3D Geometry Practice Questions - DPP for JEE

 Page 2


or 
 The locus of (x
2
, y
2
, z
2
) is
.
3. (d) For L
1
,
...(i)
...(ii)
From (i) and (ii)
...(iii)
The equation (iii) is the equation of line L
1
.
Similarly equation of line L
2
 is
Since L
1
 ? L
2
 , therefore
?  ?  ? = µ
4. (c) Let the variable point be  then according to question
?  .
So, the locus of the point is 
5. (c) Let P(a, ß, ?) be the foot of the perpendicular from the origin O(0,
0, 0) to the plane So, the plane passes throughP(a,ß, ?) and is
perpendicular to OP. Clearly direction ratios of OP i.e., normal to
the plane are a, ß, ? . Therefore, equation of the plane is a (x – a) +
ß (y – ß) + ? (z – ?) = 0
This plane passes through the fixed point (1, 2, 3), so
a (1 – a) + ß (2 – ß) + ? (3– ?) = 0
or a
2
 + ß
2
 +?
2
 – a – 2ß – 3? = 0
Generalizing a, ß and ?, locus of P (a, ß, ?) is
x
2
 + y
2
 + z
2
 – x – 2y – 3z = 0
Page 3


or 
 The locus of (x
2
, y
2
, z
2
) is
.
3. (d) For L
1
,
...(i)
...(ii)
From (i) and (ii)
...(iii)
The equation (iii) is the equation of line L
1
.
Similarly equation of line L
2
 is
Since L
1
 ? L
2
 , therefore
?  ?  ? = µ
4. (c) Let the variable point be  then according to question
?  .
So, the locus of the point is 
5. (c) Let P(a, ß, ?) be the foot of the perpendicular from the origin O(0,
0, 0) to the plane So, the plane passes throughP(a,ß, ?) and is
perpendicular to OP. Clearly direction ratios of OP i.e., normal to
the plane are a, ß, ? . Therefore, equation of the plane is a (x – a) +
ß (y – ß) + ? (z – ?) = 0
This plane passes through the fixed point (1, 2, 3), so
a (1 – a) + ß (2 – ß) + ? (3– ?) = 0
or a
2
 + ß
2
 +?
2
 – a – 2ß – 3? = 0
Generalizing a, ß and ?, locus of P (a, ß, ?) is
x
2
 + y
2
 + z
2
 – x – 2y – 3z = 0
6. (a) From the first relation, l = 5m – 3n. Putting this value of l in
second relation
Note that it, being quadratic in m, n,  gives two sets of values of m, n,
and hence gives the d.r.s. of two lines.
Now, factorising it, we get
Taking 2m – n = 0 we get 2m = n.
Also putting m = n/2 in l = 5m – 3n, we get
l = (5n/2) – 3n  l = – n/2 n = – 2l
Thus, we get, –2l = 2m = n     or    
d.r.s. of one line are –1, 1, 2.
Hence, the d,c,s. of one line are
 or  
Taking 3m – 2n = 0, we get
3m = 2n   or m = .
Putting this value in l = 5m – 3n, we obtain
Thus 
? the d.r’.ss of the second line are 1, 2, 3; and hence d.c.s. of
Page 4


or 
 The locus of (x
2
, y
2
, z
2
) is
.
3. (d) For L
1
,
...(i)
...(ii)
From (i) and (ii)
...(iii)
The equation (iii) is the equation of line L
1
.
Similarly equation of line L
2
 is
Since L
1
 ? L
2
 , therefore
?  ?  ? = µ
4. (c) Let the variable point be  then according to question
?  .
So, the locus of the point is 
5. (c) Let P(a, ß, ?) be the foot of the perpendicular from the origin O(0,
0, 0) to the plane So, the plane passes throughP(a,ß, ?) and is
perpendicular to OP. Clearly direction ratios of OP i.e., normal to
the plane are a, ß, ? . Therefore, equation of the plane is a (x – a) +
ß (y – ß) + ? (z – ?) = 0
This plane passes through the fixed point (1, 2, 3), so
a (1 – a) + ß (2 – ß) + ? (3– ?) = 0
or a
2
 + ß
2
 +?
2
 – a – 2ß – 3? = 0
Generalizing a, ß and ?, locus of P (a, ß, ?) is
x
2
 + y
2
 + z
2
 – x – 2y – 3z = 0
6. (a) From the first relation, l = 5m – 3n. Putting this value of l in
second relation
Note that it, being quadratic in m, n,  gives two sets of values of m, n,
and hence gives the d.r.s. of two lines.
Now, factorising it, we get
Taking 2m – n = 0 we get 2m = n.
Also putting m = n/2 in l = 5m – 3n, we get
l = (5n/2) – 3n  l = – n/2 n = – 2l
Thus, we get, –2l = 2m = n     or    
d.r.s. of one line are –1, 1, 2.
Hence, the d,c,s. of one line are
 or  
Taking 3m – 2n = 0, we get
3m = 2n   or m = .
Putting this value in l = 5m – 3n, we obtain
Thus 
? the d.r’.ss of the second line are 1, 2, 3; and hence d.c.s. of
second line are 
or 
7. (a) Since l
2
 + m
2
 + n
2
 = 1
? cos
2
 a + cos
2
 a + cos
2
 ? = 1       .......... (i)
( A line makes the same angle a with x and y-axes and ? with z-axis)
Also, sin
2
 ? = 2 sin
2
 a
? 1 – cos
2 
? = 2(1 – cos
2 
a) ( sin
2
 A + cos
2
 A = 1)
? cos
2
 ? = 2cos
2
 a –1 .......... (ii)
? From Eq. (i) and (ii)
2 cos
2
 a + 2 cos
2
 a – 1 = 1
?  4 cos
2
 a = 2 ? cos
2
 a = 
?    ? 
8. (b) Let Q be the image of the point P(2, 3, 4) in the plane x – 2y + 5z
= 6, then PQ is normal to the plane
? direction ratios of PQ are <1, –2, 5 >
Since PQ passes through P(2, 3, 4) and has direction ratios 1, –2, 5
? Equation of PQ is 
9. (a) Given equation of line is
or 
x = ? – 5, y = 4? – 3, z = – 9? + 6
(x, y, z) = (? – 5, 4? – 3, –9? + 6) …(i)
Let it is foot of perpendicualr
So, d.r.’s of ? line is
(? – 5 – 2, 4 ? – 3 – 4, – 9? + 6 + 1)
= (? – 7, 4?– 7, – 9? + 7)
Page 5


