Answer to Question #214397 in Analytic Geometry for prince

Question

(2.1) Find the components of a unit vector satisfying ~v· < 3, −1 >= 0.

(2.2) Show that there are infinitely many vectors in R (4) 3 with Euclidean norm 1 whose Euclidean inner product with < −1, 3, −5 > is zero.

(2.3) Determine all values of k so that ~u =< −3, 2k, −k > is orthogonal to ~v =< 2, (3) 5 2 , −k >

Answer 1

2.1) Let the vector v = <x, y>

Now v . <3, -1> = 0

<x, y> . <3, -1> = 0

3x - y = 0

One values of x, y satisfying the above equation is (2, 6)

Hence, v = <2, 6>

Now let the unit vector in the direction of v be w.

Therefore,

w = $\dfrac{v}{|v|}$ = $\dfrac{2i + 6j}{\sqrt{(2)^2 + (6)^2}}$

w = $\dfrac{2i + 6j}{\sqrt{40}}$

2.2) The given vector is a = <-1, 3, -5>

Let there be a vector v = <x, y, z>

Let the normalized vector of v be V

V = $\dfrac{v}{|v|}$

V = $\dfrac{x}{\sqrt{x^2+y^2+z^2}}, \dfrac{y}{\sqrt{x^2+y^2+z^2}}, \dfrac{z}{\sqrt{x^2+y^2+z^2}}$ ....................Equation(1)

Now as per the question, a . V = 0

Hence on taking the dot product we have

-x + 3y - 5z = 0

x = 3y - 5z

On substituting the value of x in equation (1)

V = $\dfrac{3y - 5z}{\sqrt{(3y - 5z)^2+y^2+z^2}}, \dfrac{y}{\sqrt{(3y - 5z)^2+y^2+z^2}}, \dfrac{z}{\sqrt{(3y-5z)^2+y^2+z^2}}$

V is a unitary vector orthogonal to a. Since it still depends on 2 variables we can conclude that there are infinitely many vectors whose inner dot product with a is zero.

2.3) u =< −3, 2k, −k > 

v =< 2, 52 , −k >

Since, u and v are orthogonal so

u . v = 0

<-3, 2k, -k> . <2, 52, -k> = 0

-6 + 104k + k² = 0

On solving we get

k = 0.0576603

k = −104.058