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NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced PDF Download

Short Answer Type Questions

Q.1. Find the position vector of a point A in space such thatNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedis inclined at 60º to OX and at 45° to OY andNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced= 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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
(Rejecting cos γ =NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the position vector of A isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

Q.2. Find the vector equation of the line which is parallel to the vector 
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedand which passes through the point (1,–2,3).
Ans.
We know that the equation of line is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Here, NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
 Hence, the required equation is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

Q.3. Show that the lines
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced andNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedintersect.
Also, find their point of intersection.
Ans.
The given equations are
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
LetNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ x = 2λ + 1, y = 3λ + 2 and z = 4λ + 3
andNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced

Q.4. Find the angle between the lines
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Ans.
Here,NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the required angle isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

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 | Mathematics (Maths) for JEE Main & Advanced
and its vector form isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Similarly, equation of CD is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
and its vector form isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Now, here
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Shortest distance between AB and CD
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
and z = ry + s
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ the equation becomes
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedin which d’ratios are a1 = p, b1 = 1, c1 = r
Similarly x = p′y + q′
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
and z = r′y + s′
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ the equation becomes
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedin 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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced

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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
So, the normal isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ Equation of the plane isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ (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 | Mathematics (Maths) for JEE Main & Advancedat angles of π/3 each.
Ans.
Any point on the given line is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

⇒ 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 | Mathematics (Maths) for JEE Main & Advanced

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 | Mathematics (Maths) for JEE Main & Advanced
∴ 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 | Mathematics (Maths) for JEE Main & Advanced
LetNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedbe the unit vectors along a line with d’cosines l, m, n and
(l + δl), (m + δm), (n + δn).
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & AdvancedNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ (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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
Ans.
Let OX, OY, OZ and ox, oy, oz be two rectangular systems
∴ Equations of two planes are
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced...(ii)
Length of perpendicular from origin to plane (i) is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Length of perpendicular from origin to plane (ii)
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
As per the condition of the question
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence,NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced


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 | Mathematics (Maths) for JEE Main & AdvancedAlso, find the perpendicular distance from the given point 
to the line.
Ans.
Given that:NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedis the equation of line
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ 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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
Ans.
The given equation of line is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced 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 | Mathematics (Maths) for JEE Main & Advanced 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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advancedto the plane 2x – 2y + 4z + 5 = 0.
Ans.
Given plane is 2x – 2y + 4z + 5 = 0 and given point isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced 
D’ratios of the normal to the plane are 2, – 2, 4
So, the equation of the line passing throughNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
∴ Any point in the plane is 2λ + 1,- 2λ +NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced4λ + 2
Since, the point lies in the plane, then
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ 4λ + 2 + 4λ - 3 + 16λ + 8 + 5 = 0
⇒ 24λ + 12 = 0NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
So, the coordinates of the point in the plane are
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the foot of the perpendicular isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedand the 

required lengthNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
∴ a = – 2λ, b = λ, c = 3λ
So, the equation of line is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the required equation is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
or in vector form is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given equations of lines are
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced  ...(i)
andNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced  ...(ii)
Equation (i) can be re-written as
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced ...(iii)
Here,NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ Shortest distance, SD =NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
Putting the value of λ in eq. (i), we get
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ – 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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advancedwe get
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ Direction cosines of the normal to the plane are
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
and the direction cosines of the plane (i) are
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Since, α is the angle between the planes (i) and (iii), we get
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

⇒ (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 | Mathematics (Maths) for JEE Main & Advanced
Putting the value of k in eq. (iii) we get
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedwhich 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 | Mathematics (Maths) for JEE Main & Advancedwhose perpendicular distance from origin is unity.
Ans.
Given planes are;
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced ...(i)
andNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced ...(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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced[Squaring both sides]
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advancedare equidistant from the plane NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced and lies on opposite side of it.
Ans.
 Given points areNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedand the planeNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Perpendicular distance ofNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedfrom the plane
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
and perpendicular distance ofNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedfrom the plane
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advancedare two vectors. The position vectors of the points A and C areNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedrespectively. 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 | Mathematics (Maths) for JEE Main & Advancedis perpendicular toNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedboth.
Ans.
Position vector of A isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
So, equation of any line passing through A and parallel toNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced  ...(i)
Now any point P onNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced= (6 + 3λ, 7 – λ, 4 + λ)
Similarly, position vector of C isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
andNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
So, equation of any line passing through C and parallel to NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedis
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced ...(ii)
Any point Q onNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced= (– 3μ, – 9 + 2μ, 2 + 4μ)
d’ratios ofNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
(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 | Mathematics (Maths) for JEE Main & Advancedto 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 | Mathematics (Maths) for JEE Main & Advanced 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 | Mathematics (Maths) for JEE Main & Advanced
∴ 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 | Mathematics (Maths) for JEE Main & Advanced

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 | Mathematics (Maths) for JEE Main & Advanced[from (i)]
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
Therefore, the direction ratios are proportional to l, – 2l, –2l
and l,NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advancedare such that
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
andNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
∴ 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 | Mathematics (Maths) for JEE Main & Advanced
(c)NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced 
(d)NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (d).

