Respuesta :
Answer:(x-3)=-(y-5)=z
Step-by-step explanation:
Given
the point through which line passes is (3,5,0)
so we need a vector along the line to get the equation of line
It is given that line is perpendicular to both i+j & j+k
therefore their cross product will give us the vector perpendicular to both
v=(i+j)\times (j+k)=i-j+k
therefore we get direction vector of line so we can write
[tex]\frac{x-3}{1}=\frac{y-5}{-1} =\frac{z-0}{1}[/tex]=t
i.e.
x=t+3,y=-t+5,z=t
The parametric form of the equation is;
[tex]\rm x=x_1+at, \ x=3+1t, \ x=3+t\\\\y=y_1+at, \ y=5+(-1)t, \ y=5-t\\\\z=z_1+at, \ z=0+a(1), \ z=a[/tex]
The symmetric form of the equation is [tex]\rm x - 3 = -(y -5) = z[/tex].
Given
The line through (3, 5, 0) and perpendicular to both i + j and j + k
The symmetric form of the equation of the line is given by;
[tex]\rm \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}=t[/tex]
Where the value of [tex]\rm x_1=3, \ y_1=5\ and \ z_1=0[/tex].
To find a, b, c by evaluating the product of ( i + j) and ( j + k ).
[tex]\rm= (i+j)\times (j+k)\\\\= i-j+k[/tex]
The value of a = 1, b = -1 and c = 1.
Substitute all the values in the equation.
[tex]\rm \dfrac{x-x_1}{a}=\dfrac{y-y_1}{b}=\dfrac{z-z_1}{c}\\\\\dfrac{x-3}{1}=\dfrac{y-5}{-1}=\dfrac{z-0}{1}\\\\(x-3)=-(y-5)=z[/tex]
Therefore,
The parametric form of the equation is;
[tex]\rm x=x_1+at, \ x=3+1t, \ x=3+t\\\\y=y_1+at, \ y=5+(-1)t, \ y=5-t\\\\z=z_1+at, \ z=0+a(1), \ z=a[/tex]
To know more about symmetric equations click the link given below.
https://brainly.com/question/14701215
