Table of contents  
Short Answer Type Questions  
Long Answer Type Questions  
Objective Type Questions  
Fill in the blanks  
State True or False for the statements 
Q.1. Find the position vector of a point A in space such thatis inclined at 60º to OX and at 45° to OY and= 10 units.
Ans.
Let α = 60°, β = 45° and the angle inclined to OZ axis be γ
We know that
cos^{2} α + cos^{2} β + cos^{2} γ = 1
⇒ cos^{2} 60° + cos^{2} 45° + cos^{2} γ = 1
⇒
⇒
∴
(Rejecting cos γ =
∴
Hence, the position vector of A is
Q.2. Find the vector equation of the line which is parallel to the vector
and which passes through the point (1,–2,3).
Ans.
We know that the equation of line is
Here,
Hence, the required equation is
Q.3. Show that the lines
andintersect.
Also, find their point of intersection.
Ans.
The given equations are
Let
∴ x = 2λ + 1, y = 3λ + 2 and z = 4λ + 3
and
⇒ x = 5μ + 4, y = 2μ + 1 and z = μ
If the two lines intersect each other at one point,
then 2λ + 1 = 5μ + 4
⇒ 2λ – 5μ = 3 ...(i)
3λ + 2 = 2μ + 1
⇒ 3λ – 2μ = – 1 ...(ii)
and 4λ + 3 = μ
⇒ 4λ – μ = – 3 ...(iii)
Solving eqns. (i) and (ii) we get
2λ – 5μ = 3 [multiply by 3]
⇒3λ – 2μ = – 1 [multiply by 2]
Putting the value of m in eq. (i) we get,
2λ – 5(– 1) = 3
⇒ 2λ + 5 = 3
⇒ 2λ = – 2 ∴ λ = – 1
Now putting the value of λ and m in eq. (iii) then
4(– 1) – (– 1) = – 3
– 4 + 1 = – 3
– 3 = – 3 (satisfied)
∴ Coordinates of the point of intersection are
x = 5 (– 1) + 4 = – 5 + 4 = – 1
y = 2(– 1) + 1 = – 2 + 1 = – 1
z = – 1
Hence, the given lines intersect each other at (– 1, – 1, – 1).
Alternately: If two lines intersect each other at a point, then the shortest distance between them is equal to 0.
For this we will use SD =
Q.4. Find the angle between the lines
Ans.
Here,
∴
∴
Hence, the required angle is
Q.5. Prove that the line through A (0, –1, –1) and B (4, 5, 1) intersects the line through C (3, 9, 4) and D (– 4, 4, 4).
Ans.
Given points are A(0, – 1, – 1) and B(4, 5, 1) C(3, 9, 4) and D(– 4, 4, 4)
Cartesian form of equation AB is
and its vector form is
Similarly, equation of CD is
and its vector form is
Now, here
Shortest distance between AB and CD
∴
Hence, the two lines intersect each other.
Q.6. Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.
Ans.
Given that: x = py + q
and z = ry + s
∴ the equation becomes
in which d’ratios are a_{1} = p, b_{1} = 1, c_{1} = r
Similarly x = p′y + q′
and z = r′y + s′
∴ the equation becomes
in which a_{2} = p′, b_{2 }= 1, c_{2} = r′
If the lines are perpendicular to each other, then
a_{1}a_{2} + b_{1}b_{2} + c_{1}c_{2} = 0
pp′ + 1.1 + rr′ = 0
Hence, pp′ + rr′ + 1 = 0 is the required condition.
Q.7. Find the equation of a plane which bisects perpendicularly the line joining the points A (2, 3, 4) and B (4, 5, 8) at right angles.
Ans.
