Free Objective Test 02 Practice Test - 11th and 12th
Question 1
If the perpendicular distance of a point other than the origin from the plane x+y+z=p is equal to the distance of the plane from the origin, then the coordinates of the point are
SOLUTION
Solution : C
The perpendicular distance of the origin (0, 0, 0) from the plane x + y + z = p is ∣∣∣−p√1+1+1∣∣∣=|p|√3
If the coordinates of P are (x, y, z), then we must have
∣∣∣x+y+z−p√3∣∣∣=|p|√3⇒|x+y+z−p|=|p|
Which is satisfied by (c)
Question 2
The shortest distance from the plane 12x + 4y + 3z = 327 to the sphere x2+y2+z2+4x−2y−6z=155 is
SOLUTION
Solution : B
The centre of the sphere is (-2, 1, 3) and its radius is √4+1+9+155=13.
Length of the perpendicular from the centre of the sphere on the plane is ∣∣∣−24+4+9−327√144+16+9∣∣∣=33813=26
So the plane is outside the sphere and the required distance is equal to 26 - 13 = 13.
Question 3
The ratio in which yz- plane divides the segment joining the points (-2, 4, 7) and (3, -5, 8) is
SOLUTION
Solution : A
let YZ-plane divide the segment joining (-2, 4, 7) and (3, -5, 8) in the ratio λ:1. Then ⇒3λ−2λ+1=0⇒λ=23 and the required ratio is 2 : 3
Question 4
If (l1,m1,n1) and (l2,m2,n2,) are d.c.'s of ¯¯¯¯¯¯¯¯¯¯OA, ¯¯¯¯¯¯¯¯OB such that ∠AOB=θ where ‘O’ is the origin, then the d.c.’s of the internal bisector of the angle ∠AOB are
SOLUTION
Solution : B
Let OA and OB be two lines with d.c’s l1,m1,n1 and l2,m2,n2,. Let OA = OB = 1. Then, the coordinates of A and B are (l1,m1,n1) and (l2,m2,n2), respectively. Let OC be the bisector of ∠AOB .Then, C is the mid point of AB and so its coordinates are (l1+l22,m1+m22,n1+n22).
∴ d.r's of OC are l1+l22,m1+m22,n1+n22
We have, OC=√(l1+l22)2+(m1+m22)2+(n1+n22)2=12√(l21+m21+n21)+(l22+m22+n22)+(l1l2+m1m2+n1n2)=12√2+2cosθ[Q cos θ=l1l2+m1m2+n1n2]=12√2(1+cosθ)=cos(θ2)
∴ d.c's of OC are
l1+l22(OC),m1+m22(OC),n1+n22(OC)
Question 5
The projection of any line on co-ordinate axes be respectively 3, 4, 5 then its length is
SOLUTION
Solution : C
Let the line segment be AB, then as given
ABcosα=3,ABcosβ=4,ABcosγ=5⇒AB2(cos2α+cos2β+cos2γ)=32+42+52AB=√9+16+25=5√2
where α,β and γ are the angles made by the line with the axes.
Question 6
A variable plane which remains at a constant distance p from the origin cuts the coordinate axes in A, B, C. The locus of the centroid of the tetrahedron OABC is y2z2+z2x2+x2y2=kx2y2z2, where k is equal to
SOLUTION
Solution : D
Let equation of plane is
lx+my+nz=p
or x(pl)+y(pm)+z(pn)=1
Coordinates of A, B, C are (pl,0,0)(0,pm,0) and
(0,0,pn) respectively.
∴ Centroid of OABC is (p4l,p4m,p4n)
x1=p4l,y1=p4m,z1=p4n∵l2+m2+n2=1⇒p216x21+p216y21+p216z21=1
or x21y21+y21z21+z21x21=16/p2x21y21z21
∴ Locus is x2y2+y2z2+z2x2=16p2x2y2z2
∴k=16p2
Question 7
The line joining the points (1, 1, 2) and (3, -2, 1) meets the plane 3x + 2y + z = 6 at the point
SOLUTION
Solution : B
The straight line joining the points (1, 1, 2) and (3, -2, 1) is x−12=y−1−3=z−2−1=r (say)
∴ Point is (2r + 1, 1 -3r, 2 –r )
Which lies on 3x + 2y + z = 6
∴3(2r+1)+2(1−3r)+2−r=6
∴r=1
Required point is (3, -2, 1).
