Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

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Kerala Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Plus Two Maths Three Dimensional Geometry Three Mark Questions and Answers

Question 1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 1

  1. Write the Cartesian equation. (1)
  2. Find the angle between the line. (2)

Answer:
1.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 2

2. cosθ
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 3

Question 2.
Find the vector equation of the plane passing through the intersection of the planes \(\bar{r}\).(i + j + k) = 6 and \(\bar{r}\).(2i + 3 j + 4k) = -5 at the point (1,1,1).
Answer:
The Cartesian equation of the planes are x + y + z = 6 and 2x + 3y + 4z = – 5. Therefore the equation of the plane passing through the intersection of these planes is
x + y + z – 6 + k(2x + 3y + 4z + 5) = 0
Since it pass through (1, 1, 1) we get,
1 + 1 + 1 – 6 + k(2 + 3 + 4 + 5) = 0 ⇒ -3 + k14 ⇒ k = \(\frac{3}{14}\)
∴ the equation is
x + y + z + -6 + \(\frac{3}{14}\) (2x + 3 y + 4z + 5) = 0
14x + 14y + 14z – 84 + 6x + 9y + 12z + 15 = 0
20x + 23y + 26z = 69
Vector equation is \(\bar{r}\). (20i + 23j + 26k) = 69.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 3.
Find the equation of the plane passing through the intersection of the planes x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 and contains the point (1, 0, 0).
Answer:
The equation of the planes passing through the intersection of the planes
x + y + 4z + 5 = 0 and 2x – y + 3z + 6 = 0 is
x + y + 4z + 5 + k(2x – y + 3z + 6) = 0 ____(1)
Since (1) pass through (1, 0, 0)
⇒ 1 + 0 + 0 + 5 + k(2 – 0 + 0 + 6) = 0
⇒ 6 + 8k = 0 ⇒ k = –\(\frac{3}{4}\); Then (1)
⇒ x + y + z + 5 + \(\frac{3}{4}\)(2x – y + 3z + 6) = 0
⇒ 4x + 4y + 16z + 20 – 6x + 3y – 9z – 18 = 0
⇒ 2x – 7y – 7z = 2.

Plus Two Maths Three Dimensional Geometry Four Mark Questions and Answers

Question 1.
Consider the point (-1, -2, -3).

  1. In which octant, the above point lies.(1)
  2. Find the direction cosines of the line joining (-1, -2, -3) and (3, 4, 5). (1)
  3. If P is any point such that OP = \(\sqrt{50}\) and direction cosines of OP are \(\frac{3}{\sqrt{50}}\), \(\frac{4}{\sqrt{50}}\) and \(\frac{5}{\sqrt{50}}\), then find the co-ordinate of P. (2)

Answer:
1. The point lies in the octant X’OY’Z’.

2. Direction ratios of the line joining (-1, -2, -3) and (3, 4, 5) are (3 + 1), (4 + 2), (5 + 3) ⇒ 4, 6, 8 ⇒ 2, 3, 4.
Therefore direction cosines are
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 4

3. Given, OP = \(\sqrt{50}\)
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 5
Therefore the point is (3, 4, 5).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 2.
Consider a cube of side ‘a’ unit has one vertex at the origin O.
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 6

  1. Write down the co-ordinate of 0, 0′, A and A’ (1)
  2. Find the direction ratios of OO’ and AA’. (2)
  3. Show that the angle between the main diagonals of the above cube is cos-1\(\left(\frac{1}{3}\right)\) (1)

Answer:
1. O(0, 0, 0), O'(a, a, a), A(a, 0, 0) and A'(0, a, a).

2. Direction ratios along OO’ is a – 0, a – 0, a – 0
⇒ a, a ,a ⇒ 1, 1, 1

3. cosθ
Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry 7

Question 3.
Consider two points A and B and a line L as shown in the figure.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 8

  1. Find \(\overline{A B}\) (1)
  2. Find the Cartesian equation of the line L.  (1)
  3. Find the foot of the perpendicular drawn from ( 2, 3, 4 ) to the line L. (2)

Answer:
1. \(\overline{A B}\) = (3 – 1)i + (- 3 – 3)j + (3 – 0)k = 2i – 6j + 3k.