Given that A(2, 3, 4) and B(4, 5, 8)
Coordinates of midpoint C are
Now direction ratios of the normal to the plane
= direction ratios of AB
= 4 – 2, 5 – 3, 8 – 4 = (2, 2, 4)
Equation of the plane is
a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0
⇒ 2(x – 3) + 2(y – 4) + 4(z – 6) = 0
⇒ 2x – 6 + 2y – 8 + 4z – 24 = 0
⇒ 2x + 2y + 4z = 38 ⇒ x + y + 2z = 19
Hence, the required equation of plane is
x + y + 2z = 19 or
Q.8. Find the equation of a plane which is at a distance 3 √3 units from origin and the normal to which is equally inclined to coordinate axis.
Ans.
Since, the normal to the plane is equally inclined to the axes
∴ cos α = cos β = cos γ
⇒ cos^{2} α + cos^{2} α + cos^{2} α = 1
⇒ 3 cos^{2} α = 1 ⇒ cos a =
⇒
So, the normal is
∴ Equation of the plane is
⇒
⇒ x + y+ z = 3√3.√3
⇒ x + y+z = 9
Hence, the required equation of plane is x + y + z = 9.
Q.9. If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.
Ans.
Direction ratios of the normal to the plane are
(1 + 2, – 3 + 1, 3 + 3) Þ (3,  2, 6)
Equation of plane passing through one point (x_{1}, y_{1}, z_{1}) is
a(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0
⇒ 3(x – 1) – 2(y + 3) + 6(z – 3) = 0
⇒ 3x – 3 – 2y – 6 + 6z – 18 = 0
⇒ 3x – 2y + 6z – 27 = 0 ⇒ 3x  2y + 6z = 27
Hence, the required equation is 3x  2y + 6z = 27.
Q.10. Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
Ans.
Since, the equation of the plane passing through the points (x_{1}, y_{1}, z_{1}), (x_{2}, y_{2},z_{2}) and (x_{3}, y_{3}, z_{3}) is
⇒ (x – 2) (– 21) – (y –1)(7 + 2) + z(3) = 0
⇒ – 21(x –2) – 9(y – 1) + 3z = 0
⇒ – 21x + 42 – 9y + 9 + 3z = 0
⇒ – 21x – 9y + 3z + 51 = 0 ⇒ 7x + 3y – z – 17 = 0
Hence, the required equation is 7x + 3y – z – 17 = 0.
Q.11. Find the equations of the two lines through the origin which intersect the lineat angles of π/3 each.
Ans.
Any point on the given line is
⇒ x = 2λ + 3, y = λ + 3 and z = λ
Let it be the coordinates of P
∴ Direction ratios of OP are
(2λ + 3 – 0), (λ + 3 – 0) and (λ – 0) ⇒ 2λ + 3, λ + 3, λ
But the direction ratios of the line PQ are 2, 1, 1
∴
⇒
⇒
⇒
⇒
⇒ λ^{2} + 3λ + 3 = 4λ^{2} + 9 + 12λ (Squaring both sides)
⇒ 3λ^{2} + 9λ + 6 = 0 ⇒ λ^{2} + 3λ + 2 = 0
⇒ (λ + 1)(λ + 2) = 0
∴ λ = – 1, λ = – 2
∴ Direction ratios are [2(– 1) + 3, – 1 + 3, – 1] i.e., 1, 2, – 1 when λ = – 1 and [2(– 2) + 3, – 2 + 3, – 2] i.e., – 1, 1, – 2 when l = – 2.
Hence, the required equations are
Q.12. Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l^{2} + m^{2} – n^{2} = 0.
Ans.
The given equations are
l + m + n = 0 ...(i)
l^{2} + m^{2} – n^{2} = 0 ...(ii)
From equation (i) n = – (l + m)
Putting the value of n in eq. (ii) we get
l^{2} + m^{2} + [– (l + m)^{2}] = 0
⇒ l^{2} + m^{2} – l^{2} – m^{2} – 2lm = 0
⇒ – 2lm = 0
⇒ lm = 0 ⇒ (– m – n)m = 0 [∵ l = – m – n]
⇒ (m + n)m = 0 ⇒ m = 0 or m = – n
⇒ l = 0 or l = – n
∴ Direction cosines of the two lines are
0, – n, n and – n, 0, n ⇒ 0, 1, 1 and  1, 0, 1
∴
∴ θ = π/3
Hence, the required angle is π/3.
