3k ̂)
The required line passes through the point i ̂+3j ̂+2k ̂ and is perpendicular to the lines r ⃗=(i ̂+2j ̂-k ̂ )+λ(2i ̂+j ̂+k ̂) and r ⃗=(2i ̂+6j ̂+k ̂ )+μ(i ̂+2j ̂+3k ̂); therefore it is parallel to the vector b ⃗=(2i ̂+6j ̂+k ̂ )×μ(i ̂+2j ̂+3k ̂ )=(i ̂-5j ̂+3k ̂) Hence, the equation of the required line is r ⃗=(i ̂+3j ̂+2k ̂ )+λ(i ̂-5j ̂+3k ̂)
Q9.Equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and ⊥ to the plane 2x+6y+6z-1=0 is
Solution
(b)
Any plane through (2, 2, 1) is
a(x-2)+b(y-2)+c(z-1)=0 (i)
It passes through (9, 3, 6) if 7a+b+5c=0 (ii)
Also (i) is perpendicular to 2x+6y+6z-1=0, we have
2a+6b+6c=0
∴ a+3b+3c=0 (iii)
∴ a/(-12)=b/(-16)=c/20 or a/3=b/4=c/(-5) (from (ii) and (iii))
Therefore, the required plane is 3(x-2)+4(y-2)-5(z–1)=0 or 3x+4y-5z-9=0
Q10. The intersection of the spheres x^2+y^2+z^2+7x-2y-z=13 and x^2+y^2+z^2-3x+3y+4z=8 is the same as the intersection of one of the spheres and the plane
Solution
The given sphere are
x^2+y^2+z^2+7x-2y-z-13=0 (i)
and x^2+y^2+z^2-3x+3y+4z-8=0 (ii)
Subtracting (ii) from (i), we get
10x-5y-5z-5=0
⇒ 2x-y-z=1