or 
 The locus of (x
2
, y
2
, z
2
) is
.
3. (d) For L
1
,
...(i)
...(ii)
From (i) and (ii)
...(iii)
The equation (iii) is the equation of line L
1
.
Similarly equation of line L
2
 is
Since L
1
 ? L
2
 , therefore
?  ?  ? = µ
4. (c) Let the variable point be  then according to question
?  .
So, the locus of the point is 
5. (c) Let P(a, ß, ?) be the foot of the perpendicular from the origin O(0,
0, 0) to the plane So, the plane passes throughP(a,ß, ?) and is
perpendicular to OP. Clearly direction ratios of OP i.e., normal to
the plane are a, ß, ? . Therefore, equation of the plane is a (x – a) +
ß (y – ß) + ? (z – ?) = 0
This plane passes through the fixed point (1, 2, 3), so
a (1 – a) + ß (2 – ß) + ? (3– ?) = 0
or a
2
 + ß
2
 +?
2
 – a – 2ß – 3? = 0
Generalizing a, ß and ?, locus of P (a, ß, ?) is
x
2
 + y
2
 + z
2
 – x – 2y – 3z = 0
6. (a) From the first relation, l = 5m – 3n. Putting this value of l in
second relation
Note that it, being quadratic in m, n,  gives two sets of values of m, n,
and hence gives the d.r.s. of two lines.
Now, factorising it, we get
Taking 2m – n = 0 we get 2m = n.
Also putting m = n/2 in l = 5m – 3n, we get
l = (5n/2) – 3n  l = – n/2 n = – 2l
Thus, we get, –2l = 2m = n     or    
d.r.s. of one line are –1, 1, 2.
Hence, the d,c,s. of one line are
 or  
Taking 3m – 2n = 0, we get
3m = 2n   or m = .
Putting this value in l = 5m – 3n, we obtain
Thus 
? the d.r’.ss of the second line are 1, 2, 3; and hence d.c.s. of
second line are 
or 
7. (a) Since l
2
 + m
2
 + n
2
 = 1
? cos
2
 a + cos
2
 a + cos
2
 ? = 1       .......... (i)
( A line makes the same angle a with x and y-axes and ? with z-axis)
Also, sin
2
 ? = 2 sin
2
 a
? 1 – cos
2 
? = 2(1 – cos
2 
a) ( sin
2
 A + cos
2
 A = 1)
? cos
2
 ? = 2cos
2
 a –1 .......... (ii)
? From Eq. (i) and (ii)
2 cos
2
 a + 2 cos
2
 a – 1 = 1
?  4 cos
2
 a = 2 ? cos
2
 a = 
?    ? 
8. (b) Let Q be the image of the point P(2, 3, 4) in the plane x – 2y + 5z
= 6, then PQ is normal to the plane
? direction ratios of PQ are <1, –2, 5 >
Since PQ passes through P(2, 3, 4) and has direction ratios 1, –2, 5
? Equation of PQ is 
9. (a) Given equation of line is
or 
x = ? – 5, y = 4? – 3, z = – 9? + 6
(x, y, z) = (? – 5, 4? – 3, –9? + 6) …(i)
Let it is foot of perpendicualr
So, d.r.’s of ? line is
(? – 5 – 2, 4 ? – 3 – 4, – 9? + 6 + 1)
= (? – 7, 4?– 7, – 9? + 7)
D.r.’s of given line is (1, 4, – 9) and both lines are ?
? (?–7). 1 + (4? – 7). 4 + (– 9 ?+ 7) (–9) = 0
? 98? = 98 ? ? = 1
? Point is (– 4, 1, – 3). [Substituting ? = 1 in (i)]
10. (b) Given equation of line is
? DR’s of the given line are 2, 1, 1
? DC’s of the given line are 
Since, required lines make an angle  with the given line
The DC’s of the required lines are
 respectively.
Also, both the required lines pass through the origin.
? Equation of required lines are
11. (d) Let {l, m, n} be the direction -cosines of PQ, then 
3l – m + n = 0 and 5l + m + 3n = 0
?  
Now a plane ? to PQ will have l, m , n as the coefficients of x, y and z
Hence the plane ? to PQ is  x + y – 2z = ?
It passes through (2, 1, 4); ? 2 + 1 –2.4 = ? i.e ? = – 5
Hence the required plane is  x + y – 2z = – 5
12. (c) D.R. of given line are 1, –2, 3 and the d.r. of normal to the given
plane are 1, 2, 1.
Since 1 × 1 + (–2) × 2 + 3 × 1= 0, therefore, the line is parallel to the
plane, Also, the base point of the line (1, 2, 1) lies in the given
plane.
(1 + 2 × 2 + 1 = 6 is true)
Hence, the given line lies in the given plane.