NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

Mathematics (Maths) Class 12

JEE : NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

The document NCERT Exemplar - Three Dimensional Geometry Notes | EduRev is a part of the JEE Course Mathematics (Maths) Class 12.
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Q.1. Find the position vector of a point A in space such thatNCERT Exemplar - Three Dimensional Geometry Notes | EduRevis inclined at 60º to OX and at 45° to OY andNCERT Exemplar - Three Dimensional Geometry Notes | EduRev= 10 units.
Ans.
Let α = 60°, β = 45° and the angle inclined to OZ axis be γ
We know that
cos2 α + cos2 β + cos2 γ = 1
⇒ cos2 60° + cos2 45° + cos2 γ = 1
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
(Rejecting cos γ =NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the position vector of A isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev

Q.2. Find the vector equation of the line which is parallel to the vector 
NCERT Exemplar - Three Dimensional Geometry Notes | EduRevand which passes through the point (1,–2,3).
Ans.
We know that the equation of line is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Here, NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
 Hence, the required equation is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

Q.3. Show that the lines
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev andNCERT Exemplar - Three Dimensional Geometry Notes | EduRevintersect.
Also, find their point of intersection.
Ans.
The given equations are
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
LetNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ x = 2λ + 1, y = 3λ + 2 and z = 4λ + 3
andNCERT Exemplar - Three Dimensional Geometry Notes | EduRev

⇒ 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]
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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 =NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

Q.4. Find the angle between the lines
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
Here,NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the required angle isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev

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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
and its vector form isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Similarly, equation of CD is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
and its vector form isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Now, here
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Shortest distance between AB and CD
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
and z = ry + s
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ the equation becomes
NCERT Exemplar - Three Dimensional Geometry Notes | EduRevin which d’ratios are a1 = p, b1 = 1, c1 = r
Similarly x = p′y + q′
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
and z = r′y + s′
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ the equation becomes
NCERT Exemplar - Three Dimensional Geometry Notes | EduRevin which a2 = p′, b= 1, c2 = r′
If the lines are perpendicular to each other, then
a1a2 + b1b2 + c1c2 = 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 mid-point C areNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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 – x1) + b(y – y1) + c(z – z1) = 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 orNCERT Exemplar - Three Dimensional Geometry Notes | EduRev

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
⇒ cos2 α + cos2 α + cos2 α = 1
⇒ 3 cos2 α = 1 ⇒ cos a = NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
So, the normal isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ Equation of the plane isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ 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 (x1, y1, z1) is
 a(x – x1) + b(y – y1) + c(z – z1) = 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 (x1, y1, z1), (x2, y2,z2) and (x3, y3, z3) is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ (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 lineNCERT Exemplar - Three Dimensional Geometry Notes | EduRevat angles of π/3 each.
Ans.
Any point on the given line is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev⇒ 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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

⇒ λ2 + 3λ + 3 = 4λ2 + 9 + 12λ (Squaring both sides)
⇒ 3λ2 + 9λ + 6 = 0 ⇒ λ2 + 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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

Q.12. Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0.
Ans.
The given equations are
l + m + n = 0 ...(i)
l2 + m2 – n2 = 0 ...(ii)
From equation (i) n = – (l + m)
Putting the value of n in eq. (ii) we get
l2 + m2 + [– (l + m)2] = 0
⇒ l2 + m2 – l2 – m2 – 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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ 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 δθ2 = δl2 + δm2 + δn2
Ans.
Given that l, m, n and l + δl, m + δm, n + δn, are the direction cosines of a variable line in two positions
∴ l2 + m2 + n2 = 1    ...(i)
and (l + δl)2 + (m + δm)2 + (n + δn)2 = 1    ...(ii)
⇒ l2 + δl2 + 2l.δl + m2 + δm2 + 2m.δm + n2 + δn2 + 2n.δn = 1
⇒ (l2 + m2 + n2) + (δl2 + δm2 + δn2) + 2(l.δl + m.δm + n.δn) = 1
⇒ 1 + (δl2 + δm2 + δn2) + 2(l.δl + m.δm + n.δn) = 1
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
LetNCERT Exemplar - Three Dimensional Geometry Notes | EduRevbe the unit vectors along a line with d’cosines l, m, n and
(l + δl), (m + δm), (n + δn).
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ cos δθ = l(l + δl) + m(m + δm) + n(n + δn)
⇒ cos δθ = l2 + l.δl + m2 + m.δm + n2 + n.δn
⇒ cos δθ = (l2 + m2 + n2) + (l.δl + m.δm + n.δn)
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRevNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ (δq)2 = δl2 + δm2 + δn2 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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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 – a2 + by – b2 + cz – c2 = 0
⇒ ax + by + cz = a2 + b2 + c2
Hence, the required equation is ax + by + cz = a2 + b2 + c2.