Question 8
The equation of the plane passing through the line of intersection of the planes 3x-y-4z=0 and x+3y+6=0 whose distance from the origin is 1, is
SOLUTION
Solution : A
Equation of planes passing through intersection of the planes 3x-y-4z=0 and x+3y+6=0 is, (3x−y−4z)+λ(x+3y+6)=0.....(i)
Given, distance of plane (i) from origin is 1.
∴6λ√(3+λ)2+(3λ−1)2+42=1
or 36λ2=10λ2+26 or λ=±1
Put the value of λ in (i),
∴(3x−y−4z)±(x+3y+6)=0
or 4x+2y -4z+6=0 i.e 2x+y-2z+3 =0
and 2x-4y-4z-6=0
Thus the required planes are x-2y-2z-3=0 and 2x+y-2z+3=0.
Question 9
The equation of the plane passing through the points (1,-3,-2) and perpendicular to planes x +2y+2z=5 and 3x+3y+2z=8, is
SOLUTION
Solution : A
l+2m+2n=0, 3l+3m+2n=0, l2+m2+n2=1, we get l, m, n from these equations and then putting the values in l(x-1)+m(y+3) +n(z+2)=0, we get the required result.
Trick: Checking conversely,
2(x)-4(y)+3(z)-8 =0,
So, it passes through given point.
1(2)+2(-4)+2(3) =0,
So, it is perpendicular to x+2y+2z=5.
3(2)+3(-4)+2(3)=0,
So, it is perpendicular to 3x+3y +2z=8.
Question 10
The distance between two points P and Q is d and the length of their projections of PQ on the coordinate planes are d1,d2,d3. Then d21+d22+d23=kd2, where k is
SOLUTION
Solution : D
Here, d1=d cos(90∘−α)
d2=d cos(90∘−β)
and d3=d cos(90∘−γ)
∴d1=dsinα
d2=dsinβ
and d3=dsinγ
∴d21+d22+d23=kd2
⇒d2(sin2α+sin2β+sin2γ)=k(d)2
∴k=2
Question 11
If a line makes the angle α,β,γ with three dimensional co-ordinate axes respectively, then cos2α+cos2β+cos2γ
SOLUTION
Solution : B
cos2α+cos2β+cos2γ=2cos2α−1+2cos2β−1+2cos2γ−1=2(cos2α+cos2β+cos2γ)−3=2−3=−1
Question 12
Perpendicular distance of the point (3, 4, 5) from the y-axis, is
SOLUTION
Solution : A
Perpendicular distance of (x,y,z) from y axis is given by =√x2+z2.
=> Required distance =√32+52=√34.
Question 13
The perpendicular distance of the origin form the plane which makes intercepts 12, 3 and 4 on x, y, z axes respectively, is
SOLUTION
Solution : D
Let equation of plane be lx + my + nz = p, where p is the perpendicular distance of the plane from origin and (l,m,n) are the direction cosines of the normal to the plane
Or x(pl)+y(pm)+z(pn)=1
According to question,
pl=12,pm=3,pn=4
or p12=l,p3=m,p4=n
or p2144+p29+p216=l2+m2+n2=1
⇒p2(1144+19+116)=1p2(1+16+9144)=1
Or p2=14426=7213
∴p=6√2√13
Question 14
The equation of the plane passing through the points (2, –4, 0), (3, –4, 5) and parallel to the line 2x = 3y = 4z is :
SOLUTION
Solution : D
Equation of the plane through the point (2, –1, 0) is
a(x – 2) + b(y + 1) + c(z – 0) = 0 . . . (1)
It passes through the point (3, –4, 5)
∴ a – 3b + 5c = 0 . . . (2)
The plane (1) is parallel to the line
2x = 3y = 4z i.e.,x6=y4=z3...(3)
∴ Normal to the plane (1) is perpendicular to the line (3).
∴ 6a + 4b + 3c = 0 . . .(4)
From (2) and (4), we get
a−29=b27=c22
∴ Equation of the required plane is
– 29(x – 2) + 27(y + 1) + 22(z) = 0
⇒−29x+27y+22z+85=0
⇒29x−27y−22z=85
Question 15
The distance of the point (1, –2, 3) from the plane x – y + z = 5 measured parallel to the line (12)x=(13)y=(−16)z is :
SOLUTION
Solution : A
Line parallel to the line x2=y2=z−6 and passing through (1,-2, 3) is
x−12=y+23=z−3−6=r
A = (2r + 1, 3r – 2, – 6r + 3) is a point on this line.
A lies in the plane x – y + z = 5
∴2r+1−3r+2−6r+3=5⇒r=17∴A=(97,−117,157)
Distance of A from (1, –2, 3) is
√(97−1)2+(−117+2)2+(157−3)2=17√4+9+36=77=1