Q.13. If a variable line in two adjacent positions has direction cosines l, m, n and l + δl, m + δm, n + δn, show that the small angle δθ between the two positions is given by δθ^{2} = δl^{2} + δm^{2} + δn^{2}
Ans.
Given that l, m, n and l + δl, m + δm, n + δn, are the direction cosines of a variable line in two positions
∴ l^{2} + m^{2} + n^{2} = 1 ...(i)
and (l + δl)^{2} + (m + δm)^{2} + (n + δn)^{2} = 1 ...(ii)
⇒ l^{2} + δl^{2} + 2l.δl + m^{2} + δm^{2} + 2m.δm + n^{2} + δn^{2} + 2n.δn = 1
⇒ (l^{2} + m^{2} + n^{2}) + (δl^{2} + δm^{2} + δn^{2}) + 2(l.δl + m.δm + n.δn) = 1
⇒ 1 + (δl^{2} + δm^{2} + δn^{2}) + 2(l.δl + m.δm + n.δn) = 1
Letbe the unit vectors along a line with d’cosines l, m, n and
(l + δl), (m + δm), (n + δn).
⇒ cos δθ = l(l + δl) + m(m + δm) + n(n + δn)
⇒ cos δθ = l^{2} + l.δl + m^{2} + m.δm + n^{2} + n.δn
⇒ cos δθ = (l^{2} + m^{2} + n^{2}) + (l.δl + m.δm + n.δn)
⇒ (δq)^{2} = δl^{2} + δm^{2} + δn^{2} Hence proved.
Q.14. O is the origin and A is (a, b, c).Find the direction cosines of the line OA and the equation of plane through A at right angle to OA
Ans.
We have A(a, b, c) and O(0, 0, 0)
∴ direction ratios of OA = a – 0, b – 0, c – 0
= a, b, c
∴ direction cosines of line OA
Now direction ratios of the normal to the plane are (a, b, c).
∴ Equation of the plane passing through the point
A(a, b, c) is a(x – a) + b(y – b) + c(z – c) = 0
⇒ ax – a^{2} + by – b^{2} + cz – c^{2} = 0
⇒ ax + by + cz = a^{2} + b2 + c^{2}
Hence, the required equation is ax + by + cz = a^{2} + b^{2} + c^{2}.
Q.15. Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that
Ans.
Let OX, OY, OZ and ox, oy, oz be two rectangular systems
∴ Equations of two planes are
...(ii)
Length of perpendicular from origin to plane (i) is
Length of perpendicular from origin to plane (ii)
As per the condition of the question
Hence,
Q.16. Find the foot of perpendicular from the point (2,3,–8) to the line
Also, find the perpendicular distance from the given point
to the line.
Ans.
Given that:is the equation of line
⇒
∴ Coordinates of any point Q on the line are
x = – 2λ + 4, y = 6λ and z = – 3λ + 1
and the given point is P(2, 3, – 8)
Direction ratios of PQ are – 2λ + 4 – 2, 6λ – 3, – 3λ +1 + 8
i.e., – 2λ + 2, 6λ – 3, – 3l + 9
and the D’ratios of the given line are – 2, 6, – 3.
If PQ ⊥ line
then – 2(– 2λ + 2) + 6(6λ – 3) – 3(– 3λ + 9) = 0
⇒ 4λ – 4 + 36λ – 18 + 9λ – 27 = 0
⇒ 49λ – 49 = 0
⇒ λ = 1
∴ The foot of the perpendicular is – 2(1) + 4, 6(1), – 3(1) + 1
i.e., 2, 6, – 2
Now, distance PQ =
Hence, the required coordinates of the foot of perpendicular are 2, 6, – 2 and
the required distance is 3√5 units.
