- #1
ttpp1124
- 110
- 4
- Homework Statement
- Calculus and Vectors - Vector and Parametric Equations
I solved it, can someone concur?
- Relevant Equations
- Not Available
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ttpp1124 said:When I sub in 0 for t, I get the values corresponding to the direction vector. That's how I know it's correct
how is it wrong?PeroK said:Looks wrong to me!
Is that ##x = -2t, \ y = -1 + 3t, \ z = 6 + 5t##?ttpp1124 said:how is it wrong?
YesPeroK said:Is that ##x = -2t, \ y = -1 + 3t, \ z = 6 + 5t##?
That's not right. What values of ##t## give you ##P## and ##Q##?ttpp1124 said:Yes
Is it the direction vector, 0,-1,6?PeroK said:That's not right. What values of ##t## give you ##P## and ##Q##?
The vector equation is correct. That's the direction ##\vec{QP}##.ttpp1124 said:Is it the direction vector, 0,-1,6?
Right, so I think I didn't multiply the t value and do vector addition. When I do multiply, I get r = (-2,3,5)+(0t,-t,6t).PeroK said:The vector equation is correct. That's the direction ##\vec{QP}##.
So, what value of ##t## gives you ##Q##?ttpp1124 said:Right, so I think I didn't multiply the t value and do vector addition. When I do multiply, I get r = (-2,3,5)+(0t,-t,6t).
Then I do vector addition by components, so then my parametric equations should be -2+0t, 3-t, and 5+6t
-1PeroK said:So, what value of ##t## gives you ##Q##?
My question is, how does the value of t have anything to do with the actual question? They just wanted to know the vector and parametric equations.ttpp1124 said:-1
ttpp1124 said:My question is, how does the value of t have anything to do with the actual question? They just wanted to know the vector and parametric equations.
Also, are my new parametric equations correct?
A vector in calculus is a mathematical object that has both magnitude (size) and direction. It is commonly represented by an arrow with a length and a direction. In parametric equations, a vector is represented by two or three components, depending on whether it is in two or three dimensions.
Parametric equations are a set of equations that express the coordinates of a point in terms of one or more independent variables, called parameters. They are used in calculus to describe the motion of a point or object in space, and to find the derivatives and integrals of vector functions.
To find the derivative of a vector function, you can use the chain rule. First, find the derivative of each component of the vector function with respect to the independent variable. Then, combine these derivatives to form a new vector, which is the derivative of the original vector function.
The dot product of two vectors is a scalar quantity that represents the projection of one vector onto the other. It is calculated by multiplying the magnitudes of the two vectors and the cosine of the angle between them.
The cross product of two vectors is a vector that is perpendicular to both of the original vectors. It is calculated by taking the determinant of a 3x3 matrix formed by the components of the two vectors. The magnitude of the cross product is equal to the product of the magnitudes of the two vectors and the sine of the angle between them.