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23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N?
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23. Express the vector in the Cartesian form if the angle made by it w...
To express a vector in Cartesian form, we need to find its components along the x, y, and z axes. We are given that the vector makes an angle of 60° with the y-axis and 45° with the z-axis. Let's break down the problem step by step.
Magnitude of the force: 200N
Angle with the y-axis: 60°
Angle with the z-axis: 45°
Finding the x-component:
The x-component of the vector can be found using trigonometry. Since the angle with the x-axis is not given, let's assume it to be 'a'. The x-component can be found using the formula:
x = magnitude * cos(angle)
x = 200N * cos(a)
Finding the y-component:
The y-component can be found using trigonometry. The angle with the y-axis is given as 60°. The y-component can be found using the formula:
y = magnitude * sin(angle)
y = 200N * sin(60°)
y = 200N * √3/2
y = 100√3 N
Finding the z-component:
The z-component can be found using trigonometry. The angle with the z-axis is given as 45°. The z-component can be found using the formula:
z = magnitude * sin(angle)
z = 200N * sin(45°)
z = 200N * √2/2
z = 100√2 N
Expressing the vector in Cartesian form:
The vector can now be expressed in Cartesian form as the sum of its components along the x, y, and z axes:
Vector = x î + y ĵ + z k̂
Substituting the values:
Vector = (200N * cos(a)) î + (100√3 N) ĵ + (100√2 N) k̂
Simplifying the expression:
Vector = 200 cos(a) î + 100√3 ĵ + 100√2 k̂
Therefore, the vector can be expressed in Cartesian form as 200 cos(a) î + 100√3 ĵ + 100√2 k̂.
Magnitude of the force: 200N
Angle with the y-axis: 60°
Angle with the z-axis: 45°
Finding the x-component:
The x-component of the vector can be found using trigonometry. Since the angle with the x-axis is not given, let's assume it to be 'a'. The x-component can be found using the formula:
x = magnitude * cos(angle)
x = 200N * cos(a)
Finding the y-component:
The y-component can be found using trigonometry. The angle with the y-axis is given as 60°. The y-component can be found using the formula:
y = magnitude * sin(angle)
y = 200N * sin(60°)
y = 200N * √3/2
y = 100√3 N
Finding the z-component:
The z-component can be found using trigonometry. The angle with the z-axis is given as 45°. The z-component can be found using the formula:
z = magnitude * sin(angle)
z = 200N * sin(45°)
z = 200N * √2/2
z = 100√2 N
Expressing the vector in Cartesian form:
The vector can now be expressed in Cartesian form as the sum of its components along the x, y, and z axes:
Vector = x î + y ĵ + z k̂
Substituting the values:
Vector = (200N * cos(a)) î + (100√3 N) ĵ + (100√2 N) k̂
Simplifying the expression:
Vector = 200 cos(a) î + 100√3 ĵ + 100√2 k̂
Therefore, the vector can be expressed in Cartesian form as 200 cos(a) î + 100√3 ĵ + 100√2 k̂.
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23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N?
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23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about 23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N?.
23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about 23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for 23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N?.
Solutions for 23. Express the vector in the Cartesian form if the angle made by it with y-axis and z-axis are 60° and 45° respectively. Also, it makes an angle of a with the x-axis. The magnitude of the force is 200N. (a) 100 - 100; 141.4k N (b) 100 100j 141.4k N (c) 100 - 100 - 141.4k N (d) 100i 100; - 141.4k N? in English & in Hindi are available as part of our courses for Class 10.
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