Q.17. Find the distance of a point (2,4,–1) from the line
Ans.
The given equation of line is
and any point P(2, 4, – 1)
Let Q be any point on the given line
∴ Coordinates of Q are x = λ – 5, y = 4λ – 3, z = – 9λ + 6
D’ratios of PQ are λ – 5 – 2, 4λ – 3 – 4, – 9λ + 6 + 1
i.e., λ – 7, 4λ – 7, – 9λ + 7
and the d’ratios of the line are 1, 4, – 9
If line then
1(λ  7) + 4(4λ  7)  9( 9λ + 7)
= 0 λ  7 + 16λ  28 + 81λ  63 = 0
⇒ 98λ – 98 = 0 ∴ λ = 1
So, the coordinates of Q are 1 – 5, 4 × 1 – 3, – 9 × 1 + 6
i.e., – 4, 1, – 3
Hence, the required distance is 7 units.
Q.18. Find the length and the foot of perpendicular from the point
to the plane 2x – 2y + 4z + 5 = 0.
Ans.
Given plane is 2x – 2y + 4z + 5 = 0 and given point is
D’ratios of the normal to the plane are 2, – 2, 4
So, the equation of the line passing through
and whose d’ratios are equal to the d’ratios of the normal to the plane
i.e., 2, – 2, 4 is
∴ Any point in the plane is 2λ + 1, 2λ +4λ + 2
Since, the point lies in the plane, then
⇒ 4λ + 2 + 4λ  3 + 16λ + 8 + 5 = 0
⇒ 24λ + 12 = 0
So, the coordinates of the point in the plane are
Hence, the foot of the perpendicular isand the
required length
Q.19. Find the equations of the line passing through the point (3,0,1) and parallel to the planes x + 2y = 0 and 3y – z = 0.
Ans.
Given point is (3, 0, 1) and the equation of planes are
x + 2y = 0 ...(i)
and 3y – z = 0 ...(ii)
Equation of any line l passing through (3, 0, 1) is
Direction ratios of the normal to plane (i) and plane (ii) are
(1, 2, 0) and (0, 3, – 1) Since the line is parallel to both the planes.
∴ 1.a + 2.b + 0.c = 0 ⇒ a + 2b + 0c = 0
and 0.a + 3.b – 1.c = 0 ⇒ 0.a + 3b  c = 0
So
∴ a = – 2λ, b = λ, c = 3λ
So, the equation of line is
Hence, the required equation is
or in vector form is
Q.20. Find the equation of the plane through the points (2,1,–1) and (–1,3,4), and perpendicular to the plane x – 2y + 4z = 10.
Ans.
Equation of the plane passing through two points (x_{1}, y_{1}, z_{1}) and (x_{2}, y_{2}, z_{2}) with its normal’s d’ratios is a
(x – x_{1}) + b(y – y_{1}) + c(z – z_{1}) = 0 ..(i)
If the plane is passing through the given points (2, 1, – 1) and
(– 1, 3, 4) then
a(x_{2} – x_{1}) + b(y_{2} – y_{1}) + c(z_{2} – z_{1}) = 0
⇒ a(– 1 – 2) + b(3 – 1) + c(4 + 1) = 0
⇒ – 3a + 2b + 5c = 0 ...(ii)
Since the required plane is perpendicular to the given plane
x – 2y + 4z = 10, then
1.a – 2.b + 4.c = 10 ...(iii)
Solving (ii) and (iii) we get,
a = 18λ, b = 17λ, c = 4λ
Hence, the required plane is
18λ(x – 2) + 17λ(y – 1) + 4λ(z + 1) = 0
⇒ 18x – 36 + 17y – 17 + 4z + 4 = 0
⇒ 18x + 17y + 4z – 49 = 0
Q.21. Find the shortest distance between the lines given by
Ans.
Given equations of lines are
...(i)
and ...(ii)
Equation (i) can be rewritten as
...(iii)
Here,
∴ Shortest distance, SD =
Hence, the required distance is 14 units.
