If the vectors `vecA=2hati+4hatj and vecB=5hati-phatj` are parallel to each other, the magnitude of `vecB` is

Options

A. `5sqrt5`
B. 10
C. 15
D. `2sqrt5`

Answer

A

Explanation

Here, ` vecA = 2 hat I + 4 hat j , vec B = 5 veci - vecj`
`:. A = sqrt(2^2+4^2)=sqrt20 and B = sqrt(5^2+p^2)`
`vecA .vecB = 10 - 4p` If `vecA ||vecB` , then `vecA.vecB=AB cos 0^@ = AB`
or `10 - 4p = sqrt(20) sqrt(25+p^2)`
On squaring both sides, we get
`100 + 16 p^(2) - 80 p = 20 (25 +p^2) = 500 + 20 p^2`
or`20 p^(2) - 16 p^(2) + 80 p + 400 = 0 `
or `p^(2) + 20 p + 100 = 0 " or " (p+10)^2 = 0 " or " p = - 10 `
`:. vecB = 5hati + 10 hatj` and its magnitude is
`B= sqrt(5^(2)+10^(2))=sqrt(125)=5sqrt5`