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Q.1. Which octant do the point (−5, 4, 3) lie?
(a) Octant I
(b) Octant II
(c) Octant III
(d) Octant IV
Correct Answer is option (b)
Given (−5,4,3) is the point.
Here, the xcoordinate is negative but y and z coordinates are positive. Therefore, (−5,4,3) lie in octant II.
Q.2. The direction cosines of the yaxis are:
(a) (9, 0, 0)
(b) (1, 0, 0)
(c) (0, 1, 0)
(d) (0, 0, 1)
Correct Answer is option (c)
The Direction cosines of the yaxis are (0, 1, 0)
Q.3. A point is on the xaxis. Which of the following represent the point?
(a) (0, x, 0)
(b) (0, 0, x)
(c) (x, 0, 0)
(d) None of the above
Correct Answer is option (c)
At xaxis, y and z coordinates are zero.
Q.4. Find the equation of the plane passing through the points P(1, 1, 1), Q(3, 1, 2), R(3, 5, 4).
(a) x + 2y = 0
(b) x – y – 2 = 0
(c) x + 2y – 2 = 0
(d) x + y – 2 = 0
Correct Answer is option (d)
Given three points, P(1, 1, 1), Q(3, 1, 2) and R(3, 5, 4).
Equation will be =
On solving we get;
⇒ x + y – 2 = 0
Q.5. Coordinate planes divide the space into ______ octants.
(a) 4
(b) 6
(c) 8
(d) 10
Correct Answer is option (c)
The coordinate planes divide the three dimensional space into eight octants.
Q.6. The equation x² – x – 2 = 0 in threedimensional space is represented by:
(a) A pair of parallel planes
(b) A pair of straight lines
(c) A pair of the perpendicular plane
(d) None of these
Correct Answer is option (a)
x^{2} −x−2=0⇔(x−2)(x+1)=0
⇔x=2,x=−1 which are the planes (both parallel to YOZ plane).
Q.7. What is the distance between the points (2, –1, 3) and (–2, 1, 3)?
(a) 2√5 units
(b) 25 units
(c) 4√5 units
(d) √5 units
Correct Answer is option (a)
Let the points be P (2, – 1, 3) and Q (– 2, 1, 3)
By using the distance formula,
PQ = √[(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} + (z_{2} – z_{1})^{2}]
So here,
x_{1} = 2, y_{1} = – 1, z_{1} = 3
x_{2} = – 2, y_{2} = 1, z_{2} = 3
PQ = √[(2 – 2)^{2} + (1 – (1))^{2} + (3 – 3)^{2}]
= √[(4)^{2} + (2)^{2} + (0)^{2}]
= √[16 + 4 + 0]
= √20
= 2√5
Therefore, the required distance is 2√5 units.
Q.8. The direction ratios of the normal to the plane 7x + 4y – 2z + 5 = 0 are:
(a) 7, 4,2
(b) 7, 4, 5
(c) 7, 4, 2
(d) 4, 2, 5
Correct Answer is option (a)
The direction ratios of the normal to the plane 7x + 4y – 2z + 5 = 0 are(7, 4, 2)
Q.9. The maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is:
(a) 3 units
(b) 4 units
(c) 5 units
(d) Cannot be determined
Correct Answer is option (c)
Let the two points be P (3sin θ, 0, 0) and Q (4cos θ, 0, 0)
Now by distance formula,
PQ = √{(4cos θ – 3sin θ)² + (0 – 0)² + (0 – 0)²}
PQ = √{(4cos θ – 3sin θ)²}
PQ = 4cos θ – 3sin θ
Now, maximum value of 4cos θ – 3sin θ;
= √{(4² + (3)²}
= √(16 + 9)
= √25
= 5
Thus, PQ = 5 unitsSo, the maximum distance between points (3sin θ, 0, 0) and (4cos θ, 0, 0) is 5.
Q.10. If l, m, n are the direction cosines of a line, then;
(a) l^{2}+ m^{2}+ 2n^{2} = 1
(b) l^{2}+ 2m^{2}+ n^{2} = 1
(c) 2l^{2}+ m^{2}+ n^{2} = 1
(d) l^{2}+ m^{2}+ n^{2} = 1
Correct Answer is option (d)
Let vector OP = r, r is the position vector of point P(x, y, z)
And we know that if a line segment of magnitude “r” makes angles α β and γ with x,y and z axes
then x=r cosα,y=r cosβ cosγ are nothing but the direction cosines, which are given as l, m and n, so x = l.r, y = m.r and z = n.r
So, we’ll have  (using (1))
(l.r)^{2 }+ (m.r)^{2 }+ (n.r)^{2 }= (r)^{2}
or r^{2}(l^{2 }+ m^{2 }+ n^{2}) = r^{2}
or l^{2} + m^{2} + n^{2 }= 1
Q.11. The locus represented by xy + yz = 0 is:
(a) A pair of perpendicular lines
(b) A pair of parallel lines
(c) A pair of parallel planes
(d) A pair of perpendicular planes
Correct Answer is option (d)
Locus represented by xy + yz = 0
⇒ y(x + z)=0
The planes y = 0 and x + z = 0 are perpendicular
Q.12. Direction ratio of line joining (2, 3, 4) and (−1, −2, 1), are:
(a) (−3, −5, −3)
(b) (−3, 1, −3)
(c) (−1, −5, −3)
(d) (−3, −5, 5)
Correct Answer is option (a)
The direction ratio of the line joining A (2, 3, 4) and B (−1, −2, 1), are:
(−1−2), (−2−3), (1−4)
= (−3, −5, −3)
Q.13. The perpendicular distance of the point P(6, 7, 8) from the XY – Plane is:
(a) 8
(b) 7
(c) 6
(d) None of the above
Correct Answer is option (a)
Let Q be the foot of perpendicular drawn from the point P (6, 7, 8) to the XY plane.
Thus, the distance of this foot Q from P is zcoordinate of P, i.e., 8 units
Q.14. The vector equation for the line passing through the points (–1, 0, 2) and (3, 4, 6) is:
(a) i + 2k + λ(4i + 4j + 4k)
(b) i – 2k + λ(4i + 4j + 4k)
(c) i+2k+ λ(4i + 4j + 4k)
(d) i+2k+ λ(4i – 4j – 4k)
Correct Answer is option (c)
The vector equation of the line is given by:
r = a + λ (b – a), λ ∈ R
Let a = i + 2k
And b = 3i + 4j + 6k
b – a = 4i + 4j + 4k
Let the vector equation be r, then;
r = i + 2k + λ (4i + 4j + 4k)
Q.15. The image of the point P(1,3,4) in the plane 2x – y + z = 0 is:
(a) (3, 5, 2)
(b) (3, 5, 2)
(c) (3, 5, 2)
(d) (3, 5, 2)
Correct Answer is option (a)
Let the image of the point P(1, 3, 4) is Q.
The equation of the line through P and normal to the given plane is:
(x – 1)/2 = (y – 3)/1 = (z – 4)/1
Since the line passes through Q, so let the coordinate of Q are (2k + 1, k + 3, k + 4)
Now, the coordinate of the midpoint of PQ is:
(k + 1, k/2 + 3, k/2 + 4)
Now, this midpoint lies in the given plane.
2(k + 1) – (k/2 + 3) + (k/2 + 4) + 3 = 0
⇒ 2k + 2 + k/2 – 3 + k/2 + 4 + 3 = 0
⇒ 3k + 6 = 0
⇒ k = 2
Hence, the coordinate of Q is (2k + 1, k + 3, k + 4) = (4 + 1, 2 + 3, 2 + 4)
= (3, 5, 2)
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