Q.22. Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
Ans.
The given planes are
P_{1} : 5x + 3y + 6z + 8 = 0
P_{2} : x + 2y + 3z – 4 = 0
P_{3} : 2x + y – z + 5 = 0
Equation of the plane passing through the line of intersection of P_{2} and P_{3} is (x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0
⇒ (1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0 ...(i)
Plane (i) is perpendicular to P_{1}, then
5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0
⇒ 5 + 10λ + 6 + 3λ + 18 – 6λ = 0
⇒ 7λ + 29 = 0
∴
Putting the value of λ in eq. (i), we get
⇒ – 15x – 15y + 50z – 28 – 145 = 0
⇒ – 15x – 15y + 50z – 173 = 0
⇒ 51x + 15y – 50z + 173 = 0
Q.23. The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is ax + by ±
Ans.
Given planes are:
ax + by = 0 ...(i)
z = 0 ...(ii)
Equation of any plane passing through the line of intersection of plane (i) and (ii) is
(ax + by) + kz = 0 ⇒ ax + by + kz = 0 ...(iii)
Dividing both sides bywe get
∴ Direction cosines of the normal to the plane are
and the direction cosines of the plane (i) are
Since, α is the angle between the planes (i) and (iii), we get
⇒
⇒
⇒ (a^{2} + b^{2} + k^{2}) cos^{2} α = a^{2} + b^{2}
⇒ a^{2} cos^{2} α + b^{2} cos^{2} α + k^{2} cos^{2} α = a^{2} + b^{2}
⇒ k^{2} cos^{2} α = a^{2} – a^{2} cos^{2} α + b^{2} – b^{2} cos^{2} α
⇒ k^{2} cos^{2} α = a^{2}(1 – cos^{2} α) + b^{2}(1 – cos^{2} α)
⇒ k^{2} cos^{2} α = a^{2} sin^{2} α + b^{2} sin^{2} α
⇒ k^{2} cos^{2} α = (a^{2} + b^{2}) sin^{2} α
⇒
Putting the value of k in eq. (iii) we get
which is the required equation of plane.
Hence proved.
Q.24. Find the equation of the plane through the intersection of the planeswhose perpendicular distance from origin is unity.
Ans.
Given planes are;
...(i)
and ...(ii)
Equation of the plane passing through the line of intersection of plane (i) and (ii) is
(x + 3y – 6) + k(3x – y – 4z) = 0 ...(iii)
(1 + 3k)x + (3 – k)y – 4kz – 6 = 0
Perpendicular distance from origin
[Squaring both sides]
⇒
⇒ 26k^{2} = 26
⇒ k^{2} = 1
∴ k = ± 1
Putting the value of k in eq. (iii) we get,
(x + 3y  6) ± (3x  y  4z) = 0
⇒ x + 3y  6 + 3x  y  4z = 0 and x + 3y  6  3x + y + 4z = 0
⇒ 4x + 2y  4z  6 = 0 and  2x + 4y + 4z  6 = 0
Hence, the required equations are:
4x + 2y  4z  6 = 0 and  2x + 4y + 4z  6 = 0.
Q.25. Show that the pointsare equidistant from the plane and lies on opposite side of it.
Ans.
Given points areand the plane
Perpendicular distance offrom the plane
and perpendicular distance offrom the plane
Hence, the two points are equidistant from the given plane. Opposite sign shows that they lie on either side of the plane.
Q.26. are two vectors. The position vectors of the points A and C arerespectively. Find the position vector of a point P on the line AB and a point Q on the line CD such thatis perpendicular toboth.
Ans.
