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. If the lines
(x-1)/2=(y+1)/3=(z-1)/4
and (x-3)/1=(y-k)/2=z/1
interested, then the value of k is
Solution
(b) Given,(x-1)/2=(y+1)/3=(z-1)/4=λ and (x-3)/1=(y-k)/2=z/1=μ ⟹x=2λ+1,y=3λ-1,z=4λ+1 and x=μ+3,y=2μ+k,z=μ As the lines intersect they must have a point in common. ∴2λ+1=μ+3,3λ-1=2μ+k,4λ+1=μ ⟹ λ=-3/2 and μ=-5 ∴k=3λ-2μ-1 ⟹k=3(-3/2)-2(-5)-1 ⟹k=9/2
(b) Given,(x-1)/2=(y+1)/3=(z-1)/4=λ and (x-3)/1=(y-k)/2=z/1=μ ⟹x=2λ+1,y=3λ-1,z=4λ+1 and x=μ+3,y=2μ+k,z=μ As the lines intersect they must have a point in common. ∴2λ+1=μ+3,3λ-1=2μ+k,4λ+1=μ ⟹ λ=-3/2 and μ=-5 ∴k=3λ-2μ-1 ⟹k=3(-3/2)-2(-5)-1 ⟹k=9/2
Q2.In a three-dimensional xyz space, the equation x^2-5x+6=0 represents
Solution
(b) x^2-5x+6=0 ⇒x-2=0,x-3=0 Which represents planes
(b) x^2-5x+6=0 ⇒x-2=0,x-3=0 Which represents planes
Q3. The direction ratios of a normal to the plane through (1, 0,0) and (0, 1, 0), which makes an angle of π/4 with the plane x+y=3 are
Solution
(b)
Any plane through (1, 0, 0) is a(x-1)+by+cz=0 (i)
It passes through (0, 1, 0)
∴ a(0-1)+b(1)+c(0)=0 ⇒-a+b=0 (ii)
(i) makes an angle of π/4 with x+y=3, therefore
cos〖Ï€/4=(a(1)+b(1)+c(0))/(√(a^2+b^2+c^2 ) √(1+1+0))〗
⇒ 1/√2=(a+b)/(√2 √(a^2+b^2+c^2 ))
⇒a+b=√(a^2+b^2+c^2 )
Squaring, we get
a^2+b^2+2ab=a^2+b^2+c^2
⇒2ab=c^2 ⇒2a^2=c^2
⇒c=√2 a (using (ii))
Hence, a:b:c=a:a:√2 a
=1:1:√2
Q4.
Q5.A plane passes through a fixed point (a,b,c). The locus of the foot of the perpendicular to it from the origin is a sphere of radius
Solution
(a)
Let the foot of the perpendicular from the origin on the given plane be P(α,β,γ). Since the plane passes through A(a,b,c)
AP⊥OP ⇒ (AP) ⃗.(OP) ⃗=0
⇒[(α-a) i ̂+(β-b) j ̂+(γ-c) k ̂ ].(αi ̂+βj ̂+γk ̂ )=0
⇒ α(α-a)+β(β-b)+γ(γ-c)=0
Hence, the locus of (α,β,γ) is
x(x-a)+y(y-b)+z(z-c)=0
x^2+y^2+z^2-ax-by-cz=0
Which is a sphere of radius 1/2 √(a^2+b^2+c^2 )
Q6. Line r =a +λb will not meet the plane r ∙n =q, if
Q7.Let A(a ) and B(b ) be points on two skew lines r =a +λ p and r =b +uq ⃗ and the shortest distance between the skew lines is 1, where p and q are unit vectors forming adjacent sides of a parallelogram enclosing an area of 1/2 units. If an angle between AB and the line of shortest distance is 60°, then AB=
Q8.Consider triangle AOB in the x-y plane, where A≡(1,0,0);B≡(0,2,0); and O≡(0,0,0). The new position of O, when triangle is rotated about side AB by 90° can be
Solution
(c)
Equation of a line AB is (x-1)/1=y/(-2)=z/0=λ
Now AB⊥OC⇒ 1(λ+1)+(-2λ)(-2)=0 ⇒5λ=-1 ⇒ λ=-1/5
⇒C is (4/5,2/5,0). Now
x_1^2+(y_1-2)^2+z_1^2=4 (i)
and (x_1-1)^2+y_1^2+z_1^2=1 (ii)
Now OC⊥CD
⇒ (x_1-4/5) 4/5+(y_1-2/5) 2/5+(z_1-0)0=0 (iii)
From (i) and (ii), we get
-4y_1+2x_1=0 ⇒ x_1=2y_1
From (iii), putting x_1=2y_1⇒2y_1=4/5 ⇒y_1=2/5 ⇒x_1=4/5. Putting this value of x_1 and y_1 in (i), we get
z_1=±2/√5
Q9.Let A(1,1,1),B(2,3,5) and C(-1,0,2) be three points, then equation of a plane parallel to the plane ABC which is at distance 2 is
Solution
(a)
A(1,1,1),B(2,3,5),C(-1,0,2) direction ratios of AB are <1,2,4>
Direction ratios of AC are <-2,-1,1>
Therefore, direction ratios of normal to plane ABC are <2,-3,1>
As a result, equation of the plane ABC is 2x-3y+z=0
Let the equation of the required plane is 2x-3y+z=k, then |k/√(4+9+1)|=2
k=±2√14
Hence, equation of the required plane is 2x-3y+z+2√14=0
Q10. Value of λ such that the line (x-1)/2=(y-1)/3=(z-1)/λ is ⊥ to normal to the plane r ∙(2i + 3j + 4k )=0 is