2. The Cartesian equation of a line passing through the point (4, 0, -1 ) and parallel to the vector \(\overline{A B}\) is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 9

3. We take, \(\frac{x-4}{2}=\frac{y}{-6}=\frac{z+1}{3}\) = r then any point of the line can be taken as (2r + 4, -6r, 3r – 1). Assume that this point be the foot of the perpendicular drawn from (2, 3, 4 ). The dr’s of the line is 2 : – 6 : 3 and dr’s of the perpendicular line L is
2r + 4 – 2 : -6r – 3 : 3r – 1 – 4 ⇒ 2r + 2: -6r – 3 : 3r – 5
Since perpendicular,
2(2 r + 2) – 6(-6 r -3) + 3(3r – 5) = 0
49r = -7 ⇒ r = \(\frac{-7}{49}\) = –\(\frac{1}{7}\). Therefore the foot of the
perpendicular is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 10

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 4.
Cartesian equation of two lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 11
(i) Write the vector equation of the lines. (2)
(ii) Shortest distance between the lines. (2)
Answer:
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 12
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 13
Question 5.
Consider the points (1, 3, 4) & (-3, 5, 2)

  1. Find the equation of the line through P and Q. (1)
  2. At which point that the above line cuts the plane 2x + y + z + 3 = 0. (3)

Answer:
1. Equation of a line passing through( 1, 3, 4) and (-3, 5, 2) is given by,
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 14
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 15

2. Let \(\frac{x-1}{-2}=\frac{y-3}{1}=\frac{z-4}{-1}\) = λ
Then any point on the line is (-2λ + 1, λ + 3, -λ + 4)
Since the plane 2x + y + z + 3 = 0 cuts the aboveline. We have,
⇒ 2(-λ + 1) + λ + 3 – λ + 4 + 3 = 0
⇒ -2λ + 2 + λ + 3 – λ + 4 + 3 = 0
⇒ -2λ = -12 ⇒ λ = 6
∴ point of intersection is (-2 × 6 + 1, 6 + 3, -6 + 4)
⇒ (-11, 9, -2).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 6.
Let the equation of a plane be \(\bar{r}\). (2i – 3j + 5k) = 7, then

  1. Find the Cartesian equation of the plane. (1)
  2. Find the equation of a plane passing through the point (3, 4, -1) and parallel to the given plane. (2)
  3. Find the distance between the parallel planes. (1)

Answer:
1. Given, \(\bar{r}\).(2i – 3j + 5k) = 7 and if we substitute \(\bar{r}\) = xi + yj + zk Then we get the Cartesian equation as 2x – 3y + 5z – 7 = 0.

2. The equation of a plane parallel to the above plane differ only by a constant, therefore let the equation be 2x – 3y + 5z + k = 0.
⇒ 6 – 12 – 5 + k = 0 ⇒ k = 11
Therefore the equation is 2x – 3y + 5z + 11 = 0

3. The distance between the parallel planes
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 16

Question 7.

  1. State the condition for the line \(\bar{r}\) = \(\bar{a}\) + λ \(\bar{b}\) is parallel to the plane \(\bar{r}\).\(\bar{n}\) = d. (2)
  2. Show that the line \(\bar{r}\) = i + j + λ(2i + j + 4k) is parallel to the plane \(\bar{r}\). (-2i + k) = 5. (1)
  3. Find the distance between the line and The Plane in (ii). (1)

Answer:
1. The line \(\bar{r}\) = \(\bar{a}\) + λ \(\bar{b}\) is parallel to the plane \(\bar{r}\).\(\bar{n}\) = d, if the normal of the plane is perpendicularto the line.
∴ \(\bar{b}\).\(\bar{n}\) = 0.

2. Given,
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 17
The line \(\bar{r}\) = i + j + λ(2i + j + 4k) is parallel to the plane \(\bar{r}\). (-2i + 4k) = 5 ⇒ -2x + 4y = 5.