Position vector of A is
So, equation of any line passing through A and parallel to
...(i)
Now any point P on= (6 + 3λ, 7 – λ, 4 + λ)
Similarly, position vector of C is
and
So, equation of any line passing through C and parallel to is
...(ii)
Any point Q on= (– 3μ, – 9 + 2μ, 2 + 4μ)
d’ratios of
(6 + 3λ + 3μ, 7 – λ + 9 – 2μ, 4 + λ – 2 – 4μ)
⇒ (6 + 3λ + 3μ), (16 – λ – 2μ), (2 + λ – 4μ)
Nowto eq. (i),then
3(6 + 3λ + 3μ) – 1(16 – λ – 2μ) + 1(2 + λ – 4μ) = 0
⇒ 18 + 9λ + 9μ – 16 + λ + 2μ + 2 + λ – 4μ = 0
⇒ 11λ + 7μ + 4 = 0 ...(iii)
to eq. (ii), then
 3(6 + 3λ + 3μ) + 2(16  λ  2μ) + 4(2 + λ  4μ) = 0
⇒  18  9λ  9μ + 32  2λ  4μ + 8 + 4λ  16μ = 0
⇒ – 7λ – 29μ + 22 = 0
⇒ 7λ + 29μ – 22 = 0 ...(iv)
Solving eq. (iii) and (iv) we get
∴ μ = 1
Now using μ = 1 in eq. (iv) we get
7λ + 29 – 22 = 0
⇒ λ = – 1
∴ Position vector of P = [6 + 3(– 1), 7 + 1, 4 – 1] = (3, 8, 3)
and position vector of Q = [– 3(1), –9 + 2(1), 2 + 4(1)] = (– 3, –7, 6)
Hence, the position vectors of
Q.27. Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
Ans.
Given that 2l + 2m – n = 0 ...(i)
and mn + nl + lm = 0 ...(ii)
Eliminating m from eq. (i) and (ii) we get,
[from (i)]
⇒ n^{2} + nl – 2l^{2} = 0
⇒ n^{2} + 2nl – nl – 2l^{2} = 0
⇒ n(n + 2l) – l(n + 2l) = 0
⇒ (n – l)(n + 2l) = 0
⇒ n = – 2l and n = l
Therefore, the direction ratios are proportional to l, – 2l, –2l
and l,
⇒ 1, – 2, – 2 and 2, – 1, 2
If the two lines are perpendicular to each other then 1(2) – 2(– 1) – 2 × 2 = 0
2 + 2 – 4 = 0
0 = 0
Hence, the two lines are perpendicular.
Q.28. If l_{1}, m_{1}, n_{1}; l_{2}, m_{2}, n_{2}; l_{3}, m_{3}, n_{3} are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l_{1} + l_{2} + l_{3}, m_{1} + m_{2} + m_{3}, n_{1} + n_{2} + n_{3} makes equal angles with them.
Ans.
Letare such that
and
Since the given d’cosines are mutually perpendicular then
l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2} = 0
l_{2}l_{3} + m_{2}m_{3} + n_{2}n_{3} = 0
l_{1}l_{3} + m_{1}m_{3} + n_{1}n_{3} = 0
Let α , β and γ be the angles between
∴ cos α = l_{1}(l_{1} + l_{2} + l_{3}) + m_{1}(m_{1} + m_{2} + m_{3}) + n_{1}(n_{1} + n_{2} + n_{3})
= l_{1}^{2} + l_{1}l_{2} + l_{1}l_{3 }+ m_{1}^{2} + m_{1}m_{2} + m_{1}m_{3} + n_{1}^{2}+ n_{1}n_{2} + n_{1}n_{3}
= (l_{1}^{2} + m_{1}^{2}+n_{1}^{2}) + (l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}) + (l_{1}l_{3} + m_{1}m_{3} + n_{1}n_{3})
= 1 + 0 + 0 = 1
∴ cos β = l_{2}(l_{1} + l_{2} + l_{3}) + m_{2}(m_{1} + m_{2} + m_{3}) + n_{2}(n_{1} + n_{2 }+ n_{3})
= l_{1}l_{2} + l_{2}^{2}+ l_{2} l_{3} + m_{1}m_{2} + m_{2}^{2} + m_{2 }m_{3} + n_{1}n_{2} + n_{2}^{2} + n_{2} n_{3}
