Balbharti12th Maharashtra State Board Maths Solutions BookPdf Chapter 5 Vectors Ex 5.1 Questions and Answers.
Maharashtra State Board 12th Maths Solutions Chapter 5 Vectors Ex 5.1
Solution & Step-by-Step Answer:
Let = , = Then = + = a + b Given : = = l(AB) = 24 and = = l(BC) = 7 ∴ ∠ABC = 90° ∴ [l(AC)]2 = [l(AB)]2 + [l(BC)]2 = (24)2 + (7)2 = 625 ∴ l(AC) = 25 ∴ = 25 ∴ = = 25.
Solution & Step-by-Step Answer:
Given : = 2, = 2 (i) = + = 2 + 2.
(ii)
Solution:
∵ M is the midpoint of PR
∴ = = [2 + 2]
= + .
(iii)
Solution:
= = = (2 + 2)
= – –
∴ = +
= 2 – –
= – .
Solution & Step-by-Step Answer:
Given : = , = Let AD, BE, OC meet at M. Then M bisects AD, BE, OC. = + = – + = – + = – ∵ OABM is a parallelogram Hence, the position vectors of C, D and E are 2 – 2, 2 – 3 and – 2 respectively.
Solution & Step-by-Step Answer:
ABCDEF is a regular hexagon. ∴ = and = ∴ by the triangle law of addition of vectors,
Solution & Step-by-Step Answer:
Let, if possible, the three vectors form a triangle ABC with = , = , = Now, + = + = = Hence, the three vectors do not form a triangle.
Solution & Step-by-Step Answer:
= + ∴ = – … (1) = + ∴ = + … (2) Adding equations (1) and (2), we get + = ( – ) + ( + ) = 2
Solution & Step-by-Step Answer:
Let the point A be (4, -2, 6). Then, The distance of A from XY-plane = |z| = 6
(b) The YZ-plane
Solution:
The distance of A from YZ-plane = |x| = 4
(c) The XZ-plane
Solution:
The distance of A from ZX-plane = |y| = 2
(d) The X-axis
Solution:
The distance of A from X-axis
= = = =
(e) The Y-axis
Solution:
The distance of A from Y-axis
= = = =
(f) The Z-axis
Solution:
The distance of A from Z-axis
= = = =
Solution & Step-by-Step Answer:
Let the coordinates of the point be (x, y, z). Since the point is located 3 units behind the YZ- j plane, 4 units to the right of XZ-plane and 5 units, above the XY-plane, x = -3, y = 4 and z = 5 Hence, coordinates of the required point are (-3, 4, 5)
(b) In the YZ-plane, one unit to the right of the XZ-plane and six units above the XY-plane.
Solution:
Let the coordinates of the point be (x, y, z).
Since the point is located in the YZ plane, x = 0. Also, the point is one unit to the right of XZ-plane and six units above the XY-plane.
∴ y = 1, z = 6.
Hence, coordinates of the required point are (0, 1, 6).
Solution & Step-by-Step Answer:
Let A = (1, 1, 0), B = (1, 0, 1), C = (0, 1, 1)
Solution & Step-by-Step Answer:
Let and be the position vectors of A and B. Given : A = (1, 5, 0).’. = Now, = ∴ – = ∴ = + = + = Hence, the terminal point B = (3, 1, 7).
Solution & Step-by-Step Answer:
Let , , be the position vectors of the points. A = (3, 2, -4), B = (9, 8, -10) and C = (-2, -3, 1) respectively. ∴ is a non-zero scalar multiple of ∴ they are parallel to each other. But they have the point B in common. ∴ and are collinear vectors. Hence, the points A, B and C are collinear.
(ii) P (4, 5, 2), Q (3, 2, 4), R (5, 8, 0).
Solution:
Let , , be the position vectors of the points.
P = (4, 5, 2), Q = (3, 2, 4), R = (5, 8, 0) respectively.
= 2. …[By (1)]
∴ is a non-zero scalar multiple of
∴ they are parallel to each other.
But they have the point B in common.
∴ and are collinear vectors.
Hence, the points A, B and C are collinear.
Solution & Step-by-Step Answer:
The vectors and are collinear ∴ the coefficients of are proportional
Solution & Step-by-Step Answer:
Solution: The position vectors , , , of the points A, B, C, D are By equality of vectors, y = -2 ….(1) 2x – 2y = 2 … (2) 3y = 0 … (3) From (1), y = -2 From (3), y = 0 This is not possible. Hence, the points A, B, C, D are not coplanar.
Solution & Step-by-Step Answer:
By equality of vectors, 2x + 2y + 3 = -1 x – y + z = -3 -4x + 3y – 2z = 4 We have to solve these equations by using Cramer’s Rule D = = 2(2 – 3) – 2(-2 + 4) + 3(3 – 4) = -2 – 4 – 3 = -9 ≠ 0 = 2(-4 + 9) – 2(4 – 12) – 1(3 – 4) = 10 + 16 + 1 = 27