Q.31. The distance of the planeNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced 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 | Mathematics (Maths) for JEE Main & Advanced
So, the distance of the given plane from the origin is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the correct option is (a).

Q.32. The sine of the angle between the straight lineNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced and the plane 2x – 2y + z = 5 is
(a)NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
(b)NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
(c)NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
(d)NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Ans. (d)
Solution.

Given that: l :NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced∴ 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 | Mathematics (Maths) for JEE Main & Advanced is a parallelogram.
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ Area of parallelogram ABCD =NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
Hence, the equation of plane isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

Q.38. The direction cosines of the vectorNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedare ______.

Ans.
LetNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
direction ratios of NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced are 2,2,-1
So, the direction cosines areNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the direction cosines of the given vector areNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

Q.39. The vector equation of the lineNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedis _______.

Ans.
The given equation is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced 
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Equation of the line isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the vector equation of the given line is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

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 | Mathematics (Maths) for JEE Main & Advanced
Equation of the line isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the vector equation of the line is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced

Q.41. The cartesian equation of the planeNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedis ________.
Ans.
Given equation is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ 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 | Mathematics (Maths) for JEE Main & Advanced
Ans.
Given plane is x + 2y + 3z – 6 = 0
Vector normal to the planeNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
Ans.
 Equation of the plane is 2x – 3y + 5z + 4 = 0
⇒ 2x – 3y + 5z = – 4
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
So, the required intercepts areNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the given statement is ‘true’.

Q.44. The angle between the lineNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedand the plane
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & AdvancedisNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Ans.
Equation of line isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedand the equation of the plane isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the given statement is ‘false’.

Q.45. The angle between the planesNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedis
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Ans.
The given planes areNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
So,NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the given statement is ‘false’.

Q.46. The lineNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedlies in the plane NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Ans.
Direction ratios of the lineNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Direction ratios of the normal to the plane areNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Therefore, the line is parallel to the plane.
Now point through which the line is passing NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
If line lies in the plane then
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advancedis
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Ans.
The Cartesian form of the equation is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
∴ Here x= 5, y1 = – 4, z= 6, a = 3, b = 7, c = 2
So, the vector equation isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
Hence, the given statement is ‘true’.

Q.48. The equation of a line, which is parallel toNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advancedand which passes through the point (5,–2,4), isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
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 | Mathematics (Maths) for JEE Main & Advanced
Ans.
The given equation of the plane isNCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced 
If the foot of the perpendicular to this plane is
NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced
⇒ 25 + 9 + 4 = 38
38 = 38 (satisfied)
Hence, the given statement is ‘true’.

The document NCERT Exemplar: Three Dimensional Geometry | Mathematics (Maths) for JEE Main & Advanced is a part of the JEE Course Mathematics (Maths) for JEE Main & Advanced.
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FAQs on NCERT Exemplar: Three Dimensional Geometry - Mathematics (Maths) for JEE Main & Advanced

1. How do you find the distance between two points in three-dimensional space?
Ans. To find the distance between two points (x1, y1, z1) and (x2, y2, z2) in three-dimensional space, you can use the distance formula: Distance = √((x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2).
2. What are the coordinates of the midpoint of a line segment in three-dimensional space?
Ans. The coordinates of the midpoint of a line segment with endpoints (x1, y1, z1) and (x2, y2, z2) in three-dimensional space are: ((x1 + x2)/2, (y1 + y2)/2, (z1 + z2)/2).
3. How do you determine if two lines in three-dimensional space are parallel or perpendicular?
Ans. Two lines in three-dimensional space are parallel if their direction vectors are proportional. They are perpendicular if the dot product of their direction vectors is zero.
4. How can you find the equation of a plane passing through a given point with a given normal vector in three-dimensional space?
Ans. The equation of a plane passing through a point (x0, y0, z0) with a normal vector (a, b, c) is given by: a(x - x0) + b(y - y0) + c(z - z0) = 0.
5. What is the angle between two planes in three-dimensional space?
Ans. The angle between two planes with normal vectors (a1, b1, c1) and (a2, b2, c2) is given by the formula: cos θ = |a1a2 + b1b2 + c1c2| / √(a1^2 + b1^2 + c1^2) √(a2^2 + b2^2 + c2^2).
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