Mathametics 06 – 3-Dimensional Co-ordinate geometry

1) The shortest distance between the lines 2x+ y+z=1, 3x + y + 2z = 2 and x = y = z.

(a) 1/√2               (b) √2                (c) 3/√2               (d) √3/2

(a) Any plane passing through first line 2x + y + z – 1 + λ(3x + y + 2z – 2) = 0,

if it is parallel to second line (2 + 3λ)1 + (1 + λ)1 + (1 + 2λ)1 = 0

⇒ λ = – 2/3.

Plane is y – z + 1 = 0

Distance from (0, 0, 0) = 1/√2 .

2) A mirror and a source of light are situated at the origin O and at a point on the line OX respectively. A ray of light from the source strikes the mirror at O and is reflected. The direction ratios of the normal to the plane of the mirror are (1, -1, 1); then the direction cosines of the reflected ray are

(a) 1/3 , -2/3 , 2/3               (b) -1/3 , 2/3 , 2/3                (c) -1/3 , -2/3 , -2/3               (d) 1/3 , 2/3 , -2/3

Screenshot 20200810 114529 2

(d) The d.c’s of incident ray are (–1, 0, 0). Let the d.c’s of reflected ray be (l, m, n), then the direction ratios of the normal to the plane of mirror will be (l – 1, m , n).

So, (l-1)/1 = m/-1 = n/1 = k (say)

⇒ l = k + 1, m = –k, n = k

⇒ (k + 1)2 + (–k)2 + (k)2 = l2 + m2 + n2 = 1

⇒ k = –2/3.

So, (l, m, n) is (1/3 , 2/3 , -2/3)

3) A variable plane at a distance of 1 unit from the origin cuts the co-ordinate axes at A, B and C. If the centroid D (x, y, z) of triangle ABC satisfies the relation (1/x2)+(1/y2)+(1/z2) = k, then the value of k is

(a) 3               (b) 1                (c) 1/3               (d) 9

(d) Let (x/a)+(y/b)+(z/c) = 1 be the variable plane so that |1/√(1/a2)+(1/b2)+(1/c2)| = 1 .

The plane meets the coordinate axes at A (a, 0, 0), B (0, b, 0), C (0, 0, c).

The centroid D of the triangle ABC is (a/3 , b/3 , c/3)

⇒ x = a/3 , y = b/3 , z = c/3 and (1/a2)+(1/b2)+(1/c2) = 1

⇒ (1/x2)+(1/y2)+(1/z2) = 9 . 

4) A straight line passes through the point (2, –1, –1). It is parallel to the plane 4x + y + z + 2 = 0 and is perpendicular to the line x/1 = y/-2 = (z-5)/1 . The equation of the straight line is

(a) (x-2)/4 = (y+1)/1 = (z+1)/1                                (b) (x+2)/4 = (y-1)/1 = (z-1)/1                

(c) (x-2)/-1 = (y+1)/1 = (z+1)/3                               (d) (x+2)/-1 = (y-1)/1 = (z-1)/3

(c) Let direction cosines of straight line be l, m, n ∴ 4l + m + n = 0

l – 2m + n = 0 ⇒ l/3 = m/-3 = n/-9 ⇒ l/-1 = m/1 = n/3

∴ Equation of straight line is (x-2)/-1 = (y+1)/1 = (z+1)/3 .

5) The equation of the straight line through the origin parallel to the line (b + c)x + (c + a)y + (a + b)z = k = (b – c)x + (c – a)y + (a – b)z is

(a) x/(b2-c2) = y/(c2-a2) = z/a2-b2)                            (b) x/b = y/c = z/a                

(c) x/(a2-bc) = y/(b2-ca) = z/c2-ab)                           (d) none of these

(c) Equation of straight line through origin is (x-0)/l = (y-0)/m = (z-0)/n

where l((b+c) + m(c+a) + n(a + b) = 0

and l(b – c) + m(c -a) + n(a – b) = 0

On solving, l/2(a2-bc) = m/2(b2-ca) = n/2(c2-ab)

∴ Equation of the straight line is x/(a2-bc) = y/(b2-ca) = z/(c2-ab)

6) The angle between two lines whose direction cosines are given by the equation l + m + n = 0, l2 + m2 + n2 = 0 is

(a) π/3               (b) 2π/3               (c) π/4               (d) none of these

(b) Eliminating n between the two relations, we have l2 + m2 – (l + m)2 = 0 or 2lm = 0 ⇒ either l = 0 or m = 0 if l = 0, then m + n = 0 i,e. m = – n

⇒ l/0 = m/1 = n/-1 , giving the direction ratios of one line.

