Question

How do you find the unit vector perpendicular to the vector (4i - 3j)?

Answer 1

Therefore, the reqd vectors are `(3i/5+4j/5), or, (-3/5i-4/5j)`

Explanation:

Let us name the given vctr. `(4i-3j)=(4,-3)=vecu,` and, suppose that, `vecv =(x,y)` is the reqd. unit vector perp. to `vecu`.

Since, `vecv` is unit vctr., `||vecv||=1 rArr sqrt(x^2+y^2)=1`

`:. x^2+y^2=1...............(1)`

Next, `vecv` is perp. to `vecu`, `vecv.vecu=0`

`:. (x,y)*(4,-3)=0 rArr 4x-3y=0.` or, `x=3/4y............(2)`

We solve `(1) & (2) : (3/4y)^2+y^2=1 rArr y=+-4/5`, giving, `x=+-3/5`

Therefore, the reqd vectors are `(3i/5+4j/5), or,(-3i/5-4j/5)`

Enjoy Maths.!

Answer 2

`+-1/5(3i+4j)`

Explanation:

Any unit vector is of the form (cos a, sin a ) = i cos a + j sin a..

This has to be perpendicular to the given vector.

So, the dot product of the two, (4, -3).( cos a, sin a)

`4 cos a - 3 sin a = 0`..

And so,, tan a = 4/3 > 0.

Both sin and cos have the same sign.

Thus, `(cos a, sin a)=+-(3/5, 4/5)`.