= (l_{2}^{2} + m_{2}^{2}+n_{2}^{2}) + (l_{1}l_{2} + m_{1}m_{2} + n_{1}n_{2}) + (l_{2}l_{3} + m_{2}m_{3} + n_{2}n_{3})
= 1 + 0 + 0 = 1
Similarly,
∴ cos γ = l_{3}(l_{1} + l_{2} + l_{3}) + m_{3}(m_{1} + m_{2} + m_{3}) + n_{3}(n_{1} + n_{2} + n_{3})
= l_{1}l_{3} + l_{2} l_{3} + l_{3}^{2}+ m_{1}m_{3} + m_{2} m_{3} + m_{3}^{2} + n_{1}n_{3} + n_{2} n_{3} + n_{3}^{2}
= (l_{3}^{2} + m_{3}^{2}+n_{3}^{2}) + (l_{1}l_{3} + m_{1}m_{3} + n_{1}n_{3}) + (l_{2}l_{3} + m_{2}m_{3} + n_{2}n_{3})
= 1 + 0 + 0 = 1
∴ cos α = cos β = cos γ = 1
⇒ α = β = γ which is the required result.
Q.29. Distance of the point (α,β,γ) from yaxis is
(a) β
(b)
(c)
(d)
Ans. (d)
Solution.
The given point is (α, β, γ)
Any point on yaxis = (0, β, 0)
∴ Required distance =
Hence, the correct option is (d).
Q.30. If the directions cosines of a line are k,k,k, then
(a) k > 0
(b) 0 < k < 1
(c) k = 1
(d)
Ans. (d)
Solution.
If l , m, n are the direction cosines of a line, then
l^{2} + m^{2} + n^{2} = 1
So, k^{2} + k^{2} + k^{2} = 1
⇒ 3k^{2} = 1 ⇒
Hence, the correct option is (d).
Q.31. The distance of the plane from the origin is
(a) 1
(b) 7
(c) 1/7
(d) None of these
Ans. (a)
Solution.
Given that:
So, the distance of the given plane from the origin is
Hence, the correct option is (a).
Q.32. The sine of the angle between the straight line and the plane 2x – 2y + z = 5 is
(a)
(b)
(c)
(d)
Ans. (d)
Solution.
Given that: l :
and P : 2x – 2y + z = 5
d’ratios of the line are 3, 4, 5
and d’ratios of the normal to the plane are 2, – 2, 1
∴
⇒
Hence, the correct option is (d).
Q.33. The reflection of the point (α,β,γ) in the xy– plane is
(a) (α,β,0)
(b) (0,0,γ)
(c) (–α,–β,γ)
(d) (α,β,–γ)
Ans. (d)
Solution.
Reflection of point (α, β, γ) in xyplane is (a,b,–g).
Hence, the correct option is (d).
Q.34. The area of the quadrilateral ABCD, where A(0,4,1), B (2, 3, –1), C(4, 5, 0) and D (2, 6, 2), is equal to
(a) 9 sq. units
(b) 18 sq. units
(c) 27 sq. units
(d) 81 sq. units
Ans. (a)
Solution.
Given points are
A(0, 4, 1), B(2,3,– 1), C(4, 5, 0) and D(2,6,2)
d’ratios of AB = 2,–1 –2
and d’ratios of DC = 2,–1,–2
∴ AB ║DC
Similarly, d’ratios of AD = 2, 2, 1
and d’ratios of BC = 2, 2, 1
∴ AD ║BC
So is a parallelogram.
∴ Area of parallelogram ABCD =
Hence, the correct option is (a).
Q.35. 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
Ans. (d)
Solution.
Given that:
xy + yz = 0
y.(x + z) = 0
y = 0 or x + z = 0
Here y = 0 is one plane and x + z = 0 is another plane. So, it is a pair of perpendicular planes.