3. Distance = Distance between – 2x + 4y = 5 and point (1, 1, 0) on the line
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 19

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 8.
Choose the correct answer from the bracket,
(i) If a line in the space makes angle α, β and γ with the coordinates axes, then cos2α + cos2β + cos2γ is equal to (1)
(a) 1
(b) 2
(c) 0
(d) 3
(ii) The direction ratios of the line are \(\frac{x-6}{1}=\frac{2-y}{2}=\frac{z-2}{2}\) (1)
(a) 6, -2, -2
(b) 1, 2, 2
(c) 6, 1, -2
(d) 0, 0, 0
(iii) If the vector equation of a line is \(\bar{r}\) = i + j + k + µ(2i – 3j – 4k), then the Cartesian equation of the line
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 20
(iv) If the Cartesian equation of a plane is x + y + z =12, then the vector equation of the line is (1)
(a) \(\bar{r}\).(2i + j + k) = 12
(b) \(\bar{r}\).(i + j + k) = 12
(C) \(\bar{r}\).(i + y + 2k) = 12
(d) \(\bar{r}\).(i + 3j + k) = 12
Answer:
(i) (a) 1

(ii) (b) 1, 2, 2

(iii) (b) \(\frac{x-I}{2}=\frac{y-1}{-3}=\frac{z-1}{-4}\)

(iv) (b) \(\bar{r}\).(i + j + k) = 12.

Question 9.
Consider the lines \(\bar{r}\) = (i + 2j – 2k) + λ(i + 2 j) and \(\bar{r}\) = (i + 2j – 2k) + µ(2j – k)

  1. Find the angle between the lines.
  2. Find a vector perpendicular to both the lines.
  3. Find the equation of the line passing through the point of intersection of lines and perpendicular to both the lines.

Answer:
1. \(\bar{b}_{1}\) = i + 2j; \(\bar{b}_{2}\) = 2j – k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 21

2. Perpendicular vector = \(\bar{b}_{1}\) × \(\bar{b}_{2}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 22
= i(-2 – 0) -j(-1 – 0) + k(2 – 0)
= -2i + j + k.

3. Equation of line is \(\bar{r}\) = (i + 2j – k) + µ(-2i + j + 2k).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 10.
Consider the line \(\bar{r}\) = (2i – j + k) + λ(i + 2j + 3k)

  1. Find the Cartesian equation of the line.
  2. Find the vector equation of the line passing through A (1, 0, 2) and parallel to the above line.
  3. Write two points on the line obtained in (ii) which are equidistant from A.

Answer:
1. \(\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-1}{3}\).

2. \(\bar{r}\) = (i + 2k) + λ(i + 2j + 3k).

3. Put λ = a and λ = -a for any real value ‘a’.
Let us put λ = 1 and λ = -1
\(\bar{r}\) = (i + 2k) + 1(i + 2j + 3k) = 2i + 2j + 5k
⇒ (2, 2, 5)
\(\bar{r}\) = (i + 2k) – 1(i + 2j + 3k) = 0i – 2j – k
⇒ (0, -2, -1)
The equidistant points are(2, 2, 5) and (0, -2, -1).

Question 11.

  1. Find the equation of the plane through the point(1, 2, 3) and perpendicular to the plane x – y + z = 2 and 2x + y – 3z = 5 (2)
  2. Find the distance between the planes x – 2y + 2z – 8 = 0 and 6y – 3x – 6z = 57 (2)

Answer:
1. Required equation is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 23
(x – 1)(3 – 1) – (y – 2)(-3 – 2) + (z – 3)(1 + 2) = 0
2(x – 1) + 5(y – 2) + 3(z – 3) = 0
2x + 5y + 3z – 2 – 10 – 9 = 0
2x + 5y + 3z – 21 = 0

2. The planes are
x – 2y + 2z – 8 = 0 and 3x – 6y + 6z + 57 = 0
ie, 3x – 6y + 6z – 24 = 0 and 3x – 6y + 6z + 57 = 0
Distance
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 24

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 12.
Consider the Cartesian equation of a line \(\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-5}{-2}\)

  1. Find the vector equation of the line. (1)
  2. Find its intersecting point with the plane 5x + 2y – 6z – 7 = 0 (2)
  3. Find the angle made by the line with the plane 5x + 2y – 6z – 7 = 0 (1)

Answer:
1. The vector equation is \(\bar{r}\) = (3i – j + 5k) + λ(2i + 3 j – 2k).