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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
Let OX, OY, OZ and ox, oy, oz be two rectangular systems
∴ Equations of two planes are
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev...(ii)
Length of perpendicular from origin to plane (i) is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Length of perpendicular from origin to plane (ii)
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
As per the condition of the question
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence,NCERT Exemplar - Three Dimensional Geometry Notes | EduRev


LONG ANSWER TYPE QUESTIONS

Q.16. Find the foot of perpendicular from the point (2,3,–8) to the line
NCERT Exemplar - Three Dimensional Geometry Notes | EduRevAlso, find the perpendicular distance from the given point 
to the line.
Ans.
Given that:NCERT Exemplar - Three Dimensional Geometry Notes | EduRevis the equation of line
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ 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 =NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
The given equation of line is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev 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 NCERT Exemplar - Three Dimensional Geometry Notes | EduRev 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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the required distance is 7 units.

Q.18. Find the length and the foot of perpendicular from the point
NCERT Exemplar - Three Dimensional Geometry Notes | EduRevto the plane 2x – 2y + 4z + 5 = 0.
Ans.
Given plane is 2x – 2y + 4z + 5 = 0 and given point isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev 
D’ratios of the normal to the plane are 2, – 2, 4
So, the equation of the line passing throughNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
and whose d’ratios are equal to the d’ratios of the normal to the plane
i.e., 2, – 2, 4 isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ Any point in the plane is 2λ + 1,- 2λ +NCERT Exemplar - Three Dimensional Geometry Notes | EduRev4λ + 2
Since, the point lies in the plane, then
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ 4λ + 2 + 4λ - 3 + 16λ + 8 + 5 = 0
⇒ 24λ + 12 = 0NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
So, the coordinates of the point in the plane are
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the foot of the perpendicular isNCERT Exemplar - Three Dimensional Geometry Notes | EduRevand the 

required lengthNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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
SoNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ a = – 2λ, b = λ, c = 3λ
So, the equation of line is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the required equation is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
or in vector form is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

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 (x1, y1, z1) and (x2, y2, z2) with its normal’s d’ratios is a
(x – x1) + b(y – y1) + c(z – z1) = 0    ..(i)
If the plane is passing through the given points (2, 1, – 1) and
(– 1, 3, 4) then
a(x2 – x1) + b(y2 – y1) + c(z2 – z1) = 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,
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
Given equations of lines are
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev  ...(i)
andNCERT Exemplar - Three Dimensional Geometry Notes | EduRev  ...(ii)
Equation (i) can be re-written as
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev ...(iii)
Here,NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ Shortest distance, SD =NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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
P1 : 5x + 3y + 6z + 8 = 0
P2 : x + 2y + 3z – 4 = 0
P3 : 2x + y – z + 5 = 0
Equation of the plane passing through the line of intersection of P2 and P3 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 P1, then
5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0
⇒ 5 + 10λ + 6 + 3λ + 18 – 6λ = 0
⇒ 7λ + 29 = 0
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Putting the value of λ in eq. (i), we get
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ – 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 ±NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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 byNCERT Exemplar - Three Dimensional Geometry Notes | EduRevwe get
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ Direction cosines of the normal to the plane are
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
and the direction cosines of the plane (i) are
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Since, α is the angle between the planes (i) and (iii), we get
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

⇒ (a2 + b2 + k2) cos2 α = a2 + b2
⇒ a2 cos2 α + b2 cos2 α + k2 cos2 α = a2 + b2
⇒ k2 cos2 α = a2 – a2 cos2 α + b2 – b2 cos2 α
⇒ k2 cos2 α = a2(1 – cos2 α) + b2(1 – cos2 α)
⇒ k2 cos2 α = a2 sin2 α + b2 sin2 α
⇒ k2 cos2 α = (a2 + b2) sin2 α
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Putting the value of k in eq. (iii) we get
NCERT Exemplar - Three Dimensional Geometry Notes | EduRevwhich is the required equation of plane.
Hence proved.

