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

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Answer 1 · 2016-07-10T10:13:19

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

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 · 2016-07-10T11:45:55

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

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)#.