Hence, the correct option is (d).
Q.36. The plane 2x – 3y + 6z – 11 = 0 makes an angle sin^{–1}(α) with xaxis. The value of α is equal to
(a) √3/2
(b) √2/3
(c) 2/7
(d) 3/7
Ans. (c)
Solution.
Direction ratios of the normal to the plane 2x – 3y + 6z – 11 = 0 are 2, – 3, 6
Direction ratios of xaxis are 1, 0, 0
∴ angle between plane and line is
Hence, the correct option is (c).
Q.37. A plane passes through the points (2,0,0) (0,3,0) and (0,0,4). The equation of plane is _______.
Ans.
Given points are (2, 0, 0), (0, 3, 0) and (0, 0, 4).
So, the intercepts cut by the plane on the axes are 2, 3, 4
Equation of the plane (intercept form) is
Hence, the equation of plane is
Q.38. The direction cosines of the vectorare ______.
Ans.
Let
direction ratios of are 2,2,1
So, the direction cosines are
Hence, the direction cosines of the given vector are
Q.39. The vector equation of the lineis _______.
Ans.
The given equation is
Equation of the line is
Hence, the vector equation of the given line is
Q.40. The vector equation of the line through the points (3,4,–7) and (1,–1,6) is _______.
Ans.
Given the points (3, 4, –7) and (1, –1, 6)
Equation of the line is
Hence, the vector equation of the line is
Q.41. The cartesian equation of the planeis ________.
Ans.
Given equation is
⇒
⇒ x + y – z = 2
Hence, the Cartesian equation of the plane is x + y – z = 2.
Q.42. The unit vector normal to the plane x + 2y + 3z – 6 = 0 is
Ans.
Given plane is x + 2y + 3z – 6 = 0
Vector normal to the plane
Hence, the given statement is ‘true’
Q.43. The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the coordinate axis are
Ans.
Equation of the plane is 2x – 3y + 5z + 4 = 0
⇒ 2x – 3y + 5z = – 4
So, the required intercepts are
Hence, the given statement is ‘true’.
Q.44. The angle between the lineand the plane
is
Ans.
Equation of line isand the equation of the plane is
∴
⇒
⇒
Hence, the given statement is ‘false’.
Q.45. The angle between the planesis
Ans.
The given planes are
So,
⇒
∴
Hence, the given statement is ‘false’.
Q.46. The linelies in the plane
Ans.
Direction ratios of the line
Direction ratios of the normal to the plane are
Therefore, the line is parallel to the plane.
Now point through which the line is passing
If line lies in the plane then
6 – 3 + 1 + 2 ≠ 0
So, the line does not lie in the plane.
Hence, the given statement is ‘false’.
Q.47. The vector equation of the lineis
Ans.
The Cartesian form of the equation is
∴ Here x_{1 }= 5, y_{1} = – 4, z_{1 }= 6, a = 3, b = 7, c = 2
So, the vector equation is
Hence, the given statement is ‘true’.
Q.48. The equation of a line, which is parallel toand which passes through the point (5,–2,4), is
Ans.
Here, x_{1} = 5, y_{1} = – 2, z_{1} = 4; a = 2, b = 1, c = 3
We know that the equation of line is
⇒
Hence, the given statement is ‘false’.
Q.49. If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is
Ans.
The given equation of the plane is
If the foot of the perpendicular to this plane is
⇒ 25 + 9 + 4 = 38
38 = 38 (satisfied)
Hence, the given statement is ‘true’.
209 videos443 docs143 tests

1. How do you find the distance between two points in threedimensional space? 
2. What are the coordinates of the midpoint of a line segment in threedimensional space? 
3. How do you determine if two lines in threedimensional space are parallel or perpendicular? 
4. How can you find the equation of a plane passing through a given point with a given normal vector in threedimensional space? 
5. What is the angle between two planes in threedimensional space? 
209 videos443 docs143 tests


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