Question

Find Cartesian and vector equation of the line which passes through the point `(-2,4,-5)` and parallel to the line given by `(x+3)/3=(y-4)/5=(z+8)/6` .

Answer 1

Equation of given line is `(x+3)/(3)=(y-4)/(5)=(z+8)/(6)" "`(i)
Direction ratios of line (i) are 3, 5, 6
Therefore, direction ratios of required line are also 3, 5, 6. ltBrgt Required line passes through the point (-2, 4, -5).
Hence, equation of required line is `(x+2)/(3)=(y-4)/(5)=(z+5)/(6)`
Its vector form is `vecr=(-2hati+4hatj-5hatk)+lamda(3hati+5hatj+6hatk)`