2. Any point on the line is
\(\frac{x-3}{2}=\frac{y+1}{3}=\frac{z-5}{-2}\) = λ
x = 2λ + 3, y = 3, λ – 1, z = -2λ + 5
Since this lies on the plane ,it satisfies the plane
5(2λ + 3) + 2(3λ – 1) -6(-2λ + 5) – 7 = 0
10λ + 6λ + 12λ + 15 – 2 – 30 – 7 = 0
28λ = 24
λ = 6/7
The point of intersection is \(\left[\frac{33}{7}, \frac{11}{7}, \frac{23}{7}\right]\).

3. Let θ be the angle between the line and the plane. The direction of the line and the plane
\(\bar{b}\) = 2i + 3j + k; \(\bar{m}\) = 5i + 2j – 6k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 25

Question 13.
From the following figure
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 26

  1. Find \(\overline{A B}\). (1)
  2. Find the vector equation of line L. (1)
  3. Find a point on line L other than C. (2)

Answer:
1. P.v of A = i – j + 4k,
P.v. of B = 2i + j + 2k
\(\overline{A B}\) = p. v. of B – p. v. of A
= 2i + j + 2k -(i – j + 4k) = i + 2j – 2k.

2. The line L passes through (1, -2, -3) and parallel to \(\overline{A B}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 28
∴ Vector equation of line L is \(\bar{r}=\bar{a}+\lambda \bar{m}\)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 29

3. From (1) of part (ii), we have
xi + yj + zk = (l + λ)i + (-2 + 2λ)j + (-3 – 2λ)k
Put λ = 1
⇒ xi +yj + zk = (1 +1)i + (-2 + 2)j + (-3 – 2 )k
⇒ xi + yj + zk = 2i + 0j – 5k
Therefore a point on line L is (2, 0, -5).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 14.
Find the vector equation of the plane which is at a distance of \(\frac{6}{\sqrt{29}}\) from the origin with perpendicular vector 2i – 3j + 4k. Convert into Cartesian form. Also, find the foot of the perpendicular drawn from the origin to the Plane.
Answer:
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 30
Perpendicular distance from origin = d = \(\frac{6}{\sqrt{29}}\)
The equation of the Plane is \(\bar{r}\).\(\hat{n}\) = d
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 31
Cartesian equation is 2x – 3y + 4z = 6
The direction cosines perpendicular to the Plane is \(\frac{2}{\sqrt{29}},-\frac{3}{\sqrt{29}}, \frac{4}{\sqrt{29}}\).
Perpendicular distance to the Plane is as \(\frac{6}{\sqrt{29}}\)
Hence the foot of the perpendicular is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 32

Question 15.
Consider the Plane \(\bar{r}\).(-6i -3j – 2k) + 1 = 0, find the direction cosines perpendicular to the Plane and perpendicular distance from the origin.
Answer:
Convert the equation of the plane into normal form
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 33
Direction cosines perpendicular to the Plane is \(\frac{6}{7}, \frac{3}{7}, \frac{2}{7}\)
Perpendicular distance from the origin is \(\frac{1}{7}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 16.
Consider three points (6, -1, 1), (5, 1, 2) and (1, – 5, -4) on space.

  1. Find the Cartesian equation of the plane passing through these points. (2)
  2. Find direction ratios normal to the Plane.(1)
  3. Find a unit vector normal to the Plane. (1)

Answer:
1. Equation of a plane passing through the points (6, -1, 1),(5, 1, 2) and (1, -5, 4)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 34
⇒ (x – 6)(-10 + 4) – (y + 1)(5 + 5) + (z – 1)(4 + 10) = 0
⇒ (x – 6)(-6) – (y + 1)(10) + (z – 1)(14) = 0
⇒ -6x + 36 – 10y – 10 + 14z – 14 = 0
⇒ 6x +10y – 14z -12 = 0.

2. Dr’s normal to the plane are 6 : 10 : -14 ⇒ 3 : 5 : -7.

3. Since the dr’s normal to the plane are 3 : 5 : -7, a unit vector in this direction is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 35

Question 17.
Consider a straight line through a fixed point with position vector 2i – 2j + 3k and parallel to i – j + 4k.

