Find the equation of the line (vector and Cartesian both) which is parallel to the vector `2hati-hatj+3hatk` and which passes through the point (5,-2,4)

Options

A. `r =( 3hati + 2hatj - 8 hatk ) + lamda ( 5hati + 2 hatj - 4 hatk ) , ( x - 3 ) / ( 5 ) = ( y - 2 ) /( 2 ) = ( z + 8 ) /( - 4 ) `
B. ` r = ( 5 hati + 2hatj - 4hatk) + lamda ( 3 hati + 2 hatj - 8 hatk ) , ( x - 3 )/ ( 5 ) = ( y - 2 ) /( 2 ) = ( z + 8 )/ ( - 1) `
C. ` r = ( 5 hati + 2 hatj - 4 hatk ) + lamda ( 3hati + 2 hatj - 8 hatk ) , ( x - 5 ) / ( 3 ) = ( y - 2 ) / ( 2 ) = ( z + 4 ) / ( - 8 ) `
D. ` r = ( 3 hati + 2 hatj - 8 hatk ) + lamda ( 5 hati + 2 hatj - 4hatk ) , ( x - 5 ) / ( 3 ) = ( y - 2 ) / ( 2 ) = ( z - 4 ) / ( - 8 ) `

Answer

C

Explanation

We have,
` a = 5 hati + 2hatj - 4 hatk and b = 3hati + 2hatj - 8 hatk `
Therefore, the vector equation of the line is
` r= 5hati + 2 hatj - 4hatk + lamda ( 3hati + 2hatj - 8 hatk ) `
Now, r is the position vector of any point ` P ( x, y, z ) ` on the line.
Therefore,
` x hati + yhatj + z hatk = 5 hati + 2hatj - 4hatk + lamda ( 3 hati + 2 hatj - 8 hatk ) `
` = ( 5 + 3lamda ) hati+ ( 2 + 2 lamda ) hatj + ( - 4 - 8 lamda ) hatk `
Eliminating ` lamda ` , we get
` ( x - 5 ) /( 3) = ( y -2 ) / (2 ) = ( z + 4 ) / ( - 8 ) `
which is the equuation of the line in cartesian form.