Q.24. Find the equation of the plane through the intersection of the planesNCERT Exemplar - Three Dimensional Geometry Notes | EduRevwhose perpendicular distance from origin is unity.
Ans.
Given planes are;
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev ...(i)
andNCERT Exemplar - Three Dimensional Geometry Notes | EduRev ...(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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev[Squaring both sides]
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ 26k2 = 26
⇒ k2 = 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 pointsNCERT Exemplar - Three Dimensional Geometry Notes | EduRevare equidistant from the plane NCERT Exemplar - Three Dimensional Geometry Notes | EduRev and lies on opposite side of it.
Ans.
 Given points areNCERT Exemplar - Three Dimensional Geometry Notes | EduRevand the planeNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Perpendicular distance ofNCERT Exemplar - Three Dimensional Geometry Notes | EduRevfrom the plane
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
and perpendicular distance ofNCERT Exemplar - Three Dimensional Geometry Notes | EduRevfrom the plane
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the two points are equidistant from the given plane. Opposite sign shows that they lie on either side of the plane.

Q.26. NCERT Exemplar - Three Dimensional Geometry Notes | EduRevare two vectors. The position vectors of the points A and C areNCERT Exemplar - Three Dimensional Geometry Notes | EduRevrespectively. Find the position vector of a point P on the line AB and a point Q on the line CD such thatNCERT Exemplar - Three Dimensional Geometry Notes | EduRevis perpendicular toNCERT Exemplar - Three Dimensional Geometry Notes | EduRevboth.
Ans.
Position vector of A isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
So, equation of any line passing through A and parallel toNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev  ...(i)
Now any point P onNCERT Exemplar - Three Dimensional Geometry Notes | EduRev= (6 + 3λ, 7 – λ, 4 + λ)
Similarly, position vector of C isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
andNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
So, equation of any line passing through C and parallel to NCERT Exemplar - Three Dimensional Geometry Notes | EduRevis
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev ...(ii)
Any point Q onNCERT Exemplar - Three Dimensional Geometry Notes | EduRev= (– 3μ, – 9 + 2μ, 2 + 4μ)
d’ratios ofNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
(6 + 3λ + 3μ, 7 – λ + 9 – 2μ, 4 + λ – 2 – 4μ)
⇒ (6 + 3λ + 3μ), (16 – λ – 2μ), (2 + λ – 4μ)
NowNCERT Exemplar - Three Dimensional Geometry Notes | EduRevto 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)
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev 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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ 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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

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,
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev[from (i)]
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ n2 + nl – 2l2 = 0
⇒ n2 + 2nl – nl – 2l2 = 0
⇒ n(n + 2l) – l(n + 2l) = 0
⇒ (n – l)(n + 2l) = 0
⇒ n = – 2l and n = l
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Therefore, the direction ratios are proportional to l, – 2l, –2l
and l,NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ 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 l1, m1, n1; l2, m2, n2; l3, m3, n3 are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l1 + l2 + l3, m1 + m2 + m3, n1 + n2 + n3 makes equal angles with them.
Ans.
LetNCERT Exemplar - Three Dimensional Geometry Notes | EduRevare such that
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
andNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Since the given d’cosines are mutually perpendicular then
l1l2 + m1m2 + n1n2 = 0
l2l3 + m2m3 + n2n3 = 0
l1l3 + m1m3 + n1n3 = 0
Let α , β and γ be the angles betweenNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ cos α = l1(l1 + l2 + l3) + m1(m1 + m2 + m3) + n1(n1 + n2 + n3)
= l12 + l1l2 + l1l+ m12 + m1m2 + m1m3 + n12+ n1n2 + n1n3
= (l12 + m12+n12) + (l1l2 + m1m2 + n1n2) + (l1l3 + m1m3 + n1n3)
= 1 + 0 + 0 = 1
∴ cos β = l2(l1 + l2 + l3) + m2(m1 + m2 + m3) + n2(n1 + n+ n3)
= l1l2 + l22+ l2 l3 + m1m2 + m22 + mm3 + n1n2 + n22 + n2 n3
= (l22 + m22+n22) + (l1l2 + m1m2 + n1n2) + (l2l3 + m2m3 + n2n3)
= 1 + 0 + 0 = 1
Similarly,
∴ cos γ = l3(l1 + l2 + l3) + m3(m1 + m2 + m3) + n3(n1 + n2 + n3)
= l1l3 + l2 l3 + l32+ m1m3 + m2 m3 + m32 + n1n3 + n2 n3 + n32
= (l32 + m32+n32) + (l1l3 + m1m3 + n1n3) + (l2l3 + m2m3 + n2n3)
= 1 + 0 + 0 = 1
∴ cos α = cos β = cos γ = 1
⇒ α = β = γ which is the required result.


OBJECTIVE TYPE QUESTIONS

Q.29. Distance of the point (α,β,γ) from y-axis is
(a) β 
(b)NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
(c)NCERT Exemplar - Three Dimensional Geometry Notes | EduRev 
(d)NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans. (d)
Solution.

