Free Three Dimensional Geometry 01 Practice Test - 12th Grade - Commerce
Question 1
Points (1, 1, 1), (–2, 4, 1), (–1, 5, 5) and (2, 2, 5) are the vertices of a
SOLUTION
Solution : B
Let A=(1,1,1);B=(-2,4,1);C=(-1,5,5) & D=(2,2,5)
AB=√9+9+0=3√2,BC=√1+1+16=3√2 and CD=3√2 and AD=3√2.
also AB.CD=0. Hence it is a square.
Question 2
A straight line, makes an angle of 60∘ with each of y and z-axis,then the inclination of the line with x-axis is
SOLUTION
Solution : C
If α is a single at which the straight line is inclined to x-axis, then cos2α+cos260∘+cos260∘=1
⇒cos2α=12
⇒α=45∘or135∘.
Question 3
If the projections of a line on the axes are 9, 12 and 8. Then the length of the line is
SOLUTION
Solution : B
Given r cosα=9,r cosβ=12 and r cosγ=8
∴r2(cos2α+cos2β+cos2γ)=81+144+64
r2.1=289
r=17
Question 4
If the direction cosines of a variable line in two adjacent positions be l, m, n and l + a, m + b, n + c and the small angle between the two positions be θ, then :
SOLUTION
Solution : B
l2+m2+n2=1,(l+a)2+(m+b)2+(n+c)2=1⇒al+bm+cn=−12(a2+b2+c2)
cosθ=(a+l)l+m(b+m)+n(c+n)
= 1 + al + bm + cn
⇒a2+b2+c2=2(1−cosθ)=4sin2(θ2)
Since θ is small, sin(θ2)≈θ2
∴θ2=a2+b2+c2[∵sin2(θ2)≈θ24]
Question 5
Direction cosines of the line which is perpendicular to the lines whose direction ratios are 1, –1, 2 and 2, 1, –1 are given by :
SOLUTION
Solution : A
If l, m, n are the d.c.’s of the line which is perpendicular to the given lines, then
l – m + 2n = 0 and 2l + m –n = 0
∴l1−2=m4+1=n1+2⇒l−1=m5=n3.
so d.r's =(−1,5,3)
Question 6
The three lines drawn from O with direction ratios, (1, –1, k), (2, –3, 0) and (1, 0, 3) are coplanar. Then k =
SOLUTION
Solution : A
If lines are coplanar and l, m, n are d.c.’s of the normal to the plane, then
l – 1.m + k.n = 0
l – 3.m + 0.n = 0
l – 0.m + 3.n = 0
⇒∣∣ ∣∣1−1k2−30103∣∣ ∣∣=0⇒k=1
Question 7
The equation of the plane containing the two lines of intersection of the two pairs of planes x + 2y – z – 3 = 0 and 3x – y + 2z – 1 = 0, 2x – 2y + 3z = 0 and x – y + z + 1 =0 is :
SOLUTION
Solution : D
The planes through given lines of intersection are
(x+2y−z−3)+λ(3x−y+2z−1)=0....(1)
and (2x−2y+3z)+μ(x−y+z+1)=0....(2)
Planes (1) and (2) are same. Comparing the coefficients,
we get, 1+3λ2+μ=2−λ−2−μ=−1+2λ3+μ=−3−λμ
Solving for λ,μ, these equations are inconsistent. Therefore, there exists no such plane i.e., the lines are neither parallel nor intersecting.
Question 8
Equation of the plane perpendicular to the plane x – 2y + 5z + 1 = 0 which passes through the points (2, –3, 1) and (–1, 1, –7) is given by :
SOLUTION
Solution : B
Equation of the plane through (2, –3, 1) is
a(x – 2) + b(y + 3) + c(z – 1) = 0
It passes through (–1, 1, –7) and perpendicular to the plane x – 2y + 5z + 1 = 0
∴ –3a+ 4b – 8c = 0 and a – 2b + 5c = 0
∴a4=b7=c2
Hence, equation of required plane is
4x + 7y + 2z + 11 = 0
Question 9
The foot of the perpendicular from (0, 2, 3) to the line x+35=y−12=z+43 is :
SOLUTION
Solution : D
x+35=y−12=z+43=λ
Let P(5λ−3,2λ+1,3λ−4) be the foot of perpendicular of A(0, 2, 3). Then AP and given line are perpendicular to each other.
∴(5λ−3)5+(2λ−1)2+(3λ−7)3=0
⇒38λ−38=0⇒λ=1.
Hence, P = (2, 3, –1)
Question 10
Reflection of the line x−1−1=y−23=z−41 in the plane x +y +z =7 is :
SOLUTION
Solution : C
Given line passes through the point A(1, 2, 4) and this point also lies in the plane. To find the reflection of the line, we need one more point of the line. Clearly P(0, 5, 5) also lies in the line.
Let Q(α,β,γ)be the reflection of P in the plane x + y + z = 7
Then α2=β+52=γ+52=7⇒α+β+γ=4
Also PQ⊥ to the plane, i.e., parallel to the normal of the plane.
∴α−01=β−51=γ−51=λ⇒α=λ,β=λ+5,γ=λ+5∴λ+λ+5λ+5=4⇒λ=−2
∴ Q is (-2,3,3)
Now AQ will be the reflection of the given line, equation of AQ is
x−1−3=y−21=z−4−1