  1. Write down the vector equation of the straight line. (1)
  2. Show that the straight line is parallel to the plane \(\bar{r}\).(i + 5y + k) = 5 (1)
  3. Find the distance between the line and plane. (2)

Answer:
1. Vector equation of a straight line is \(\bar{r}=\bar{a}+\lambda \bar{b}\) where a is \(\bar{a}\) fixed point and \(\bar{b}\) is a vector parallel to the line. Here \(\bar{a}\) = 2i – 2y + 3 it and \(\bar{b}\) = i – j + 4k. Therefore vector equation of the line \(\bar{r}\) = 2i – 2j + 3k + λ(i – j + 4k).

2. The vector parallel to the line is i – j + 4k and vector normal to the plane is i + 5j + k.
Then, (i – j + 4k). (i + 5j + k) = 1 – 5 + 4 = 0
implies that straight line and plane are parallel.

3. A point on the line is 2i – 2j + 3k. Then the distance of 2i – 2j + 3k to the given plane
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 36

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 18.
Consider the vector equation of two planes \(\bar{r}\).(2i + j + k) = 3, \(\bar{r}\).(i – j – k) = 4

  1. Find the vector equation of any plane through the intersection of the above two planes. (2)
  2. Find the vector equation of the plane through the intersection of the above planes and the point (1, 2, -1 ) (2)

Answer:
1. The cartesian equation are 2x + y + z – 3 = 0 and x – y – z – 4 = 0 Required equation of the plane is
(2x + y + z – 3) + λ(x – y – z – 4) = 0
(2+ λ)x + (1 – λ)y + (1 – λ)z + (-3 – 4λ) = 0.

2. The above plane passes through (1, 2, -1)
(2+ λ)1 + (1 – λ)2 + (1 – λ)(-1) + (-3 – 4λ) = 0
3 – 3 + 4λ = 0
λ = 0
Equation of the plane is 2x + y + z – 3 = 0
\(\bar{r}\).(2i + j + k) = 3.

Question 19.
(i) Distance of the point(0, 0, 1) from the plane x + y + z = 3
(a) \(\frac{1}{\sqrt{3}}\) units
(b) \(\frac{2}{\sqrt{3}}\) units
(c) \(\sqrt{3}\) units
(d) \(\frac{\sqrt{3}}{2}\) units
(ii) Find the equation of the plane through the line of intersection of the planes x + y + z = 1 and 2x + 3y + 4z = 5 which is perpendicular to x – y + z = 0 (3)
Answer:
(i) (b) \(\frac{2}{\sqrt{3}}\) units.

(ii) Equation of the plane passing through the intersection is of the form
x + y + z – 1 + λ(2x + 3y + 4z – 5) = 0 _____(1)
(1 + 2λ)x + (1 + 3λ)j + (1 + 4λ)z – 1 – 5λ = 0
Thr Dr’s of the required plane is
(1 + 2λ), (1 + 3λ), (1 + 4λ)
Thr Dr’s of the Perpendicular plane is 1, -1, 1
⇒ (1 + 2λ)(1) + (1 + 3λ)(-1) + (1 + 4λ)(1) = 0
⇒ 1 + 2λ – 1 – 3λ + 1 + 4λ = 0
⇒ 3λ + 1 = 0 ⇒ λ = \(\frac{-1}{3}\)
(1) ⇒ x + y + z – \(\frac{1}{3}\)(2x + 3y + 4z – 5) = 0
⇒ 3x + 3y + 3z – 2x – 3y – 4z + 5 = 0
⇒ x – z + 2 = 0.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 20.
Consider a plane \(\bar{r}\).(6i – 3j – 2k) + 1 = 0

  1. Find dc’s perpendicular to the plane. (2)
  2. Find a vector of magnitude 14 units perpendicular to given plane. (1)
  3. Find the equation of a line parallel to the above vector and passing through the point (1, 2, 1 ). (1)

Answer:
1. Given, \(\bar{r}\).(6i – 3j – 2k) + 1 = 0 ____(1)
Now, |6i – 3j – 2k| = \(\sqrt{36+9+4}\) = 7
∴ \(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\) is a unit perpendicular to the plane (1)
⇒ the dc’s perpendicular to the plane (1) are \(\frac{6}{7},-\frac{3}{7},-\frac{2}{7}\).

