Question

Let O be the origin, c a given number, and u a given direction (i.e., a unit vector).


Describe geometrically the locus of all points P in space that satisfy the vector equation
-

OP · u = c|OP| .


In particular, tell for what value(s) of c the locus will be (Hint: divide through by |OP|):


a) a plane

b) a ray (i.e., a half-line)

c) empty​

Answer 1

We're given to discuss about locus of point P in space satisfying the equation,

$\longrightarrow\mathbf{OP\cdot \^u}=c\,\mathbf{|OP|}\quad\quad\dots(1)$

Let $\bf{OP=v.}$ Let $\theta$ be the angle between vectors $\bf{u}$ and $\bf{v.}$

Then (1) becomes,

$\longrightarrow\mathbf{v\cdot \^u}=c\,\mathbf{|v|}$

$\longrightarrow\mathbf{\dfrac{v}{|v|}\cdot \^u}=c$

$\longrightarrow\mathbf{\^v\cdot \^u}=c$

$\longrightarrow\mathbf{|\^v|\cdot|\^u|\,\cos\theta}=c$

Since $\bf{\^u}$ and $\bf{\^v}$ are unit vectors, their magnitude is 1 each. So,

$\longrightarrow c=\cos\theta$

For the locus being a plane, the angle between vectors should be $\sf{90^o.}$

$\longrightarrow\theta=90^o$

Therefore,

$\longrightarrow c=\cos90^o$

$\longrightarrow\underline{\underline{c=0}}$

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For the locus being a ray, the angle between vectors should be $\sf{0^o}$ or $\sf{180^o.}$

$\longrightarrow\theta=0^o\quad OR\quad\theta=180^o$

Therefore,

$\longrightarrow c=\cos0^o\quad OR\quad c=\cos180^o$

$\longrightarrow\underline{\underline{c=\pm1}}$

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For the locus being empty, it can be possible that,

$\longrightarrow\underline{\underline{c\in\mathbb{R}-[-1,\ 1]}}$

because $c=\cos\theta$ is not defined under this set.

Answer 2

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Explanation:

Refer to the attachment dear

hope this helps you