If m = 0, then l + n = 0 i.e. l = – n

⇒ l/1 = m/1 = n/-1 , giving direction ratios of the other lines.

The angles between these lines is cos-1 {± [0.1+0.1+(-1)(-1)]/√[02+12+(-1)2]√[12+02+(-1)2]} = cos-1 (±1/2) = π/3 or 2π/3 .

7) The equation of the plane which contains the line of intersection of the planes x + y + z – 6 = 0 and 2x + 3y + z + 5 = 0 and perpendicular to the xy plane is:

(a) x – 2y + 11 = 0               (b) x + 2y + 11 = 0               (c) x + 2y – 11 = 0               (d) x – 2y – 11 = 0

(b) Equation of the required plane is (x + y + z – 6) + λ(2x + 3y + z + 5) = 0

i.e. (1 + 2λ)x + (1 + 3λ)y + (1 + λ)z + (–6 + 5λ) = 0

This plane is perpendicular to xy plane whose equation is z = 0

i.e. 0 . x + 0 . y + z = 0

∴ By condition of perpendicularity

0.(1 + 2λ) + 0. (1 + 3λ) + (1 + λ) .1 = 0 i.e. λ = –1

∴ Equation of required plane is (1 – 2)x + (1 – 3)y + (1 – 1)z + (–6 – 5) = 0

or x + 2y + 11 = 0.

8) The coordinates of the foot of the perpendicular drawn from the origin to the plane 3x + 4y – 6z + 1 = 0 are :

(a) (-3/61 , 4/61 , 6/61)                                            (b) (3/61 , -4/61 , 6/61)               

(c) (-3/61 , -4/61 , 6/61)                                           (d) (3/61 , 4/61 , 6/61)

(c) The equation of the plane is 3x + 4y – 6z + 1 = 0 ……(1)

The direction ratios of the normal to the plane (1) are 3, 4, –6. So equation of the line through (0, 0, 0) and perpendicular to the plane (1) are

x/3 = y/4 = z/-6 = r (say) ……(2)

The coordinates of any point P on (2) are (3r, 4r, –6r). If this point lie on the plane (1), then

3(3r) + 4(4r) – 6(–6r) + 1 = 0 i.e. r = – 1/61

Putting the value of r coordinates of the foot of the perpendicular P are (-3/61 , -4/61 , 6/61) .

9) If equation of a sphere is 2(x2 + y2 + z2) – 4x – 8y + 12z – 7 = 0 and one extremity of its diameter is (2, –1, 1), then the other extremity of diameter of the sphere will be

(a) (2,9,-13)               (b) (0,9,7)               (c) (0,5,7)               (d) (2,5,-13)

(c) The centre of the sphere is (1, 2, –3) so if other extremity of diameter is (x1, y1, z1), then

(x1+2)/2 = 1

(y1-1)/2 = 2

(z1+1)/2 = -3

∴ Required point is (0, 5, 7).

10) The image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0 is :

(a) (-3,5,2)               (b) (3,2,5)               (c) (-5,3,2)               (d) (-2,5,3)

Screenshot 20200810 114741 2

(a) As it is clear from the figure that PQ will be perpendicular to the plane and foot of this perpendicular is mid point

of PQ i.e. N.