2. We have, \(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\) is a unit perpendicular to the Plane (1). Therefore, a vector of magnitude 14 units perpendicular to the Plane (1) is 14(\(\frac{6}{7} i-\frac{3}{7} j-\frac{2}{7} k\))
⇒ 12i – 6j – 4k.

3. Equation of a line parallel to the vector 12i – 6j – 4k and passing through the point (1, 2, 1 )is given by
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 37

Plus Two Maths Three Dimensional Geometry Six Mark Questions and Answers

Question 1.
Consider the pair of lines whose equations are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 38

  1. Write the direction ratios of the lines. (1)
  2. Find the shortest distance between the above skew lines. (4)
  3. Find the angle between these two lines. (1)

Answer:
1. The direction ratios are 2, 5, – 3 and – 1, 8, 4.

2. The given lines are \(\bar{r}\) = (2i + j – 3k) + λ(2i + 5j – 3k)
i.e. \(\bar{r}\) = \(\overline{a_{1}}+\lambda \overline{b_{1}}\),
where \(\overline{a_{1}}\) = 2i + j – 3k) + λ(2i + 5j – 3k)
and \(\bar{r}\) =(-i + 4j + 5k) + µ(-i + 8j + 4k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 39

3. cosθ
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 40

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 2.
Consider the pair of lines \(\bar{r}\) = 3i + 4j – 2k + λ(-i + 2j + k) ——L1, \(\bar{r}\) = i – 7j – 2k + µ(i + 3j + 2k) ——L2

  1. Find one point each on lines L1 and L2. (1)
  2. Find the distance between those points. (2)
  3. Find the shortest distance between L1 and L2. (3)

Answer:
1. By putting λ = 0 in line L1 and µ = 0 in L2 we get the required points. L1 ⇒ \(\bar{r}\) = 3i + 4j – 2k
∴ Co-ordinate is (3, 4, -2)
L2 ⇒ \(\bar{r}\) = i – 7j – 2k
∴ Co-ordinate is (1, -7, -2).

2. Distance between (3, 4, -2) and (1, -7, -2)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 41

3. Let L1 ⇒ \(\bar{r}\) = 3i + 4j – 2k + λ(-i + 2j + k) is of the form \(\bar{r}=\overline{a_{1}}+\lambda \overline{b_{1}}\) where
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 42

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 3.
Consider the points A (2, 2, -1), B (3, 4, 2) and C (7, 0, 6).

  1. Are A, B, and C collinear? Explain.
  2. Find the vector and Cartesian equation of the plane passing these three points. (2)
  3. Find the angle between the above plane and the line \(\bar{r}\) = (i + 2j – k) + λ(i – j + k) (2)

Answer:
1. Direction ratios along A and B is 3 -2, 4 -2, 2 + 1 ⇒ 1, 2, 3
Direction ratios along B and C is
7 -3, 0 -4, 6 -2 ⇒ 4, -4, 4
Since the direction ratios are not proportional they are not collinear.

2. Cartesian equation of the Plane is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 43
⇒ (x – 2)(14 + 6) -(y – 2)(7 – 15) + (z + 1)(-2 -10) = 0
⇒ 20(x – 2) + 8(y – 2) – 12(z + 1) = 0
⇒ 20x – 40 + 8y – 16 – 12z – 12 = 0
⇒ 20x + 8y – 12z = 68
⇒ 5x + 2y – 3z = 17
Vector Equation is \(\bar{r}\).(5i + 2j – 3k) = 17.

3. Angle between the Plane and the Line
\(\bar{r}\) = (i + 2j – k) + λ(i – j + k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 44

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 4.
Consider three points on space ( 2, 1, 0 ), (3, -2, -2)and(3, 1, 7)

  1. Find the Cartesian equation of the plane passing through the above points. (2)
  2. Convert the above equation into vector form.
  3. Hence, find a unit vector perpendicular to the above plane and also find the perpendicular distance of the plane from the origin. (2)

Answer:
1. Equation of the plane is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 45
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 46
⇒ (x – 2)(-21) – (y – 1)(7 + 2) + z(0 + 3) = 0
⇒ 21x + 42 – 9y + 9 + 3z = 0 ⇒ -21x – 9y + 3z + 51 = 0
⇒ 7x + 3y – z = 17.

