Question

Let $P=(2,3,0,-1)$and $Q=(3,-2,1,6)$be points in four-dimensional space.

(a) Find $\overrightarrow{PQ}$.

(b) Find $\overrightarrow{PQO}$.

(c) Find the unit vector in the direction of $\overrightarrow{PQ}$.

Short answer

Part A:

The value of $\overrightarrow{PQ}$is $⟨1,-5,1,7⟩$

Part B:

The norm of the vector $\Vert \overrightarrow{PQ}\Vert$is $\sqrt{76}$.

Part C:

The unit vector in the direction of $\overrightarrow{PQ}$is

$\frac{1}{\sqrt{76}}⟨1,-5,1,7⟩$

Step-by-step solution

Introduction (Part A)

• Consider the vector $P=(2,3,0,-1)$and $Q=(3,-2,1,6)$be the points in four-dimensional space.
• The Target is to find $\overrightarrow{PQ}$.

Given Information (Part A)

If there are two points to consider

$P=\left(x_0,y_0,z_0,w_0\right)$and $Q=\left(x_1,y_1,z_1,w_1\right)$,

then

$\overrightarrow{PQ}=\left(x_1-x_0,y_1-y_0,z_1-z_0,w_1-w_0\right).$

Explanation (Part A)

Now,

$P=(2,3,0,-1)$and $Q=(3,-2,1,6)$

As a result , $\overrightarrow{PQ}=⟨3-2,-2-3,1-0,6-(-1)⟩$

Hence, $\overrightarrow{PQ}=⟨1,-5,1,7⟩$

As a result, the value of $\overrightarrow{PQ}$is $⟨1,-5,1,7⟩$

Introduction (Part B)

• Consider the points $P=(2,3,0,-1)$and $Q=(3,-2,1,6)$be the points in four-dimensional space.
• The aim is to find the norm $\Vert \overrightarrow{PQ}\Vert$.

Given Information (Part B)

If

$v$is the vector in $R^4$such that $v=⟨a,b,c,d⟩$,

then

norm of the vector is given by

$\Vert v\Vert =\sqrt{a^2+b^2+c^2+d^2}$

Explanation (Part B)

Now, $P=(2,3,0,-1)$and$Q=(3,-2,1,6)$

Therefore,$\overrightarrow{PQ}=⟨3-2,-2-3,1-0,6-(-1)⟩$

Hence,$\overrightarrow{PQ}=⟨1,-5,1,7⟩$

Therefore,

$\Vert \overrightarrow{PQ}\Vert =\sqrt{1^2+(-5{)}^2+1^2+7^2}$

$=\sqrt{1+25+1+49}$

$=\sqrt{76}$

Therefore, the norm of the vector$\Vert \overrightarrow{PQ}\Vert$is$\sqrt{76}$

Introduction (Part C)

• Consider the points $P=(2,3,0,-1)$and $Q=(3,-2,1,6)$be the points in four - dimensional space.
• The objective is to find the unit vector in the direction of$\overrightarrow{PQ}$

Given Information(Part C)

Now, $P=(2,3,0,-1)$and $Q=(3,-2,1,6)$

Therefore,$\overrightarrow{PQ}=⟨3-2,-2-3,1-0,6-(-1)⟩$

Hence, $\overrightarrow{PQ}=⟨1,-5,1,7⟩$.

Explanation (Part C)

$\Vert \overrightarrow{PQ}\Vert =\sqrt{1^2+(-5{)}^2+1^2+7^2}$

$=\sqrt{1+25+1+49}$

$=\sqrt{76}$

Explanation (Part C)

The unit vector in the direction of $\overline{PQ}$is given by $\frac{1}{\Vert \overrightarrow{PQ}\Vert }\overrightarrow{PQ}$.

Substitute the values,

$\overrightarrow{PQ}=\frac{1}{\sqrt{76}}⟨1,-5,1,7⟩$

As a result, the unit vector in the direction of

$\overrightarrow{PQ}$is $\frac{1}{\sqrt{76}}⟨1,-5,1,7⟩$