So, direction ratios of line PQ is 2, –1, 1

⇒ Equation of line PQ = (x-1)/2 = (y-3)/-1 = (z-4)/1 = r (say)

Any point on line PQ is (2r + 1, –r + 3, r + 4)

If this point lies on the plane, then 2(2r + 1) – (–r + 3) + (r + 4) + 3 = 0 ⇒ r = –1

∴ coordinate of foot of perpendicular N = (–1, 4, 3)

As N is middle point of PQ

∴ –1 = (1+x1)/2

4 = (3+y1)/2

3 = (4+z1)/2

⇒ x1 = –3, y1 = 5, z1 = 2

∴ Image of point P (1, 3, 4) is the point Q (–3, 5, 2).

11) The equation of the sphere which passes through the points (1, 0, 0), (0, 1, 0) and (0, 0, 1) and has its radius as small as possible is :

(a) 3(x2 + y2 + z2) + 2(x + y + z) – 1= 0                    (b) 3(x2 + y2 + z2) – 2(x + y + z) – 1= 0              

(c) (x2 + y2 + z2) – 2(x + y + z) – 1= 0                      (d) (x2 + y2 + z2) + 2(x + y + z) – 1= 0

(b) Let equation of sphere be given by x2 + y2 + z2 + 2ux + 2vy + 2wz + d = 0 ……(1)

As sphere passes through points (1, 0, 0), (0, 1, 0) and (0, 0, 1).

So we have 1 + 2u + d = 0, 1 + 2v + d = 0, 1 + 2w + d = 0

On solving u = v = w = – 1/2 (d + 1)

If r is the radius of the sphere, then r = √(u2+v2+w2-d)

r2 = 3/4(d + 1)2 – d = μ (say)

for r to be minimum

dμ/dd = 0 ⇒ 3/4 .2(d + 1) – 1 = 0 or d = –1/3

Also, d2 μ/dd2 3/2 = positive at d = –1/3

Hence μ is minimum at d = –1/3 

So, substituting value of d we have u = v = w = –1/3

⇔ equation of the sphere x2 + y2 + z2 – 2/3(x + y + z) – 1/3 = 0

⇒ 3(x2 + y2 + z2) – 2(x + y + z) – 1= 0.

12) A point moves so that the ratio of its distances from two fixed points is constant. Its locus is a:

(a) plane               (b) straight lines               (c) circle               (d) sphere

(d) Let the coordinates of moving point P be (x, y, z). Let A (a, 0, 0) and B (–a, 0, 0) be two fixed points.

According to given condition

AP/BP = constant = k (say)

⇒ AP2 = k2BP2

or, (x – a)2 + (y – 0)2 + (z – 0)2 = k2{(x + a)2 + (y – 0)2 + (z – 0)2}

⇒ (1 – k2)(x2 + y2 + z2) – 2ax(1 + k2) + a2(1 – k2) = 0

∴ required locus is x2 + y2 + z2 – {[2a(1+k2)]/(1-k2)}x + a2 = 0.

Which is a sphere.

13) A line makes angles α, β, γ, δ with the four diagonals of a cube then cos2α + cos2β + cos2γ + cos2δ =

(a)              (b) 4/3               (c) 3/4               (d) 4/5

(b) The direction ratios of the diagonal OR (1, 1, 1)

Direction cosine are (1/√3 , 1/√3 , 1/√3)

Similarly, direction cosine of AS are (1/√3 , -1/√3 , -1/√3)

BP are (1/√3 , 1/√3 , -1/√3)

CQ are (1/√3 , -1/√3 , 1/√3)

Let l, m, n be direction cosines of the line cosα = (l+m+n)/√3 , cosβ = (l-m-n)/√3 , cosγ = (l+m-n)/√3 ,

cosδ = (l-m+n)/√3

cos2α + cos2β + cos2γ + cos2δ = [4(l2+m2+n2)]/3 = 4/3

(Since l2 + m2 + n2 = 1)

14) The points (0, -1, -1), (-4, 4, 4), (4, 5, 1) and (3, 9, 4) are

(a) collinear              (b) coplanar             (c) forming a square              (d) none of these

(b) Equation of the plane passing through the points (0, -1, -1),

 

(-4, 4, 4) and (4, 5, 1) is          x      (y+1)      (z+1)

                                           -4         5             5       = 0   …. (1)                              

                                                 4         6             2

 

The point (3, 9, 4) satisfies the equation (1).

