Q. 52 - Calculus

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 $\overrightarrow{PQ}$.

Step-by-Step Solution

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 1: 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 $\overrightarrow{PQ}$.

Step 2: 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

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

Step 3: 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 $⟨1,-5,1,7⟩$

Step 4: 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 $\Vert \overrightarrow{PQ}\Vert$.

Step 5: 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

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

Step 6: 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 $\sqrt{76}$

Step 7: 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 $\overrightarrow{PQ}$

Step 8: 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, $\overrightarrow{PQ}=⟨1,-5,1,7⟩$.

Step 9: Explanation (Part C)

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

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

$=\sqrt{76}$

Step 10: 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 $\frac{1}{\sqrt{76}}⟨1,-5,1,7⟩$