Question

For any vector `vec"a"`, the value of `(vec"a" xx hat"i")^2 + (vec"a" xx hat"j")^2 + (vec"a" xx hat"k")^2` is equal to ______.

पर्याय

• `vec"a"^2`

• `3vec"a"^2`

• `4vec"a"^2`

• `2vec"a"^2`

Answer 1

For any vector `vec"a"`, the value of `(vec"a" xx hat"i")^2 + (vec"a" xx hat"j")^2 + (vec"a" xx hat"k")^2` is equal to `2vec"a"^2`.

Explanation:

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

∴ `vec"a"^2 = "a"_1^2 + "a"_2^2 + "a"_3^2`

Now, `vec"a" xx hat"i" = ("a"_1hat"i" + "a"2hat"j" + "a"_3hat"k") xx hat"i"`

= `|(hat"i", hat"j", hat"k"),("a"_1, "a"_2, "a"_3),(1, 0, 0)|`

= `hat"i"(0 - 0) - hat"j"(0 - "a"_3) + hat"k"(0 - "a"_2)`

= `"a"_3hat"j" - "a"_2hat"k"`

∴ `(vec"a" xx hat"i")^2 = ("a"_3hat"j" - "a"_2hat"k") * ("a"_3hat"j" - "a"_2hat"k")`

= `"a"_3^2 + "a"_2^2`

Similarly `(vec"a" xx hat"i")^2 = "a"_1^2 + "a"_3^2`

And `(vec"a" xx hat"k")^2 = "a"_1^2 + "a"_2^2`

∴ `(vec"a" xx hat"i")^2 + (vec"a" xx hat"j")^2 + (vec"a" xx hat"k")^2 = "a"_3^2 + "a"_2^2 + "a"_1^2 + "a"_3^2 + "a"_1^2 + "a"_2^2`

= `2("a"_1^2 + "a"_2^2 + "a"_3^2)`

= `2vec"a"^2`