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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.
Solution
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")`.
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