Four lines `x+3y-10=0, x+3y-20=0, 3x-y + 5 = 0` and `3x - y - 5 = 0` form a figure which is.

Answer

` a to p ,r; b to q, ; c, to p, q ,s; d to p`

Explanation

a. vector `-3hati = 3hatj + 4hatk and hati + hatj` are coplanar with `veca and vecb`.
b. `vecaxxvecb=|{:(hati,hatj,hatk),(-1,2,2),(-2,1,2):}|`
`2hati-2hatj+3hatk`
c. if `vecc` is equally inclined to `veca and vecb` , then we must have `veca.vecc= vecb.vecc`.
which is true for vectors in options p,q,s
d. vector is forming a triangle with `veca and vecb`. thus
`vecc= veca + vecb = -3hati + 3hatj + 4hatk`