cosine angle to Cartesian equation.
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Dr.Peterson
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Do you know that the direction cosines are the components of a unit vector in the direction of the line?
Do you know the general equation for a line in terms of a point (in this case, the origin) and the direction vector? See the last part of this page: https://tutorial.math.lamar.edu/classes/calciii/eqnsoflines.aspx
Do you know the general equation for a line in terms of a point (in this case, the origin) and the direction vector? See the last part of this page: https://tutorial.math.lamar.edu/classes/calciii/eqnsoflines.aspx
I think I got it but I cant explain how ?. I think the inverse would give you the ratios of their magnitude.
Dr.Peterson
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Please show some actual work, even if you know it's wrong or incomplete, so I can be sure what you're thinking. Ignore what you see in the given solution; just try writing some equations, using what you know. The best way to understand something is to do it yourself.
Dr.Peterson
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You seem to have started with what they wrote, not with the information in the problem; that is the answer to the problem, apart from simplification, so it can't be where you are having trouble!
Then you wrote a single equation, in vector form and then in rectangular form (though you dropped the radical), not for the line, but for a plane normal to it. That makes me wonder whether you know what the equations of a line look like in the first place, which could be the source of your difficulty. A plane has a single equation; a line requires two.
Please look again (if you did at all) at the link I gave, https://tutorial.math.lamar.edu/classes/calciii/eqnsoflines.aspx. At the bottom, it gives the form we need: [MATH]\frac{x−x_0}{a}=\frac{y−y_0}{b}=\frac{z−z_0}{c}[/MATH]. Observe that this is two equations, not one.
Since your line has to go through the origin, [MATH](0,0,0)[/MATH], this becomes [MATH]\frac{x}{a}=\frac{y}{b}=\frac{z}{c}[/MATH]. And a, b, c are the components of a vector in the direction of the line, such as the direction cosines (called l, m, and n in your book).
I'll finish for you, doing the important step they left out and which is most likely confusing you: Given that [MATH]l=\frac{\sqrt{2}}{2}[/MATH], [MATH]m=\frac{1}{2}[/MATH], and [MATH]n=\frac{1}{2}[/MATH], the equation becomes [MATH]\frac{x}{\frac{\sqrt{2}}{2}}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{1}{2}}[/MATH], which simplifies to the equation they wrote.
Does that make it clearer?
Then you wrote a single equation, in vector form and then in rectangular form (though you dropped the radical), not for the line, but for a plane normal to it. That makes me wonder whether you know what the equations of a line look like in the first place, which could be the source of your difficulty. A plane has a single equation; a line requires two.
Please look again (if you did at all) at the link I gave, https://tutorial.math.lamar.edu/classes/calciii/eqnsoflines.aspx. At the bottom, it gives the form we need: [MATH]\frac{x−x_0}{a}=\frac{y−y_0}{b}=\frac{z−z_0}{c}[/MATH]. Observe that this is two equations, not one.
Since your line has to go through the origin, [MATH](0,0,0)[/MATH], this becomes [MATH]\frac{x}{a}=\frac{y}{b}=\frac{z}{c}[/MATH]. And a, b, c are the components of a vector in the direction of the line, such as the direction cosines (called l, m, and n in your book).
I'll finish for you, doing the important step they left out and which is most likely confusing you: Given that [MATH]l=\frac{\sqrt{2}}{2}[/MATH], [MATH]m=\frac{1}{2}[/MATH], and [MATH]n=\frac{1}{2}[/MATH], the equation becomes [MATH]\frac{x}{\frac{\sqrt{2}}{2}}=\frac{y}{\frac{1}{2}}=\frac{z}{\frac{1}{2}}[/MATH], which simplifies to the equation they wrote.
Does that make it clearer?
now i get it. I wondered where x0, y0 and z0 diappeared.