Question

If `veca=2hati+hatj-hatk, vecb=4hati-7hatj+hatk`, find a vector \[\vec{c}\] such that \[\vec{a} \times \vec{c} = \vec{b} \text { and }\vec{a} \cdot \vec{c} = 6\].

Answer 1

Let `vecc= xhati + yhatj + zhatk .`The,

`veca xx vecc = vecb` can be written as `[[hati,hatj,hatk],[2,1, -1],[x,y,z]] = hat4i - hat7j +hatk`

⇒ `(z+y)hati - (2z - x ) hatj +(2y - x)hatk =4hati - 7hatj + k`

\[ \Rightarrow z + y = 4, x - 2z = - 7, 2y - x = 1 . . . \left( 1 \right)\]

Also, 

\[\vec{a} \cdot \vec{c} = 6\]

`(2hati + hatj - hatk) (xhat i + yhatj + zhatk) = 6`

\[ \Rightarrow 2x + y - z = 6\]

\[ \Rightarrow 4x + 2y - 2z = 12\]

\[ \Rightarrow 3x + 2y + x - 2z = 12\]

\[\Rightarrow 3x + 2y - 7 = 12 \left[ From \left( 1 \right) \right]\]

\[ \Rightarrow 3x + 2y = 19 . . . \left( 2 \right)\]

From (1) and (2) we get,
2y − x = 1
2y + 3x = 19

On solving these two equations we get the value of x = \[\frac{9}{2}\] ,y = \[\frac{11}{4}\]

Using the values of x and y we get the value of z as \[\frac{5}{4}\].

So, 

`vec c=9/2hati+11/4hatj+5/4hatk`.