Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1). If `vec'a' + vec'b' + vec'c'` = 0, show that `vec'a' xx vec'b' = vec'b' xx vec'c' = vec'c' xx vec'a'`. Using vectors find the area of triangle ABC with vertices A(1, 2, 3), B(2, −1, 4) and C(4, 5, −1).įind the angle between the vectors `vec'a' + vec'b' and vec'a' -vec'b' if vec'a' = 2hat'i'-hat'j'+3hat'k' and vec'b' = 3hat'i' + hat'j'-2hat'k', and'hence find a vector perpendicular to both' vec'a' + vec'b' and vec'a' - vec'b'`.įind the angle between the vectors `2hat'i' - hat'j' + hat'k'` and `3hat'i' + 4hat'j' - hat'k'`. If `vecr=xhati+yhatj+zhatk` ,find `(vecrxxhati).(vecrxxhatj)+xy`įind `veca.(vecbxxvecc), ' if ' veca=2hati+hatj+3hatk, vecb=-hati+2hatj+hatk ' and ' vecc=3hati+hatj+2hatk` If `|vecaxxvecb|^2+|veca.vecb|^2=400 ` and `|vec a| = 5`, then write the value of `|vecb|`

Hence, the required area is `1/2 sqrt(62)` sq.units Now area of parallelogram whose diagonals are `2hat'i' - hat'j' + hat'k'` and `hat'i' + 3hat'j' - hat'k'` So, the area of the parallelogram ABCD = `|vec'p' xx vec'q'| = 1/2|vec'a' xx vec'b'|` Subtracting equation (ii) from equation (i) we get

In parallelogram abcd find mâˆ a