Question

v) Find a vector which is parallel to v=i-2j
and has a magnitude 10.​

Answer 1

Explanation:

Answer the following question.

Find a vector which is parallel to `vec"v" = hat"i" - 2hat"j"` and has a magnitude 10.

Let the vector be `vec"w" = "w"_"x" hat"i" + "w"_"y"hat"j"`

`|vec"w"| = sqrt("w"_"x"^2 + "w"_"y"^2) = 10` ....(Given)

∴ `"w"_"x"^2 + "w"_"y"^2 = 100` ...(i)

Also, `vec"v"*vec"w"` = vw ....(∵ `|vec"v"| and |vec"w"|`are parallel vectors)

`=> (hat"i" - 2hat"j")*("w"_"x" hat"i" + "w"_"y"hat"j") = sqrt((1)^2 + (-2)^2) xx 10` ....`(∵ |vec"v"| = sqrt((1)^2 + (-2)^2))`

∴ `"w"_"x" - 2"w"_"y" = 10sqrt5` ...(ii)

Substituting for wx in (i) using equation (ii),

`(10sqrt5 + 2"w"_"y")^2 + "w"_"y"^2 = 100`

∴ `500 + 40sqrt5 "w"_"y" + 4"w"_"y"^2 - 100 = 0`

∴ `5"w"_"y"^2 + 40sqrt5"w"_"y" + 400 = 0`

∴ `"w"_"y"^2 + 8sqrt5"w"_"y" + 80 = 0`

Using factorisation formula,

x = `(-"b" +- sqrt("b"^2 - 4"ac"))/"2a"`

`"w"_"y" = (-8sqrt5+-sqrt((8sqrt5)^2 - 4 xx 1 xx 80))/(2xx1)`

`= "w"_"y" = (-8sqrt5 +- 0)/2 = - 4sqrt5 = (-20)/sqrt5`

Using equation (ii),

`"w"_"x" = 10sqrt5 + 2((-20)/sqrt5)`

`= 10sqrt5 - 40/sqrt5`

`= ((10sqrt5 xx sqrt5) - 40)/sqrt5 = (50 - 40)/sqrt5 = 10/sqrt5`

∴ `vec"w" = "w"_"x"hat"i" + "w"_"y"hat"j" = 10/sqrt5 hat"i" - 20/sqrt5hat"j"`

Required vector is `10/sqrt5 hat"i" - 20/sqrt5hat"j"`

I hope this answer is helpful for you