Find a unit vector that is perpendicular to both of the vectors 4i+2j-3k and 2i-3j+k.?

Answer 1

call it a*i+b*j+c*k
(When multiplied with another vector called a1*i+b1*j+c1*k, the product is:
a*a1+b*b1+c*c1)

It is perpendicular to the 1st vector then you have the first equation:
4*a+2*b-3*c = 0 (1)

Similarly you have second equation:

2*a-3*b+c = 0 (2)

It is unit vector, then you have: a^2+b^2+c^2 = 1 (3);

3 equations and 3 unknowns; I think you can solve it.

(By the way, the answer by Mathematica(R) is: two vectors (a,b,c) and (-a,-b,-c) with
a = 7/(9*sqrt(5))
b = 2*sqrt(5)/9
c = 16/(9*sqrt(5))
)

Hope it helps.

Answer 2

Find a unit vector that is perpendicular to both of the vectors 4i + 2j - 3k and 2i - 3j + k.

Take the cross product. The result will be a vector n, that is perpendicular to both of the given vectors.

n = <4, 2, -3> X <2, -3, 1> = <-7, -10, -16>

We can multiply this by any non-zero scalar and it will still be perpendicular. Multiply by -1.

n = <7, 10, 16>

Calculate the magnitude of n.

|| n || = √(7² + 10² + 16²) = √(49 + 100 + 256) = √405 = 9√5

Divide n by its magnitude to get a unit vector perpendicular to both given vectors.

n / || n || = <7, 10, 16> / (9√5)

n / || n || = <7 / (9√5), 2√5 / 9, 16 / (9√5)>

Answer 3

First, take the bypass product, v, of the two those vectors to get a vector perpendicular to the two: this is the determinant of i j ok 4 2 -3 2 -3 a million which fits out to -7i -10j -16k. yet -v additionally satisfies the situations of the priority. So we choose a unit vector interior the different direction of v. as a effect we would desire to divide via its length which fits out to be ?405. So the suitable answer is a million/?405(7i + 10 j + 16k). So Sin