2. Vector form is \(\bar{r}\).(7i + 3j – k) = 17 _____(1)

3. Now, |7i + 3j – k| = \(\sqrt{49+9+1}=\sqrt{59}\)
Dividing equation (1) by \(\sqrt{59}\), we get
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 47
Therefore the above equation is the normal form of the plane. Then \(\frac{7 i+3 j-k}{\sqrt{59}}\) is the unit vector perpendicular to the plane and \(\frac{17}{\sqrt{59}}\) is the perpendicular distance from the origin.

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 5.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 48
\(\overline{O A}\) = i + 2j + 3k
\(\overline{O B}\) = i – 2j + 4k
\(\overline{O C}\) = 2i + 3j + k
are adjacent sides of the parallelopiped.

  1. Find the base area of the parallelopiped. (2)
    (Base determined by \(\overline{O A}\) and \(\overline{O B}\))
  2. Find the volume of the parallelopiped. (2)
  3. Find the height of the parallelopiped. (2)

Answer:
1. \(\overline{O A}\) × \(\overline{O B}\) =
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 49
= 14i – j – 4k
Base area = |l4i – j – 4k|
\(=\sqrt{196+1+16}=\sqrt{213}\)

2. Volume of the parallelopiped is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 50
= (14i – j – 4k).(2i + 3 j + k)
= 28 – 3 – 4 = 21.

3. Height = \(\frac{\text {volume}}{\text {base area}}=\frac{21}{\sqrt{213}}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 6.

  1. Find the equation of the line passing through the point (2, 1, 0) and (3, 2, -1) (3)
  2. Find the shortest distance of the above line from the line \(\bar{r}\) = (i – j + 2k) + λ(2i + j – 3k) (3)

Answer:
1.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 51

2.
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 52
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 53
= i(-3 + 1) – j(-3 + 2) + k(1 – 2)
= -2i + j – k
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 54

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 7.
The equation of two lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 55

  1. Find the dr’s of the given lines. (2)
  2. Find the angle between the given lines. (2)
  3. Find the equation of the line passing through (2, 1, 3) and perpendicular to the given lines. (2)

Answer:
1. The given lines are
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 56
The dr’s of (1) are 2, 2, 3 and dr’s of (2) are -3, 2, 5.

2. The angle between (1) and (2) is given by
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 57

3. Let a, b, c be the dr’s of the line perpendicular to lines (1) and (2).
∴ 2a + 2b + 3c = 0, -3a + 2b + 5c = 0
Solving by the rule of cross-multiplication, we get
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 58
∴ dr’s of the required line are 4, -19, 10 and the line passes through (2, 1, 3).
∴ Equation of the required line is
\(\frac{x-2}{4}=\frac{y-1}{-19}=\frac{z-3}{10}\).

Plus Two Maths Chapter Wise Questions and Answers Chapter 11 Three Dimensional Geometry

Question 8.

  1. Find the direction cosines of the vector 2i + 2j – k. (1)
  2. Find the distance of the point (2, 3, 4) from the plane \(\bar{r}\).(3i – 6j + 2 k) = -11. (2)
  3. Find the shortest distance between the lines \(\bar{r}\) = (2i – j – k)+ λ(3i – 5 j + 2k) an \(\bar{r}\) = (i+ 2 j + k)+ µ(i – j + k) (3)

Answer:
1. Direction ratios of the vector 2i + 2j – k is 2, 2, -1
Direction cosines of the vector 2i + 2j – k is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 59

2. The equation of the plane in the Cartesian form is 3x – 6y + 2z + 11 = 0 . Then distance from the point (2, 3, 4) is
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 60

3. The given lines are \(\bar{r}\) = (2i – j – k) + λ(3i – 5j + 2k)
Plus Two Maths Three Dimensional Geometry 4 Mark Questions and Answers 61

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