How to find the parametric equation of line with vector equation
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Answer: Normal vector of the plane (1)x + ( -3 )y + ( -5 )z = 10
is n = ( 1, -3, -5 )
The given line is parallel to this normal and passes thru' ( 3, -1, 0).
Hence, its equations in Symmetric Form are
( x - 3 ) / 1 = ( y + 1 ) / ( -3 ) = ( z - 0 ) / ( -5 ) = t, say.
Then
x - 3 = t, y + 1 = -3t, z = -5t.
Hence, the parametric equations of the line are :
x = t + 3, y = - 3t - 1, z = - 5t,................................Ans.
where t is a parameter.
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