Answer:
Explanation:
Suppose u = i + j =
& v = j + k =
![\Big \langle 0,1,1 \Big\rangle](https://img.qammunity.org/2021/formulas/mathematics/college/mgxy4eooa7vjvrm5gcl2tk8shim2ua7li7.png)
The direction vector for the line u*v is:
![u* v = \left |\begin{array}{ccc}i&j&k\\1&1&0\\0&1&1\end {array} \right|](https://img.qammunity.org/2021/formulas/mathematics/college/rknb9a1kam79e9a8wzbahq0g47cejewof2.png)
= (1-0) i - (1 - 0)j + ( 1- 0) k
= i - j - k
=
![\Big \langle 1,-1,1 \Big\rangle](https://img.qammunity.org/2021/formulas/mathematics/college/23f1ccra46gmezsk9c33ppftxlkhfyezm6.png)
Hence, the equation of the line via the point (5,4,0 ) and the direction vector
is as follows:
r(t) = (5,4,0) + t
![\Big \langle 1,-1,1 \Big\rangle](https://img.qammunity.org/2021/formulas/mathematics/college/23f1ccra46gmezsk9c33ppftxlkhfyezm6.png)
r(t) = (5+t, 4-t, t)
The symmetric equations of the line are:
![(x-5)/(1)= (y-4)/(-1) = (z)/(1)](https://img.qammunity.org/2021/formulas/mathematics/college/rcjg1pfdvbaffug6yywchztyzi0fep0cdx.png)
x - 5 = -(y-4) = z
The parametric equation of the line is:
![(x-5)/(1)= (y-4)/(-1) = (z)/(1)= t](https://img.qammunity.org/2021/formulas/mathematics/college/dab4yj72kgwbkxpq5sad2t2ogoo2thkb1v.png)
x = 5 + t , y = 4 - t , z = t