15) A variable plane passes through a fixed point (a, b, c) and meets the coordinate axes in A, B, C. The locus of the point common to plane through A, B, C parallel to coordinate planes is

(a) ayz + bzx + cxy = xyz                            (b) axy + byz + czx = xyz            

(c) axy + byz + czx = abc                            (d) bcx + acy + abz = abc

(a) Let the equation to the plane be (x/α)+(y/β)+(z/γ) = 1

⇒ (a/α)+(b/β)+(c/γ) = 1 (the plane passes through a, b, c)

Now the points of intersection of the plane with the coordinate axes are A(α, 0, 0), B(0, β, 0) & C(0, 0, γ)

⇒ Equation to planes parallel to the coordinate planes and passing through A, B & C are

x = α, y = β and z = γ.

∴ The locus of the common point is (a/x)+(b/y)+(c/z) = 1 (by eliminating α, β, γ from above equation)

16) The plane (x/a)+(y/b)+(z/c) = 1 meets the coordinate axes in A, B, C. Then area of ΔABC is

(a) 1/2√(a2b2+b2c2+c2a2                           (b) 1/2√(ab+bc+ca)            

(c) 1/4√(a2b2+b2c2+c2a2)                            (d) 1/2√(a+b+c)

(a) A is (a, 0, 0) B is (0, b, 0) C is (0, 0, c)

The area of the projection on xy plane i.e. z = 0

= Area of the triangle formed by vertices (a, 0), (0, b), (0, 0)

Δ1 = 1/2[ab] ; Δ12 = 1/4(a2b2)

Similarly Δ22 = 1/4(b2c2) , Δ32 = 1/4(c2a2)

Required area √(Δ122232) = √[(1/4(a2b2)+1/4(b2c2)+1/4(c2a2)] = 1/2√(a2b2+b2c2+c2a2)

17) The internal bisector of angle A of ΔABC meets BC in D where A, B, C are (3, 2, 0), (5, 3, 2) and (-9,6,-3) . Then D is

(a) (19/8,-57/16,-17/16)                                (b) (19/8,57/16,17/16)             

(c) (19/8,-57/16,17/16)                                  (d) (-19/8,57/16,-17/16)

(b) AB = 3 , AC = 13

D divides BC in the ratio 3 : 13 

D = [(3×-9+13×5)/(3+13) , (3×6+13×3)/(3+13) , (3×-3+13×2)/(3+13)]

⇒(19/8 , 57/16 , 17/16)

18) Distance between the planes 2x+y+2z = 8 , 4x+2y+4z+5 = 0 is

(a) 3/2              (b) 5/2             (c) 7/2              (d) 9/2

(c) Note that the two planes are parallel. Let (x1,y1,z1) be a point on the I plane 2x1+y1+2z1 = 8

Perpendicular distance (x1,y1,z1) from the II plane

= (4x1+2y1+4z1+5)/6 = [(2×8)+5]/6 = 21/6 = 7/2

19) The vector equation of straight line passing through a and perpendicular to b and c is

(a) r – a = t(b×c)               (b) r = a×(b×c)             (c) r – b = t(a×c)              (d) r = b×(a×c)

(a) Since the line is perpendicular to both b and c , the line is parallel to b×c

Line passes through a .

∴ Equation is r = a+t(b×c)

r – a = t(b×c) where t is a scalar.

20) The centroid of the triangle whose vertices are (5,7,1), (1,1,3) is given by (1,3,5), then the 3rd vertex is

(a) (-3,1,11)               (b) (3,-1,11)             (c) (3,1,-11)             (d) none of these

(a) Let (x1,y1,z1) be the 3rd vertex.

Then centroid is [(5+1+x1)/3 , (7+1+y1)/3 , (1+3+z1)/3] = (1,3,5)

(6+x1)/3 = 1 ⇒ 6+x1 = 3 ⇒ x1 = -3

(8+y1)/3 = 3 ⇒ 8+y1 = 9 ⇒ y1 = 1

(4+z1)/3 = 5 ⇒ 4+z1 = 15 ⇒ z1 = 11

(-3,1,11)

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