The given point is (α, β, γ)
Any point on y-axis = (0, β, 0)
∴ Required distance =NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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)NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans. (d)
Solution.

If l , m, n are the direction cosines of a line, then
l2 + m2 + n2 = 1
So, k2 + k2 + k2 = 1
⇒ 3k2 = 1 ⇒NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the correct option is (d).

Q.31. The distance of the planeNCERT Exemplar - Three Dimensional Geometry Notes | EduRev from the origin is
(a) 1 
(b) 7 
(c) 1/7
(d) None of these
Ans. (a)
Solution.

Given that:NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
So, the distance of the given plane from the origin is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the correct option is (a).

Q.32. The sine of the angle between the straight lineNCERT Exemplar - Three Dimensional Geometry Notes | EduRev and the plane 2x – 2y + z = 5 is
(a)NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
(b)NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
(c)NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
(d)NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans. (d)
Solution.

Given that: l :NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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 xy-plane 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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev∴ AB ║DC
Similarly, d’ratios of AD = 2, 2, 1
and d’ratios of BC = 2, 2, 1
∴ AD ║BC
So NCERT Exemplar - Three Dimensional Geometry Notes | EduRev is a parallelogram.
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ Area of parallelogram ABCD =NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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 x-axis. 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 x-axis are 1, 0, 0
∴ angle between plane and line is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the correct option is (c).

Fill in the blanks
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
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the equation of plane isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev

Q.38. The direction cosines of the vectorNCERT Exemplar - Three Dimensional Geometry Notes | EduRevare ______.

Ans.
LetNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
direction ratios of NCERT Exemplar - Three Dimensional Geometry Notes | EduRev are 2,2,-1
So, the direction cosines areNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the direction cosines of the given vector areNCERT Exemplar - Three Dimensional Geometry Notes | EduRev

Q.39. The vector equation of the lineNCERT Exemplar - Three Dimensional Geometry Notes | EduRevis _______.

Ans.
The given equation is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev 
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Equation of the line isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the vector equation of the given line is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

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)
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Equation of the line isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the vector equation of the line is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev

Q.41. The cartesian equation of the planeNCERT Exemplar - Three Dimensional Geometry Notes | EduRevis ________.
Ans.
Given equation is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ x + y – z = 2
Hence, the Cartesian equation of the plane is x + y – z = 2.

State True or False for the statements
Q.42. The unit vector normal to the plane x + 2y + 3z – 6 = 0 is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
Given plane is x + 2y + 3z – 6 = 0
Vector normal to the planeNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the given statement is ‘true’

Q.43. The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis areNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
 Equation of the plane is 2x – 3y + 5z + 4 = 0
⇒ 2x – 3y + 5z = – 4
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
So, the required intercepts areNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the given statement is ‘true’.

Q.44. The angle between the lineNCERT Exemplar - Three Dimensional Geometry Notes | EduRevand the plane
NCERT Exemplar - Three Dimensional Geometry Notes | EduRevisNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
Equation of line isNCERT Exemplar - Three Dimensional Geometry Notes | EduRevand the equation of the plane isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the given statement is ‘false’.

Q.45. The angle between the planesNCERT Exemplar - Three Dimensional Geometry Notes | EduRevis
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
The given planes areNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
So,NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the given statement is ‘false’.

Q.46. The lineNCERT Exemplar - Three Dimensional Geometry Notes | EduRevlies in the plane NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
Direction ratios of the lineNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Direction ratios of the normal to the plane areNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Therefore, the line is parallel to the plane.
Now point through which the line is passing NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
If line lies in the plane then
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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 lineNCERT Exemplar - Three Dimensional Geometry Notes | EduRevis
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
The Cartesian form of the equation is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
∴ Here x= 5, y1 = – 4, z= 6, a = 3, b = 7, c = 2
So, the vector equation isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Hence, the given statement is ‘true’.

Q.48. The equation of a line, which is parallel toNCERT Exemplar - Three Dimensional Geometry Notes | EduRevand which passes through the point (5,–2,4), isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
Here, x1 = 5, y1 = – 2, z1 = 4; a = 2, b = 1, c = 3
We know that the equation of line isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
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 isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev
Ans.
The given equation of the plane isNCERT Exemplar - Three Dimensional Geometry Notes | EduRev 
If the foot of the perpendicular to this plane is
NCERT Exemplar - Three Dimensional Geometry Notes | EduRev
⇒ 25 + 9 + 4 = 38
38 = 38 (satisfied)
Hence, the given statement is ‘true’.

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