Question

If `vec"a" = hat"i" + hat"j" + hat"k"` and `vec"b" = hat"j" - hat"k"`, find a vector `vec"c"` such that `vec"a" xx vec"c" = vec"b"` and `vec"a"*vec"c"` = 3.

Answer 1

Let `vec"c" = "c"_1hat"i" + "c"_2hat"j" + "c"_3hat"k"`

Also given that `vec"a" = hat"i" + hat"j" + hat"k"` and `vec"b" = hat"j" - hat"k"`

Since, `vec"a" xx vec"c" = vec"b"`

∴ `|(hat"i", hat"j", hat"k"),(1, 1, 1),("c"_1, "c"_2, "c"_3)| = hat"j" - hat"k"`

= `hat"i"("c"_3 - "c"_2) - hat"j"("c"_3 - "c"_1) + hat"k"("c"_2 - "c"_1)`

= `hat"j" - hat"k"`

On comparing the like terms, we get

c₃ – c₂ = 0  ......(i)

c₁ – c₃ = 1  ....(ii)

And c₂ – c₁ = –1  ....(iii)

Now for `vec"a"*vec"c"` = 3  

`(hat"i" + hat"j" + hat"k") * ("c"_1hat"i" + "c"_2hat"j" + "c"_3hat"k")` = 3

∴ c₁ + c₂ + c₃ = 3   ......(iv)

Adding equation (ii) and equation (iii) we get,

c₂ – c₃ = 0   ......(iv)

From (iv) and (v) we get

c₁ + 2c₂ = 3   .....(vi)

From (iii) and (vi) we get

              c₁ + 2c₂ = 3
             – c₁ + c₂ = – 1
Adding          3c₂ = 2

∴ c₂ = `2/3`

c₃ – c₂ = 0

⇒ `"c"_3 - 2/3` = 0

∴ c₃ = `2/3`

Now c₂ – c₁ = –1

⇒ `2/3 - "c"_1` = –1

⇒ c₁ = `1 + 2/3 = 5/3`

∴ `vec"c" = 5/3 hat"i" + 2/3hat"j" + 2/3hat"k"`

Hence, `vec"c" = 1/3(5hat"i" + 2hat"j" + 2hat"k")`.